JEE Main + AdvancedClass XIHeat & ThermodynamicsMolecular picture

Kinetic Theory

How the ceaseless random motion of molecules explains pressure, temperature, specific heats and the transport of gases — the molecular picture behind the gas laws

🎯 Overview Chapter hero + roadmap

🔬 Interactive 3D · One view of the whole chapter: a box of molecules in ceaseless random motion drums on the walls to produce pressure $P=\tfrac{1}{3}nm\overline{v^2}$; raise the temperature and the molecules speed up, because their mean kinetic energy is $\tfrac{3}{2}k_BT$. Drag the temperature slider to watch the speed distribution broaden and the wall pressure climb, and follow one molecule's zig-zag path between collisions — its mean free path.

Kinetic theory explains the behaviour of gases by picturing them as they really are: enormous numbers of tiny molecules in ceaseless, random motion, colliding with one another and with the walls of their container. Where thermodynamics describes a gas from the outside, through bulk quantities such as pressure, volume and temperature, kinetic theory works from the inside, deriving those same bulk quantities as averages over the motion of the molecules. It is the bridge between the microscopic world of atoms and the macroscopic world of the gas laws, and it was one of the first great triumphs of the atomic hypothesis, developed in the nineteenth century by Maxwell, Boltzmann and others. 🔉⇢

The chapter opens with the molecular nature of matter — the long road from the speculations of Kanada and Democritus, through Dalton's chemical evidence for atoms, to the modern estimate of molecular sizes of about a few angstroms. The key physical distinction is one of spacing: in a solid or liquid the molecules are packed almost as closely as they can be, whereas in a gas at ordinary pressure they are separated by distances tens of times their own size, so that atoms are much freer in gases and can travel long distances without colliding. This diluteness is exactly what makes a gas simple enough to treat by averaging, and it is why kinetic theory succeeds so well for gases. 🔉⇢

Before building the molecular model, the chapter revisits the behaviour of gases and the ideal-gas equation. Experiment shows that at low pressures and not-too-low temperatures all gases obey, to a good approximation, the single relation $PV=\mu RT$, where $\mu$ is the number of moles and $R$ the universal gas constant. This one equation contains Boyle's law (constant $T$), Charles' law (constant $P$) and Avogadro's hypothesis (equal volumes of gases at the same $T$ and $P$ contain equal numbers of molecules). Dalton's law of partial pressures — that the total pressure of a mixture of ideal gases is the sum of partial pressures — follows immediately once the molecules are assumed not to interact. 🔉⇢

The heart of the chapter is the kinetic-theory derivation of pressure. Treating the molecules of a gas as being in incessant random motion, making perfectly elastic collisions with the walls, one computes the momentum they deliver to a wall per second and finds the pressure $P=\tfrac{1}{3}nm\overline{v^2}$, where $n$ is the number of molecules per unit volume, $m$ the mass of a molecule and $\overline{v^2}$ the mean of the squared speeds. This single result, derived from Newtonian mechanics plus a statistical average, is the foundation for everything that follows in the chapter. 🔉⇢

Rewriting the pressure result gives $PV=\tfrac{2}{3}N\left(\tfrac{1}{2}m\overline{v^2}\right)$, so that the product $PV$ is proportional to the total translational kinetic energy of the molecules. Comparing this with the ideal-gas equation $PV=\mu RT=Nk_BT$ yields the kinetic interpretation of temperature: the average kinetic energy of a molecule is proportional to the absolute temperature of the gas, specifically $\tfrac{1}{2}m\overline{v^2}=\tfrac{3}{2}k_BT$. Temperature, mysterious in thermodynamics, is here revealed as nothing but a measure of the mean translational kinetic energy of the molecules, and this at once gives the root-mean-square speed $v_{rms}=\sqrt{3k_BT/m}=\sqrt{3RT/M}$. 🔉⇢

That the mean kinetic energy depends only on temperature, and not on the mass of the molecule, has a striking consequence: in a mixture of gases at a common temperature, the heavier molecules move more slowly and the lighter ones more quickly, but every species has the same average kinetic energy. This is why the RMS speed of a light gas such as hydrogen far exceeds that of a heavy gas such as oxygen at the same temperature, and it underlies phenomena from the escape of light gases from the atmosphere to the separation of uranium isotopes by gaseous diffusion. 🔉⇢

The law of equipartition of energy generalises the picture from translation to all the ways a molecule can store energy. A molecule has several degrees of freedom — three of translation for a point mass, plus rotational and (at high temperature) vibrational ones for molecules with structure. The law states that in thermal equilibrium the energy is shared equally among all the degrees of freedom, each quadratic term contributing an average energy of $\tfrac{1}{2}k_BT$ per molecule. Counting the active degrees of freedom of a gas therefore fixes its internal energy. 🔉⇢

From equipartition the specific heat capacities follow at once. A monatomic gas, with three translational degrees of freedom, has $C_v=\tfrac{3}{2}R$; a diatomic gas at ordinary temperatures, with three translational and two rotational degrees of freedom, has $C_v=\tfrac{5}{2}R$; and adding the constant $R$ of Mayer's relation gives $C_p$ in each case. Indeed the relation $C_p-C_v=R$ is true for any ideal gas, whatever its atomicity, and the ratio $\gamma=C_p/C_v$ takes the values $5/3$, $7/5$ and (for a typical polyatomic gas) $4/3$ that recur throughout thermodynamics. 🔉⇢

The chapter closes with the mean free path — the average distance a molecule travels between successive collisions. Although molecules move at speeds of hundreds of metres per second, they do not travel far in a straight line before striking another molecule, and this is why a smell diffuses across a room only slowly. A short calculation gives $l=1/(\sqrt{2}\,n\pi d^2)$, so the mean free path depends inversely on the number density and the size of the molecules. The mean free path is the quantity that controls the transport properties of a gas — diffusion, viscosity and thermal conduction — and it ties the molecular picture back to measurable, everyday behaviour. 🔉⇢

For the JEE, kinetic theory is a reliable source of one or two marks every year in Main, and it is the molecular backbone of the heat-and-thermodynamics block. The examiners return again and again to a small set of ideas: the pressure result and the RMS speed; the kinetic meaning of temperature and the mass-independence of the mean kinetic energy; the degrees-of-freedom count and the specific heats and $\gamma$ that follow; and mean-free-path and Avogadro-number estimates. Master these, keep the three molecular speeds distinct, and remember that a diatomic gas has five active degrees of freedom at ordinary temperatures, and the chapter is largely won. 🔉⇢

A recurring theme, and the one the examiners most love to test, is the difference between per-molecule and per-mole book-keeping. The mean energy of a molecule is written with Boltzmann's constant $k_B$; the energy or specific heat of a mole is written with the gas constant $R=N_A k_B$. Slipping between the two, or forgetting the factor of Avogadro's number, is the single commonest slip in the chapter. Keeping the two levels — molecule and mole — explicitly separate, and always working temperature in kelvin, prevents most of the errors students make here. 🔉⇢

How to use this page: the Concepts map links each idea to a deep-dive, and the interactive 3D scenes let you watch molecules drum on the walls to make pressure, see the speed distribution broaden as you raise the temperature, and trace one molecule's zig-zag path between collisions. The Worked Examples and Question Bank build problem-solving fluency, the PYQ tab shows exactly how the ideas have been examined across fourteen years, and the Mock Test rehearses them under time. Work the scenes and examples actively — predict the RMS speed or the specific heat before you check it — and the formulas will attach themselves to a physical picture rather than floating free as symbols to be memorised. 🔉⇢

What you will master 🔉⇢

🧠 Concepts Map of the deep dives

This chapter breaks into the concepts below. Each has its own tab with interactive 3D, a full derivation, and a JEE worked example. The Contents tab has the complete prose.

Molecular Nature of 1,=10^-10,mBehaviour of Gases &PV= RT=Nk_BTKinetic Theory of anP=13nm v^2▶Kinetic InterpretatiP=13nm v^2▶Law of EquipartitionTSpecific Heat CapaciC_v=dUdTMean Free Pathd▶
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What you are looking at

A map of the whole chapter. Each box is one concept with its governing relation, and the arrows are the recommended order to learn them in.

Click any box to jump straight to that concept’s tab.

🗺️ Concept map — click any concept to open its tab. Each box shows the concept and its governing formula; the path is the recommended learning order.

Molecular Nature of Matter 🔉⇢

Matter is made up of atoms and molecules — little particles in perpetual motion that attract one another at a few angstroms and repel when squeezed closer. An atom is about one angstrom ($1\,\text{\AA}=10^{-10}\,\text{m}$) across; a mole of any substance contains Avogadro's number $N_A=6.02\times10^{23}$ molecules. In gases the molecules are far apart, so their mutual interaction is negligible except during collisions.

Behaviour of Gases & the Ideal-Gas Equation 🔉⇢

At low pressures and high temperatures a gas obeys the ideal-gas equation $PV=\mu RT=Nk_BT$, where $\mu$ is the number of moles, $R=8.314\,\text{J mol}^{-1}\text{K}^{-1}$, $N$ is the number of molecules and $k_B=1.38\times10^{-23}\,\text{J K}^{-1}$ is the Boltzmann constant. An ideal gas is one that obeys this relation exactly at all $P$ and $T$; real gases approach it as interactions become negligible.

Kinetic Theory of an Ideal Gas — Pressure 🔉⇢

Modelling a gas as a large number of molecules in incessant random motion undergoing elastic collisions, the pressure on the walls arises from the momentum transferred by molecular impacts. Averaging over all molecules gives $P=\tfrac{1}{3}nm\langle v^2\rangle$, where $n$ is the number density, $m$ the molecular mass and $\langle v^2\rangle$ the mean square speed. Pressure is thus a statistical, bulk result of microscopic motion.

Kinetic Interpretation of Temperature 🔉⇢

Combining the kinetic pressure $P=\tfrac{1}{3}nm\langle v^2\rangle$ with the ideal-gas law gives $\tfrac{1}{2}m\langle v^2\rangle=\tfrac{3}{2}k_BT$: the average translational kinetic energy of a molecule is proportional to the absolute temperature and independent of pressure, volume or the nature of the gas. Hence the root-mean-square speed is $v_{\text{rms}}=\sqrt{3k_BT/m}=\sqrt{3RT/M}$.

Law of Equipartition of Energy 🔉⇢

The law of equipartition of energy states that in thermal equilibrium at absolute temperature $T$, the total energy of a molecule is shared equally among every independent quadratic (square) term in its energy, each such term carrying an average of $\tfrac12 k_B T$. Each translational and rotational degree of freedom contributes $\tfrac12 k_B T$, while each vibrational mode contributes $k_B T$ because it stores both kinetic and potential energy.

Specific Heat Capacity of Gases 🔉⇢

The molar specific heat of a gas is the heat needed to raise the temperature of one mole by one kelvin. Equipartition fixes the internal energy, so the molar heat at constant volume is $C_v=\tfrac{dU}{dT}$ and, for an ideal gas, $C_p-C_v=R$ (Mayer's relation). Their ratio $\gamma=C_p/C_v$ is $\tfrac53$ for monatomic, $\tfrac75$ for rigid diatomic and about $\tfrac43$ for polyatomic gases.

Mean Free Path 🔉⇢

The mean free path is the average distance a molecule travels between two successive collisions. Modelling molecules as spheres of diameter $d$ with number density $n$, a molecule sweeps a collision cross-section $\pi d^2$, giving $l=\dfrac{1}{\sqrt{2}\,n\pi d^2}$ once the motion of all molecules is accounted for. For air at STP, $l\sim10^{-7}\,\text{m}$, about a hundred times the interatomic spacing.

📖 Contents Full chapter — read straight through

The complete chapter in reading order. Each section is narratable; the concept tabs add interactive 3D and worked examples on top of this text.

Kinetic Theory
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What you are looking at

The chapter banner — the physics of this chapter in a single moving picture, with live symbols and values.

Every diagram below it is interactive: drag the controls and the numbers move with the drawing.

Molecular Nature of Matter 🔉⇢

🎯 All matter is built from molecules of finite size d ~ 10^-10 m. In a solid or liquid they touch; in a gas the average spacing is ten or more diameters, so each molecule travels freely between rare collisions. Kinetic theory rests on exactly this picture. Raise the temperature and the molecules vibrate more violently and drift apart.
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matter is molecular — molecular size d ≈ 10−10 m; in a gas the mean spacing ≫ d, so molecules are almost always free. Raise T and the molecules jiggle harder and sit farther apart.
What this shows

All matter is built from molecules of finite size d ~ 10^-10 m. In a solid or liquid they touch; in a gas the average spacing is ten or more diameters, so each molecule travels freely between rare collisions. Kinetic theory rests on exactly this picture. Raise the temperature and the molecules vibrate more violently and drift apart.

Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.

Definition: Matter is made up of atoms and molecules — little particles in perpetual motion that attract one another at a few angstroms and repel when squeezed closer. An atom is about one angstrom ($1\,\text{\AA}=10^{-10}\,\text{m}$) across; a mole of any substance contains Avogadro's number $N_A=6.02\times10^{23}$ molecules. In gases the molecules are far apart, so their mutual interaction is negligible except during collisions. 🔉⇢

Kinetic theory explains the behaviour of gases by picturing a gas as an enormous number of tiny atoms or molecules in rapid, ceaseless motion. Before we can build that picture into quantitative laws, we must be convinced of the underlying idea itself: that matter is not continuous but is made up of discrete particles. Richard Feynman, one of the great physicists of the twentieth century, considered the discovery that 'matter is made up of atoms' to be the single most significant piece of scientific knowledge — the one sentence he would choose to pass to a future civilisation if all other knowledge were lost. His compact statement of the atomic hypothesis is worth memorising: all things are made of atoms, little particles that move around in perpetual motion, attracting each other when they are a little distance apart, but repelling upon being squeezed into one another. 🔉⇢

The idea that matter may not be continuous is very old. In India, the Vaiseshika school of thought founded by Kanada in the sixth century B.C. developed an atomic picture in considerable detail: atoms were thought to be eternal, indivisible, infinitesimal and the ultimate parts of matter. The argument was elegant — if matter could be subdivided without end, there would be no difference between a mustard seed and the Meru mountain. Four kinds of Paramanu (the Sanskrit word for the smallest particle) were postulated — Bhoomi (earth), Ap (water), Tejas (fire) and Vayu (air) — while Akasa (space) was held to be continuous and structureless. In ancient Greece, Democritus of the fourth century B.C. is best known for the word 'atom' itself, which means 'indivisible'. These were fascinating conjectures, but because they were never tested and modified by quantitative experiment — the hallmark of modern science — they could not evolve much further. 🔉⇢

The scientific atomic theory is usually credited to John Dalton, who proposed it about two hundred years ago to explain the laws of definite and multiple proportions obeyed by elements when they combine into compounds. The law of definite proportions says that any given compound has a fixed proportion by mass of its constituents. The law of multiple proportions says that when two elements form more than one compound, for a fixed mass of one element the masses of the other element are in the ratio of small integers. To explain these laws Dalton suggested that the smallest constituents of an element are atoms; that atoms of one element are identical but differ from those of other elements; and that a small number of atoms of each element combine to form a molecule of the compound. This is the cornerstone on which the whole kinetic picture rests. 🔉⇢

Two further early-nineteenth-century laws complete the chemical evidence for atoms. Gay-Lussac's law of combining volumes states that when gases combine chemically to yield another gas, their volumes are in the ratios of small integers. Avogadro's law (or hypothesis) states that equal volumes of all gases at equal temperature and pressure contain the same number of molecules. Avogadro's law, when combined with Dalton's theory, neatly explains Gay-Lussac's law of combining volumes, because equal volumes carrying equal numbers of molecules turn integer molecule-ratios into integer volume-ratios. Since elements are often present in the form of molecules rather than free atoms, Dalton's atomic theory is also referred to as the molecular theory of matter. The theory is now universally accepted, although even at the end of the nineteenth century there were still famous scientists who did not believe in atoms. 🔉⇢

In modern times we have direct evidence that molecules — made up of one or more atoms — constitute matter. Electron microscopes and scanning tunnelling microscopes let us effectively 'see' individual atoms and molecules on a surface, so the atomic hypothesis is no longer an inference but an observation. This closes a long historical arc: what began as philosophical speculation in ancient India and Greece, was made quantitative by Dalton, Gay-Lussac and Avogadro, and is today imaged directly. For the purposes of kinetic theory, the crucial point is simply that a gas is a very large collection of identical molecules whose individual behaviour can be described mechanically and whose collective behaviour gives the pressure, temperature and volume that we measure in the laboratory. 🔉⇢

The characteristic length scale of the atomic world is the angstrom. The size of an atom is about one angstrom, that is $1\,\text{\AA}=10^{-10}\,\text{m}$; a typical molecule is one to two angstroms across. This is extraordinarily small — about a hundred-millionth of a centimetre — which is why the discreteness of matter is completely hidden from our senses and why matter appears smooth and continuous in everyday life. Fixing this scale in mind is essential, because almost every estimate in this chapter — of molecular volumes, of intermolecular separations, of the mean free path — is expressed as a multiple of the angstrom, and the relative sizes of these numbers are exactly what distinguish the solid, liquid and gaseous states from one another. 🔉⇢

The interatomic spacing tells us how the three states of matter differ. In solids, which are tightly packed, atoms are spaced only a few angstroms apart — about two angstroms — and are held rigidly in place, which is why solids keep their shape. In liquids the separation between atoms is also about the same as in solids, but the atoms are not rigidly fixed; they can move around each other, and this mobility is what enables a liquid to flow while remaining nearly incompressible. In gases, by contrast, the interatomic distances are in the tens of angstroms — roughly ten times or more the molecular size — so a gas is mostly empty space. This one geometric fact, that gases are dilute, is the reason a gas expands to fill any container and is easily compressed, unlike a solid or a liquid. 🔉⇢

Because the molecules of a gas are on average about ten times their own size apart, the volume actually occupied by the molecules is a tiny fraction of the volume of the container. A famous NCERT estimate makes this concrete: for water vapour at 100 degrees Celsius and one atmosphere, the ratio of the molecular volume to the total volume occupied by the vapour is only about $6\times10^{-4}$. In other words, well over 99.9 percent of a gas is empty space. This is why the molecules travel freely in straight lines for most of the time and only occasionally come near enough to interact, and it is precisely this dilution that makes the ideal-gas idealisation — non-interacting point-like molecules — such a good approximation for real gases at low pressure. 🔉⇢

It is illuminating to trace where the numbers come from. Treating liquid water and a water molecule as having roughly the same density, one mole of water has a mass of about $18\,\text{g}=0.018\,\text{kg}$ and contains about $6\times10^{23}$ molecules, so a single molecule has a mass of about $3\times10^{-26}\,\text{kg}$. Dividing by the density of water ($1000\,\text{kg m}^{-3}$) gives a molecular volume of about $3\times10^{-29}\,\text{m}^3$; setting this equal to $\tfrac{4}{3}\pi r^3$ yields a radius of about $2\,\text{\AA}$. Thus the abstract statement 'a molecule is a couple of angstroms across' is a direct consequence of the measured density of water and the value of Avogadro's number, and requires no microscope at all. 🔉⇢

Avogadro's number, $N_A=6.02\times10^{23}$, is the bridge between the microscopic and macroscopic worlds. It is defined so that the mass of $N_A$ molecules of a substance, expressed in grams, equals the molecular weight; equivalently, the mass of 22.4 litres of any gas at standard temperature and pressure (273 K and 1 atm) equals its molecular weight in grams. That amount of substance — $N_A$ entities — is called one mole. Avogadro originally guessed the equality of the number of molecules in equal volumes of different gases at fixed temperature and pressure purely from chemical reactions, and kinetic theory later justified this hypothesis on mechanical grounds. Because $N_A$ is so enormous, even a small, everyday quantity of gas contains an astronomical number of molecules, which is exactly why statistical averages over molecules are so sharply defined. 🔉⇢

Continuing the chain of estimates, one can also work out how far apart the molecules of a gas are. Since a given mass of water in the vapour state occupies about $1.67\times10^3$ times the volume it occupies as a liquid, the volume available to each molecule increases by the same factor; because volume scales as the cube of a length, the average spacing increases by roughly the cube root, about ten times. Starting from a molecular radius of about $2\,\text{\AA}$, this gives an average intermolecular distance of the order of $40\,\text{\AA}$ in the vapour. So the rule of thumb — that in a gas the average distance between molecules is about ten times the molecular size — is not a guess but a calculated result, and it is the geometric foundation of the whole idea that gas molecules move almost independently. 🔉⇢

Closely related to intermolecular spacing is the mean free path: the average distance a molecule travels without colliding with another molecule. In gases the mean free path is of the order of thousands of angstroms — very much larger than the molecular size — so molecules are quite free and can travel long distances in straight lines between collisions. This is why, if a gas is not enclosed, its molecules simply disperse away and the gas spreads out to fill whatever space is available. In solids and liquids, by contrast, the closeness of the atoms makes the interatomic force important at all times, so the atoms cannot wander off; this is the mechanical reason solids and liquids have a definite volume while a gas does not. 🔉⇢

The interatomic force that governs all of this has a characteristic shape: a long-range attraction and a short-range repulsion. Atoms attract one another when they are a few angstroms apart, which is what binds solids and liquids together, but they repel strongly when pushed closer, which is what makes condensed matter nearly incompressible. In a gas the molecules are, on average, far outside the range where the attraction matters, so for most of the time they feel essentially no force and move in straight lines according to Newton's first law. Only during the brief moments of a collision do the forces act. This is the physical justification for the central simplification of kinetic theory: that intermolecular forces can be ignored except during collisions. 🔉⇢

The apparently static appearance of a gas at rest is deeply misleading. The gas is in fact full of activity, and its equilibrium is a dynamic one. In this dynamic equilibrium the molecules collide constantly and change their speeds at every collision; what remains constant is not the state of any individual molecule but only the average properties of the whole collection — the number density, the average speed, the pressure and the temperature. Recognising that equilibrium is dynamic rather than static is essential for the rest of the chapter, because the pressure a gas exerts and the temperature it possesses are both averages over the ceaseless microscopic motion, not fixed properties of stationary particles. 🔉⇢

It is worth stressing why gases are so much easier to treat than solids and liquids. In a gas the molecules are far from one another and their mutual interactions are negligible except during the brief collisions, so to a very good approximation each molecule moves freely and independently. In a solid or a liquid the molecules are always within range of their neighbours' forces, so their motions are strongly coupled and a simple free-particle description fails. This is exactly why kinetic theory begins with gases: the diluteness of the gas — the very fact that the average molecular separation is about ten times the molecular size — is what makes a clean, calculable model possible, and it is the launch point for deriving the pressure and the meaning of temperature in the sections that follow. 🔉⇢

Finally, the atomic hypothesis is a beginning, not an end. We now know that atoms are neither indivisible nor truly elementary: an atom consists of a nucleus surrounded by electrons, the nucleus is made of protons and neutrons, and these are themselves built from quarks — and even quarks may not be the final word, with string-like entities among the possibilities. For the purposes of this chapter, however, none of that inner structure matters. A gas molecule can be treated as a single mechanical particle of definite mass, and it is the number, mass and motion of these particles — set against the measured length scales of roughly one angstrom for size and tens of angstroms for separation — that will let us derive the macroscopic gas laws from Newtonian mechanics in the sections that follow. 🔉⇢

Kinetic theory is spectacularly successful precisely because it connects these microscopic parameters to measurable bulk properties. It gives a molecular interpretation of the pressure and the temperature of a gas, it is consistent with the gas laws and with Avogadro's hypothesis, and it correctly explains the specific heat capacities of many gases. Beyond such equilibrium properties, it relates the transport properties of gases — their viscosity, thermal conduction and diffusion — to molecular parameters such as the mean free path and the molecular speed, and from measurements of these it yields estimates of molecular sizes and masses. The theory was developed in the nineteenth century by Maxwell, Boltzmann and others, more than a hundred and fifty years after Boyle discovered his gas law in 1661 and long before atoms could be seen directly; that it works so well is among the strongest confirmations of the atomic hypothesis, and it is the reason the molecular picture is now beyond dispute. 🔉⇢

The number density $n$ — the number of molecules per unit volume — is the single most useful quantity carried forward from this section into the derivations that follow. It packages the atomic-scale information (how many molecules there are and how closely they are spaced) into one macroscopic-looking symbol. From the molecular size of about two angstroms and the average separation of tens of angstroms one can estimate $n$; conversely, measuring $n$ through the pressure and temperature of a gas lets one work back to molecular quantities. The interplay between the microscopic length scales established in this card and the number density is exactly what will make the pressure formula $P=\tfrac{1}{3}nm\langle v^2\rangle$ of the next card both meaningful and calculable. 🔉⇢

The estimate of the interatomic distance in a gas is worth restating, because it fixes the diluteness numerically. A given mass of water in the vapour state occupies about $1.67\times10^3$ times the volume it occupies as a liquid, so each molecule has that many times more room. Because volume scales as the cube of a length, the average spacing grows only by the cube root of about $10^3$, that is roughly ten times; starting from a molecular radius of about $2\,\text{\AA}$ this gives an average intermolecular distance of the order of $40\,\text{\AA}$ in the vapour. Thus in a gas the molecules sit, on the average, about ten to twenty times their own diameter apart — a vast and mostly empty arena within which they move almost entirely freely, colliding only occasionally. 🔉⇢

It is worth being explicit about why gases disperse away when unconfined while solids and liquids hold together. In a gas the mean free path is of the order of thousands of angstroms and the intermolecular attraction is negligible at those distances, so once a molecule heads outward there is nothing to pull it back and the gas expands without limit to fill its container. In a liquid or a solid the molecules are always within the few-angstrom range of the attractive part of the interatomic force, which supplies the cohesion that gives condensed matter a definite volume. The same force law — long-range attraction, short-range repulsion — thus explains both the near-incompressibility of solids and liquids and the expansiveness of gases, simply through the very different average spacings involved in each state. 🔉⇢

To gather the section together: matter is made of atoms and molecules; the chemical evidence of Dalton, Gay-Lussac and Avogadro established this, and modern microscopy confirms it directly. Atoms are about one angstrom in size, a mole contains $N_A=6.02\times10^{23}$ molecules, and the three states of matter are distinguished by their interatomic spacing — a couple of angstroms in solids and liquids, tens of angstroms in gases. Because a gas is so dilute, its molecules move almost freely, interacting only in brief elastic collisions, and its equilibrium is dynamic. These facts, expressed in the language of number density, molecular mass, mean free path and Avogadro's number, are the raw material from which the quantitative kinetic theory of the next two sections is constructed. 🔉⇢

Derivation 🔉⇢

  1. Goal: estimate the size of a molecule from bulk data. Take water: density $\rho=1000\,\text{kg m}^{-3}$ and molar mass $M_0=18\,\text{g}=0.018\,\text{kg}$.
  2. Mass of one molecule: $m=M_0/N_A=0.018/(6\times10^{23})\approx3\times10^{-26}\,\text{kg}$, using Avogadro's number $N_A=6.02\times10^{23}$.
  3. Volume of one molecule (liquid density $\approx$ molecular density): $V=m/\rho=(3\times10^{-26})/1000=3\times10^{-29}\,\text{m}^3$.
  4. Radius from $V=\tfrac{4}{3}\pi r^3$: $r=\left(3V/4\pi\right)^{1/3}\approx2\times10^{-10}\,\text{m}=2\,\text{\AA}$ — of the order of one angstrom, as claimed.
  5. Intermolecular spacing in vapour: vapour volume is about $1.67\times10^3$ times the liquid volume, so the linear spacing grows by $(10^3)^{1/3}\approx10$, giving an average separation of order $10\times2\,\text{\AA}\times2\approx40\,\text{\AA}$.
  6. Conclusion: molecular size $\sim1$–$2\,\text{\AA}$ while average separation in a gas $\sim$ tens of angstroms; the gas is $\sim10\times$ dilute, so intermolecular interaction is negligible except during collisions.
⚠️ JEE trap: A common error is to imagine a gas as densely packed with molecules touching, or to think of equilibrium as a static, frozen arrangement. In fact a gas is overwhelmingly empty space — the molecular volume is only about $6\times10^{-4}$ of the total — and the average molecular separation is roughly ten times the molecular size. Equally, equilibrium is dynamic, not static: molecules collide incessantly and change speed at every collision, and only the averages (number density, pressure, temperature) stay constant. Confusing 'nothing changes macroscopically' with 'nothing moves' misses the entire mechanism of kinetic theory. 🔉⇢

Behaviour of Gases & the Ideal-Gas Equation 🔉⇢

🎯 An ideal gas obeys PV = nRT, tying pressure, volume and temperature to the amount of gas. Squeeze the volume at fixed temperature and the pressure rises (Boyle); heat it at fixed pressure and it expands (Charles). Drag V and T and read P off the gauge as the piston and the molecular bombardment respond.
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PV = nRT   Boyle (T fixed: P ∝ 1/V), Charles (P fixed: V ∝ T), Gay-Lussac (V fixed: P ∝ T). Here n = 1 mol.
What this shows

An ideal gas obeys PV = nRT, tying pressure, volume and temperature to the amount of gas. Squeeze the volume at fixed temperature and the pressure rises (Boyle); heat it at fixed pressure and it expands (Charles). Drag V and T and read P off the gauge as the piston and the molecular bombardment respond.

Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.

Definition: At low pressures and high temperatures a gas obeys the ideal-gas equation $PV=\mu RT=Nk_BT$, where $\mu$ is the number of moles, $R=8.314\,\text{J mol}^{-1}\text{K}^{-1}$, $N$ is the number of molecules and $k_B=1.38\times10^{-23}\,\text{J K}^{-1}$ is the Boltzmann constant. An ideal gas is one that obeys this relation exactly at all $P$ and $T$; real gases approach it as interactions become negligible. 🔉⇢

The properties of gases are far easier to understand than those of solids and liquids, and the reason is geometric: in a gas the molecules are far from one another and their mutual interactions are negligible except when two molecules collide. Because of this, gases at low pressures and high temperatures — well above the conditions at which they would liquefy or solidify — obey a remarkably simple relation among their pressure, volume and temperature. Historically this relation was pieced together from experiment long before the molecular picture was secure: Boyle discovered his law in 1661, and Charles, Gay-Lussac and Avogadro added further pieces in the following century and a half. Kinetic theory would eventually explain all of them from a single molecular model. 🔉⇢

The empirical starting point is that for a given sample of gas the product of pressure and volume, divided by the absolute temperature, is a constant: $PV=KT$, where $T$ is the temperature on the kelvin or absolute scale. Here $K$ is a constant for a given sample of gas but it varies with the amount of gas — specifically with the volume, or better, with the number of molecules present. Introducing the molecular picture, we write $K=Nk_B$, where $N$ is the number of molecules in the sample. The astonishing experimental observation is that the constant $k_B$ is the same for all gases, whatever their chemical identity. This universal constant is called the Boltzmann constant, denoted $k_B$, with the value $k_B=1.38\times10^{-23}\,\text{J K}^{-1}$ in SI units. 🔉⇢

Because $k_B$ is universal, we can immediately compare two states of the same or different gases. For two samples we have $P_1V_1/(N_1T_1)=P_2V_2/(N_2T_2)=k_B$. A direct consequence follows: if $P$, $V$ and $T$ are the same for two gases, then $N$ must also be the same. This is precisely Avogadro's hypothesis — that equal volumes of all gases at the same temperature and pressure contain the same number of molecules, or equivalently that the number of molecules per unit volume is the same for all gases at a fixed temperature and pressure. What Avogadro originally guessed from chemical reactions, kinetic theory thus derives as a mechanical necessity, and this is one of the early triumphs of the theory. 🔉⇢

Counting molecules directly is impractical, so it is convenient to group them into moles. The number of molecules in one mole is Avogadro's number, $N_A=6.02\times10^{23}$. The mass of 22.4 litres of any gas at standard temperature and pressure (a standard temperature of 273 K and a pressure of 1 atm) equals its molecular weight in grams, and this amount of substance is called a mole. If the sample contains $N$ molecules, the number of moles is $\mu=N/N_A$; equivalently, if the sample has mass $M$ and the molar mass is $M_0$, then $\mu=M/M_0=N/N_A$. This double expression for $\mu$ is the practical link between the mass of gas we weigh out in the laboratory and the number of molecules that the kinetic picture deals with. 🔉⇢

Rewriting the perfect-gas relation in terms of moles gives the most familiar form of the ideal-gas equation, $PV=\mu RT$, where $R=N_A k_B$ is a universal constant called the universal gas constant. In the kelvin scale its value is $R=8.314\,\text{J mol}^{-1}\text{K}^{-1}$. Because $R=N_A k_B$, the two constants carry the same physics at different bookkeeping levels: $k_B$ works per molecule, $R$ works per mole. The molecular form $PV=Nk_BT$ and the molar form $PV=\mu RT$ are therefore completely equivalent, and being fluent in switching between them — using $\mu=N/N_A$ and $R=N_A k_B$ — is essential for solving JEE problems quickly and without error. 🔉⇢

A third equivalent form expresses the equation through the number density. Writing $n=N/V$ for the number of molecules per unit volume, the relation $PV=Nk_BT$ becomes $P=nk_BT$. This form is especially useful in kinetic theory itself, because the pressure derived from molecular collisions comes out naturally in terms of the number density $n$ rather than the total number $N$. A fourth form uses the mass density: since $\mu=M/M_0$ and the density is $\rho=M/V$, the ideal-gas equation can be written $P=\rho RT/M_0$. Each of these versions — $PV=Nk_BT$, $PV=\mu RT$, $P=nk_BT$, $P=\rho RT/M_0$ — is the same physical law, and choosing the right one for the quantities given in a problem is half the battle. 🔉⇢

An ideal gas is defined precisely as a gas that satisfies the relation $PV=\mu RT$ exactly at all pressures and temperatures. It is a simple theoretical model; no real gas is truly ideal. The value of the model is that real gases approach it closely under the right conditions, and the ideal-gas equation is the reference against which real-gas behaviour is measured. It is important to be clear that 'ideal' is an idealisation — a limiting behaviour — and not a description of any actual substance. Nonetheless the idealisation is extraordinarily useful, because for the low pressures and moderate-to-high temperatures of most laboratory and exam situations, the error made in treating a common gas as ideal is small. 🔉⇢

Experiments show clear and systematic departures from ideal behaviour for real gases. A plot of $PV/\mu T$ against pressure for a real gas is not the horizontal line $R$ that the ideal-gas law predicts; instead it curves, and it does so differently at different temperatures. The key regularity is that all such curves approach the ideal-gas value as the pressure is lowered and the temperature raised. The physical reason is exactly the one from the previous section: at low pressures or high temperatures the molecules are far apart and molecular interactions are negligible, and without interactions the gas behaves like an ideal one. Deviations grow as the gas is compressed or cooled toward the point where it would liquefy, because then the neglected intermolecular attractions and the finite molecular volume start to matter. 🔉⇢

Boyle's law is recovered from the ideal-gas equation by fixing the amount of gas and the temperature. Setting $\mu$ and $T$ constant in $PV=\mu RT$ gives $PV=\text{constant}$: at fixed temperature, the pressure of a given mass of gas varies inversely with its volume. Compressing a fixed quantity of gas at constant temperature therefore raises its pressure in exact inverse proportion. Experimental pressure-volume curves confirm Boyle's law, and once again the agreement is best at high temperatures and low pressures, where the ideal-gas idealisation is most nearly exact. Boyle's law is thus not an independent postulate but a special case of the ideal-gas equation, obtained by holding two of its variables fixed. 🔉⇢

Charles' law is recovered by fixing the pressure instead. From $PV=\mu RT$, holding $P$ (and $\mu$) constant gives $V\propto T$: at fixed pressure the volume of a gas is directly proportional to its absolute temperature. Heating a gas at constant pressure makes it expand in direct proportion to the kelvin temperature, and cooling it makes it contract. Experimental temperature-volume curves for real gases follow Charles' law closely under the usual conditions and deviate as the gas approaches liquefaction. Boyle's law and Charles' law together capture the two most common controlled experiments on gases, and both fall out of the single ideal-gas equation as soon as one variable is held fixed — a good illustration of the economy of the equation. 🔉⇢

The use of the absolute (kelvin) temperature scale in all of these relations is not a matter of convenience but of necessity. The proportionalities $V\propto T$ and $P\propto T$ hold only when $T$ is measured from absolute zero; they fail badly if a Celsius temperature is used, because Celsius has an arbitrary zero. The ideal-gas law itself defines a natural temperature scale: the constant-volume gas thermometer, using the pressure of a low-density gas held at fixed volume as its thermometric property, reads a temperature that in the low-density limit is independent of which gas is used. This is the perfect-gas or ideal-gas temperature, and it coincides with the absolute thermodynamic temperature, which is why kelvin is the mandatory unit in every formula of this chapter. 🔉⇢

When several gases that do not react are mixed in one vessel, the ideal-gas equation extends in a simple additive way. For $\mu_1$ moles of gas 1, $\mu_2$ moles of gas 2, and so on, held in a vessel of volume $V$ at temperature $T$, the equation of state of the mixture is $PV=(\mu_1+\mu_2+\dots)RT$. Rearranging, the total pressure is $P=\mu_1 RT/V+\mu_2 RT/V+\dots=P_1+P_2+\dots$, where each term $P_i=\mu_i RT/V$ is the pressure that gas $i$ alone would exert if it occupied the vessel by itself at the same volume and temperature. This term is called the partial pressure of that gas, and the mixture simply adds them. 🔉⇢

The statement just derived is Dalton's law of partial pressures: the total pressure of a mixture of ideal gases is the sum of the partial pressures the individual gases would exert on their own. It follows directly from the fact that non-interacting molecules of different species contribute to the pressure independently, each obeying the ideal-gas law as if the others were not there. Dalton's law is enormously useful for mixtures such as air, and it is a favourite of examiners because it links directly to the molar composition: since $P_i/P=\mu_i/\mu$, the ratio of partial pressures equals the ratio of the numbers of moles, and hence the ratio of the numbers of molecules, of the components. 🔉⇢

A short NCERT example shows the law in action. A vessel contains two non-reactive gases, neon (monatomic) and oxygen (diatomic), whose partial pressures are in the ratio 3:2. Since each gas separately obeys $P_iV=\mu_i RT$ with common $V$ and $T$, the ratio of partial pressures equals the ratio of moles, $\mu_1/\mu_2=3/2$, and hence the ratio of the numbers of molecules $N_1/N_2=\mu_1/\mu_2=3/2$ as well. The mass-density ratio then follows by weighting the mole ratio with the molar masses ($20.2\,\text{u}$ for neon, $32.0\,\text{u}$ for oxygen), giving $\rho_1/\rho_2=(3/2)(20.2/32.0)\approx0.947$. The example shows how partial pressures translate cleanly into molecule counts and densities. 🔉⇢

It is worth pausing on what the ideal-gas law does and does not assume. It treats the gas as a collection of molecules whose sizes and mutual forces are negligible; it does not distinguish monatomic from diatomic gases, because $PV=\mu RT$ contains no reference to internal molecular structure. That is why neon and oxygen, so different chemically, obey exactly the same equation of state. The internal structure of the molecules does matter for other properties — notably the specific heat capacities, through the law of equipartition of energy studied later in the chapter — but for the relation among pressure, volume, temperature and molecule number, all ideal gases behave identically. Keeping this scope in mind prevents the common mistake of over-thinking simple gas-law problems. 🔉⇢

It is useful to have the numerical values firmly in mind. The universal gas constant is $R=8.314\,\text{J mol}^{-1}\text{K}^{-1}$, the Boltzmann constant is $k_B=1.38\times10^{-23}\,\text{J K}^{-1}$, and Avogadro's number is $N_A=6.02\times10^{23}\,\text{mol}^{-1}$, with the three linked by $R=N_A k_B$. At standard temperature and pressure — a temperature of 273 K and a pressure of 1 atm — one mole of any ideal gas occupies 22.4 litres, a result that follows directly from $PV=\mu RT$ with $\mu=1$. These constants recur throughout the chapter, and having them ready avoids arithmetic slips in numerical problems. 🔉⇢

Avogadro's hypothesis, originally a bold guess made purely from the integer volume-ratios of chemical reactions, is not an independent postulate within kinetic theory but a derived result. Because the constant $k_B$ in $PV=Nk_BT$ is the same for every gas, equal values of $P$, $V$ and $T$ necessarily force equal $N$ — which is precisely Avogadro's statement that equal volumes at equal temperature and pressure contain equal numbers of molecules. Kinetic theory thus explains from mechanics what chemistry had only inferred, and this convergence of independent lines of evidence is a large part of why the molecular picture of gases became universally accepted. 🔉⇢

The perfect-gas or ideal-gas temperature deserves a fuller word, because it is the operational backbone of the kelvin scale. A constant-volume gas thermometer holds a fixed quantity of a low-density gas at fixed volume and uses its pressure as the thermometric property; since $PV=\mu RT$ gives $P\propto T$ at fixed $V$ and $\mu$, the pressure reads the absolute temperature directly. Crucially, as the amount of gas is reduced toward the low-density limit, the temperature inferred becomes independent of which gas is used, converging on a single universal scale. This ideal-gas temperature coincides with the absolute thermodynamic temperature defined later through the Carnot cycle, which is why every formula in this chapter is written in kelvin. 🔉⇢

The systematic study of departures from ideal behaviour is instructive. If one plots the quantity $PV/\mu T$ against pressure for a real gas at several fixed temperatures, the ideal-gas law predicts a single horizontal line at the value $R$. Real gases instead give curves that dip or rise and differ from one temperature to another, but all of them approach the ideal value $R$ as the pressure tends to zero and as the temperature is raised. The deviations are largest near the conditions at which the gas would liquefy, because there the two neglected effects — the finite volume of the molecules and the attractive forces between them — become significant; far from liquefaction they are negligible and the gas is very nearly ideal. 🔉⇢

The molecular reason for the low-pressure, high-temperature rule is exactly the diluteness discussed earlier. At low pressure the number density is small, so the molecules are far apart and spend almost all their time outside the range of one another's forces; at high temperature the molecules move fast, so even when they do approach, the brief attractive tug barely deflects them. In both limits the molecules behave as free, non-interacting particles — the very definition of an ideal gas — and $PV=\mu RT$ holds accurately. Compression or cooling reverses this, crowding the molecules and slowing them so that the intermolecular forces reassert themselves and measurable deviations appear. 🔉⇢

The mass-density form of the equation, $P=\rho RT/M_0$, is often the most convenient in practice, because density is directly measurable. It shows that at a given pressure and temperature the density of a gas is proportional to its molar mass, which is why carbon dioxide (molar mass 44) is denser than air and tends to pool in low-lying spaces, while hydrogen and helium rise. Rearranged, the same relation lets one determine an unknown molar mass from a measured density, pressure and temperature — a standard laboratory technique. All four forms of the ideal-gas equation are simply this one physical law viewed through different measured quantities. 🔉⇢

Dalton's law of partial pressures is the natural tool for real gaseous mixtures such as the atmosphere. Air is chiefly nitrogen and oxygen, and each component contributes a partial pressure in proportion to its mole fraction: the partial pressure of oxygen is about 21 percent of the total atmospheric pressure, that of nitrogen about 78 percent, and so on, with the partial pressures summing to the total. Because $P_i/P=\mu_i/\mu=N_i/N$, measuring partial pressures is equivalent to measuring the molar composition of the mixture. This additivity works precisely because the different species, being ideal, do not interact, and each fills the whole volume independently as though the others were absent. 🔉⇢

Gathering the section: at low pressures and high temperatures a gas obeys the ideal-gas equation, which can be written equivalently as $PV=\mu RT$, $PV=Nk_BT$, $P=nk_BT$ or $P=\rho RT/M_0$, with the universal constants linked by $R=N_A k_B$. Boyle's law and Charles' law are the constant-temperature and constant-pressure special cases; Avogadro's hypothesis is the statement that equal volumes hold equal numbers of molecules; and Dalton's law adds the partial pressures of a non-reacting mixture. An ideal gas is one that obeys this relation exactly at all $P$ and $T$, a limit that real gases approach as interactions vanish. These relations, all resting on the universality of $k_B$, are the macroscopic facts that kinetic theory must reproduce from the motion of molecules — the task of the next two cards. 🔉⇢

Derivation 🔉⇢

  1. Empirical law for a fixed sample: $PV=KT$ with $T$ in kelvin. Bring in molecules: $K=Nk_B$, with $k_B$ found experimentally to be the same for all gases (the Boltzmann constant).
  2. Hence $PV=Nk_BT$. Comparing two states, $P_1V_1/(N_1T_1)=P_2V_2/(N_2T_2)=k_B$; so equal $P,V,T$ force equal $N$ (Avogadro's hypothesis).
  3. Group molecules into moles: $\mu=N/N_A=M/M_0$, and define $R=N_A k_B=8.314\,\text{J mol}^{-1}\text{K}^{-1}$. Then $PV=Nk_BT=\mu N_A k_B T=\mu RT$.
  4. Special cases: fix $\mu,T\Rightarrow PV=\text{const}$ (Boyle's law); fix $\mu,P\Rightarrow V\propto T$ (Charles' law). Other forms: $P=nk_BT$ with $n=N/V$, and $P=\rho RT/M_0$ with $\rho=M/V$.
  5. Mixture of non-reacting ideal gases: $PV=(\mu_1+\mu_2+\dots)RT$, so $P=\mu_1 RT/V+\mu_2 RT/V+\dots=P_1+P_2+\dots$, where $P_i=\mu_i RT/V$ is the partial pressure.
  6. Therefore the total pressure of a mixture of ideal gases is the sum of the partial pressures (Dalton's law); and $P_i/P=\mu_i/\mu=N_i/N$ links partial pressures to molecule counts.
⚠️ JEE trap: Two errors recur. First, using Celsius instead of kelvin: relations like $V\propto T$ and $P\propto T$ and the whole equation $PV=\mu RT$ require the absolute scale, since Celsius has an arbitrary zero. Second, thinking the ideal-gas equation depends on the type of gas: it does not — monatomic neon and diatomic oxygen obey the identical $PV=\mu RT$, because the equation contains no reference to molecular structure. Molecular structure affects specific heats (via equipartition), not the equation of state. Always convert temperatures to kelvin and never insert a factor for 'diatomic' into the gas law itself. 🔉⇢

Kinetic Theory of an Ideal Gas — Pressure 🔉⇢

🎯 Pressure is not a property of a single molecule — it is the steady drum-roll of countless molecules striking the wall and reversing their momentum. Averaging over all directions gives P = (1/3)(N/V) m v^2. Add molecules (N) or speed them up (v) and the wall is hit harder and more often, so the pressure climbs.
🔉⇢
P = ⅓ (N/V) m v̅2   pressure is the time-averaged bombardment of the walls by molecules; more molecules or faster molecules → higher P.
What this shows

Pressure is not a property of a single molecule — it is the steady drum-roll of countless molecules striking the wall and reversing their momentum. Averaging over all directions gives P = (1/3)(N/V) m v^2. Add molecules (N) or speed them up (v) and the wall is hit harder and more often, so the pressure climbs.

Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.

Definition: Modelling a gas as a large number of molecules in incessant random motion undergoing elastic collisions, the pressure on the walls arises from the momentum transferred by molecular impacts. Averaging over all molecules gives $P=\tfrac{1}{3}nm\langle v^2\rangle$, where $n$ is the number density, $m$ the molecular mass and $\langle v^2\rangle$ the mean square speed. Pressure is thus a statistical, bulk result of microscopic motion. 🔉⇢

The kinetic theory of gases is built entirely on the molecular picture of matter established earlier. A given amount of gas is regarded as a collection of a very large number of molecules — typically of the order of Avogadro's number — that are in incessant random motion. This single phrase, 'incessant random motion', encapsulates the whole model: the molecules never stop moving, and there is no preferred direction to their velocities. Everything measurable about the gas — its pressure, its temperature, its internal energy — will be shown to be an average over this microscopic motion. The aim of this section is the first great result of the theory: a formula for the pressure of a gas in terms of the mass, number density and mean square speed of its molecules. 🔉⇢

Full derivation, worked example and interactive 3D on the Kinetic Theory of an Ideal Gas — Pressure tab →

Kinetic Interpretation of Temperature 🔉⇢

🎯 Temperature has a molecular meaning: it is a direct measure of the average translational kinetic energy of the molecules, (3/2)k_B T per molecule, independent of the gas. Heating raises v_rms as the square root of T. Drag T and watch the single molecule speed up (v_rms shown for N2, m = 4.65e-26 kg).
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½ m vrms2 = ₂₃ kBT   so vrms = √(3kBT/m). Temperature is mean molecular kinetic energy.
What this shows

Temperature has a molecular meaning: it is a direct measure of the average translational kinetic energy of the molecules, (3/2)k_B T per molecule, independent of the gas. Heating raises v_rms as the square root of T. Drag T and watch the single molecule speed up (v_rms shown for N2, m = 4.65e-26 kg).

Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.

Definition: Combining the kinetic pressure $P=\tfrac{1}{3}nm\langle v^2\rangle$ with the ideal-gas law gives $\tfrac{1}{2}m\langle v^2\rangle=\tfrac{3}{2}k_BT$: the average translational kinetic energy of a molecule is proportional to the absolute temperature and independent of pressure, volume or the nature of the gas. Hence the root-mean-square speed is $v_{\text{rms}}=\sqrt{3k_BT/m}=\sqrt{3RT/M}$. 🔉⇢

The pressure formula derived in the previous card, $P=\tfrac{1}{3}nm\langle v^2\rangle$, is a purely mechanical result: it contains no reference to temperature at all. The great achievement of this section is to bring temperature into the molecular picture, and thereby to give a molecular meaning to a quantity that thermodynamics had treated as primitive. The route is short but profound — we simply compare the kinetic pressure formula with the experimentally established ideal-gas law — and the result, that the average kinetic energy of a molecule is proportional to the absolute temperature, is one of the most important single equations in all of physics. 🔉⇢

Full derivation, worked example and interactive 3D on the Kinetic Interpretation of Temperature tab →

Law of Equipartition of Energy 🔉⇢

🎯 In thermal equilibrium energy is shared equally among every quadratic degree of freedom, each getting (1/2)k_B T. A monatomic molecule has only 3 translational modes (f=3); a diatomic adds 2 rotational modes (f=5), and vibration adds more at high T. Slide f and watch the active energy arrows switch on.
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U = (f/2) kBT per molecule   each quadratic degree of freedom (translation, rotation, vibration) carries ½kBT. Monatomic f = 3, diatomic f = 5.
What this shows

In thermal equilibrium energy is shared equally among every quadratic degree of freedom, each getting (1/2)k_B T. A monatomic molecule has only 3 translational modes (f=3); a diatomic adds 2 rotational modes (f=5), and vibration adds more at high T. Slide f and watch the active energy arrows switch on.

Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.

Definition: The law of equipartition of energy states that in thermal equilibrium at absolute temperature $T$, the total energy of a molecule is shared equally among every independent quadratic (square) term in its energy, each such term carrying an average of $\tfrac12 k_B T$. Each translational and rotational degree of freedom contributes $\tfrac12 k_B T$, while each vibrational mode contributes $k_B T$ because it stores both kinetic and potential energy. 🔉⇢

The law of equipartition of energy is the bridge that carries us from the microscopic picture of molecules in ceaseless motion to the measurable thermal properties of bulk matter. In the kinetic theory of an ideal gas we already found that the pressure is $P=\tfrac13 n m \overline{v^2}$, and combining this with the ideal gas equation $PV=\mu R T = k_B N T$ gives the celebrated result that the average translational kinetic energy of a molecule is $\tfrac12 m \overline{v^2}=\tfrac32 k_B T$. This single relation is the kinetic interpretation of temperature: the temperature of a gas is a direct measure of the average kinetic energy of random molecular motion, independent of the nature of the gas or the molecule. Equipartition generalises this insight, telling us how energy is shared not just among the three translational modes but among every independent mode a molecule possesses. 🔉⇢

To make the idea precise we must first count degrees of freedom. A molecule free to move in space needs three coordinates to specify its location; if it is constrained to move in a plane it needs two, and if constrained to a line it needs just one. We therefore say it has one degree of freedom for motion along a line, two for motion in a plane and three for motion in space. Motion of a body as a whole from one point to another is called translation, so a molecule free to move in space has exactly three translational degrees of freedom. These are the coordinates and velocity components $v_x$, $v_y$ and $v_z$ along the three Cartesian axes, and each one is an independent avenue along which the molecule can carry kinetic energy. 🔉⇢

Now examine the structure of the kinetic energy of a single molecule. For a point-like monatomic particle the energy is purely translational, $\varepsilon_t=\tfrac12 m v_x^2+\tfrac12 m v_y^2+\tfrac12 m v_z^2$. The crucial feature is that each translational degree of freedom contributes a term containing the square of some variable of motion. For a gas in thermal equilibrium at temperature $T$, the average value of the whole translational energy is $\langle\varepsilon_t\rangle=\tfrac32 k_B T$. Because space has no preferred direction, this energy must be shared equally among the three axes, so $\tfrac12 m\overline{v_x^2}=\tfrac12 m\overline{v_y^2}=\tfrac12 m\overline{v_z^2}=\tfrac12 k_B T$. Each translational degree of freedom therefore carries, on the average, an energy of exactly $\tfrac12 k_B T$, and this equal sharing among the three is the seed of the general law. 🔉⇢

Molecules of a monatomic gas like argon, helium or neon have only translational degrees of freedom, so three numbers completely describe their motion. But a diatomic gas such as oxygen or nitrogen is richer. A molecule of $\mathrm{O_2}$ has the same three translational degrees of freedom, but in addition it can rotate about its centre of mass. There are two independent axes of rotation, both perpendicular to the line joining the two atoms, and rotation about each contributes a term $\tfrac12 I_1\omega_1^2$ or $\tfrac12 I_2\omega_2^2$ to the energy, where $I_1$ and $I_2$ are the moments of inertia and $\omega_1$, $\omega_2$ the angular speeds. Rotation about the third axis, the line joining the atoms itself, has a negligibly small moment of inertia and, for quantum mechanical reasons, simply does not come into play. So a rigid diatomic molecule has two rotational degrees of freedom in addition to its three translational ones. 🔉⇢

Notice the recurring pattern: each rotational degree of freedom, exactly like each translational one, contributes a term to the energy that contains the square of a variable of motion, here an angular speed. This is not a coincidence but the organising principle of the whole discussion. Whenever the energy of a system can be written as a sum of independent squared terms, each such term is an independent way in which the molecule can absorb and store energy. It is these quadratic terms, not the geometric degrees of freedom in a naive sense, that equipartition counts. For translational and rotational motion the two happen to coincide, one squared term per degree of freedom, but for vibration, as we shall see, one physical mode brings two squared terms. 🔉⇢

We assumed above that the $\mathrm{O_2}$ molecule is a rigid rotator, that is, that the two atoms are held at a fixed separation like a rigid dumbbell. This assumption is found to be true at moderate temperatures for oxygen and nitrogen, but it is not always valid. Molecules such as carbon monoxide, even at moderate temperatures, have a mode of vibration in which the two atoms oscillate back and forth along the interatomic axis like a one-dimensional harmonic oscillator. The vibrational energy of such a mode is $\varepsilon_v=\tfrac12 m\left(\dfrac{dy}{dt}\right)^2+\tfrac12 k y^2$, where $k$ is the force constant of the oscillator and $y$ is the vibrational coordinate measured from equilibrium. The full molecular energy then reads $\varepsilon=\varepsilon_t+\varepsilon_r+\varepsilon_v$. 🔉⇢

The vibrational energy deserves special attention because it breaks the simple one-term-per-degree counting. The expression $\varepsilon_v$ contains two squared terms, one for the kinetic energy $\tfrac12 m(dy/dt)^2$ and one for the potential energy of the spring $\tfrac12 k y^2$. While each translational and rotational degree of freedom contributes only one squared term to the energy, one vibrational mode contributes two. This is the physical reason a vibrational mode is worth twice a translational or rotational one when we come to divide up the energy. Keeping this distinction firmly in mind is the single most reliable way to avoid errors when counting the thermal energy of a polyatomic molecule. 🔉⇢

We can now state the law itself. Each quadratic term occurring in the expression for the energy is a mode of absorption of energy by the molecule. In thermal equilibrium at absolute temperature $T$ we have already seen that for each translational mode the average energy is $\tfrac12 k_B T$. The most elegant principle of classical statistical mechanics, first proved by Maxwell, states that this is so for every mode of energy alike, whether translational, rotational or vibrational. That is, in equilibrium the total energy is distributed equally among all the possible energy modes, with each mode carrying an average energy of $\tfrac12 k_B T$. This is known as the law of equipartition of energy, and its reach extends far beyond gases to any classical system in thermal equilibrium. 🔉⇢

Reading off the consequences: each translational and each rotational degree of freedom of a molecule contributes $\tfrac12 k_B T$ to the average energy, while each vibrational mode contributes $2\times\tfrac12 k_B T=k_B T$, precisely because a vibrational mode carries both a kinetic and a potential energy term. The rigorous proof of the law lies beyond the scope of an introductory treatment, but its use is straightforward and enormously powerful: we shall apply it to predict the specific heats of gases theoretically, and later to the specific heats of solids. Wherever the energy is a sum of squared terms and the temperature is high enough for classical physics to hold, equipartition tells us the average energy at once. 🔉⇢

Let us apply the law to compute the internal energy of a mole of gas, degree of freedom by degree of freedom. For a monatomic gas the molecule has only its three translational degrees of freedom, so the average energy of one molecule at temperature $T$ is $\tfrac32 k_B T$. Multiplying by Avogadro's number $N_A$ gives the internal energy of one mole, $U=\tfrac32 k_B T\times N_A=\tfrac32 R T$, since $R=k_B N_A$. This is the exact result we anticipated from the kinetic interpretation of temperature, now derived cleanly from equipartition. It says that for a monatomic ideal gas the entire internal energy is the translational kinetic energy of random molecular motion, with three modes each holding $\tfrac12 R T$ per mole. 🔉⇢

For a diatomic gas treated as a rigid rotator, like a dumbbell, there are five degrees of freedom: three translational and two rotational. Using equipartition, the total internal energy of one mole is $U=\tfrac52 k_B T\times N_A=\tfrac52 R T$. Each of the five modes again carries $\tfrac12 R T$ per mole, so the diatomic gas stores more internal energy at a given temperature than a monatomic gas simply because it has more places to put that energy. This is the microscopic origin of the fact, familiar from the specific heats, that diatomic gases are 'harder to heat' per degree than monatomic ones: the same input of heat must be spread over five reservoirs rather than three. 🔉⇢

If the diatomic molecule is not rigid but also vibrates, we must add the vibrational contribution. A single vibrational mode carries two squared terms and therefore an average energy $k_B T$, so the molecular energy becomes $\tfrac52 k_B T+k_B T=\tfrac72 k_B T$, and the molar internal energy is $U=\left(\tfrac52 k_B T+k_B T\right)N_A=\tfrac72 R T$. Counting squared terms rather than physical modes, a vibrating diatomic molecule behaves as though it had seven degrees of freedom: three translational, two rotational and two from the single vibration. Whether vibration is active depends on temperature, a subtlety that classical equipartition cannot itself explain and that we return to when discussing specific heats. 🔉⇢

A polyatomic molecule is the most general case. In general it has three translational and three rotational degrees of freedom, together with a certain number $f$ of vibrational modes that depends on its structure. According to the law of equipartition of energy, one mole of such a gas has internal energy $U=\left(\tfrac32 k_B T+\tfrac32 k_B T+f k_B T\right)N_A$. The first term is the translational energy from three modes, the second is the rotational energy from three modes, and the third is the vibrational energy, each of the $f$ vibrational modes carrying its full $k_B T$. This compact formula contains the monatomic and diatomic results as special cases and shows how systematic the bookkeeping becomes once the law is trusted. 🔉⇢

It is worth dwelling on a deep and useful consequence: for an ideal gas the internal energy is a function of temperature alone. Every energy term we have counted, translational, rotational and vibrational, is proportional to $T$, and none depends on the volume or the pressure separately. Physically this is because in an ideal gas the molecules do not interact except during instantaneous collisions, so there is no potential energy of intermolecular forces to store. The internal energy is entirely the microscopic mechanical energy of the molecules, and it changes only when the temperature changes. This is exactly the assumption used when we write $\Delta U=\mu C_v\Delta T$ for any process of an ideal gas, and equipartition is what justifies it from first principles. 🔉⇢

The law also sharpens the meaning of 'purely kinetic'. For a monatomic ideal gas the internal energy is literally the translational kinetic energy of the atoms, with no rotational or vibrational store at all. For diatomic and polyatomic gases some of the internal energy resides in rotational kinetic energy and, when vibration is active, half of the vibrational energy is potential energy of the interatomic bond. So the loose statement that 'the internal energy of an ideal gas is purely kinetic' is exactly true only for a monatomic gas; for molecular gases it is more accurate to say that the internal energy is a temperature-dependent sum of kinetic and, through vibration, some potential contributions, all of them microscopic and none of them due to intermolecular forces. 🔉⇢

Equipartition is remarkable for how little it assumes and how much it delivers. It does not require us to know the detailed shape of the molecular velocity distribution, only that the system is in thermal equilibrium and that its energy is a sum of independent quadratic terms. From this alone it assigns $\tfrac12 k_B T$ to every such term. Because $k_B=R/N_A$ is a universal constant, the energy per mode per mole, $\tfrac12 R T$, is the same for every gas at a given temperature. This universality is why gases as different as helium and nitrogen have specific heats that fall into a small number of predictable families, distinguished only by how many active modes each molecule carries. 🔉⇢

It is instructive to see how the counting of degrees of freedom depends on the shape of the molecule, because this is where careless bookkeeping most often goes astray. A monatomic gas such as argon is treated as a structureless point and carries only three translational degrees of freedom. A diatomic or any linear molecule, such as oxygen, nitrogen or carbon dioxide, has three translational and only two rotational degrees of freedom, because rotation about the axis passing through the atoms themselves involves a vanishingly small moment of inertia and does not store appreciable energy. A nonlinear polyatomic molecule such as water or ammonia, by contrast, has three translational and three rotational degrees of freedom, since it can tumble independently about three distinct axes. Getting the rotational count right, two for a linear molecule and three for a nonlinear one, is essential before any vibrational modes are added. 🔉⇢

The exclusion of rotation about the internuclear axis of a diatomic molecule deserves a closer look, because it is a first glimpse of quantum ideas intruding on a classical argument. Classically one might expect three rotational degrees of freedom for any object, yet experiment insists on only two for a diatomic gas. The reason is that the moment of inertia about the line joining the two atoms is extremely small, the mass being concentrated on that very axis, and quantum mechanics then places the first excited rotational level of that mode enormously high in energy. At ordinary temperatures the thermal energy $k_B T$ is far too small to excite it, so the mode is frozen and contributes nothing. Equipartition, being a purely classical statement, cannot by itself predict this exclusion; it must be told which modes are active, and only then does it distribute the energy among them. 🔉⇢

Why does the constant that appears is $k_B$, and why is the result universal across all gases? The Boltzmann constant $k_B=R/N_A\approx 1.38\times10^{-23}\,\text{J K}^{-1}$ is the fundamental bridge between temperature, a macroscopic quantity, and energy at the level of a single molecule. Equipartition assigns the same $\tfrac12 k_B T$ to every quadratic mode regardless of the mass, size or chemical identity of the molecule, so the average energy per mode is a property of the temperature alone. This is why, in a mixture of gases at a common temperature, every species has the same average energy per degree of freedom even though the heavier molecules move more slowly. It is the same universality that made the kinetic interpretation of temperature so compelling: temperature is nothing but a measure of the average microscopic energy per mode. 🔉⇢

The fact that the internal energy of an ideal gas is a function of temperature alone is worth restating as a working principle, because it is used constantly in thermodynamics. Whenever an ideal gas changes state, its change in internal energy is $\Delta U=\mu C_v\Delta T$ no matter how the change is brought about, whether at constant volume, constant pressure, or along some curved path on the pressure-volume diagram. The internal energy does not care about the path, only about the endpoints' temperatures, precisely because equipartition ties $U$ to $T$ through the fixed count of active modes. This is what allows one to compute $\Delta U$ for a complicated process by imagining a simpler constant-volume path with the same temperature change, a device that recurs throughout the study of the first law and the special processes of a gas. 🔉⇢

Finally, a word on the limits of the classical law prepares the ground for what follows. Classical equipartition predicts that every mode, once it exists, always carries its full $\tfrac12 k_B T$, regardless of temperature. Experiment shows this is not quite true: at ordinary temperatures the vibrational modes of many diatomic gases are effectively dormant, and even rotation freezes out at very low temperatures. The resolution lies in quantum mechanics, which allows a mode to be excited only when $k_B T$ is comparable to the spacing of its energy levels. Until then the mode is 'frozen' and does not share in the energy. Equipartition is thus the correct classical, high-temperature limit, and it is against this benchmark that the measured specific heats of gases are best understood, as the next card develops in detail. 🔉⇢

Derivation 🔉⇢

  1. Start from the kinetic-theory pressure of an ideal gas, $P=\tfrac13 n m\overline{v^2}$, where $n$ is the number density, $m$ the molecular mass and $\overline{v^2}$ the mean square speed.
  2. Combine with the ideal gas law $PV=\mu R T=k_B N T$ to eliminate $P$; using $n=N/V$ gives $\tfrac13 m\overline{v^2}=k_B T$, hence the mean translational kinetic energy $\tfrac12 m\overline{v^2}=\tfrac32 k_B T$.
  3. Because space is isotropic, $\overline{v_x^2}=\overline{v_y^2}=\overline{v_z^2}$, so each of the three translational squared terms carries $\tfrac12 m\overline{v_x^2}=\tfrac12 k_B T$: equal sharing among the three modes.
  4. Generalise (Maxwell): every independent quadratic term in the molecular energy — translational $\tfrac12 m v_i^2$, rotational $\tfrac12 I\omega_i^2$, vibrational $\tfrac12 m(dy/dt)^2$ and $\tfrac12 k y^2$ — carries average energy $\tfrac12 k_B T$ in equilibrium.
  5. Count terms per molecule: monatomic $=3$ (translation); rigid diatomic $=5$ (3 translation + 2 rotation); vibrating diatomic $=7$ (3 + 2 + 2, since one vibration = two squared terms); polyatomic $=3+3+2f$ (with $f$ vibrational modes).
  6. Average molecular energy $=(\text{number of squared terms})\times\tfrac12 k_B T$; multiply by $N_A$ and use $R=k_B N_A$ to obtain the molar internal energy.
  7. Results: monatomic $U=\tfrac32 R T$; rigid diatomic $U=\tfrac52 R T$; vibrating diatomic $U=\tfrac72 R T$; polyatomic $U=\left(3+f\right)R T$, each valid for one mole of ideal gas and depending on temperature alone.
⚠️ JEE trap: A very common error is to count physical degrees of freedom and forget that a vibrational mode contributes two squared terms, not one. Students write the vibrational energy as $\tfrac12 k_B T$ per vibration and obtain a wrong internal energy. The law of equipartition assigns $\tfrac12 k_B T$ to each quadratic (squared) term in the energy expression; a single vibration has both a kinetic term $\tfrac12 m(dy/dt)^2$ and a potential term $\tfrac12 k y^2$, so it carries $k_B T$, twice as much as a translational or rotational mode. A second, related error is assuming vibration is always active: classically it is, but experimentally vibrational (and at very low temperature even rotational) modes freeze out, so the effective number of active modes — and hence $U$ — depends on temperature. 🔉⇢

Specific Heat Capacity of Gases 🔉⇢

🎯 Heating a gas at constant volume raises only its internal energy, C_V = (f/2)R; at constant pressure the gas also does expansion work, so C_P = C_V + R (Mayer's relation). Their ratio gamma = 1 + 2/f sets the adiabatic behaviour. Slide the degrees of freedom f and watch both molar heats and gamma respond.
🔉⇢
CV = (f/2)R,   CP = CV + R,   γ = CP/CV = 1 + 2/f   CP exceeds CV by R (the work of expansion).
What this shows

Heating a gas at constant volume raises only its internal energy, C_V = (f/2)R; at constant pressure the gas also does expansion work, so C_P = C_V + R (Mayer's relation). Their ratio gamma = 1 + 2/f sets the adiabatic behaviour. Slide the degrees of freedom f and watch both molar heats and gamma respond.

Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.

Definition: The molar specific heat of a gas is the heat needed to raise the temperature of one mole by one kelvin. Equipartition fixes the internal energy, so the molar heat at constant volume is $C_v=\tfrac{dU}{dT}$ and, for an ideal gas, $C_p-C_v=R$ (Mayer's relation). Their ratio $\gamma=C_p/C_v$ is $\tfrac53$ for monatomic, $\tfrac75$ for rigid diatomic and about $\tfrac43$ for polyatomic gases. 🔉⇢

The specific heat capacity of a gas measures how much heat is required to change its temperature, and it is here that the law of equipartition of energy pays its most striking experimental dividend. Because equipartition fixes the internal energy $U$ of an ideal gas as a definite multiple of $R T$ per mole, and because the internal energy of an ideal gas depends only on temperature, we can predict the specific heats of whole families of gases from nothing more than a count of molecular degrees of freedom. The agreement between these theoretical predictions and the measured values, and the instructive places where the agreement breaks down, together form one of the great early successes and signposts of the molecular picture of matter. 🔉⇢

First we must be careful about what specific heat we mean, because a gas is unusual in having more than one. When we supply heat to a gas we can hold its volume fixed, in which case all the heat goes into raising the internal energy, or we can hold its pressure fixed, in which case the gas also expands and does work on its surroundings, so more heat is needed for the same temperature rise. This is why a gas has two principal molar specific heats: $C_v$, the molar specific heat at constant volume, and $C_p$, the molar specific heat at constant pressure. For a solid or a liquid the distinction is unimportant because expansion is negligible, but for a gas it is central and must always be respected. 🔉⇢

Consider first heating at constant volume. Since the volume does not change, the gas does no work, $W=0$, and the first law of thermodynamics $\Delta Q=\Delta U+\Delta W$ reduces to $\Delta Q=\Delta U$. All the heat supplied goes into internal energy. By definition the molar specific heat at constant volume is the heat per mole per unit temperature rise, so $C_v=\dfrac{dU}{dT}$. This simple relation is the hinge of the whole subject: once equipartition hands us $U(T)$, differentiating gives $C_v$ immediately, and then Mayer's relation gives $C_p$. Everything about the thermal capacity of an ideal gas flows from the single function $U(T)$. 🔉⇢

For a monatomic gas the molecule has only three translational degrees of freedom, so the internal energy of one mole is $U=\tfrac32 R T$. Differentiating, $C_v=\dfrac{dU}{dT}=\tfrac32 R$. This is the smallest specific heat any gas can have, because a monatomic atom has the fewest ways to store energy. Numerically $C_v=\tfrac32\times 8.31\approx 12.5\,\text{J mol}^{-1}\text{K}^{-1}$, a figure confirmed to good accuracy by measurements on helium, neon, argon and the other noble gases. That so simple a count reproduces the observed heat capacity of real gases is a powerful vindication of the kinetic theory and of equipartition. 🔉⇢

To pass from $C_v$ to $C_p$ we need Mayer's relation. For an ideal gas the equation of state is $PV=\mu R T$, so for one mole at constant pressure $P\,dV=R\,dT$. Heating at constant pressure, the first law gives $\Delta Q=\Delta U+P\Delta V$, and dividing by $dT$ for one mole yields $C_p=\dfrac{dU}{dT}+P\dfrac{dV}{dT}=C_v+R$. Rearranged, this is $C_p-C_v=R$, Mayer's relation, one of the tidiest results in all of thermal physics. The extra $R$ is precisely the work done per mole per kelvin as the gas expands to keep its pressure constant; it is the same universal gas constant that appears in the equation of state. 🔉⇢

The most important feature of Mayer's relation is its universality. The step $P\,dV=R\,dT$ used only the ideal gas equation and nothing about the internal structure of the molecule, so the conclusion holds regardless of how many degrees of freedom the molecule has. In the words of the text, $C_p-C_v=R$ is true for any ideal gas, whether mono, di or polyatomic. The molecular complexity changes $C_v$ and $C_p$ individually, raising both by the same amount as more modes are added, but their difference is pinned to the constant $R$ for every ideal gas alike. This is a beautiful example of a robust thermodynamic result surviving intact across very different microscopic pictures. 🔉⇢

For a monatomic gas, therefore, $C_p=C_v+R=\tfrac32 R+R=\tfrac52 R\approx 20.8\,\text{J mol}^{-1}\text{K}^{-1}$. The ratio of the specific heats, a quantity that appears everywhere from the speed of sound to the adiabatic gas law, is $\gamma=\dfrac{C_p}{C_v}=\dfrac{\tfrac52 R}{\tfrac32 R}=\dfrac53\approx 1.67$. This value is the fingerprint of a monatomic gas, and measuring $\gamma$ for an unknown gas is a standard way to infer how many degrees of freedom its molecules possess. A high $\gamma$ near $1.67$ signals a simple, structureless molecule with only translational modes. 🔉⇢

A diatomic gas treated as a rigid rotator has five degrees of freedom, so $U=\tfrac52 R T$ and $C_v=\dfrac{dU}{dT}=\tfrac52 R\approx 20.8\,\text{J mol}^{-1}\text{K}^{-1}$. Mayer's relation then gives $C_p=\tfrac52 R+R=\tfrac72 R\approx 29.1\,\text{J mol}^{-1}\text{K}^{-1}$, and the ratio is $\gamma=\dfrac{\tfrac72 R}{\tfrac52 R}=\dfrac75=1.40$. This prediction matches the measured room-temperature specific heats of hydrogen, oxygen, nitrogen and the other common diatomic gases with impressive accuracy, and the value $\gamma=1.4$ is so characteristic that it is worth committing to memory as the signature of a diatomic gas behaving as a rigid rotator. 🔉⇢

If the diatomic molecule also vibrates, an extra vibrational mode adds two squared terms and hence $k_B T$ per molecule, giving $U=\tfrac72 R T$ per mole. Then $C_v=\tfrac72 R$, $C_p=\tfrac92 R$ and $\gamma=\dfrac{\tfrac92 R}{\tfrac72 R}=\dfrac97\approx 1.29$. Whether a given diatomic gas shows the rigid-rotator value $C_v=\tfrac52 R$ or the higher vibrating value $C_v=\tfrac72 R$ depends on temperature, because vibration switches on only at high temperature. At ordinary temperatures most diatomic gases sit at $\tfrac52 R$; as they are heated strongly their specific heats climb towards $\tfrac72 R$, a temperature dependence that classical equipartition alone cannot explain and that first pointed physicists towards quantum ideas. 🔉⇢

For a polyatomic gas the counting is completely systematic. In general such a molecule has three translational, three rotational degrees of freedom and $f$ vibrational modes, so equipartition gives $U=\left(\tfrac32+\tfrac32+f\right)R T=(3+f)R T$ per mole. Differentiating, $C_v=(3+f)R$ and, by Mayer's relation, $C_p=(4+f)R$, so that $\gamma=\dfrac{C_p}{C_v}=\dfrac{4+f}{3+f}$. When vibrational modes are ignored, $f=0$, a nonlinear polyatomic molecule has six active modes with $C_v=3R$, $C_p=4R$ and $\gamma=\dfrac43\approx 1.33$. This lower $\gamma$ near $1.33$ is the hallmark of a complex, many-atom molecule with many ways to store energy. 🔉⇢

Gathering the three families side by side makes the pattern vivid. The monatomic gas has $C_v=\tfrac32 R$, $C_p=\tfrac52 R$ and $\gamma=\tfrac53$; the rigid diatomic gas has $C_v=\tfrac52 R$, $C_p=\tfrac72 R$ and $\gamma=\tfrac75$; the polyatomic gas has $C_v=3R$, $C_p=4R$ and $\gamma=\tfrac43$. As the molecule grows more complex, both $C_v$ and $C_p$ increase because there are more modes to fill, while $\gamma$ falls steadily towards one. Textbook tables of predicted specific heats, computed on exactly this basis and ignoring vibration, agree well with measured values for many gases, confirming that the number of active modes is what governs the thermal capacity. 🔉⇢

The agreement is good but not perfect, and the discrepancies are as instructive as the successes. The predicted values assume vibrational modes are inactive; for gases such as chlorine, ethane and many other polyatomic species the measured specific heats are larger than the simple prediction, which is exactly what we expect if vibrational modes have begun to contribute. Including those modes in the count improves the agreement. Thus the systematic excess of measured over predicted specific heat is not a failure of the theory but a signal, telling us which internal modes have woken up at the temperature of measurement. The law of equipartition of energy is, in this way, well verified experimentally at ordinary temperatures. 🔉⇢

The same equipartition machinery extends naturally to solids, giving the Dulong and Petit law. Model a solid as $N$ atoms each vibrating about a fixed lattice site. An oscillation in one dimension has average energy $2\times\tfrac12 k_B T=k_B T$, again because a vibration carries both a kinetic and a potential squared term. Each atom oscillates in three dimensions, so its average energy is $3k_B T$, and for one mole the internal energy is $U=3k_B T\times N_A=3RT$. At constant pressure the expansion of a solid is negligible, so $\Delta Q\approx\Delta U$ and the molar specific heat is $C=\dfrac{\Delta Q}{\Delta T}=\dfrac{dU}{dT}=3R\approx 25\,\text{J mol}^{-1}\text{K}^{-1}$. This constant value is the law of Dulong and Petit, and it agrees well with the measured specific heats of most solids at ordinary temperature. 🔉⇢

The Dulong and Petit prediction is not universal, however, and the exceptions again point to quantum physics. Light, tightly bound solids such as diamond have specific heats well below $3R$ at room temperature, because their high vibrational frequencies mean the modes are only partly excited until much higher temperatures. As with the vibrational modes of gases, the classical result $3R$ is the high-temperature limit that a solid approaches once all its lattice vibrations are fully active. Water, treated crudely as a collection of atoms each contributing $3R$, gives a molar specific heat of roughly $3\times 3R\approx 9R$ for its three atoms, which is qualitatively in line with the large specific heat of water that makes it such an effective coolant and heat store. 🔉⇢

It is worth making the temperature dependence of specific heats fully explicit, because it is the chapter's deepest lesson. Classical equipartition says every mode always carries its full share, so specific heats should be constant, independent of temperature. Experiment flatly contradicts this: the specific heat of hydrogen, for instance, is about $\tfrac32 R$ at very low temperatures (only translation active), rises to $\tfrac52 R$ at ordinary temperatures (rotation switched on) and climbs further towards $\tfrac72 R$ at high temperatures (vibration switched on). The modes appear to switch on one after another as the temperature rises, a staircase that classical physics cannot reproduce and that demanded a new theory. 🔉⇢

The resolution is quantum: a mode can absorb energy only in discrete quanta whose size is set by the spacing of its energy levels. A mode contributes its classical $\tfrac12 k_B T$ per squared term only when $k_B T$ is large compared with that spacing; when $k_B T$ is much smaller, the mode is effectively 'frozen out' and takes almost no energy. Vibrational levels are widely spaced, so vibration freezes at ordinary temperatures; rotational levels are closer, so rotation freezes only at very low temperatures; translational energy is essentially continuous and never freezes. This qualitative picture of the freezing of degrees of freedom explains, without any fitting, why specific heats rise in steps as temperature increases. 🔉⇢

It is helpful to hold the three families in mind as a simple ladder of complexity. A monatomic gas, the simplest, has the lowest heat capacity because it can store energy only in translation; a diatomic gas can also rotate, so it holds more energy at the same temperature; a polyatomic gas, richer still, can rotate about more axes and vibrate in more ways, and so has the largest heat capacity of the three. Reading the trend the other way, the ratio $\gamma$ falls steadily from $\tfrac53$ to $\tfrac75$ to $\tfrac43$ as complexity grows, because adding modes raises $C_v$ faster than it raises the fixed gap $C_p-C_v=R$. Measuring $\gamma$ for an unfamiliar gas is therefore a quick experimental route to the number of degrees of freedom its molecules bring into play, and hence to their structure. 🔉⇢

The comparison between predicted and measured specific heats, laid out in the standard tables of the kinetic theory, is a small triumph worth appreciating in full. For the noble gases the monatomic prediction $C_v=\tfrac32 R$ is confirmed almost exactly. For hydrogen, oxygen, nitrogen and other common diatomic gases at room temperature, the rigid-rotator prediction $C_v=\tfrac52 R$ and $\gamma=1.40$ agree closely with experiment. The predicted values for triatomic and larger molecules, computed while ignoring vibration, are also in reasonable agreement. Where measured specific heats exceed the simple predictions, as they do for chlorine and many heavier polyatomic gases, the excess is systematically upward, exactly the direction expected if vibrational modes have started to absorb energy. The pattern of both agreement and controlled disagreement is what gives the theory its authority. 🔉⇢

Turning to solids, the law of Dulong and Petit is a striking illustration of how far equipartition reaches beyond gases. Historically, Dulong and Petit found experimentally, long before the theory was understood, that the molar specific heats of most solid elements at ordinary temperature cluster around the same value, close to $3R$ or about $25\,\text{J mol}^{-1}\text{K}^{-1}$. The kinetic theory explains this at once: each atom in the solid is bound to its lattice site and oscillates in three dimensions like three independent one-dimensional oscillators, each carrying $k_B T$ of energy, so the molar internal energy is $3RT$ and the specific heat is $3R$. That a law discovered empirically should fall straight out of counting squared terms is a persuasive sign that the molecular picture underlying it is correct. 🔉⇢

The whole account rests on the kinetic picture of a gas as a vast number of molecules in random motion, colliding elastically with one another and with the walls of the container. Maxwell and Boltzmann showed by statistical mechanics that in equilibrium the average kinetic energy per molecule is fixed by the absolute temperature, and Avogadro's number relates the molar quantities to the behaviour of a single molecule. Each quadratic term in the energy, whether translational along the three axes, rotational about the axes of a diatomic dumbbell, or vibrational, absorbs its equal share. The macroscopic specific heat capacities $C_v$ and $C_p$ are simply the sums of these microscopic absorptions of energy, converted to a per-mole, per-kelvin basis through the universal gas constant $R$, so a bulk laboratory measurement reaches all the way down to the moments of inertia and the momentum of individual molecules. 🔉⇢

In practical units the gas constant is $R=8.31\,\text{J mol}^{-1}\text{K}^{-1}$, so the monatomic $C_v=\tfrac32 R$ is about $12.5$, the rigid diatomic $C_v=\tfrac52 R$ about $20.8$, and the triatomic value about $24.9$, each rising by $8.31$ from $C_v$ to $C_p$ exactly as Mayer's relation demands. These theoretical predictions, summarised in the standard tables, are compared against measured values for argon, helium, hydrogen, oxygen, nitrogen and carbon dioxide, and the agreement is close wherever vibration is negligible. The discrepancies for heavier compounds, whose measured capacities are greater than the predicted values, are traced to vibrational modes that the simple count ignored; including those modes improves the agreement and confirms that the number of active quadratic terms is what governs the specific heat capacity of a gas. 🔉⇢

In summary, the specific heats of gases are among the sharpest tests of the molecular theory of matter. Equipartition together with the first law gives $C_v=\dfrac{dU}{dT}$ and the universal Mayer relation $C_p-C_v=R$, from which the ratios $\gamma=\tfrac53$, $\tfrac75$ and $\tfrac43$ follow for monatomic, diatomic and polyatomic gases. The same reasoning gives the Dulong and Petit value $3R$ for solids. Where the classical predictions succeed they confirm the reality of molecules and their degrees of freedom; where they fail, systematically and in the direction of frozen modes, they were the very clues that forced the birth of quantum theory. Few topics in introductory physics reward careful bookkeeping so richly. 🔉⇢

Derivation 🔉⇢

  1. Constant volume: with $W=0$, the first law $\Delta Q=\Delta U+\Delta W$ gives $\Delta Q=\Delta U$, so by definition $C_v=\dfrac{dU}{dT}$ (heat per mole per kelvin).
  2. Insert equipartition energies: monatomic $U=\tfrac32 R T\Rightarrow C_v=\tfrac32 R$; rigid diatomic $U=\tfrac52 R T\Rightarrow C_v=\tfrac52 R$; polyatomic $U=(3+f)R T\Rightarrow C_v=(3+f)R$.
  3. Constant pressure, one mole of ideal gas: $PV=RT\Rightarrow P\,dV=R\,dT$; the first law $\Delta Q=\Delta U+P\Delta V$ divided by $dT$ gives $C_p=\dfrac{dU}{dT}+R=C_v+R$.
  4. Hence Mayer's relation $C_p-C_v=R$, which uses only the equation of state and therefore holds for any ideal gas — mono, di or polyatomic.
  5. Form the ratio $\gamma=\dfrac{C_p}{C_v}=\dfrac{C_v+R}{C_v}$: monatomic $\gamma=\dfrac{\tfrac52 R}{\tfrac32 R}=\tfrac53$; rigid diatomic $\gamma=\dfrac{\tfrac72 R}{\tfrac52 R}=\tfrac75$; polyatomic $\gamma=\dfrac{4+f}{3+f}\to\tfrac43$ when $f=0$.
  6. Solids (Dulong–Petit): each atom is a 3-D oscillator with energy $3\times k_B T=3k_B T$ (six squared terms); one mole gives $U=3RT$, and since $\Delta V\approx 0$, $C=\dfrac{dU}{dT}=3R\approx 25\,\text{J mol}^{-1}\text{K}^{-1}$.
  7. Quantum caveat: a mode contributes its classical share only when $k_B T$ exceeds its level spacing; otherwise it is frozen, so measured specific heats rise in steps with temperature toward the classical limits above.
⚠️ JEE trap: Students often speak of 'the' specific heat of a gas as if it had only one, the way a solid or liquid effectively does. A gas has two principal molar specific heats, $C_v$ and $C_p$, and they differ by exactly $R$ because at constant pressure the gas also does expansion work $P\Delta V$ that must be supplied on top of the internal-energy change. Confusing the two, or forgetting the work term, is a frequent source of wrong answers. The correction is to always ask whether volume or pressure is held constant: at constant volume $\Delta Q=\Delta U$ so use $C_v=\tfrac{dU}{dT}$; at constant pressure $\Delta Q=\Delta U+P\Delta V$ so use $C_p=C_v+R$. A second misconception is that specific heats are fixed constants; in reality vibrational (and at low temperature rotational) modes freeze out, so the effective $C_v$ depends on temperature. 🔉⇢

Mean Free Path 🔉⇢

🎯 Between collisions a molecule flies in a straight line; the average of these free flights is the mean free path l = 1/(sqrt2 pi n d^2). It shrinks when the gas is denser (more targets, n) or the molecules are larger (bigger cross-section, d). Slide n and d and watch the zig-zag tighten as l drops.
🔉⇢
l = 1 / (√2 π n d2)   the mean distance a molecule travels between collisions: denser gas (n) or fatter molecules (d) → shorter free path.
What this shows

Between collisions a molecule flies in a straight line; the average of these free flights is the mean free path l = 1/(sqrt2 pi n d^2). It shrinks when the gas is denser (more targets, n) or the molecules are larger (bigger cross-section, d). Slide n and d and watch the zig-zag tighten as l drops.

Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.

Definition: The mean free path is the average distance a molecule travels between two successive collisions. Modelling molecules as spheres of diameter $d$ with number density $n$, a molecule sweeps a collision cross-section $\pi d^2$, giving $l=\dfrac{1}{\sqrt{2}\,n\pi d^2}$ once the motion of all molecules is accounted for. For air at STP, $l\sim10^{-7}\,\text{m}$, about a hundred times the interatomic spacing. 🔉⇢

Molecules in a gas have rather large speeds, of the order of the speed of sound, hundreds of metres per second at ordinary temperatures. Yet a gas leaking from a cylinder in a kitchen takes a considerable time to diffuse to the other corners of the room, and the top of a cloud of smoke can hold together for hours. If molecules truly flew in straight lines at their thermal speeds, such slowness would be inexplicable. The resolution is that molecules in a gas have a finite, though small, size, so they are bound to undergo collisions. As a result they cannot move straight unhindered; their paths keep getting incessantly deflected, and the molecule executes a random zig-zag rather than a straight dash across the room. 🔉⇢

Full derivation, worked example and interactive 3D on the Mean Free Path tab →

Kinetic Theory of an Ideal Gas — Pressure 🔉⇢deep concept

Definition: Modelling a gas as a large number of molecules in incessant random motion undergoing elastic collisions, the pressure on the walls arises from the momentum transferred by molecular impacts. Averaging over all molecules gives $P=\tfrac{1}{3}nm\langle v^2\rangle$, where $n$ is the number density, $m$ the molecular mass and $\langle v^2\rangle$ the mean square speed. Pressure is thus a statistical, bulk result of microscopic motion. 🔉⇢

🔬 Interactive 3D · Molecules bouncing elastically in a cube; one wall collision shows momentum change 2mv_x building pressure P=(1/3)nm⟨v²⟩.

The kinetic theory of gases is built entirely on the molecular picture of matter established earlier. A given amount of gas is regarded as a collection of a very large number of molecules — typically of the order of Avogadro's number — that are in incessant random motion. This single phrase, 'incessant random motion', encapsulates the whole model: the molecules never stop moving, and there is no preferred direction to their velocities. Everything measurable about the gas — its pressure, its temperature, its internal energy — will be shown to be an average over this microscopic motion. The aim of this section is the first great result of the theory: a formula for the pressure of a gas in terms of the mass, number density and mean square speed of its molecules. 🔉⇢

The model rests on a small number of clearly stated assumptions, and it is worth setting them out explicitly because every step of the derivation uses one of them. First, a gas consists of a very large number of identical molecules moving in random directions with a distribution of speeds. Second, at ordinary pressure and temperature the average distance between molecules is a factor of ten or more larger than the typical molecular size of about two angstroms, so the molecules are, for most of the time, far apart. Third, because they are so far apart, the interaction between molecules is negligible except during collisions, and between collisions each molecule moves freely in a straight line according to Newton's first law. 🔉⇢

The remaining assumptions concern the collisions themselves. The molecules collide incessantly, both against one another and against the walls of the container, and in each such collision their velocities change. These collisions are taken to be perfectly elastic, which has two immediate consequences: the total kinetic energy of the colliding molecules is conserved, and (as always in mechanics) the total momentum is conserved. A further idealisation is that the time spent during a collision is negligible compared with the time a molecule spends travelling freely between collisions, and that the molecules themselves are so small that their own volume is negligible compared with the volume of the container. These are exactly the conditions under which a real gas behaves ideally. 🔉⇢

It is important to appreciate that these assumptions are not arbitrary; they are the mechanical translation of the facts established in the section on the molecular nature of matter. The diluteness of the gas — average separation about ten times the molecular size — justifies neglecting both the molecular volume and the intermolecular forces between collisions. The observed dynamic equilibrium justifies the picture of incessant motion with a steady distribution of velocities. And the elasticity of the collisions is what keeps the gas from simply losing energy and settling to the floor. With these assumptions in place, the problem of finding the pressure becomes a straightforward, if careful, exercise in Newtonian mechanics applied to one wall of a container. 🔉⇢

To make the calculation concrete, consider a gas enclosed in a cube of side $l$, with the coordinate axes taken parallel to the sides of the cube. Focus on one molecule of mass $m$ with velocity components $(v_x,v_y,v_z)$ that strikes the wall parallel to the $yz$-plane, a wall of area $A=l^2$. Because the collision with the wall is elastic and the wall is smooth, the molecule rebounds with its speed unchanged: the $y$- and $z$-components of its velocity are unaffected, but the $x$-component simply reverses sign. So the velocity after the collision is $(-v_x,v_y,v_z)$. This clean reversal of only the perpendicular component is the microscopic heart of the pressure calculation. 🔉⇢

The change in the molecule's momentum in this single collision is therefore confined to the $x$-direction. Its $x$-momentum changes from $mv_x$ to $-mv_x$, a change of $-mv_x-(mv_x)=-2mv_x$. By the principle of conservation of momentum, whatever momentum the molecule loses the wall gains, so the momentum imparted to the wall in one collision is $2mv_x$. This quantity — twice the perpendicular momentum of the molecule — is the elementary packet of momentum that a single impact delivers to the wall. Pressure will emerge from adding up an enormous number of such packets per unit area per unit time, so the next task is to count how many molecules strike the wall in a given interval. 🔉⇢

To find the force, and hence the pressure, we need the rate at which momentum is delivered to the wall. Consider a small time interval $\Delta t$. A molecule with $x$-component of velocity $v_x$ can reach the wall in this interval only if it lies within a distance $v_x\Delta t$ of it; that is, only molecules inside the slab of volume $Av_x\Delta t$ adjacent to the wall are candidates to strike it. But of the molecules in that slab, on the average only half are moving towards the wall (the other half are moving away). If $n$ is the number of molecules per unit volume, the number striking the wall in time $\Delta t$ with this velocity is therefore $\tfrac{1}{2}nAv_x\Delta t$. 🔉⇢

The total momentum $Q$ transferred to the wall in time $\Delta t$ by this group of molecules is the momentum per collision times the number of collisions: $Q=(2mv_x)\times(\tfrac{1}{2}nAv_x\Delta t)=nmAv_x^2\Delta t$. The force on the wall is the rate of momentum transfer, $Q/\Delta t$, and the pressure is the force per unit area. Dividing by the area $A$ gives the pressure contributed by this group of molecules: $P=Q/(A\Delta t)=nmv_x^2$. Notice already that both the area $A$ and the time interval $\Delta t$ have cancelled out of the final expression — the pressure does not depend on the size of the patch of wall or on the interval chosen, as it should not. 🔉⇢

In reality, not all molecules share the same velocity; there is a whole distribution of velocities. The expression $P=nmv_x^2$ therefore represents only the pressure due to the group of molecules whose $x$-velocity is $v_x$, with $n$ standing for the number density of that particular group. To obtain the total pressure we must sum the contributions of all such groups, which amounts to replacing $v_x^2$ by its average over all molecules. Writing $\langle v_x^2\rangle$ (also written $\overline{v_x^2}$) for the average of the squared $x$-velocity, the total pressure becomes $P=nm\langle v_x^2\rangle$. This is the point at which the mechanical calculation for a single molecule turns into a statistical statement about the whole gas. 🔉⇢

Now the assumption of randomness is invoked. The gas is isotropic: there is no preferred direction for the molecular velocities inside the vessel, so on the average the motion is shared equally among the three axes. By this symmetry the averages of the squared components are equal, $\langle v_x^2\rangle=\langle v_y^2\rangle=\langle v_z^2\rangle$. Since the speed satisfies $v^2=v_x^2+v_y^2+v_z^2$, taking averages gives $\langle v^2\rangle=\langle v_x^2\rangle+\langle v_y^2\rangle+\langle v_z^2\rangle=3\langle v_x^2\rangle$, and therefore $\langle v_x^2\rangle=\tfrac{1}{3}\langle v^2\rangle$. This is the crucial step that removes the special role of the $x$-direction and expresses everything in terms of the mean square speed $\langle v^2\rangle$. 🔉⇢

Substituting this result gives the central formula of the section: $P=\tfrac{1}{3}nm\langle v^2\rangle$, where $n$ is the number density of molecules, $m$ is the mass of one molecule and $\langle v^2\rangle$ is the mean of the squared molecular speeds. This compact expression relates a directly measurable macroscopic quantity — the pressure — to microscopic molecular properties. The product $nm$ is just the mass density $\rho$ of the gas, so the formula can equally be written $P=\tfrac{1}{3}\rho\langle v^2\rangle$. It says that pressure is proportional both to how many molecules there are per unit volume and to how vigorously, on average, they are moving. 🔉⇢

The role played by $\langle v^2\rangle$, the mean square speed, deserves emphasis. Pressure depends not on the average velocity of the molecules — which is zero, since the motion is random and as many molecules move one way as the other — but on the average of the square of the speed, which is positive. Squaring removes the cancellation of directions and gives extra weight to the faster molecules. The square root of $\langle v^2\rangle$ is a natural measure of molecular speed called the root-mean-square speed, and it is this quantity, not the ordinary average of the velocity, that governs the pressure. Recognising that pressure is tied to $\langle v^2\rangle$ prepares the ground for the next card, where $\langle v^2\rangle$ is linked directly to the temperature. 🔉⇢

Several features of the derivation deserve comment. First, although we chose a cubical container, the shape of the vessel is actually immaterial: for a vessel of arbitrary shape one can always select a small planar element of wall and carry through exactly the same steps. This is consistent with Pascal's law, according to which the pressure in a gas in equilibrium is the same everywhere. The fact that the area $A$ and time $\Delta t$ dropped out of the answer is the mathematical reflection of this physical truth: pressure is a local, direction-independent property of the gas, not an artefact of the particular wall we chose to examine. 🔉⇢

Second, the derivation quietly ignored collisions between the molecules themselves; only collisions with the wall were counted. This can be justified qualitatively. Because the collisions are random and the gas is in a steady state, whenever one molecule with velocity $(v_x,v_y,v_z)$ is knocked into a different velocity by an intermolecular collision, there is always, on the average, another molecule that is knocked into the velocity $(v_x,v_y,v_z)$. The distribution of velocities therefore stays steady, and since the pressure formula involves only the average $\langle v_x^2\rangle$, the reshuffling of individual molecules among velocities does not affect the result — provided the collisions are not too frequent and the time spent in a collision is negligible compared with the time between collisions. 🔉⇢

The deepest lesson of this section is that pressure is a statistical, bulk result. No single molecule 'has' a pressure; pressure emerges only when the momentum delivered by an astronomical number of individual impacts is averaged over an area and over time. The steadiness we perceive as a constant pressure on the wall is really the smoothing-out of a fantastically rapid succession of tiny, discrete momentum kicks, made smooth by the sheer number of molecules involved (of order Avogadro's number). This is why kinetic theory is fundamentally a statistical theory: it connects the definite, reproducible macroscopic quantities of thermodynamics to averages over the chaotic microscopic motion of the molecules. 🔉⇢

Although the derivation uses only the mean square speed, it is important to remember that the molecules do not all move at the same speed. There is a distribution of molecular speeds: at any instant some molecules move slowly, some very fast and the rest in between, and the collisions constantly reshuffle individual speeds while leaving the distribution as a whole unchanged. The kinetic pressure formula deliberately averages over this distribution, replacing $v_x^2$ by its mean $\langle v_x^2\rangle$, so the final result depends only on the average and not on the detailed shape of the distribution. This is why a single number, the mean square speed, suffices to determine the pressure of the whole gas. 🔉⇢

The assumption that the collisions are elastic is doing essential work and deserves careful statement. An elastic collision is one in which the total kinetic energy is conserved, momentum being conserved as always. For a collision with the smooth wall this means the molecule rebounds with unchanged speed, only its perpendicular velocity component reversing; for a collision between two molecules it means the total kinetic energy of the pair is unchanged. If the collisions were inelastic, kinetic energy would steadily drain away into some internal form and the gas would cool and settle — contrary to the observed fact that a gas in an insulated container maintains its pressure and temperature indefinitely. Elasticity is thus what keeps the incessant molecular motion truly perpetual. 🔉⇢

The factor of one-half in the count of wall collisions repays a second look. Within the slab of thickness $v_x\Delta t$ next to the wall, the molecules with $x$-velocity of magnitude $v_x$ are moving along the $x$-axis in two senses: half toward the wall and half away from it. Only those heading toward the wall can strike it in the interval $\Delta t$. Hence the number striking is half the number in the slab, $\tfrac{1}{2}nAv_x\Delta t$, and not the full $nAv_x\Delta t$. Overlooking this factor is a common slip that produces a pressure twice too large; the factor is a direct consequence of the isotropy of the molecular motion, which shares the molecules equally between the two directions along each axis. 🔉⇢

The independence of the result from the shape of the vessel is more than a mathematical convenience; it is demanded by Pascal's law. In a gas in equilibrium the pressure is the same at every point and in every direction, so any small planar element of the wall, wherever it sits and however it is oriented, must experience the same pressure. The derivation reflects this exactly: the area $A$ and the interval $\Delta t$ both cancelled, leaving $P=\tfrac{1}{3}nm\langle v^2\rangle$ with no trace of the geometry. For a container of arbitrary shape one simply applies the argument to a small planar patch and obtains the identical formula, confirming that pressure is an intrinsic property of the gas and not of the box that holds it. 🔉⇢

The neglect of intermolecular collisions in the derivation might seem troubling, since molecules collide with one another far more often than with the walls. The justification rests on the steady state. Because the gas is in dynamic equilibrium, the velocity distribution does not change with time: whenever a collision knocks a molecule out of the velocity group $(v_x,v_y,v_z)$, some other collision, on the average, knocks a different molecule into that same group. The population of each velocity group is therefore maintained, and since the pressure depends only on the average $\langle v_x^2\rangle$ over the steady distribution, the constant reshuffling of individual molecules among groups leaves the pressure untouched — provided the collisions are neither too frequent nor too long-lasting compared with the free flights between them. 🔉⇢

It is worth translating the formula into the language of measurable quantities. The number density $n$ is the number of molecules per unit volume, so the product $nm$ is the mass per unit volume, that is the mass density $\rho$ of the gas. The pressure formula can therefore be written equivalently as $P=\tfrac{1}{3}\rho\langle v^2\rangle$, which is convenient because $\rho$ is directly measurable whereas $n$ and $m$ separately are not. Rearranged, this form gives $\langle v^2\rangle=3P/\rho$, so the mean square molecular speed of a gas can be found from nothing more than its measured pressure and density — a remarkable window onto the microscopic world opened by purely macroscopic measurements. 🔉⇢

The most striking feature of this derivation is that it uses nothing beyond Newtonian mechanics and simple averaging. The reversal of a velocity component in an elastic bounce, the conservation of momentum, the counting of molecules in a slab — these are the tools of ordinary mechanics, applied to one representative molecule and then summed over an astronomical number of them. No new force law and no thermodynamic postulate is required to obtain the pressure; the macroscopic quantity emerges entirely from the mechanics of the microscopic constituents. This is the sense in which kinetic theory explains pressure rather than merely describing it, and it is the template for the kinetic interpretation of temperature that follows in the next card. 🔉⇢

In summary, starting from the assumptions that a gas is a large number of molecules in incessant random motion, dilute enough that they interact only through negligibly brief elastic collisions, we followed a single molecule's elastic bounce off a wall (momentum transfer $2mv_x$), counted the impacts in a slab of volume $\tfrac{1}{2}nAv_x\Delta t$, formed the pressure $P=nm\langle v_x^2\rangle$, and used isotropy ($\langle v_x^2\rangle=\tfrac{1}{3}\langle v^2\rangle$) to arrive at $P=\tfrac{1}{3}nm\langle v^2\rangle=\tfrac{1}{3}\rho\langle v^2\rangle$. This formula, obtained purely from mechanics and averaging, is the bridge to the kinetic interpretation of temperature developed in the next card, where $\langle v^2\rangle$ is identified with the absolute temperature of the gas. 🔉⇢

Derivation from first principles 🔉⇢

  1. Setup: a molecule of mass $m$, velocity $(v_x,v_y,v_z)$, in a cube of side $l$, hits the wall of area $A=l^2$ parallel to the $yz$-plane. The collision is elastic, so it rebounds as $(-v_x,v_y,v_z)$.
  2. Momentum change of the molecule $=-mv_x-(mv_x)=-2mv_x$; by conservation of momentum the wall receives $2mv_x$ per impact.
  3. In time $\Delta t$, molecules within $v_x\Delta t$ of the wall can strike it; half move toward it, so the number hitting is $\tfrac{1}{2}nAv_x\Delta t$ ($n=$ number density).
  4. Momentum delivered: $Q=(2mv_x)(\tfrac{1}{2}nAv_x\Delta t)=nmAv_x^2\Delta t$. Pressure $P=Q/(A\Delta t)=nmv_x^2$; averaging over the velocity distribution, $P=nm\langle v_x^2\rangle$.
  5. Isotropy: $\langle v_x^2\rangle=\langle v_y^2\rangle=\langle v_z^2\rangle$ and $v^2=v_x^2+v_y^2+v_z^2\Rightarrow\langle v_x^2\rangle=\tfrac{1}{3}\langle v^2\rangle$.
  6. Therefore $P=\tfrac{1}{3}nm\langle v^2\rangle=\tfrac{1}{3}\rho\langle v^2\rangle$. Both $A$ and $\Delta t$ cancel, consistent with Pascal's law that pressure is the same throughout the gas.
⚠️ JEE trap: A frequent mistake is to think the pressure depends on the average molecular velocity. The average velocity of gas molecules is zero, because the motion is random and equally likely in every direction; pressure depends on the mean square speed $\langle v^2\rangle$, which is positive. Another error is to picture the pressure as coming from molecules 'pushing' steadily on the wall — in fact it is the time-averaged rate of momentum transfer from an enormous number of discrete elastic impacts, each delivering $2mv_x$. Pressure is a statistical bulk quantity; no individual molecule possesses a pressure. 🔉⇢

Worked example · JEE Advanced 🔉⇢

SITUATION Nitrogen gas ($\text{N}_2$, molar mass $28\,\text{g mol}^{-1}$) is held at temperature $T=300\,\text{K}$. The mass of one molecule is $m=M/N_A=28\times10^{-3}/(6.02\times10^{23})\approx4.65\times10^{-26}\,\text{kg}$.
TARGET Estimate the root-mean-square speed of a nitrogen molecule, illustrating how $P=\tfrac{1}{3}nm\langle v^2\rangle$ ties pressure to molecular motion.
STRATEGY Combine the pressure result $P=\tfrac{1}{3}nm\langle v^2\rangle$ with the ideal-gas law $P=nk_BT$ to eliminate $P$ and $n$, giving $\langle v^2\rangle=3k_BT/m$; then take the square root.
EXECUTE $\tfrac{1}{3}nm\langle v^2\rangle=nk_BT\Rightarrow\langle v^2\rangle=3k_BT/m=3(1.38\times10^{-23})(300)/(4.65\times10^{-26})\approx(516)^2\,\text{m}^2\text{s}^{-2}$. Hence $v_{\text{rms}}=\sqrt{\langle v^2\rangle}\approx516\,\text{m s}^{-1}$.
REFLECT The rms speed is of the order of the speed of sound in air, which is reasonable since sound propagates through molecular motion. The pressure formula thus yields not just $P$ but, together with the gas law, the actual scale of molecular speeds.

Source: NCERT Class XI Ch 12 Kinetic Theory

Kinetic Interpretation of Temperature 🔉⇢deep concept

Definition: Combining the kinetic pressure $P=\tfrac{1}{3}nm\langle v^2\rangle$ with the ideal-gas law gives $\tfrac{1}{2}m\langle v^2\rangle=\tfrac{3}{2}k_BT$: the average translational kinetic energy of a molecule is proportional to the absolute temperature and independent of pressure, volume or the nature of the gas. Hence the root-mean-square speed is $v_{\text{rms}}=\sqrt{3k_BT/m}=\sqrt{3RT/M}$. 🔉⇢

🔬 Interactive 3D · Molecules colored by speed; temperature and mass sliders drive v_rms=√(3kT/m); lighter molecules move faster.

The pressure formula derived in the previous card, $P=\tfrac{1}{3}nm\langle v^2\rangle$, is a purely mechanical result: it contains no reference to temperature at all. The great achievement of this section is to bring temperature into the molecular picture, and thereby to give a molecular meaning to a quantity that thermodynamics had treated as primitive. The route is short but profound — we simply compare the kinetic pressure formula with the experimentally established ideal-gas law — and the result, that the average kinetic energy of a molecule is proportional to the absolute temperature, is one of the most important single equations in all of physics. 🔉⇢

Begin by rewriting the pressure result in terms of the total number of molecules. Multiplying $P=\tfrac{1}{3}nm\langle v^2\rangle$ by the volume $V$ and using $n=N/V$, we get $PV=\tfrac{1}{3}Nm\langle v^2\rangle$. This can be rearranged to display the kinetic energy explicitly: $PV=\tfrac{2}{3}N\left(\tfrac{1}{2}m\langle v^2\rangle\right)$. The quantity in the bracket, $\tfrac{1}{2}m\langle v^2\rangle$, is the average translational kinetic energy of a single molecule. So the product $PV$ is directly proportional to the total translational kinetic energy of all the molecules in the sample — a first hint that pressure and volume are really about molecular energy. 🔉⇢

For an ideal gas the internal energy is purely kinetic — the molecules are treated as point masses with no potential energy of interaction between collisions — so the total internal energy of translation is $E=N\times\tfrac{1}{2}m\langle v^2\rangle$. With this, the previous relation becomes simply $PV=\tfrac{2}{3}E$. This is a striking statement in its own right: it says that the product of pressure and volume equals two-thirds of the total translational kinetic energy of the gas. We now have two independent expressions for the same $PV$ — one from mechanics, $PV=\tfrac{2}{3}E$, and one from experiment, the ideal-gas law $PV=Nk_BT$ — and equating them is the decisive step. 🔉⇢

Setting the mechanical result $PV=\tfrac{2}{3}E$ equal to the empirical ideal-gas law $PV=Nk_BT$ gives $\tfrac{2}{3}E=Nk_BT$, that is $E=\tfrac{3}{2}Nk_BT$. Dividing through by the number of molecules $N$, the average translational kinetic energy per molecule is $E/N=\tfrac{1}{2}m\langle v^2\rangle=\tfrac{3}{2}k_BT$. This is the kinetic interpretation of temperature. In words: the average kinetic energy of a molecule is proportional to the absolute temperature of the gas. Temperature, which thermodynamics introduced as an abstract label of hotness, is revealed to be nothing other than a measure of the average energy of molecular motion. 🔉⇢

Every feature of this result rewards close reading. The average kinetic energy $\tfrac{3}{2}k_BT$ depends only on the temperature $T$; it is independent of the pressure, the volume, and — most remarkably — the nature of the ideal gas. At a given temperature a molecule of hydrogen and a molecule of carbon dioxide have exactly the same average translational kinetic energy, even though their masses differ enormously. This is a fundamental result relating a macroscopic, measurable thermodynamic variable, the temperature, to a molecular quantity, the average kinetic energy of a molecule; the two domains are connected by the single universal constant $k_B$, the Boltzmann constant. It is hard to overstate how much of physics rests on this one bridge. 🔉⇢

The relation $E=\tfrac{3}{2}Nk_BT$ also settles the question of what the internal energy of an ideal gas depends on. Because the right-hand side contains only $N$ and $T$, the internal energy of an ideal gas depends only on its temperature, not on its pressure or volume. This is a fact we shall use repeatedly in thermodynamics — for instance, it is why the internal energy of an ideal gas does not change during an isothermal process, however much the gas expands or is compressed. With this interpretation of temperature in hand, kinetic theory is seen to be completely consistent with the ideal-gas equation and with all the gas laws (Boyle, Charles, Avogadro, Dalton) derived from it. 🔉⇢

It is illuminating to see how Dalton's law of partial pressures re-emerges from this energy picture. For a mixture of non-reacting ideal gases the total pressure is $P=\tfrac{1}{3}\left[n_1m_1\langle v_1^2\rangle+n_2m_2\langle v_2^2\rangle+\dots\right]$. In equilibrium the average kinetic energy of the molecules of every species is the same, $\tfrac{1}{2}m_1\langle v_1^2\rangle=\tfrac{1}{2}m_2\langle v_2^2\rangle=\tfrac{3}{2}k_BT$, because average kinetic energy depends only on temperature. Substituting, each term becomes $n_ik_BT$, so $P=(n_1+n_2+\dots)k_BT$ — which is exactly Dalton's law of partial pressures, now derived from the kinetic interpretation of temperature rather than assumed. 🔉⇢

The result immediately gives the typical speed of molecules in a gas. From $\tfrac{1}{2}m\langle v^2\rangle=\tfrac{3}{2}k_BT$ we obtain $\langle v^2\rangle=3k_BT/m$. The square root of the mean square speed is called the root-mean-square speed, written $v_{\text{rms}}=\sqrt{\langle v^2\rangle}$; it is the natural measure of how fast the molecules move. Thus $v_{\text{rms}}=\sqrt{3k_BT/m}$. Since the molecular mass is $m=M/N_A$ and $R=N_A k_B$, this can be rewritten in molar quantities as $v_{\text{rms}}=\sqrt{3RT/M}$, where $M$ is the molar mass. The two forms are equivalent; which one to use depends simply on whether the data are given per molecule or per mole. 🔉⇢

A concrete number fixes the scale. For nitrogen at $T=300\,\text{K}$, with molecular mass $m=28/(6.02\times10^{26})\approx4.65\times10^{-26}\,\text{kg}$ (mass in kilograms per molecule), the mean square speed is $\langle v^2\rangle=3k_BT/m\approx(516)^2\,\text{m}^2\text{s}^{-2}$, so the root-mean-square speed is $v_{\text{rms}}\approx516\,\text{m s}^{-1}$. This is of the order of the speed of sound in air — which makes physical sense, since sound is a disturbance carried by the molecular motion, and cannot travel much faster than the molecules themselves. That such an everyday quantity as the speed of sound emerges from the kinetic formula is a satisfying check on the whole theory. 🔉⇢

The formula $v_{\text{rms}}=\sqrt{3k_BT/m}=\sqrt{3RT/M}$ shows two dependences that are constantly tested. At a fixed temperature, the rms speed varies inversely as the square root of the molecular mass, so lighter molecules move faster than heavier ones. This is why, at the same temperature, hydrogen molecules (molar mass 2) move much faster than oxygen molecules (molar mass 32) — by a factor of $\sqrt{32/2}=4$. The comparison of rms speeds of different gases at the same temperature is therefore governed entirely by the inverse-square-root-of-mass rule, while at fixed mass the rms speed grows as the square root of the absolute temperature. 🔉⇢

The temperature dependence deserves a separate emphasis. Because $\langle v^2\rangle=3k_BT/m$, the mean square speed is directly proportional to the absolute temperature, and the rms speed grows as $\sqrt{T}$. To double the rms speed of the molecules in a gas one must raise the absolute temperature by a factor of four. This square-root behaviour is a favourite source of examination questions, and it also underlies practical facts such as the increase in the speed of sound and in reaction rates with rising temperature. It is the direct, quantitative statement of the intuitive idea that heating a gas makes its molecules move faster. 🔉⇢

The kinetic interpretation of temperature gives a beautifully simple meaning to absolute zero. Since the average kinetic energy of a molecule is $\tfrac{3}{2}k_BT$, it decreases steadily as the temperature is lowered, and it would fall to zero at $T=0$ on the kelvin scale. Absolute zero is therefore the temperature at which (in this classical picture) all translational molecular motion ceases and the kinetic energy vanishes. This explains at once why the kelvin scale, with its zero at $-273.15^\circ\text{C}$, is the natural scale for a gas: it is the scale on which molecular kinetic energy is directly proportional to temperature, with no arbitrary offset. Negative absolute temperatures are impossible in this picture because kinetic energy cannot be negative. 🔉⇢

The picture also explains diffusion — the gradual spreading and intermingling of gases — and, quantitatively, Graham's law. Since lighter molecules have a larger rms speed at a given temperature ($v_{\text{rms}}\propto1/\sqrt{M}$), they wander across a region faster, and the rate at which a gas diffuses is proportional to its molecular speed. Graham's law of diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass (or density): $r\propto1/\sqrt{M}$. So hydrogen diffuses about four times faster than oxygen, in the same ratio as their rms speeds. Graham's law is thus a direct consequence of the kinetic interpretation of temperature, and it provides a laboratory method of comparing molecular masses. 🔉⇢

It is worth clarifying the scope of the phrase 'kinetic energy' in this section. The energy $\tfrac{3}{2}k_BT$ per molecule is the average translational kinetic energy — the energy associated with the motion of the molecule's centre of mass through space. Real molecules can also rotate and vibrate, and these internal motions carry additional energy that is not counted in $\tfrac{3}{2}k_BT$. For a monatomic gas, whose molecules are effectively point masses, translation is the whole story and the internal energy per mole is $\tfrac{3}{2}RT$. For diatomic and polyatomic gases the extra rotational and vibrational degrees of freedom add to the internal energy; this is handled by the law of equipartition of energy, which is studied in the next section of the chapter and which explains the specific heat capacities of gases. 🔉⇢

The kinetic interpretation of temperature is the conceptual summit of the chapter, so it is worth stating clearly what has and has not been shown. We have shown that for an ideal gas the absolute temperature is a direct measure of the average translational kinetic energy of its molecules, $\tfrac{1}{2}m\langle v^2\rangle=\tfrac{3}{2}k_BT$, independent of pressure, volume and the identity of the gas, with $k_B$ as the universal conversion factor. This single equation unifies the thermodynamic and molecular pictures: it turns 'hotness' into 'molecular energy', explains why all ideal gases share the same average energy at a given temperature, and yields the rms speed, the meaning of absolute zero, and Graham's law of diffusion as immediate corollaries. 🔉⇢

The energy $\tfrac{3}{2}k_BT$ per molecule is specifically the translational kinetic energy — the energy of motion of the molecule as a whole through space. For a monatomic gas, whose atoms can be treated as structureless points, this translational energy is the entire internal energy, so one mole has internal energy $\tfrac{3}{2}RT$ and the gas has no other way to store thermal energy. This is the simplest instance of the deeper law of equipartition of energy, which assigns an average energy of $\tfrac{1}{2}k_BT$ to each independent quadratic degree of freedom; a point atom has three translational degrees of freedom, giving $3\times\tfrac{1}{2}k_BT=\tfrac{3}{2}k_BT$. Diatomic and polyatomic molecules have additional rotational (and, at high temperature, vibrational) degrees of freedom, and so larger internal energies and specific heats — the subject of the next section of the chapter. 🔉⇢

The fact that the average translational kinetic energy is $\tfrac{3}{2}k_BT$ for every ideal gas, whatever its mass or chemical nature, is worth dwelling on, because it is so counter-intuitive. Place hydrogen and carbon dioxide in the same vessel at the same temperature; a carbon-dioxide molecule is about twenty-two times as massive as a hydrogen molecule, yet the two have exactly the same average kinetic energy. The heavier molecule compensates for its greater mass by moving more slowly, so that $\tfrac{1}{2}m\langle v^2\rangle$ comes out the same for both. Temperature, in the kinetic view, is the great equaliser: it fixes the average energy per molecule, and the molecules adjust their speeds to their masses in order to comply. 🔉⇢

It is important to distinguish the several 'speeds' that arise. The ordinary average of the velocity vector is zero, because the motion is random and every direction is equally likely; this is why the velocity itself never appears in the pressure or the temperature. The meaningful measure is the root-mean-square speed, $v_{\text{rms}}=\sqrt{\langle v^2\rangle}=\sqrt{3k_BT/m}$, obtained by squaring the speeds first (which removes the cancellation of directions), then averaging, and finally taking the square root. It is the rms speed, not the vanishing average velocity, that carries the energy and the pressure, and it is the quantity quoted whenever one speaks of the speed of the molecules in a gas. 🔉⇢

A few numbers make the mass dependence vivid. At 300 K the rms speed of nitrogen (molar mass 28) is about $516\,\text{m s}^{-1}$; hydrogen (molar mass 2), being fourteen times lighter, moves $\sqrt{14}\approx3.7$ times faster, at roughly $1900\,\text{m s}^{-1}$; oxygen (molar mass 32) moves a little slower than nitrogen, and heavy gases such as carbon dioxide slower still. Every one of these speeds is comparable to, or greater than, the speed of sound in the respective gas, which is no coincidence: sound is a pressure wave carried by the molecular motion, so it cannot outrun the molecules that carry it. These order-of-magnitude figures are worth remembering as sanity checks in problems. 🔉⇢

The re-derivation of Dalton's law from the energy picture also illuminates why different gases in a mixture share a common temperature. When gases are mixed, collisions between unlike molecules exchange energy until the average kinetic energy per molecule is the same for every species — this common value is precisely what $\tfrac{3}{2}k_BT$ measures, and it is the statement that the mixture has a single temperature. Only once this equality of average kinetic energy among the species is reached does each gas contribute a partial pressure $n_ik_BT$, and the total pressure become the simple sum $(n_1+n_2+\dots)k_BT$. Thermal equilibrium in a mixture is thus the equalisation of average molecular kinetic energy, not of speed and not of pressure. 🔉⇢

The kinetic meaning of absolute zero also clarifies why negative kelvin temperatures are meaningless in this picture. Since kinetic energy is a sum of squares and cannot be negative, the average energy $\tfrac{3}{2}k_BT$ cannot fall below zero, so $T$ cannot be negative on the kelvin scale. As $T$ is lowered toward zero the molecular motion becomes ever more sluggish, but removing the last increments of energy becomes progressively harder — a foreshadowing of the third law of thermodynamics. The classical picture, in which all translational motion ceases at $T=0$, is modified by quantum mechanics (which leaves a residual zero-point energy), but for the ideal-gas discussion of this chapter the identification of absolute zero with the vanishing of molecular kinetic energy is the essential idea. 🔉⇢

Graham's law of diffusion follows so directly from the temperature interpretation that it serves as an experimental test of the theory. The rate at which a gas diffuses through a fine opening, or spreads through another gas, is set by how fast its molecules move, and since $v_{\text{rms}}\propto1/\sqrt{M}$ at a given temperature, the diffusion rate is inversely proportional to the square root of the molar mass: $r\propto1/\sqrt{M}$. Two gases compared under identical conditions diffuse in the inverse ratio of the square roots of their molar masses, so hydrogen ($M=2$) diffuses $\sqrt{32/2}=4$ times faster than oxygen ($M=32$). This is exactly how a lighter isotope can be separated from a heavier one by repeated diffusion, and it provides a laboratory route to relative molecular masses. 🔉⇢

Finally, the kinetic interpretation gives the deepest possible answer to the question of what a thermometer measures. A thermometer brought into contact with a gas comes to thermal equilibrium with it, meaning its own molecules acquire the same average kinetic energy as the gas molecules; the reading it displays is, at root, a measure of that shared molecular kinetic energy. Temperature is therefore not an arbitrary human construct but a direct physical property of the molecular motion, common to any two systems in thermal equilibrium. This closes the conceptual circle of the chapter: the macroscopic temperature of thermodynamics and the microscopic kinetic energy of the molecules are one and the same thing, joined by the Boltzmann constant $k_B$. 🔉⇢

The kinetic interpretation also makes the various gas laws transparent as consequences of molecular energy. Boyle's law, $PV=\text{constant}$ at fixed temperature, is simply the statement $PV=\tfrac{2}{3}E$ with $E$ fixed, since the average kinetic energy — and hence the total translational energy at fixed $N$ — depends only on temperature. Charles' law, $V\propto T$ at fixed pressure, follows because raising the temperature raises the average molecular kinetic energy and therefore the rate of momentum transfer, so the gas must expand to keep the pressure constant. Avogadro's hypothesis is recovered because equal average kinetic energies at equal temperatures, combined with equal pressures and volumes, force equal numbers of molecules. Every empirical gas law is thus seen to be a shadow of the one molecular fact $\tfrac{1}{2}m\langle v^2\rangle=\tfrac{3}{2}k_BT$. 🔉⇢

To collect the section: rewriting the kinetic pressure as $PV=\tfrac{2}{3}E$ and equating it with the ideal-gas law $PV=Nk_BT$ yields $\tfrac{1}{2}m\langle v^2\rangle=\tfrac{3}{2}k_BT$, the statement that average molecular kinetic energy is proportional to absolute temperature. From it follow the internal energy of an ideal gas ($E=\tfrac{3}{2}Nk_BT$, a function of $T$ alone), the root-mean-square speed $v_{\text{rms}}=\sqrt{3k_BT/m}=\sqrt{3RT/M}$ with lighter molecules moving faster, the interpretation of absolute zero as the vanishing of molecular kinetic energy, and Graham's law of diffusion $r\propto1/\sqrt{M}$. This completes the molecular account of pressure and temperature that the kinetic theory of an ideal gas set out to provide. 🔉⇢

Derivation from first principles 🔉⇢

  1. Start from the pressure result $P=\tfrac{1}{3}nm\langle v^2\rangle$; multiply by $V$ and use $n=N/V$: $PV=\tfrac{1}{3}Nm\langle v^2\rangle=\tfrac{2}{3}N\left(\tfrac{1}{2}m\langle v^2\rangle\right)$.
  2. For an ideal gas the internal energy is purely translational kinetic: $E=N\cdot\tfrac{1}{2}m\langle v^2\rangle$, so $PV=\tfrac{2}{3}E$.
  3. Equate with the ideal-gas law $PV=Nk_BT$: $\tfrac{2}{3}E=Nk_BT\Rightarrow E=\tfrac{3}{2}Nk_BT$.
  4. Per molecule: $E/N=\tfrac{1}{2}m\langle v^2\rangle=\tfrac{3}{2}k_BT$ — average KE $\propto$ absolute temperature, independent of $P$, $V$ and the nature of the gas.
  5. Solve for the mean square speed: $\langle v^2\rangle=3k_BT/m$; hence $v_{\text{rms}}=\sqrt{3k_BT/m}=\sqrt{3RT/M}$, using $m=M/N_A$ and $R=N_A k_B$.
  6. Corollaries: $v_{\text{rms}}\propto1/\sqrt{M}$ (lighter molecules faster) and $\propto\sqrt{T}$; Graham's law $r\propto1/\sqrt{M}$; and $E\to0$ as $T\to0$ gives the kinetic meaning of absolute zero.
⚠️ JEE trap: Students often think heavier molecules move faster because they 'have more energy'. At a given temperature all ideal-gas molecules share the same average kinetic energy $\tfrac{3}{2}k_BT$, so $\tfrac{1}{2}m\langle v^2\rangle$ is equal for all species; since the energy is fixed, a larger mass $m$ means a smaller $\langle v^2\rangle$, i.e. lighter molecules move faster ($v_{\text{rms}}\propto 1/\sqrt{M}$). A second error is using Celsius in $v_{\text{rms}}=\sqrt{3RT/M}$: only the absolute (kelvin) temperature is proportional to kinetic energy, so $T$ must be in kelvin. 🔉⇢

Worked example · JEE Advanced 🔉⇢

SITUATION A flask contains argon (atomic mass $39.9\,\text{u}$) and chlorine (molecular mass $70.9\,\text{u}$) at a common temperature of $27^\circ\text{C}$.
TARGET Find the ratio of (i) the average kinetic energy per molecule and (ii) the root-mean-square speed $v_{\text{rms}}$ of the two gases.
STRATEGY Use $\tfrac{1}{2}m\langle v^2\rangle=\tfrac{3}{2}k_BT$: average KE depends only on $T$, so at a common temperature it is equal for both gases; the rms-speed ratio then comes from $v_{\text{rms}}\propto1/\sqrt{m}$.
EXECUTE (i) Both gases are at the same temperature, so the average kinetic energy per molecule is $\tfrac{3}{2}k_BT$ for each; the ratio is $1:1$. (ii) Since $\tfrac{1}{2}m v_{\text{rms}}^2$ is equal for both, $v_{\text{rms,Ar}}/v_{\text{rms,Cl}}=\sqrt{m_{\text{Cl}}/m_{\text{Ar}}}=\sqrt{70.9/39.9}\approx1.33$.
REFLECT Equal temperature forces equal average kinetic energy regardless of the gas, but the lighter argon atoms move faster, illustrating $v_{\text{rms}}\propto1/\sqrt{M}$ — the same rule behind Graham's law of diffusion.

Source: NCERT Class XI Ch 12 Kinetic Theory

Mean Free Path 🔉⇢deep concept

Definition: The mean free path is the average distance a molecule travels between two successive collisions. Modelling molecules as spheres of diameter $d$ with number density $n$, a molecule sweeps a collision cross-section $\pi d^2$, giving $l=\dfrac{1}{\sqrt{2}\,n\pi d^2}$ once the motion of all molecules is accounted for. For air at STP, $l\sim10^{-7}\,\text{m}$, about a hundred times the interatomic spacing. 🔉⇢

🔬 Interactive 3D · One molecule traces a zig-zag path colliding with others; n and d sliders drive l=1/(√2 nπd²) and collision frequency.

Molecules in a gas have rather large speeds, of the order of the speed of sound, hundreds of metres per second at ordinary temperatures. Yet a gas leaking from a cylinder in a kitchen takes a considerable time to diffuse to the other corners of the room, and the top of a cloud of smoke can hold together for hours. If molecules truly flew in straight lines at their thermal speeds, such slowness would be inexplicable. The resolution is that molecules in a gas have a finite, though small, size, so they are bound to undergo collisions. As a result they cannot move straight unhindered; their paths keep getting incessantly deflected, and the molecule executes a random zig-zag rather than a straight dash across the room. 🔉⇢

The concept that captures this picture quantitatively is the mean free path. Between one collision and the next a molecule travels freely in a straight line; the length of that free flight varies from one interval to the next, being sometimes long and sometimes short. The average distance between two successive collisions, called the mean free path $l$, is the natural measure of how far a molecule gets before its direction is randomised. Together with the average time between collisions and the collision frequency, the mean free path is the key microscopic length that connects the size of molecules to the bulk transport properties of a gas, such as diffusion, viscosity and thermal conduction. 🔉⇢

To derive it we model the molecules of a gas as hard spheres of diameter $d$. Focus attention on a single molecule moving with the average speed $\langle v\rangle$, and imagine, as a first simplification, that all the other molecules are held at rest. Our chosen molecule will suffer a collision with any other molecule whose centre comes within a distance $d$ of its own centre, because two spheres of diameter $d$ touch when their centres are a distance $d$ apart. The moving molecule therefore behaves as though it carries a circular target of radius $d$ around its centre, and any molecule whose centre lies within that reach will be struck. 🔉⇢

This target is the collision cross-section. As the molecule advances, it sweeps out a cylinder whose circular cross-section has radius $d$ and hence area $\pi d^2$. In a time interval $\Delta t$ the molecule moves a distance $\langle v\rangle\Delta t$, so it sweeps a volume $\pi d^2\langle v\rangle\Delta t$. Any other molecule whose centre lies inside this swept volume will collide with our molecule during the interval. The quantity $\pi d^2$ is called the collision cross-section; it is the effective area a molecule presents for collisions, and it grows as the square of the molecular diameter, so even a modest increase in molecular size sharply raises the collision rate. 🔉⇢

Now bring in the number density. If $n$ is the number of molecules per unit volume, then the number of molecular centres lying inside the swept volume is $n$ times that volume, namely $n\pi d^2\langle v\rangle\Delta t$. That is the number of collisions the molecule suffers in time $\Delta t$. Dividing by $\Delta t$, the rate of collisions, or collision frequency, is $\nu=n\pi d^2\langle v\rangle$. Its reciprocal is the average time between two successive collisions, $\tau=\dfrac{1}{n\pi\langle v\rangle d^2}$. This collision time is extremely short at ordinary densities, of the order of a nanosecond for air, which is why a gas comes to internal equilibrium so quickly after a disturbance. 🔉⇢

The mean free path follows at once as the distance travelled in the collision time. Since the molecule moves at average speed $\langle v\rangle$ for a time $\tau$ between collisions, the average distance between two successive collisions is $l=\langle v\rangle\tau=\dfrac{\langle v\rangle}{n\pi\langle v\rangle d^2}=\dfrac{1}{n\pi d^2}$. A remarkable feature emerges immediately: the average speed $\langle v\rangle$ has cancelled out, so in this simplified treatment the mean free path does not depend on how fast the molecules move but only on how many there are and how big they are. It is set purely by the number density and the collision cross-section. 🔉⇢

This first estimate contains a hidden approximation, however. We pretended that all the other molecules stood still while our chosen molecule moved. In reality every molecule is moving, and what governs the collision rate is not the speed of one molecule but the average relative speed between colliding pairs. A more careful treatment that averages over the Maxwell distribution of velocities shows that the relevant speed is larger than $\langle v\rangle$ by a factor of $\sqrt{2}$. Replacing $\langle v\rangle$ by the average relative speed introduces this factor into the denominator, and the corrected, exact result for the mean free path is $l=\dfrac{1}{\sqrt{2}\,n\pi d^2}$. 🔉⇢

This is the standard formula for the mean free path, and its structure repays study. The mean free path is inversely proportional to the number density $n$: pack the molecules more tightly and they collide sooner, so $l$ shrinks. It is also inversely proportional to the square of the molecular diameter $d$, through the cross-section $\pi d^2$: larger molecules present bigger targets and collide more often, again reducing $l$. The factor $\sqrt{2}$ is a fixed numerical correction from the relative-motion averaging. Nothing else enters, which is why a single measurement of the mean free path, combined with a known number density, can be used to estimate molecular sizes. 🔉⇢

Let us put numbers to the formula for air at standard temperature and pressure. The average molecular speed is about $\langle v\rangle\approx 485\,\text{m s}^{-1}$. The number density follows from the fact that one mole occupies $22.4\,\text{litres}$ at STP, giving $n=\dfrac{6.02\times10^{23}}{22.4\times10^{-3}}\approx 2.7\times10^{25}\,\text{m}^{-3}$. Taking a molecular diameter $d\approx 2\times10^{-10}\,\text{m}$, the collision time comes out to $\tau\approx 6.1\times10^{-10}\,\text{s}$, less than a nanosecond, and the mean free path is $l\approx 2.9\times10^{-7}\,\text{m}$. So between collisions an air molecule travels only about three ten-millionths of a metre before its direction is scrambled. 🔉⇢

It is illuminating to compare this length with the spacing between molecules. The mean free path of about $2.9\times10^{-7}\,\text{m}$ works out to roughly $1500$ molecular diameters, and it is on the order of a hundred times the average interatomic distance in the gas. In other words a molecule typically flies past many neighbours, threading a long path through the crowd, before it actually strikes one. It is precisely this large ratio of mean free path to molecular size that gives a gas its characteristic behaviour: the molecules are, most of the time, far from one another and moving freely, which is exactly the assumption underlying the ideal gas model. 🔉⇢

The dependence on number density has a striking practical consequence in rarefied gases. In a highly evacuated tube the number density $n$ is very small, so the mean free path grows correspondingly large and can become as long as the length of the tube itself. When that happens, molecules travel from one wall to the other without colliding with each other at all; collisions are then almost entirely with the walls rather than between molecules. This regime, in which $l$ exceeds the dimensions of the container, underlies the operation of vacuum tubes, thermos flasks and many high-vacuum technologies, and it is a direct and testable prediction of the mean-free-path formula. 🔉⇢

The temperature and pressure dependence follows from how $n$ responds to conditions. At fixed temperature, the ideal gas law makes $n$ proportional to pressure, so the mean free path is inversely proportional to pressure: halve the pressure and you double the mean free path. At fixed pressure, $n$ is inversely proportional to absolute temperature, so heating a gas at constant pressure increases $l$. As a worked illustration, water vapour at $373\,\text{K}$ has, taking the same molecular diameter as air, a number density reduced from the STP value in the ratio $273/373$, giving $n\approx 2\times10^{25}\,\text{m}^{-3}$ and a mean free path of about $4\times10^{-7}\,\text{m}$, again roughly a hundred times the interatomic distance. 🔉⇢

The mean free path is far more than a curiosity of molecular geometry; it is the master variable of gas transport phenomena. Using the kinetic theory of gases, the bulk measurable properties like viscosity, thermal conductivity and diffusion can all be related to microscopic parameters such as the molecular size, and in every case the mean free path appears as the length over which molecules carry momentum, energy or their own identity before a collision randomises them. It is through such relations that molecular sizes were first estimated historically, long before molecules could be observed by any direct means, a triumph of the kinetic theory. 🔉⇢

Take diffusion first. A molecule wanders through the gas in a random walk, each step of average length $l$ ending in a collision that redirects it. Because the net displacement in a random walk grows only as the square root of the number of steps, a molecule spreads slowly even though each step is taken at high thermal speed. This is exactly why the kitchen gas takes so long to reach the far corner: it is not travelling in a straight line but diffusing, and the diffusion coefficient is proportional to the product $\langle v\rangle l$. A larger mean free path means faster diffusion, which is why gases at low pressure mix more readily. 🔉⇢

Viscosity has the same microscopic origin. When one layer of gas slides past another, molecules crossing between the layers carry momentum with them, and they carry it, on the average, over one mean free path before colliding and depositing it in the new layer. This transport of momentum across the flow is what we perceive macroscopically as internal friction, or viscosity. The kinetic theory shows the coefficient of viscosity is proportional to $n m\langle v\rangle l$, and because $l\propto 1/n$, the density cancels: strikingly, the viscosity of a dilute gas is nearly independent of its pressure, a counter-intuitive prediction of the theory that experiment handsomely confirms. 🔉⇢

Thermal conduction completes the trio and works by the same mechanism with energy in place of momentum. Molecules crossing from a hotter region to a cooler one carry their extra kinetic energy with them, again over about one mean free path, and give it up in a collision, so heat flows down the temperature gradient. The thermal conductivity is likewise proportional to the mean free path and the average speed. Diffusion, viscosity and conduction are thus three faces of the same idea: the transport of some quantity, be it molecules, momentum or energy, carried by moving molecules over a mean free path between randomising collisions. 🔉⇢

The random walk executed by a molecule is worth picturing concretely, because it explains the puzzle with which we began. Between collisions the molecule flies straight at its full thermal speed, but at the end of each free flight, after a distance of about one mean free path, it strikes a neighbour and sets off in a new, essentially random direction. Over many collisions these steps add up like a drunkard's walk, in which forward and backward steps largely cancel, so the net displacement grows only as the square root of the number of steps rather than in proportion to it. A molecule that covers hundreds of metres of path length in a second may therefore have wandered a net distance of only a few centimetres from where it started. This is precisely why a whiff of gas released in one corner of a room takes minutes to be smelled across it, even though the individual molecules are moving faster than sound. 🔉⇢

The mean free path also has a distinguished place in the history of physics, because it provided the first serious estimates of the sizes of molecules. In the nineteenth century, before molecules could be seen or manipulated, their reality was still doubted by many. By measuring a bulk property such as the viscosity of a gas, which the kinetic theory relates to the mean free path, and combining it with an independent estimate of the number density, Loschmidt and others were able to extract the molecular diameter $d$ from the relation $l=1/(\sqrt{2}\,n\pi d^2)$. The values that emerged, of the order of a few angstroms, were consistent across different measurements and different gases, and this internal consistency was among the strongest early evidence that molecules are real objects of a definite size. The mean free path thus served as a kind of microscope built out of thermodynamics. 🔉⇢

The collision cross-section $\pi d^2$ that lies at the heart of the derivation is itself a concept of lasting importance well beyond this chapter. It is the effective area that one particle presents to another for the purpose of interaction, and the same idea, generalised, reappears throughout physics in the scattering of light, the absorption of neutrons in a reactor and the collisions of particles in accelerators. Here it takes its simplest hard-sphere form: two molecules of diameter $d$ collide whenever their centres approach within $d$, so the target has radius $d$ and area $\pi d^2$. Because the cross-section grows as the square of the diameter, molecular size enters the mean free path very sensitively, and a gas of large molecules is far more opaque to the passage of its own members than a gas of small ones at the same density. 🔉⇢

The regime in which the mean free path becomes comparable to or larger than the size of the container has its own name and its own technology. When a tube is highly evacuated the number density falls so low that a molecule can cross the whole apparatus without meeting another; the gas is then said to be in the free-molecular or Knudsen regime, in which collisions are almost entirely with the walls. This is the operating condition of vacuum tubes and of the insulating gap in a thermos flask, where suppressing molecular collisions suppresses the conduction of heat. It is also the reason that maintaining a good vacuum is central to electron microscopes and particle accelerators, in which stray gas molecules must not be allowed to scatter the beam. All of these follow directly from the inverse dependence of the mean free path on number density. 🔉⇢

The pressure dependence of the mean free path has practical consequences that are easy to observe and to reason about. Because the number density of a gas at fixed temperature is proportional to its pressure, the mean free path is inversely proportional to pressure: pump a chamber down to a tenth of atmospheric pressure and the average distance between collisions grows tenfold. This is why processes such as thin-film coating, sputtering and the growth of semiconductor crystals are carried out in evacuated chambers, where a long mean free path lets atoms travel from source to target in straight lines without being knocked off course by collisions with residual gas. The same reasoning explains why the upper atmosphere, where the pressure is very low, has mean free paths of metres or more, so that its rarefied gas behaves quite differently from the dense air at the ground. 🔉⇢

It is worth reflecting on the orders of magnitude involved, because they give a vivid sense of the granularity of a gas. In air at ordinary conditions a molecule suffers something like a few billion collisions every second, since the collision time is under a nanosecond, yet between those collisions it travels a distance that is enormous on the molecular scale, roughly a thousand molecular diameters. The molecule is thus almost always in free flight and only very rarely in contact, which is exactly the condition under which the ideal gas approximation, treating intermolecular forces as acting only during brief collisions, is an excellent one. The mean free path quantifies this separation of scales and tells us precisely when a real gas may safely be treated as ideal and when, at high density or low temperature, the approximation must be abandoned. 🔉⇢

It helps to place the mean free path within the wider kinetic theory. A gas is a collection of molecules in random motion that collide elastically with one another and with the walls of the vessel, and in thermal equilibrium Maxwell's distribution fixes the spread of molecular velocities about the average speed. The absolute temperature sets the mean kinetic energy, while Avogadro's number ties the number density to the number of moles. Against this backdrop the mean free path is the average length of the free flight between successive elastic collisions, and the collision frequency is its reciprocal companion, both determined by the molecular diameter through the cross-section $\pi d^2$ and by the number density of the gas, quantities that also fix the pressure and temperature through the equation of state. 🔉⇢

The same molecular diameter that fixes the mean free path also governs the transport coefficients of the gas: the viscosity, the thermal conductivity and the diffusion coefficient are all proportional to the product of the average speed and the mean free path. For nitrogen, oxygen or argon at moderate pressures these theoretical predictions agree well with measurement, and it was through such quantitative relations, using a measured viscosity together with Avogadro's number, that the diameters and hence the sizes of molecules were first estimated. Whether the gas is monatomic like helium, diatomic like hydrogen and nitrogen, or polyatomic, the mean free path formula applies universally, depending only on the number density and the collision cross-section, which is why it is such a useful and general microscopic length. 🔉⇢

In summary, the mean free path $l=\dfrac{1}{\sqrt{2}\,n\pi d^2}$ is the single length that encapsulates how the finite size of molecules limits their free flight. Derived from the collision cross-section $\pi d^2$ and the number density $n$, it comes out to about $10^{-7}\,\text{m}$ for air at STP, some hundred times the interatomic distance, and it lengthens as the gas is rarefied or heated. Its reciprocal companions, the collision frequency $\nu=n\pi d^2\langle v\rangle$ and the collision time $\tau$, govern how fast a gas equilibrates. Above all, the mean free path is the microscopic length that lets the kinetic theory predict diffusion, viscosity and thermal conduction, and through them measure the sizes of molecules themselves. Historically this quantity, first estimated by nineteenth-century physicists, let them infer the molecular diameter and the value of Avogadro number; in a modern reading it sits alongside the equipartition of energy across translational, rotational and vibrational degrees of freedom that, through the count of quadratic modes, fixes the specific heat capacities of a gas. 🔉⇢

Derivation from first principles 🔉⇢

  1. Model molecules as hard spheres of diameter $d$; two collide when their centres approach within a distance $d$, so each molecule presents a circular target of radius $d$, i.e. a collision cross-section $\pi d^2$.
  2. A molecule moving at average speed $\langle v\rangle$ sweeps, in time $\Delta t$, a cylindrical volume $\pi d^2\langle v\rangle\Delta t$ within which any other molecular centre lies in its path.
  3. With number density $n$, the number of collisions in $\Delta t$ is $n\pi d^2\langle v\rangle\Delta t$, so the collision frequency is $\nu=n\pi d^2\langle v\rangle$ and the mean collision time is $\tau=\dfrac{1}{n\pi\langle v\rangle d^2}$.
  4. The mean free path is the distance covered in one collision time: $l=\langle v\rangle\tau=\dfrac{1}{n\pi d^2}$; note the speed cancels in this rest-target approximation.
  5. Correct for the fact that all molecules move: replacing $\langle v\rangle$ by the average relative speed introduces a factor $\sqrt{2}$, giving the exact result $l=\dfrac{1}{\sqrt{2}\,n\pi d^2}$.
  6. Read off the dependences: $l\propto 1/n$ (and hence $\propto T/P$ via $PV=\mu R T$) and $l\propto 1/d^2$ — denser gases and larger molecules give shorter free paths.
  7. Numerical estimate for air at STP: with $n\approx 2.7\times10^{25}\,\text{m}^{-3}$, $d\approx 2\times10^{-10}\,\text{m}$ and $\langle v\rangle\approx 485\,\text{m s}^{-1}$, $\tau\approx 6.1\times10^{-10}\,\text{s}$ and $l\approx 2.9\times10^{-7}\,\text{m}\approx 1500\,d$, about a hundred times the interatomic distance.
⚠️ JEE trap: A frequent error is to think the mean free path depends on how fast the molecules move, or to expect hot gas molecules to have a longer free path simply because they travel faster. In the derivation the average speed cancels: $l=\langle v\rangle\tau=1/(\sqrt{2}\,n\pi d^2)$ depends only on the number density and the collision cross-section, not on the speed. Temperature affects $l$ only indirectly, through its effect on the number density $n$ at fixed pressure ($n\propto 1/T$), not through the speed. A second common slip is to omit the $\sqrt{2}$ factor: the naive result $1/(n\pi d^2)$ treats the other molecules as stationary, whereas accounting for their motion through the average relative speed correctly introduces the $\sqrt{2}$ in the denominator. 🔉⇢

Worked example · JEE Advanced 🔉⇢

SITUATION Estimate the mean free path for a water molecule in water vapour at $373\,\text{K}$. Take the molecular diameter to be the same as for air, and use the air result at STP, $l_{air}\approx 2.9\times10^{-7}\,\text{m}$ with $n_{air}\approx 2.7\times10^{25}\,\text{m}^{-3}$.
TARGET Find the mean free path of the water molecule and compare it with the interatomic distance in the vapour.
STRATEGY Since $l=1/(\sqrt{2}\,n\pi d^2)$ and $d$ is unchanged, $l\propto 1/n$. At fixed pressure the number density is inversely proportional to absolute temperature, so scale $n$ from its STP value by the temperature ratio $273/373$, then scale $l$ inversely.
EXECUTE The number density falls to $n=2.7\times10^{25}\times\dfrac{273}{373}\approx 2\times10^{25}\,\text{m}^{-3}$. Because $l\propto 1/n$, the mean free path rises in the ratio $373/273$: $l\approx 2.9\times10^{-7}\times\dfrac{373}{273}\approx 4\times10^{-7}\,\text{m}$.
REFLECT This mean free path is about a hundred times the interatomic distance of roughly $4\times10^{-9}\,\text{m}$. It is this large value of the mean free path relative to molecular spacing that leads to typical gaseous behaviour and explains why gases cannot be confined without a container.

Source: NCERT Class XI Ch 12 Kinetic Theory

✍️ Worked Examples Polya 5-move · JEE tier

WE1 · Molecular Nature of Matter · JEE Advanced 🔉⇢

SITUATION The density of water is $1000\,\text{kg m}^{-3}$ and the density of water vapour at $100^\circ\text{C}$ and $1\,\text{atm}$ is $0.6\,\text{kg m}^{-3}$.
TARGET Estimate the fraction of the total volume of the vapour that is actually occupied by the molecules themselves (the molecular volume).
STRATEGY For a fixed mass, volume is inversely proportional to density. In the liquid the molecules are close-packed, so the molecular volume nearly equals the total volume; in the vapour the same molecules occupy a much larger total volume, so the occupied fraction falls by the density ratio.
EXECUTE The vapour occupies $1000/0.6=1/(6\times10^{-4})$ times the liquid volume. Taking the molecular-volume fraction in the liquid to be about 1, the fraction in the vapour is smaller by the same factor: fraction $\approx6\times10^{-4}$.
REFLECT Only about $0.06\%$ of a gas is 'solid' molecule; the rest is empty space. This quantifies why gas molecules move almost freely and why the non-interacting ideal-gas model works so well at low pressure.

Source: NCERT Class XI Ch 12 Kinetic Theory

WE2 · Behaviour of Gases & the Ideal-Gas Equation · JEE Advanced 🔉⇢

SITUATION A vessel contains two non-reactive gases, neon (monatomic) and oxygen (diatomic), whose partial pressures are in the ratio $P_{\text{Ne}}:P_{\text{O}_2}=3:2$. Atomic mass of Ne is $20.2\,\text{u}$; molecular mass of $\text{O}_2$ is $32.0\,\text{u}$.
TARGET Find the ratio of (i) the number of molecules and (ii) the mass densities of neon and oxygen in the vessel.
STRATEGY Each gas separately obeys $P_iV=\mu_i RT$ with common $V$ and $T$, so partial-pressure ratios equal mole ratios (Dalton's law). Molecule ratio equals mole ratio; density ratio weights the mole ratio by molar mass.
EXECUTE $(P_1/P_2)=(\mu_1/\mu_2)=3/2$. Since $\mu_i=N_i/N_A$, the molecule ratio is $N_1/N_2=\mu_1/\mu_2=3/2$. For densities, $\rho_i=\mu_i M_i/V$, so $\rho_1/\rho_2=(\mu_1/\mu_2)(M_1/M_2)=(3/2)(20.2/32.0)\approx0.947$.
REFLECT Partial pressure is the pressure a gas would exert alone; its ratios directly give molecule counts, and only the density comparison brings in the molar masses. The lighter gas here has the larger molecule count but not necessarily the larger density.

Source: NCERT Class XI Ch 12 Kinetic Theory

WE3 · Law of Equipartition of Energy · JEE Advanced 🔉⇢

SITUATION One mole of oxygen gas ($\mathrm{O_2}$, a diatomic molecule) is held at absolute temperature $T=300\,\text{K}$. Treat it as a rigid rotator so that vibration is not excited.
TARGET Find the internal energy of the sample and the share of that energy stored in translational versus rotational motion, using the law of equipartition of energy.
STRATEGY Count the active quadratic terms (3 translational + 2 rotational = 5), assign $\tfrac12 k_B T$ to each per molecule, multiply by $N_A$ to get the molar energy, and split the total in the ratio of the mode counts.
EXECUTE Each mode contributes $\tfrac12 R T$ per mole, so $U=5\times\tfrac12 R T=\tfrac52 R T=\tfrac52\times 8.31\times 300\approx 6.23\times10^{3}\,\text{J}$. Of the five modes, three are translational, giving $U_{trans}=\tfrac32 R T\approx 3.74\times10^{3}\,\text{J}$, and two are rotational, giving $U_{rot}=R T\approx 2.49\times10^{3}\,\text{J}$.
REFLECT The translational-to-rotational energy ratio is $3:2$, exactly the ratio of the mode counts, illustrating equal sharing per squared term. Had we allowed vibration, two more terms would raise $U$ to $\tfrac72 R T$; the fact that measured room-temperature values match $\tfrac52 R T$ shows the vibrational mode is frozen, a hint of quantum behaviour.

Source: NCERT Class XI Ch 12 Kinetic Theory

WE4 · Specific Heat Capacity of Gases · JEE Advanced 🔉⇢

SITUATION A cylinder of fixed capacity $44.8\,\text{litres}$ contains helium gas at standard temperature and pressure ($273\,\text{K}$, $1\,\text{atm}$). Take $R=8.31\,\text{J mol}^{-1}\text{K}^{-1}$.
TARGET Find the amount of heat needed to raise the temperature of the helium in the cylinder by $15.0\,^\circ\text{C}$.
STRATEGY Use the molar volume to find the number of moles, recognise helium as monatomic so $C_v=\tfrac32 R$, and because the volume is fixed use $\Delta Q=\mu C_v\Delta T$.
EXECUTE One mole of an ideal gas at STP occupies $22.4\,\text{litres}$, so the cylinder holds $\mu=44.8/22.4=2\,\text{mol}$. Helium is monatomic, $C_v=\tfrac32 R$. Since the volume is fixed, no work is done and $\Delta Q=\mu C_v\Delta T=2\times\tfrac32 R\times 15.0=45R=45\times 8.31\approx 3.7\times10^{2}\,\text{J}$.
REFLECT Had the heating been at constant pressure we would have used $C_p=\tfrac52 R$, needing $75R\approx 6.2\times10^{2}\,\text{J}$ — more heat, because some goes into expansion work. The example shows why specifying the constraint (constant $V$ versus constant $P$) is essential for a gas.

Source: NCERT Class XI Ch 12 Kinetic Theory

WE5 · Problem 1 · easy 🔉⇢

SITUATION The density of liquid water is $1000\ \text{kg m}^{-3}$, while the density of water vapour at $100^\circ\text{C}$ and $1\ \text{atm}$ is $0.6\ \text{kg m}^{-3}$.
TARGET Estimate the fraction of the total volume of the vapour that is actually occupied by the molecules themselves.
STRATEGY For a fixed mass, volume is inversely proportional to density. In the liquid the molecules are essentially close-packed, so the molecular volume there is nearly the whole volume. On vaporising, the same molecules spread over a much larger volume, so the occupied fraction shrinks by the density ratio.
EXECUTE Volume ratio $\dfrac{V_{\text{vapour}}}{V_{\text{liquid}}}=\dfrac{\rho_{\text{liquid}}}{\rho_{\text{vapour}}}=\dfrac{1000}{0.6}\approx 1.67\times10^{3}$. Taking the molecular (packed) fraction in the liquid as $\approx 1$, the fraction in the vapour is $\dfrac{1}{1.67\times10^{3}}\approx 6\times10^{-4}$.
REFLECT Only about $0.06\%$ of a gas's volume is molecules; the rest is empty space. This is precisely why the finite size of molecules can be ignored in the ideal-gas model, and why $PV^{\gamma}$ and $PV=\mu RT$ work so well at ordinary pressures.

Source: NCERT XI §12.3, Example 12.1 (derived)

WE6 · Problem 2 · medium 🔉⇢

SITUATION One mole of water has mass $18\ \text{g}$ and contains $N_A=6.02\times10^{23}$ molecules; liquid water has density $1000\ \text{kg m}^{-3}$.
TARGET Estimate the volume and radius of a single water molecule.
STRATEGY Treat the liquid as close-packed so a molecule's own density equals the bulk density. Get the mass of one molecule from the molar mass and $N_A$, divide by density for the volume, then model the molecule as a sphere.
EXECUTE Mass of one molecule $m=\dfrac{0.018}{6.02\times10^{23}}=3.0\times10^{-26}\ \text{kg}$. Volume $V=\dfrac{m}{\rho}=\dfrac{3.0\times10^{-26}}{1000}=3.0\times10^{-29}\ \text{m}^3$. With $V=\tfrac{4}{3}\pi r^3$, $r=\left(\dfrac{3V}{4\pi}\right)^{1/3}=\left(\dfrac{3\times3.0\times10^{-29}}{4\pi}\right)^{1/3}\approx 1.9\times10^{-10}\ \text{m}\approx 2\ \text{Å}$.
REFLECT The angstrom-scale radius ($\sim 2\ \text{Å}$) is the size quoted throughout the chapter and sets the molecular diameter $d\approx 2\text{-}4\ \text{Å}$ used later in the mean-free-path formula.

Source: NCERT XI §12.3, Example 12.2 (derived)

WE7 · Problem 3 · easy 🔉⇢

SITUATION A room of volume $25.0\ \text{m}^3$ holds air at $27^\circ\text{C}$ and $1\ \text{atm}=1.01\times10^{5}\ \text{Pa}$. Take $k_B=1.38\times10^{-23}\ \text{J K}^{-1}$.
TARGET Estimate the total number of gas molecules in the room.
STRATEGY Use the ideal-gas law in the molecular form $PV=Nk_BT$, so $N=PV/(k_BT)$, with $T$ in kelvin.
EXECUTE $T=300\ \text{K}$. $N=\dfrac{PV}{k_BT}=\dfrac{(1.01\times10^{5})(25.0)}{(1.38\times10^{-23})(300)}=\dfrac{2.525\times10^{6}}{4.14\times10^{-21}}\approx 6.1\times10^{26}\ \text{molecules}$.
REFLECT This is roughly $10^{3}\ \text{mol}$ of air, a staggering count that shows why bulk gas behaviour is so smooth and statistical, and why we describe it by a handful of macroscopic variables.

Source: NCERT XI §12.4, Exercise 12.6 (derived)

WE8 · Problem 4 · medium 🔉⇢

SITUATION A vessel holds a non-reacting mixture of neon (monatomic, atomic mass $20.2\ \text{u}$) and oxygen (diatomic, molecular mass $32.0\ \text{u}$). Their partial pressures are in the ratio $P_{\text{Ne}}:P_{\text{O}_2}=3:2$.
TARGET Find the ratio of (i) the number of molecules and (ii) the mass densities of neon and oxygen.
STRATEGY At common $V$ and $T$ each gas obeys $P_iV=\mu_iRT$, so partial-pressure ratio equals mole ratio equals molecule-number ratio. Mass density then follows by weighting the mole ratio with molar mass.
EXECUTE (i) $\dfrac{\mu_{\text{Ne}}}{\mu_{\text{O}_2}}=\dfrac{P_{\text{Ne}}}{P_{\text{O}_2}}=\dfrac{3}{2}$, and since $N=\mu N_A$, $\dfrac{N_{\text{Ne}}}{N_{\text{O}_2}}=\dfrac{3}{2}$. (ii) $\dfrac{\rho_{\text{Ne}}}{\rho_{\text{O}_2}}=\dfrac{\mu_{\text{Ne}}M_{\text{Ne}}}{\mu_{\text{O}_2}M_{\text{O}_2}}=\dfrac{3}{2}\times\dfrac{20.2}{32.0}=0.95$.
REFLECT Partial pressure tracks the number of molecules, not their mass; the lighter but more numerous neon ends up with almost the same mass density as the heavier oxygen. This is Dalton's law in action.

Source: NCERT XI §12.3, Example 12.4 (derived)

WE9 · Problem 5 · medium 🔉⇢

SITUATION An air bubble of volume $1.0\ \text{cm}^3$ forms at the bottom of a lake $40\ \text{m}$ deep where the temperature is $12^\circ\text{C}$. It rises to the surface, where the temperature is $35^\circ\text{C}$. Take $\rho_{\text{water}}=1000\ \text{kg m}^{-3}$, $g=9.8\ \text{m s}^{-2}$, atmospheric pressure $1.01\times10^{5}\ \text{Pa}$.
TARGET Find the volume of the bubble as it reaches the surface.
STRATEGY The trapped gas obeys $\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$. Compute the bottom pressure as atmospheric plus the water column, use surface pressure as atmospheric, and convert temperatures to kelvin.
EXECUTE $P_1=P_{\text{atm}}+\rho g h=1.01\times10^{5}+(1000)(9.8)(40)=1.01\times10^{5}+3.92\times10^{5}=4.93\times10^{5}\ \text{Pa}$; $P_2=1.01\times10^{5}\ \text{Pa}$. $T_1=285\ \text{K}$, $T_2=308\ \text{K}$. $V_2=V_1\dfrac{P_1}{P_2}\dfrac{T_2}{T_1}=1.0\times\dfrac{4.93\times10^{5}}{1.01\times10^{5}}\times\dfrac{308}{285}=1.0\times4.88\times1.081\approx 5.3\ \text{cm}^3$.
REFLECT The bubble grows more than fivefold, dominated by the pressure drop; the temperature rise contributes only an $8\%$ enhancement. Always add the hydrostatic head to atmospheric pressure at depth.

Source: NCERT XI §12.3, Exercise 12.5 (derived)

WE10 · Problem 6 · medium 🔉⇢

SITUATION Nitrogen gas ($M=28\ \text{g mol}^{-1}$) is held at $300\ \text{K}$ and $1.01\times10^{5}\ \text{Pa}$. Take $R=8.314\ \text{J mol}^{-1}\text{K}^{-1}$.
TARGET Find the mass density of the gas.
STRATEGY Use the density form of the ideal-gas law, $P=\dfrac{\rho R T}{M_0}$, so $\rho=\dfrac{P M_0}{RT}$.
EXECUTE $\rho=\dfrac{P M_0}{RT}=\dfrac{(1.01\times10^{5})(0.028)}{(8.314)(300)}=\dfrac{2828}{2494}\approx 1.13\ \text{kg m}^{-3}$.
REFLECT This matches the tabulated density of $\text{N}_2$ near room conditions, confirming the ideal-gas approximation is excellent for air at ordinary pressures and temperatures.

Source: NCERT XI §12.3 (density form, derived)

WE11 · Problem 7 · medium 🔉⇢

SITUATION Oxygen at $0^\circ\text{C}$ and $1\ \text{atm}$ has density $\rho=1.43\ \text{kg m}^{-3}$; the pressure is $P=1.01\times10^{5}\ \text{Pa}$.
TARGET Find the root-mean-square speed of the oxygen molecules using the pressure relation of kinetic theory.
STRATEGY Kinetic theory gives $P=\tfrac{1}{3}\rho\,\overline{v^2}$ (writing $nm=\rho$). Solve for $\overline{v^2}$ and take the square root.
EXECUTE $\overline{v^2}=\dfrac{3P}{\rho}=\dfrac{3(1.01\times10^{5})}{1.43}=2.12\times10^{5}\ \text{m}^2\text{s}^{-2}$. Hence $v_{\text{rms}}=\sqrt{\overline{v^2}}=\sqrt{2.12\times10^{5}}\approx 4.6\times10^{2}\ \text{m s}^{-1}$.
REFLECT The result, about $460\ \text{m s}^{-1}$, is of the order of the speed of sound in air, exactly as kinetic theory predicts. Note that only the bulk quantities $P$ and $\rho$ were needed, no microscopic data.

Source: NCERT XI §12.4 (derived)

WE12 · Problem 8 · easy 🔉⇢

SITUATION The average translational kinetic energy of the molecules in a container of gas of volume $V$ is $E$.
TARGET Show how the pressure of the gas is related to this kinetic energy, and state $P$ if $E=300\ \text{J}$ in $V=0.02\ \text{m}^3$.
STRATEGY Start from $P=\tfrac{1}{3}nm\,\overline{v^2}$ and rewrite using $E=N\cdot\tfrac{1}{2}m\overline{v^2}$, i.e. relate the pressure to the kinetic-energy density $E/V$.
EXECUTE From $PV=\tfrac{1}{3}Nm\,\overline{v^2}=\tfrac{2}{3}\left(N\cdot\tfrac{1}{2}m\overline{v^2}\right)=\tfrac{2}{3}E$, so $P=\dfrac{2}{3}\dfrac{E}{V}$. With $E=300\ \text{J}$, $V=0.02\ \text{m}^3$: $P=\dfrac{2}{3}\times\dfrac{300}{0.02}=1.0\times10^{4}\ \text{Pa}$.
REFLECT Pressure is literally two-thirds of the translational kinetic-energy density. This compact relation $PV=\tfrac{2}{3}E$ is the bridge from the microscopic picture to the macroscopic gas law.

Source: NCERT XI §12.4 (derived)

WE13 · Problem 9 · advanced 🔉⇢

SITUATION Nitrogen at STP has number density $n=2.7\times10^{25}\ \text{m}^{-3}$; each molecule has mass $m=4.65\times10^{-26}\ \text{kg}$ and the rms speed is $v_{\text{rms}}=516\ \text{m s}^{-1}$.
TARGET Compute the pressure predicted by kinetic theory from these microscopic quantities, and check it against the known STP value.
STRATEGY Apply $P=\tfrac{1}{3}nm\,\overline{v^2}$ with $\overline{v^2}=v_{\text{rms}}^2$. Then compare with $1\ \text{atm}$ to verify self-consistency of the microscopic model.
EXECUTE $\overline{v^2}=(516)^2=2.66\times10^{5}\ \text{m}^2\text{s}^{-2}$. $nm=(2.7\times10^{25})(4.65\times10^{-26})=1.256\ \text{kg m}^{-3}$ (the mass density). $P=\tfrac{1}{3}(1.256)(2.66\times10^{5})=\tfrac{1}{3}(3.34\times10^{5})\approx 1.11\times10^{5}\ \text{Pa}$, i.e. very nearly $1\ \text{atm}$.
REFLECT The microscopic formula reproduces the measured STP pressure to within a couple of percent, which is the whole triumph of kinetic theory: it derives a macroscopic law from molecular motion. The small excess comes from rounding in $v_{\text{rms}}$ and $n$.

Source: NCERT XI §12.4 (derived)

WE14 · Problem 10 · easy 🔉⇢

SITUATION Nitrogen ($M=28\ \text{g mol}^{-1}$) is at $T=300\ \text{K}$. Take $R=8.314\ \text{J mol}^{-1}\text{K}^{-1}$.
TARGET Find the root-mean-square speed of a nitrogen molecule.
STRATEGY Use the kinetic interpretation of temperature $\tfrac{1}{2}m\overline{v^2}=\tfrac{3}{2}k_BT$, which gives $v_{\text{rms}}=\sqrt{3RT/M}$ when written per mole.
EXECUTE $v_{\text{rms}}=\sqrt{\dfrac{3RT}{M}}=\sqrt{\dfrac{3(8.314)(300)}{0.028}}=\sqrt{\dfrac{7482.6}{0.028}}=\sqrt{2.67\times10^{5}}\approx 517\ \text{m s}^{-1}$.
REFLECT The rms speed depends only on temperature and molar mass, not on pressure or density. At $300\ \text{K}$ nitrogen molecules move at roughly $0.5\ \text{km s}^{-1}$, comparable to the speed of sound.

Source: NCERT XI §12.4.2 (derived)

WE15 · Problem 11 · medium 🔉⇢

SITUATION Argon (atomic mass $39.9\ \text{u}$) and chlorine (molecular mass $70.9\ \text{u}$) are held in the same flask at the same temperature.
TARGET Find the ratio of the rms speeds of the argon and chlorine molecules.
STRATEGY Average kinetic energy per molecule, $\tfrac{1}{2}m v_{\text{rms}}^2=\tfrac{3}{2}k_BT$, is the same for both gases at a common $T$, so $v_{\text{rms}}\propto 1/\sqrt{M}$.
EXECUTE $\dfrac{v_{\text{rms,Ar}}}{v_{\text{rms,Cl}_2}}=\sqrt{\dfrac{M_{\text{Cl}_2}}{M_{\text{Ar}}}}=\sqrt{\dfrac{70.9}{39.9}}=\sqrt{1.777}\approx 1.33$.
REFLECT Both gases share the same average kinetic energy (temperature is common), yet the lighter argon moves faster. The composition by mass is irrelevant; only the molar masses enter.

Source: NCERT XI §12.4.2, Example 12.5 (derived)

WE16 · Problem 12 · medium 🔉⇢

SITUATION A gas is at $T=300\ \text{K}$. Take $k_B=1.38\times10^{-23}\ \text{J K}^{-1}$ and $R=8.314\ \text{J mol}^{-1}\text{K}^{-1}$.
TARGET Find the average translational kinetic energy of one molecule and of one mole of the gas.
STRATEGY The kinetic interpretation gives average translational KE per molecule $=\tfrac{3}{2}k_BT$, independent of the gas; multiply by $N_A$ (equivalently use $R$) for one mole.
EXECUTE Per molecule: $\overline{E}=\tfrac{3}{2}k_BT=\tfrac{3}{2}(1.38\times10^{-23})(300)=6.2\times10^{-21}\ \text{J}$. Per mole: $E=\tfrac{3}{2}RT=\tfrac{3}{2}(8.314)(300)\approx 3.74\times10^{3}\ \text{J}$.
REFLECT The average kinetic energy depends only on temperature, not on the nature of the gas or its pressure. Helium and xenon molecules at the same temperature carry identical average kinetic energy.

Source: NCERT XI §12.4.2 (derived)

WE17 · Problem 13 · hard 🔉⇢

SITUATION Argon (atomic mass $39.9\ \text{u}$) and helium (atomic mass $4.0\ \text{u}$) are in separate cylinders. The helium is at $-20^\circ\text{C}$.
TARGET At what temperature is the rms speed of an argon atom equal to that of a helium atom at $-20^\circ\text{C}$?
STRATEGY Set $v_{\text{rms,Ar}}=v_{\text{rms,He}}$. Since $v_{\text{rms}}=\sqrt{3RT/M}$, equality gives $T_{\text{Ar}}/M_{\text{Ar}}=T_{\text{He}}/M_{\text{He}}$.
EXECUTE $T_{\text{He}}=253\ \text{K}$. $\dfrac{T_{\text{Ar}}}{M_{\text{Ar}}}=\dfrac{T_{\text{He}}}{M_{\text{He}}}\Rightarrow T_{\text{Ar}}=T_{\text{He}}\dfrac{M_{\text{Ar}}}{M_{\text{He}}}=253\times\dfrac{39.9}{4.0}=253\times9.975\approx 2.5\times10^{3}\ \text{K}$.
REFLECT The heavy argon must be about ten times hotter to match the nimble helium's speed. Speed equality is not energy equality here: at these different temperatures the argon atoms carry far more kinetic energy.

Source: NCERT XI §12.4.2, Exercise 12.9 (derived)

WE18 · Problem 14 · advanced 🔉⇢

SITUATION For oxygen ($M=32\ \text{g mol}^{-1}$) at $T=300\ \text{K}$, consider the most-probable speed $v_{mp}=\sqrt{2RT/M}$, the mean speed $\overline{v}=\sqrt{8RT/\pi M}$, and the rms speed $v_{\text{rms}}=\sqrt{3RT/M}$.
TARGET Compute all three speeds and state their ratio.
STRATEGY All three scale as $\sqrt{RT/M}$ with fixed numerical prefactors from the Maxwell speed distribution: $\sqrt{2}:\sqrt{8/\pi}:\sqrt{3}$. Evaluate the common factor once, then scale.
EXECUTE Common factor $\sqrt{RT/M}=\sqrt{(8.314)(300)/0.032}=\sqrt{7.79\times10^{4}}=279.2\ \text{m s}^{-1}$. Then $v_{mp}=\sqrt{2}(279.2)=395\ \text{m s}^{-1}$; $\overline{v}=\sqrt{8/\pi}(279.2)=1.596(279.2)=446\ \text{m s}^{-1}$; $v_{\text{rms}}=\sqrt{3}(279.2)=484\ \text{m s}^{-1}$. Ratio $v_{mp}:\overline{v}:v_{\text{rms}}=1.00:1.13:1.22$.
REFLECT The ordering $v_{mp}<\overline{v}<v_{\text{rms}}$ always holds because squaring weights the fast molecules more heavily; the rms speed is the largest of the three. JEE problems frequently test which speed a given formula refers to.

Source: JEE-style (authored); Maxwell distribution

WE19 · Problem 15 · hard 🔉⇢

SITUATION A sealed flask contains $2\ \text{g}$ of hydrogen gas ($M=2\ \text{g mol}^{-1}$) at $300\ \text{K}$. Take $R=8.314\ \text{J mol}^{-1}\text{K}^{-1}$.
TARGET Find the total translational kinetic energy of the gas.
STRATEGY The internal (translational) energy of an ideal gas is $E=\tfrac{3}{2}\mu RT$; find the number of moles from the mass and molar mass.
EXECUTE $\mu=\dfrac{2}{2}=1\ \text{mol}$. $E=\tfrac{3}{2}\mu RT=\tfrac{3}{2}(1)(8.314)(300)\approx 3.74\times10^{3}\ \text{J}$.
REFLECT For $\text{H}_2$ this counts only the translational part; the two rotational degrees of freedom add further internal energy (see equipartition). Translational KE alone sets the pressure and temperature.

Source: NCERT XI §12.4.2 (derived)

WE20 · Problem 16 · easy 🔉⇢

SITUATION Consider a monatomic gas (e.g. argon), a rigid diatomic gas (e.g. $\text{N}_2$ at moderate temperature), and a non-linear triatomic gas.
TARGET State the number of degrees of freedom and the internal energy per mole for each, using the law of equipartition.
STRATEGY Each quadratic energy term (degree of freedom) contributes $\tfrac{1}{2}k_BT$ per molecule, i.e. $\tfrac{1}{2}RT$ per mole. Count translational ($3$) plus rotational ($0$, $2$, $3$) modes.
EXECUTE Monatomic: $f=3$, $U=\tfrac{3}{2}RT$. Rigid diatomic: $f=5$ ($3$ translational $+2$ rotational), $U=\tfrac{5}{2}RT$. Non-linear triatomic: $f=6$ ($3+3$), $U=3RT$ (ignoring vibration).
REFLECT Equipartition converts a molecular geometry (how many ways it can store energy) directly into a macroscopic internal energy and hence a specific heat. Vibrational modes, if excited, add two degrees of freedom each.

Source: NCERT XI §12.5 (derived)

WE21 · Problem 17 · medium 🔉⇢

SITUATION A vessel contains $2\ \text{mol}$ of a rigid diatomic gas at $300\ \text{K}$. Take $R=8.314\ \text{J mol}^{-1}\text{K}^{-1}$.
TARGET Find the total internal energy of the gas.
STRATEGY A rigid diatomic molecule has $f=5$ degrees of freedom, so $U=\tfrac{f}{2}\mu RT=\tfrac{5}{2}\mu RT$.
EXECUTE $U=\tfrac{5}{2}\mu RT=\tfrac{5}{2}(2)(8.314)(300)=\tfrac{5}{2}(4988.4)\approx 1.25\times10^{4}\ \text{J}$.
REFLECT Of this, $\tfrac{3}{5}$ ($=7.5\times10^{3}\ \text{J}$) is translational and $\tfrac{2}{5}$ is rotational. Had we (wrongly) used $f=3$, we would have undercounted the internal energy by $40\%$.

Source: NCERT XI §12.5 (derived)

WE22 · Problem 18 · advanced 🔉⇢

SITUATION A container holds $2\ \text{mol}$ of helium (monatomic) and $1\ \text{mol}$ of hydrogen (rigid diatomic) in equilibrium at temperature $T=300\ \text{K}$.
TARGET Find the total internal energy of the mixture and the effective number of degrees of freedom per molecule.
STRATEGY Internal energy is additive: $U=\sum_i \tfrac{f_i}{2}\mu_i RT$ with $f_{\text{He}}=3$, $f_{\text{H}_2}=5$. The effective $f$ follows from $U=\tfrac{f_{\text{eff}}}{2}\mu_{\text{tot}}RT$.
EXECUTE $U=\tfrac{3}{2}(2)RT+\tfrac{5}{2}(1)RT=(3+2.5)RT=5.5RT=5.5(8.314)(300)\approx 1.37\times10^{4}\ \text{J}$. With $\mu_{\text{tot}}=3\ \text{mol}$: $\tfrac{f_{\text{eff}}}{2}(3)=5.5\Rightarrow f_{\text{eff}}=\dfrac{2(5.5)}{3}=3.67$.
REFLECT The effective degrees of freedom lie between $3$ and $5$, weighted by the mole fractions. This same weighting is what gives a mixture its own effective $\gamma$, a favourite JEE-Advanced twist.

Source: JEE-style (authored); NCERT XI §12.5

WE23 · Problem 19 · medium 🔉⇢

SITUATION A rigid diatomic molecule is in thermal equilibrium at $T=300\ \text{K}$. Take $k_B=1.38\times10^{-23}\ \text{J K}^{-1}$.
TARGET Find the average rotational kinetic energy of one such molecule.
STRATEGY A rigid diatomic molecule has $2$ rotational degrees of freedom, each contributing $\tfrac{1}{2}k_BT$ by equipartition.
EXECUTE $\overline{E}_{\text{rot}}=2\times\tfrac{1}{2}k_BT=k_BT=(1.38\times10^{-23})(300)=4.14\times10^{-21}\ \text{J}$.
REFLECT The rotational energy equals $k_BT$, exactly $\tfrac{2}{3}$ of the translational energy $\tfrac{3}{2}k_BT$. Rotation about the molecular axis is excluded because its moment of inertia is negligible.

Source: NCERT XI §12.5 (derived)

WE24 · Problem 20 · easy 🔉⇢

SITUATION Consider an ideal monatomic gas. Take $R=8.314\ \text{J mol}^{-1}\text{K}^{-1}$.
TARGET Find its molar specific heats $C_v$ and $C_p$ and the ratio $\gamma=C_p/C_v$.
STRATEGY A monatomic molecule has $f=3$, so $U=\tfrac{3}{2}RT$ and $C_v=dU/dT=\tfrac{3}{2}R$. Then Mayer's relation $C_p-C_v=R$ gives $C_p$.
EXECUTE $C_v=\tfrac{3}{2}R=\tfrac{3}{2}(8.314)=12.5\ \text{J mol}^{-1}\text{K}^{-1}$; $C_p=C_v+R=\tfrac{5}{2}R=20.8\ \text{J mol}^{-1}\text{K}^{-1}$; $\gamma=\dfrac{C_p}{C_v}=\dfrac{5/2}{3/2}=\dfrac{5}{3}\approx 1.67$.
REFLECT These predicted values match the measured specific heats of the noble gases closely, an early triumph of kinetic theory. $\gamma=5/3$ is the signature of a monatomic gas.

Source: NCERT XI §12.6.1 (derived)

WE25 · Problem 21 · medium 🔉⇢

SITUATION A rigid diatomic ideal gas is heated. Take $R=8.314\ \text{J mol}^{-1}\text{K}^{-1}$.
TARGET Find $C_v$, $C_p$, $\gamma$, and verify Mayer's relation.
STRATEGY A rigid diatomic molecule has $f=5$, so $C_v=\tfrac{5}{2}R$; $C_p=C_v+R$ (Mayer's relation, true for any ideal gas), and $\gamma=C_p/C_v$.
EXECUTE $C_v=\tfrac{5}{2}R=20.8\ \text{J mol}^{-1}\text{K}^{-1}$; $C_p=\tfrac{7}{2}R=29.1\ \text{J mol}^{-1}\text{K}^{-1}$; $\gamma=\dfrac{7/2}{5/2}=\dfrac{7}{5}=1.40$. Check: $C_p-C_v=\tfrac{7}{2}R-\tfrac{5}{2}R=R=8.31\ \text{J mol}^{-1}\text{K}^{-1}$. ✓
REFLECT $\gamma=1.40$ is the classic value for air (mostly $\text{N}_2$ and $\text{O}_2$). If vibration switches on at high $T$, $f\to 7$ and $\gamma$ drops toward $9/7$.

Source: NCERT XI §12.6.2 (derived)

WE26 · Problem 22 · hard 🔉⇢

SITUATION A cylinder of fixed volume $44.8\ \text{L}$ holds helium at standard temperature and pressure. Take $R=8.31\ \text{J mol}^{-1}\text{K}^{-1}$; molar volume at STP is $22.4\ \text{L}$.
TARGET Find the heat needed to raise the temperature of the gas by $15.0^\circ\text{C}$.
STRATEGY First get the number of moles from the molar volume. At constant volume no work is done, so $Q=\mu C_v\Delta T$ with $C_v=\tfrac{3}{2}R$ for monatomic helium.
EXECUTE $\mu=\dfrac{44.8}{22.4}=2\ \text{mol}$. $Q=\mu C_v\Delta T=(2)\left(\tfrac{3}{2}R\right)(15.0)=2(1.5)(8.31)(15.0)=374\ \text{J}$.
REFLECT Because the volume is fixed, every joule goes into internal energy; none is spent doing work. Had the pressure been held constant instead, we would use $C_p=\tfrac{5}{2}R$ and need more heat for the same rise.

Source: NCERT XI §12.6, Example 12.8 (derived)

WE27 · Problem 23 · advanced 🔉⇢

SITUATION A mixture contains $2\ \text{mol}$ of helium (monatomic, $C_v=\tfrac{3}{2}R$) and $1\ \text{mol}$ of nitrogen (rigid diatomic, $C_v=\tfrac{5}{2}R$).
TARGET Find the effective molar specific heat $C_v$ of the mixture and its effective $\gamma$.
STRATEGY For a mixture the molar heat capacities combine as mole-weighted averages: $C_{v,\text{mix}}=\dfrac{\mu_1 C_{v1}+\mu_2 C_{v2}}{\mu_1+\mu_2}$; then $C_{p,\text{mix}}=C_{v,\text{mix}}+R$ and $\gamma=C_p/C_v$.
EXECUTE $C_{v,\text{mix}}=\dfrac{2(\tfrac{3}{2}R)+1(\tfrac{5}{2}R)}{3}=\dfrac{3R+2.5R}{3}=\dfrac{5.5R}{3}=1.833R$. $C_{p,\text{mix}}=1.833R+R=2.833R$. $\gamma_{\text{mix}}=\dfrac{2.833R}{1.833R}=1.55$.
REFLECT The mixture's $\gamma=1.55$ sits between the monatomic $1.67$ and diatomic $1.40$ values, pulled toward helium because it supplies two of the three moles. Mole fractions, not mass fractions, do the weighting.

Source: JEE-style (authored); NCERT XI §12.6

WE28 · Problem 24 · medium 🔉⇢

SITUATION Nitrogen at STP has number density $n=2.7\times10^{25}\ \text{m}^{-3}$ and molecular diameter $d=2.0\times10^{-10}\ \text{m}$.
TARGET Estimate the mean free path of a nitrogen molecule.
STRATEGY Use the mean-free-path formula $l=\dfrac{1}{\sqrt{2}\,n\pi d^2}$, which accounts for the relative motion of colliding molecules.
EXECUTE $\pi d^2=\pi(2.0\times10^{-10})^2=1.26\times10^{-19}\ \text{m}^2$. Denominator $\sqrt{2}\,n\pi d^2=(1.414)(2.7\times10^{25})(1.26\times10^{-19})=4.8\times10^{6}\ \text{m}^{-1}$. So $l=\dfrac{1}{4.8\times10^{6}}\approx 2.1\times10^{-7}\ \text{m}$.
REFLECT The mean free path ($\sim 2000\ \text{Å}$) is about a thousand times the molecular size and a hundred times the intermolecular spacing, which is exactly why a gas behaves so differently from a liquid.

Source: NCERT XI §12.7 (derived)

WE29 · Problem 25 · advanced 🔉⇢

SITUATION The mean free path of a gas at STP ($273\ \text{K}$, $1\ \text{atm}$) is $l_0=2.1\times10^{-7}\ \text{m}$. The molecular diameter is unchanged.
TARGET Express $l$ in terms of pressure and temperature, and find $l$ when the gas is taken to $2\ \text{atm}$ and $546\ \text{K}$.
STRATEGY Write $n=P/(k_BT)$ in $l=1/(\sqrt{2}\,n\pi d^2)$ to get $l=\dfrac{k_BT}{\sqrt{2}\,\pi d^2 P}$, so $l\propto T/P$. Scale from the STP value.
EXECUTE $l=\dfrac{k_BT}{\sqrt{2}\,\pi d^2 P}\propto \dfrac{T}{P}$. Hence $l=l_0\dfrac{T/T_0}{P/P_0}=2.1\times10^{-7}\times\dfrac{546/273}{2/1}=2.1\times10^{-7}\times\dfrac{2}{2}=2.1\times10^{-7}\ \text{m}$.
REFLECT Doubling both the temperature and the pressure leaves the mean free path unchanged, because the extra thermal expansion is exactly cancelled by the extra compression. The lesson: $l$ tracks $T/P$, i.e. it is inversely proportional to number density alone.

Source: JEE-style (authored); NCERT XI §12.7

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📐 Formula Sheet Printable · every formula cited

Ideal Gas and Gas Laws

QuantityFormulaWhat it means / when to useSource
Ideal-gas equation 🔉⇢$PV=\mu RT=Nk_BT$For $\mu$ moles ($N=\mu N_A$ molecules) of an ideal gas, with $R=8.314\,\mathrm{J\,mol^{-1}K^{-1}}$ and $k_B=R/N_A=1.38\times10^{-23}\,\mathrm{J\,K^{-1}}$. Valid at low density; contains Boyle's and Charles' laws.NCERT XI Ch 12 (§12.3)
Boyle's law 🔉⇢$PV=\text{constant}$ (at fixed $T$)At constant temperature the pressure of a fixed mass of gas varies inversely with its volume.NCERT XI Ch 12 (§12.3)
Charles' law 🔉⇢$\dfrac{V}{T}=\text{constant}$ (at fixed $P$)At constant pressure the volume of a fixed mass of gas is proportional to its absolute temperature.NCERT XI Ch 12 (§12.3)
Dalton's law of partial pressures 🔉⇢$P=P_1+P_2+\cdots$The total pressure of a mixture of non-interacting ideal gases is the sum of the partial pressures each would exert alone in the same volume.NCERT XI Ch 12 (§12.3)

Kinetic Theory of Pressure and Temperature

QuantityFormulaWhat it means / when to useSource
Pressure of an ideal gas 🔉⇢$P=\dfrac{1}{3}nm\overline{v^2}=\dfrac{1}{3}\rho\overline{v^2}$$n$ = molecules per unit volume, $m$ = molecular mass, $\rho=nm$ = density, $\overline{v^2}$ = mean-square speed. Derived from elastic collisions with the walls.NCERT XI Ch 12 (§12.4)
Mean kinetic energy of a molecule 🔉⇢$\tfrac{1}{2}m\overline{v^2}=\tfrac{3}{2}k_BT$The average translational kinetic energy of a molecule depends only on the absolute temperature, not on the gas or the molecular mass.NCERT XI Ch 12 (§12.4.2)
RMS speed 🔉⇢$v_{rms}=\sqrt{\overline{v^2}}=\sqrt{\dfrac{3k_BT}{m}}=\sqrt{\dfrac{3RT}{M}}$$M$ = molar mass. At a given temperature $v_{rms}\propto1/\sqrt{M}$, so lighter molecules move faster.NCERT XI Ch 12 (§12.4.2)
Three molecular speeds 🔉⇢$v_{mp}:\bar v:v_{rms}=\sqrt{2}:\sqrt{8/\pi}:\sqrt{3}$Most-probable $<$ average $<$ RMS, in the ratio $1.41:1.60:1.73$. $\bar v=\sqrt{8k_BT/\pi m}$, $v_{mp}=\sqrt{2k_BT/m}$.NCERT XI Ch 12 (§12.6, Maxwell distribution)

Equipartition and Specific Heats

QuantityFormulaWhat it means / when to useSource
Law of equipartition 🔉⇢$\langle E\rangle=\tfrac{1}{2}k_BT$ per degree of freedomIn thermal equilibrium each quadratic degree of freedom (translational, rotational, and each vibrational mode counting two) carries an average energy $\tfrac{1}{2}k_BT$ per molecule.NCERT XI Ch 12 (§12.5)
Internal energy (f degrees of freedom) 🔉⇢$U=\tfrac{f}{2}\mu RT$With $f$ active degrees of freedom per molecule, one mole has internal energy $\tfrac{f}{2}RT$. Monatomic $f=3$, diatomic $f=5$ (ordinary $T$).NCERT XI Ch 12 (§12.5)
Molar specific heats 🔉⇢$C_v=\tfrac{f}{2}R,\quad C_p=\left(\tfrac{f}{2}+1\right)R$At constant volume all heat raises internal energy; at constant pressure the gas also does work $R\,\Delta T$ per mole.NCERT XI Ch 12 (§12.6)
Mayer's relation 🔉⇢$C_p-C_v=R$True for any ideal gas whatever its atomicity. The constant-pressure molar heat exceeds the constant-volume one by exactly the gas constant.NCERT XI Ch 12 (§12.6)
Ratio of specific heats 🔉⇢$\gamma=\dfrac{C_p}{C_v}=1+\dfrac{2}{f}$Monatomic $\gamma=5/3$; diatomic $\gamma=7/5$; typical polyatomic $\gamma=4/3$. Sets the adiabatic exponent used in thermodynamics.NCERT XI Ch 12 (§12.6)

Mean Free Path and Molecular Estimates

QuantityFormulaWhat it means / when to useSource
Mean free path 🔉⇢$l=\dfrac{1}{\sqrt{2}\,n\pi d^2}$$n$ = number density, $d$ = molecular diameter. Depends inversely on the number density and on the square of the molecular size; the $\sqrt{2}$ accounts for the motion of all molecules.NCERT XI Ch 12 (§12.7)
Mean free path vs P, T 🔉⇢$l=\dfrac{k_BT}{\sqrt{2}\,\pi d^2 P}$Using $n=P/k_BT$. At fixed temperature $l\propto1/P$; at fixed pressure $l\propto T$.NCERT XI Ch 12 (§12.7)
Collision frequency 🔉⇢$\nu=\dfrac{\bar v}{l}=\sqrt{2}\,n\pi d^2\bar v$The average number of collisions a molecule makes per second — its average speed divided by the mean free path.NCERT XI Ch 12 (§12.7)
Avogadro's number 🔉⇢$N_A=6.02\times10^{23}\,\mathrm{mol^{-1}}$The number of molecules in one mole. Equal volumes of ideal gases at the same $T$ and $P$ contain equal numbers of molecules (Avogadro's hypothesis).NCERT XI Ch 12 (§12.3)

📜 Previous-Year Questions Authentic NTA · 65 questions

Every PYQ traces to an official NTA/JAB paper (year + shift + paper). Answer keys match the official key. OCR-extracted questions are flagged for SME verification.

JEE Main 2021 · Paper 1 · August 27 Shift 1 · Q12 (official key (printed in paper)) Answer: (C) 4/T

An ideal gas is expanding such that $PT^{3}$ = constant. The coefficient of volume expansion of the gas is :

  • (A) ${1 \over T}$
  • (B) ${2 \over T}$
  • (C) ${4 \over T}$
  • (D) ${3 \over T}$
JEE Main 2021 · Paper 1 · August 31 Shift 1 · Q15 (official key (printed in paper)) Answer: (A) p_0/(aeR)

For an ideal gas the instantaneous change in pressure 'p' with volume 'v' is given by the equation ${{dp} \over {dv}} = - ap$. If p = $p_{0}$ at v =0 is the given boundary condition, then the maximum temperature one mole of gas can attain is : (Here R is the gas constant)

  • (A) ${{{p_0}} \over {aeR}}$
  • (B) ${{a{p_0}} \over {eR}}$
  • (C) infinity
  • (D) 0$^\circ$C
JEE Main 2021 · Paper 1 · August 31 Shift 2 · Q15 (official key (printed in paper)) Answer: (C) 16 : 3

A mixture of hydrogen and oxygen has volume 500 $cm^{3}$, temperature 300 K, pressure 400 kPa and mass 0.76 g. The ratio of masses of oxygen to hydrogen will be :-

  • (A) 3 : 8
  • (B) 3 : 16
  • (C) 16 : 3
  • (D) 8 : 3
JEE Main 2021 · Paper 1 · August 27 Shift 1 · Q6 (official key (printed in paper)) Answer: (D) 123.54 kg

A balloon carries a total load of 185 kg at normal pressure and temperature of 27$^\circ$C. What load will the balloon carry on rising to a height at which the barometric pressure is 45 cm of Hg and the temperature is $-$7$^\circ$C. Assuming the volume constant?

  • (A) 181.46 kg
  • (B) 214.15 kg
  • (C) 219.07 kg
  • (D) 123.54 kg
JEE Main 2021 · Paper 1 · August 27 Shift 2 · Q9 (official key (printed in paper)) Answer: (A) 640 m/s

if the rms speed of oxygen molecules at 0$^\circ$C is 160 m/s, find the rms speed of hydrogen molecules at 0$^\circ$C.

  • (A) 640 m/s
  • (B) 40 m/s
  • (C) 80 m/s
  • (D) 332 m/s
JEE Main 2021 · Paper 1 · March 16 Shift 1 · Q9 (official key (printed in paper)) Answer: (D) (5/2)RT/V

The volume V of an enclosure contains a mixture of three gases, 16 g of oxygen, 28 g of nitrogen and 44 g of carbon dioxide at absolute temperature T. Consider R as universal gas constant. The pressure of the mixture of gases is :

  • (A) ${{3RT} \over V}$
  • (B) ${{4RT} \over V}$
  • (C) ${{88RT} \over V}$
  • (D) ${5 \over 2}{{RT} \over V}$
JEE Main 2023 · Paper 1 · January 24 Shift 1 · Q17 (official key (printed in paper)) Answer: (A) Statement I is true but Statement II is false

Given below are two statements : Statement I : The temperature of a gas is $-73^\circ$C. When the gas is heated to $527^\circ$C, the root mean square speed of the molecules is doubled. Statement II : The product of pressure and volume of an ideal gas will be equal to translational kinetic energy of the molecules. In the light of the above statements, choose the correct answer from the option given below :

  • (A) Statement I is true but Statement II is false
  • (B) Both Statement I and Statement II are true
  • (C) Statement I is false but Statement II is true
  • (D) Both Statement I and Statement II are false
JEE Main 2023 · Paper 1 · February 1 Shift 1 · Q18 (official key (printed in paper)) Answer: (D) Compressibility

$\left(P+\frac{a}{V^{2}}\right)(V-b)=R T$ represents the equation of state of some gases. Where $P$ is the pressure, $V$ is the volume, $T$ is the temperature and $a, b, R$ are the constants. The physical quantity, which has dimensional formula as that of $\frac{b^{2}}{a}$, will be:

  • (A) Energy density
  • (B) Bulk modulus
  • (C) Modulus of rigidity
  • (D) Compressibility
JEE Main 2023 · Paper 1 · February 1 Shift 1 · Q19 (official key (printed in paper)) Answer: (C) proportional to absolute temperature

The average kinetic energy of a molecule of the gas is

  • (A) proportional to volume
  • (B) dependent on the nature of the gas
  • (C) proportional to absolute temperature
  • (D) proportional to pressure
JEE Main 2023 · Paper 1 · January 24 Shift 2 · Q2 (official key (printed in paper)) Answer: (B) 25/21

Let $\gamma_1$ be the ratio of molar specific heat at constant pressure and molar specific heat at constant volume of a monoatomic gas and $\gamma_2$ be the similar ratio of diatomic gas. Considering the diatomic gas molecule as a rigid rotator, the ratio, $\frac{\gamma_1}{\gamma_2}$ is :

  • (A) $\frac{35}{27}$
  • (B) $\frac{25}{21}$
  • (C) $\frac{21}{25}$
  • (D) $\frac{27}{35}$
JEE Main 2023 · Paper 1 · January 25 Shift 2 · Q2 (official key (printed in paper)) Answer: (D) 7/2 R

According to law of equipartition of energy the molar specific heat of a diatomic gas at constant volume where the molecule has one additional vibrational mode is :-

  • (A) $\frac{9}{2}R$
  • (B) $\frac{5}{2}R$
  • (C) $\frac{3}{2}R$
  • (D) $\frac{7}{2}R$
JEE Main 2023 · Paper 1 · January 30 Shift 1 · Q2 (official key (printed in paper)) Answer: (D) 3/T

The pressure $(\mathrm{P})$ and temperature ($\mathrm{T})$ relationship of an ideal gas obeys the equation $\mathrm{PT}^{2}=$ constant. The volume expansion coefficient of the gas will be :

  • (A) $3 T^{2}$
  • (B) $\frac{3}{T^2}$
  • (C) $\frac{3}{T^3}$
  • (D) $\frac{3}{T}$
JEE Main 2023 · Paper 1 · January 30 Shift 2 · Q2 (official key (printed in paper)) Answer: (A) 1 : 1

A flask contains hydrogen and oxygen in the ratio of $2: 1$ by mass at temperature $27^{\circ} \mathrm{C}$. The ratio of average kinetic energy per molecule of hydrogen and oxygen respectively is:

  • (A) 1 : 1
  • (B) 4 : 1
  • (C) 1 : 4
  • (D) 2 : 1
JEE Main 2023 · Paper 1 · January 31 Shift 2 · Q7 (official key (printed in paper)) Answer: (C) 525 J

Heat energy of $735 \mathrm{~J}$ is given to a diatomic gas allowing the gas to expand at constant pressure. Each gas molecule rotates around an internal axis but do not oscillate. The increase in the internal energy of the gas will be :

  • (A) $572 \mathrm{~J}$
  • (B) $441 \mathrm{~J}$
  • (C) $525 \mathrm{~J}$
  • (D) $735 \mathrm{~J}$
JEE Main 2023 · Paper 1 · January 25 Shift 1 · Q8 (official key (printed in paper)) Answer: (D) Proportional to square root of temperature (sqrt(T))

The root mean square velocity of molecules of gas is

  • (A) Proportional to temperature ($T$)
  • (B) Inversely proportional to square root of temperature $\left( {\sqrt {{1 \over T}} } \right)$
  • (C) Proportional to square of temperature ($T^2$)
  • (D) Proportional to square root of temperature ($\sqrt T$)
IIT-JEE 2008 · Paper 1 · Q28 (official key) Answer: C

An ideal gas is expanding such that $PT^2 = $ constant. The coefficient of volume expansion of the gas is

  • (A) $\dfrac{1}{T}$
  • (B) $\dfrac{2}{T}$
  • (C) $\dfrac{3}{T}$
  • (D) $\dfrac{4}{T}$
IIT-JEE 2009 · Paper 1 · Q52 (official key) Answer: B,D

$C_v$ and $C_p$ denote the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then

  • (A) $C_p - C_v$ is larger for a diatomic ideal gas than for a monoatomic ideal gas
  • (B) $C_p + C_v$ is larger for a diatomic ideal gas than for a monoatomic ideal gas
  • (C) $C_p / C_v$ is larger for a diatomic ideal gas than for a monoatomic ideal gas
  • (D) $C_p \cdot C_v$ is larger for a diatomic ideal gas than for a monoatomic ideal gas
IIT-JEE 2010 · Paper 1 · Q58 (official key) Answer: D

A real gas behaves like an ideal gas if its

  • (A) pressure and temperature are both high
  • (B) pressure and temperature are both low
  • (C) pressure is high and temperature is low
  • (D) pressure is low and temperature is high
IIT-JEE 2012 · Paper 1 · Q8 (official key) Answer: D

A mixture of $2$ moles of helium gas (atomic mass $=4$ amu) and $1$ mole of argon gas (atomic mass $=40$ amu) is kept at $300$ K in a container. The ratio of the rms speeds $\left(\frac{v_{rms}(\text{helium})}{v_{rms}(\text{argon})}\right)$ is

  • (A) $0.32$
  • (B) $0.45$
  • (C) $2.24$
  • (D) $3.16$
JEE Advanced 2013 · Paper 1 · Q4 (official key) Answer: D

Two non-reactive monoatomic ideal gases have their atomic masses in the ratio $2 : 3$. The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is $4 : 3$. The ratio of their densities is

  • (A) $1 : 4$
  • (B) $1 : 2$
  • (C) $6 : 9$
  • (D) $8 : 9$
JEE Advanced 2015 · Paper 1 · Q16 (official key) Answer: A,B,D

A container of fixed volume has a mixture of one mole of hydrogen and one mole of helium in equilibrium at temperature $T$. Assuming the gases are ideal, the correct statement(s) is(are)

  • (A) The average energy per mole of the gas mixture is $2RT$.
  • (B) The ratio of speed of sound in the gas mixture to that in helium gas is $\sqrt{6/5}$.
  • (C) The ratio of the rms speed of helium atoms to that of hydrogen molecules is $1/2$.
  • (D) The ratio of the rms speed of helium atoms to that of hydrogen molecules is $1/\sqrt{2}$.
JEE Advanced 2017 · Paper 1 · Q1 (official key) Answer: A, B, D

A flat plate is moving normal to its plane through a gas under the action of a constant force $F$. The gas is kept at a very low pressure. The speed of the plate $v$ is much less than the average speed $u$ of the gas molecules. Which of the following options is/are true?

  • (A) The pressure difference between the leading and trailing faces of the plate is proportional to $uv$
  • (B) The resistive force experienced by the plate is proportional to $v$
  • (C) The plate will continue to move with constant non-zero acceleration, at all times
  • (D) At a later time the external force $F$ balances the resistive force
JEE Advanced 2020 · Paper 1 · Q12 (official key) Answer: A, B, C

As shown schematically in the figure, two vessels contain water solutions (at temperature $T$) of potassium permanganate (KMnO$_4$) of different concentrations $n_1$ and $n_2$ ($n_1 > n_2$) molecules per unit volume with $\Delta n = (n_1 - n_2) \ll n_1$. When they are connected by a tube of small length $l$ and cross-sectional area $S$, KMnO$_4$ starts to diffuse from the left to the right vessel through the tube. Consider the collection of molecules to behave as dilute ideal gases and the difference in their partial pressure in the two vessels causing the diffusion. The speed $v$ of the molecules is limited by the viscous force $-\beta v$ on each molecule, where $\beta$ is a constant. Neglecting all terms of the order $(\Delta n)^2$, which of the following is/are correct? ($k_B$ is the Boltzmann constant)

  • (A) the force causing the molecules to move across the tube is $\Delta n\, k_B T S$
  • (B) force balance implies $n_1 \beta v l = \Delta n\, k_B T$
  • (C) total number of molecules going across the tube per sec is $\left(\dfrac{\Delta n}{l}\right)\left(\dfrac{k_B T}{\beta}\right) S$
  • (D) rate of molecules getting transferred through the tube does not change with time
JEE Advanced 2023 · Paper 2 · Q4 (official key) Answer: C

An ideal gas is in thermodynamic equilibrium. The number of degrees of freedom of a molecule of the gas is $n$. The internal energy of one mole of the gas is $U_n$ and the speed of sound in the gas is $v_n$. At a fixed temperature and pressure, which of the following is the correct option?

  • (A) $v_3 < v_6$ and $U_3 > U_6$
  • (B) $v_5 > v_3$ and $U_3 > U_5$
  • (C) $v_5 > v_7$ and $U_5 < U_7$
  • (D) $v_6 < v_7$ and $U_6 < U_7$
JEE Advanced 2025 · Paper 2 · Q13 (official key) Answer: 0.2

The left and right compartments of a thermally isolated container of length $L$ are separated by a thermally conducting, movable piston of area $A$. The left and right compartments are filled with $\dfrac{3}{2}$ and $1$ moles of an ideal gas, respectively. In the left compartment the piston is attached by a spring with spring constant $k$ and natural length $\dfrac{2L}{5}$. In thermodynamic equilibrium, the piston is at a distance $\dfrac{L}{2}$ from the left and right edges of the container. Under the above conditions, if the pressure in the right compartment is $P = \dfrac{kL}{A}\alpha$, then the value of $\alpha$ is ____

JEE Main 2026 · Paper 1 · April 2 Shift 2 · Q34 (official key) Answer: C

A mixture of carbon dioxide and oxygen has volume 8310 $cm^{3}$, temperature 300 K, pressure 100 kPa and mass 13.2 g. The number of moles of carbon dioxide and oxygen gases in the mixture respectively are ______. (Assume both carbon dioxide and oxygen gases behave like ideal gases) [R = 8.31 J/mol K]

  • (A) 0.15 and 0.18
  • (B) 0.25 and 0.08
  • (C) 0.21 and 0.12
  • (D) 0.13 and 0.20
JEE Main 2026 · Paper 1 · April 6 Shift 1 · Q35 (official key) Answer: A

Two closed vessels of same volume are joined through a narrow tube and both vessels are filled with air of pressure 90 kPa and temperature 400 K . Keeping the temperature of one vessel constant at 400 K the second vessel temperature is raised to 500 K . The final pressure in the vessels is $\_\_\_\_$ kPa .

  • (A) 100
  • (B) 120
  • (C) 90
  • (D) 105
JEE Main 2019 · Paper 1 · January 11 Shift 1 · Q11 (published compilation) Answer: D⚑ verify

A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. considering only translational and rotational modes, the total internal energy of the system is :

  • (A) 12 RT
  • (B) 20 RT
  • (C) 4 RT
  • (D) 15 RT
JEE Main 2020 · Paper 1 · January 9 Shift 2 · Q10 (published compilation) Answer: B⚑ verify

Two gases-argon (atomic radius 0.07 nm, atomic weight 40) and xenon (atomic radius 0.1 nm, atomic weight 140) have the same number density and are at the same temperature. The raito of their respective mean free times is closest to :

  • (A) 2.3
  • (B) 1.83
  • (C) 4.67
  • (D) 3.67
JEE Main 2020 · Paper 1 · January 8 Shift 2 · Q11 (published compilation) Answer: D⚑ verify

Consider a mixture of n moles of helium gas and 2n moles of oxygen gas (molecules taken to be rigid) as an ideal gas. Its $C_{P}$/$C_{V}$ value will be :

  • (A) 23/15
  • (B) 67/45
  • (C) 40/27
  • (D) 19/13
JEE Main 2020 · Paper 1 · September 3 Shift 2 · Q12 (published compilation) Answer: A⚑ verify

To raise the temperature of a certain mass of gas by $50^{o}$C at a constant pressure, 160 calories of heat is required. When the same mass of gas is cooled by $100^{o}$C at constant volume, 240 calories of heat is released. How many degrees of freedom does each molecule of this gas have (assume gas to be ideal)?

  • (A) 6
  • (B) 7
  • (C) 5
  • (D) 3
JEE Main 2020 · Paper 1 · September 6 Shift 1 · Q15 (published compilation) Answer: A⚑ verify

Molecules of an ideal gas are known to have three translational degrees of freedom and two rotational degrees of freedom.The gas is maintained at a temperature of T. The total internal energy, U of a mole of this gas, and the value of $\gamma \left( { = {{{C_p}} \over {{C_v}}}} \right)$ are given, respectively by:

  • (A) U = ${5 \over 2}RT$ and $\gamma = {7 \over 5}$
  • (B) U = 5RT and $\gamma = {6 \over 5}$
  • (C) U = 5RT and $\gamma = {7 \over 5}$
  • (D) U = ${5 \over 2}RT$ and $\gamma = {6 \over 5}$
JEE Main 2020 · Paper 1 · September 2 Shift 2 · Q17 (published compilation) Answer: C⚑ verify

An ideal gas in a closed container is slowly heated. As its temperature increases, which of the following statements are true? (A) the mean free path of the molecules decreases. (B) the mean collision time between the molecules decreases. (C) the mean free path remains unchanged. (D) the mean collision time remains unchanged.

  • (C) and (D)
  • (A) and (D)
  • (B) and (C)
  • (A) and (B)
JEE Main 2020 · Paper 1 · September 4 Shift 2 · Q23 (published compilation) Answer: 150⚑ verify

The change in the magnitude of the volume of an ideal gas when a small additional pressure $\Delta$P is applied at a constant temperature, is the same as the change when the temperature is reduced by a small quantity $\Delta$T at constant pressure. The initial temperature and pressure of the gas were 300 K and 2 atm. respectively. If |$\Delta$T| = C|$\Delta$P| then value of C in (K/atm.) is _________.

JEE Main 2020 · Paper 1 · September 6 Shift 1 · Q24 (published compilation) Answer: 5⚑ verify

Initially a gas of diatomic molecules is contained in a cylinder of volume $V_{1}$ at a pressure $P_{1}$ and temperature 250 K. Assuming that 25% of the molecules get dissociated causing a change in number of moles. The pressure of the resulting gas at temperature 2000 K, when contained in a volume $2V_{1}$ is given by $P_{2}$ . The ratio ${{{P_2}} \over {{P_1}}}$ is ________.

JEE Main 2020 · Paper 1 · January 7 Shift 2 · Q3 (published compilation) Answer: A⚑ verify

Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from ${\tau _1}$ to ${\tau _2}$ . If ${{{C_p}} \over {{C_v}}} = \gamma$ for this gas then a good estimate for ${{{\tau _2}} \over {{\tau _1}}}$ is given by :

  • (A) ${\left( 2 \right)^{{{1 + \gamma } \over 2}}}$
  • (B) 2
  • (C) ${\left( {{1 \over 2}} \right)^{{{1 + \gamma } \over 2}}}$
  • (D) ${\left( {{1 \over 2}} \right)^\gamma }$
JEE Main 2021 · Paper 1 · February 25 Shift 2 · Q15 (published compilation) Answer: C⚑ verify

Given below are two statements : Statement I : In a diatomic molecule, the rotational energy at a given temperature obeys Maxwell's distribution. Statement II : In a diatomic molecule, the rotational energy at a given temperature equals the translational kinetic energy for each molecule. In the light of the above statements, choose the correct answer from the options given below :

  • (A) Both Statement I and Statement II are true
  • (B) Both Statement I and Statement II are false
  • (C) Statement I is true but Statement II is false.
  • (D) Statement I is false but Statement II is true.
JEE Main 2021 · Paper 1 · March 18 Shift 1 · Q8 (published compilation) Answer: A⚑ verify

What will be the average value of energy along one degree of freedom for an ideal gas in thermal equilibrium at a temperature T? ($k_{B}$ is Boltzmann constant)

  • (A) ${1 \over 2}{k_B}T$
  • (B) ${2 \over 3}{k_B}T$
  • (C) ${3 \over 2}{k_B}T$
  • (D) ${k_B}T$
JEE Main 2022 · Paper 1 · July 26 Shift 2 · Q12 (published compilation) Answer: A⚑ verify

A gas has $n$ degrees of freedom. The ratio of specific heat of gas at constant volume to the specific heat of gas at constant pressure will be :

  • (A) $\frac{n}{n+2}$
  • (B) $\frac{n+2}{n}$
  • (C) $\frac{n}{2n+2}$
  • (D) $\frac{n}{n-2}$
JEE Main 2022 · Paper 1 · July 26 Shift 1 · Q13 (published compilation) Answer: B⚑ verify

7 mol of a certain monoatomic ideal gas undergoes a temperature increase of $40 \mathrm{~K}$ at constant pressure. The increase in the internal energy of the gas in this process is : (Given $\mathrm{R}=8.3 \,\mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ )

  • (A) 5810 J
  • (B) 3486 J
  • (C) 11620 J
  • (D) 6972 J
JEE Main 2022 · Paper 1 · June 29 Shift 2 · Q13 (published compilation) Answer: C⚑ verify

A vessel contains 16g of hydrogen and 128g of oxygen at standard temperature and pressure. The volume of the vessel in $cm^{3}$ is :

  • (A) 72 $\times 10^{5}$
  • (B) 32 $\times 10^{5}$
  • (C) 27 $\times 10^{4}$
  • (D) 54 $\times 10^{4}$
JEE Main 2022 · Paper 1 · July 25 Shift 2 · Q14 (published compilation) Answer: B⚑ verify

Sound travels in a mixture of two moles of helium and n moles of hydrogen. If rms speed of gas molecules in the mixture is $\sqrt2$ times the speed of sound, then the value of n will be :

  • (A) 1
  • (B) 2
  • (C) 3
  • (D) 4
JEE Main 2022 · Paper 1 · July 29 Shift 2 · Q19 (published compilation) Answer: C⚑ verify

The root mean square speed of smoke particles of mass $5 \times 10^{-17} \mathrm{~kg}$ in their Brownian motion in air at NTP is approximately. [Given $\mathrm{k}=1.38 \times 10^{-23} \mathrm{JK}^{-1}$]

  • (A) $60 \mathrm{~mm} \mathrm{~s}^{-1}$
  • (B) $12 \mathrm{~mm} \mathrm{~s}^{-1}$
  • (C) $15 \mathrm{~mm} \mathrm{~s}^{-1}$
  • (D) $36 \mathrm{~mm} \mathrm{~s}^{-1}$
JEE Main 2022 · Paper 1 · June 25 Shift 1 · Q7 (published compilation) Answer: B⚑ verify

The relation between root mean square speed ($v_{rms}$) and most probable sped ($v_{p}$) for the molar mass M of oxygen gas molecule at the temperature of 300 K will be :

  • (A) ${v_{rms}} = \sqrt {{2 \over 3}} {v_p}$
  • (B) ${v_{rms}} = \sqrt {{3 \over 2}} {v_p}$
  • (C) ${v_{rms}} = {v_p}$
  • (D) ${v_{rms}} = \sqrt {{1 \over 3}} {v_p}$
JEE Main 2022 · Paper 1 · June 26 Shift 2 · Q7 (published compilation) Answer: D⚑ verify

A flask contains argon and oxygen in the ratio of 3 : 2 in mass and the mixture is kept at 27$^\circ$C. The ratio of their average kinetic energy per molecule respectively will be :

  • (A) 3 : 2
  • (B) 9 : 4
  • (C) 2 : 3
  • (D) 1 : 1
JEE Main 2022 · Paper 1 · July 28 Shift 1 · Q8 (published compilation) Answer: D⚑ verify

Given below are two statements : Statement I : The average momentum of a molecule in a sample of an ideal gas depends on temperature. Statement II : The rms speed of oxygen molecules in a gas is $v$. If the temperature is doubled and the oxygen molecules dissociate into oxygen atoms, the rms speed will become $2 v$. In the light of the above statements, choose the correct answer from the options given below :

  • (A) Both Statement I and Statement II are true
  • (B) Both Statement I and Statement II are false
  • (C) Statement I is true but Statement II is false
  • (D) Statement I is false but Statement II is true
JEE Main 2022 · Paper 1 · July 28 Shift 2 · Q8 (published compilation) Answer: C⚑ verify

A vessel contains $14 \mathrm{~g}$ of nitrogen gas at a temperature of $27^{\circ} \mathrm{C}$. The amount of heat to be transferred to the gas to double the r.m.s speed of its molecules will be : Take $\mathrm{R}=8.32 \mathrm{~J} \mathrm{~mol}^{-1} \,\mathrm{k}^{-1}$.

  • (A) 2229 J
  • (B) 5616 J
  • (C) 9360 J
  • (D) 13,104 J
JEE Main 2022 · Paper 1 · June 27 Shift 2 · Q8 (published compilation) Answer: B⚑ verify

According to kinetic theory of gases, A. The motion of the gas molecules freezes at 0$^\circ$C. B. The mean free path of gas molecules decreases if the density of molecules is increased. C. The mean free path of gas molecules increases if temperature is increased keeping pressure constant. D. Average kinetic energy per molecule per degree of freedom is ${3 \over 2}{k_B}T$ (for monoatomic gases). Choose the most appropriate answer from the options given below :

  • (A) A and C only
  • (B) B and C only
  • (C) A and B only
  • (D) C and D only
JEE Main 2022 · Paper 1 · July 27 Shift 1 · Q9 (published compilation) Answer: C⚑ verify

Same gas is filled in two vessels of the same volume at the same temperature. If the ratio of the number of molecules is $1: 4$, then A. The r.m.s. velocity of gas molecules in two vessels will be the same. B. The ratio of pressure in these vessels will be $1: 4$. C. The ratio of pressure will be $1: 1$. D. The r.m.s. velocity of gas molecules in two vessels will be in the ratio of $1: 4$. Choose the correct answer from the options given below :

  • (A) A and C only
  • (B) B and D only
  • (C) A and B only
  • (D) C and D only
JEE Main 2022 · Paper 1 · July 27 Shift 2 · Q9 (published compilation) Answer: B⚑ verify

Which statements are correct about degrees of freedom ? (A) A molecule with n degrees of freedom has n$^{2}$ different ways of storing energy. (B) Each degree of freedom is associated with $\frac{1}{2}$ RT average energy per mole. (C) A monatomic gas molecule has 1 rotational degree of freedom where as diatomic molecule has 2 rotational degrees of freedom. (D) $\mathrm{CH}_{4}$ has a total of 6 degrees of freedom. Choose the correct answer from the options given below :

  • (B) and (C) only
  • (B) and (D) only
  • (A) and (B) only
  • (C) and (D) only
JEE Main 2022 · Paper 1 · June 26 Shift 1 · Q9 (published compilation) Answer: B⚑ verify

A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats 1.4. Vessel is moving with speed v and is suddenly brought to rest. Assuming no heat is lost to the surrounding and vessel temperature of the gas increases by : (R = universal gas constant)

  • (A) ${{M{v^2}} \over {7R}}$
  • (B) ${{M{v^2}} \over {5R}}$
  • (C) 2${{M{v^2}} \over {7R}}$
  • (D) 7${{M{v^2}} \over {5R}}$
JEE Main 2022 · Paper 1 · June 28 Shift 2 · Q9 (published compilation) Answer: B⚑ verify

What will be the effect on the root mean square velocity of oxygen molecules if the temperature is doubled and oxygen molecule dissociates into atomic oxygen?

  • (A) The velocity of atomic oxygen remains same
  • (B) The velocity of atomic oxygen doubles
  • (C) The velocity of atomic oxygen becomes half
  • (D) The velocity of atomic oxygen becomes four times
JEE Main 2023 · Paper 1 · April 6 Shift 2 · Q31 (published compilation) Answer: B⚑ verify

The temperature of an ideal gas is increased from $200 \mathrm{~K}$ to $800 \mathrm{~K}$. If r.m.s. speed of gas at $200 \mathrm{~K}$ is $v_{0}$. Then, r.m.s. speed of the gas at $800 \mathrm{~K}$ will be:

  • (A) $v_{0}$
  • (B) $2 v_{0}$
  • (C) $4 v_{0}$
  • (D) $\frac{v_{0}}{4}$
JEE Main 2023 · Paper 1 · April 11 Shift 1 · Q33 (published compilation) Answer: B⚑ verify

Three vessels of equal volume contain gases at the same temperature and pressure. The first vessel contains neon (monoatomic), the second contains chlorine (diatomic) and third contains uranium hexafloride (polyatomic). Arrange these on the basis of their root mean square speed $\left(v_{\mathrm{rms}}\right)$ and choose the correct answer from the options given below:

  • (A) $\mathrm{v}_{\mathrm{rms}}($ mono $)=\mathrm{v}_{\mathrm{rms}}($ dia $)=\mathrm{v}_{\mathrm{rms}}($ poly $)$
  • (B) $\mathrm{v}_{\mathrm{rms}}$ (mono) $> \mathrm{v}_{\mathrm{rms}}($ dia $) > \mathrm{v}_{\mathrm{rms}}$ (poly)
  • (C) $\mathrm{v}_{\mathrm{rms}}$ (dia) $< \mathrm{v}_{\mathrm{rms}}$ (poly) $< \mathrm{v}_{\text {rms }}$ (mono)
  • (D) $\mathrm{v}_{\mathrm{rms}}$ (mono) $< \mathrm{v}_{\mathrm{rms}}$ (dia) $< \mathrm{v}_{\mathrm{rms}}$ (poly)
JEE Main 2023 · Paper 1 · April 6 Shift 2 · Q34 (published compilation) Answer: D⚑ verify

The ratio of speed of sound in hydrogen gas to the speed of sound in oxygen gas at the same temperature is:

  • (A) $1: 1$
  • (B) $1: 2$
  • (C) $1: 4$
  • (D) $4: 1$
JEE Main 2023 · Paper 1 · April 13 Shift 2 · Q38 (published compilation) Answer: C⚑ verify

The mean free path of molecules of a certain gas at STP is $1500 \mathrm{~d}$, where $\mathrm{d}$ is the diameter of the gas molecules. While maintaining the standard pressure, the mean free path of the molecules at $373 \mathrm{~K}$ is approximately:

  • (A) $750 \mathrm{~d}$
  • (B) $1500 \mathrm{~d}$
  • (C) $\mathrm{2049~ d}$
  • (D) $1098 \mathrm{~d}$
JEE Main 2023 · Paper 1 · April 6 Shift 1 · Q41 (published compilation) Answer: B⚑ verify

The number of air molecules per cm$^3$ increased from $3\times10^{19}$ to $12\times10^{19}$. The ratio of collision frequency of air molecules before and after the increase in number respectively is:

  • (A) 1.25
  • (B) 0.25
  • (C) 0.50
  • (D) 0.75
JEE Main 2023 · Paper 1 · April 10 Shift 2 · Q42 (published compilation) Answer: D⚑ verify

A gas mixture consists of 2 moles of oxygen and 4 moles of neon at temperature T. Neglecting all vibrational modes, the total internal energy of the system will be,

  • (A) 4RT
  • (B) 16RT
  • (C) 8RT
  • (D) 11RT
JEE Main 2023 · Paper 1 · April 8 Shift 2 · Q42 (published compilation) Answer: C⚑ verify

The temperature at which the kinetic energy of oxygen molecules becomes double than its value at $27^{\circ} \mathrm{C}$ is

  • (A) $627^{\circ} \mathrm{C}$
  • (B) $927^{\circ} \mathrm{C}$
  • (C) $327^{\circ} \mathrm{C}$
  • (D) $1227^{\circ} \mathrm{C}$
JEE Main 2023 · Paper 1 · April 11 Shift 2 · Q47 (published compilation) Answer: D⚑ verify

The root mean square speed of molecules of nitrogen gas at $27^{\circ} \mathrm{C}$ is approximately : (Given mass of a nitrogen molecule $=4.6 \times 10^{-26} \mathrm{~kg}$ and take Boltzmann constant $\mathrm{k}_{\mathrm{B}}=1.4 \times 10^{-23} \mathrm{JK}^{-1}$ )

  • (A) 91 m/s
  • (B) 1260 m/s
  • (C) 27.4 m/s
  • (D) 523 m/s
JEE Main 2023 · Paper 1 · April 13 Shift 2 · Q49 (published compilation) Answer: D⚑ verify

The initial pressure and volume of an ideal gas are P$_0$ and V$_0$. The final pressure of the gas when the gas is suddenly compressed to volume $\frac{V_0}{4}$ will be : (Given $\gamma$ = ratio of specific heats at constant pressure and at constant volume)

  • (A) P$_0$(4)$^{\frac{1}{\gamma}}$
  • (B) P$_0$
  • (C) 4P$_0$
  • (D) P$_0$(4)$^{\gamma}$
JEE Main 2023 · Paper 1 · April 15 Shift 1 · Q49 (published compilation) Answer: D⚑ verify

A flask contains Hydrogen and Argon in the ratio $2: 1$ by mass. The temperature of the mixture is $30^{\circ} \mathrm{C}$. The ratio of average kinetic energy per molecule of the two gases ( $\mathrm{K}$ argon/K hydrogen) is : (Given: Atomic Weight of $\mathrm{Ar}=39.9$ )

  • (A) $\frac{39.9}{2}$
  • (B) 2
  • (C) 39.9
  • (D) 1
JEE Main 2026 · Paper 1 · April 4 Shift 2 · Q33 (published compilation) Answer: C⚑ verify

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A : If the average kinetic energy of $\mathrm{H}_2$ and $\mathrm{O}_2$ molecules, kept in two different sized containers are same, then their temperatures will be same. Reason R : The r.m.s speed of $\mathrm{H}_2$ and $\mathrm{O}_2$ molecules are same at same temperature. Choose the correct answer from the options given below

  • (A) Both $\mathbf{A}$ and $\mathbf{R}$ are true and $\mathbf{R}$ is the correct explanation of $\mathbf{A}$
  • (B) Both $\mathbf{A}$ and $\mathbf{R}$ are true but $\mathbf{R}$ is NOT the correct explanation of $\mathbf{A}$
  • (C) A is true but $\mathbf{R}$ is false
  • (D) A is false but $\mathbf{R}$ is true
JEE Main 2026 · Paper 1 · April 4 Shift 1 · Q35 (published compilation) Answer: B⚑ verify

One gas of $n_1$ mole of molecules at temperature $T_1$, volume $V_1$, and pressure $P_1$, and another gas of $n_2$ mole of molecules at temperature $T_2$, volume $V_2$, and pressure $P_2$, are mixed resulting in pressure $P$ and volume $V$ of the mixture. The temperature of the mixture is $\_\_\_\_$ .

  • (A) $\frac{\left(\mathrm{T}_1+\mathrm{T}_2\right)}{2}$
  • (B) $\frac{T_1 T_2 P V}{T_2 P_1 V_1+T_1 P_2 V_2}$
  • (C) $\frac{\left(T_2 P_1 V_1+T_1 P_2 V_2\right)}{T_1 T_2 P V}$
  • (D) $\frac{\left[\mathrm{T}_1-\mathrm{T}_2\right]}{2}$
JEE Main 2026 · Paper 1 · April 6 Shift 2 · Q35 (published compilation) Answer: D⚑ verify

If 2 mole of an ideal monoatomic gas at temperature $T$, is mixed with 6 mole of another ideal monoatomic gas at temperature $2 T$ then the temperature of mixture is:

  • (A) $\frac{5}{2} T$
  • (B) $\frac{5}{4} T$
  • (C) $\frac{7}{2} T$
  • (D) $\frac{7}{4} T$

🎯 Question Bank 105 MCQs · graded

Distribution — advanced: 16 · easy: 39 · hard: 18 · medium: 32. Every question carries a source trace; each ends in an SME-verify solution.

Q1 Matter is made up of easy
Step solution + source
Kinetic theory begins from the atomic picture: molecules (made up of one or more atoms) constitute matter. Bulk properties like pressure and temperature emerge from the motion of these $\sim 10^{23}$ molecules, not from a continuous jelly. 🔉⇢

Source: NCERT-derived

Q2 A typical atomic size is of the order of easy
Step solution + source
Atoms are about an angstrom across, $1\,\text{\AA}=10^{-10}\,\text{m}$. This tiny size means a millimetre of matter stacks roughly $10^{7}$ atoms, which is why matter looks continuous at human scales. 🔉⇢

Source: NCERT-derived

Q3 Compared with solids and liquids, molecules in a gas are easy
Step solution + source
In a gas the atoms are much freer than in solids or liquids and can travel long distances without colliding. The mean spacing is $d\approx n^{-1/3}$, about ten times the molecular size at ordinary conditions. 🔉⇢

Source: NCERT-derived

Q4 Avogadro's number $N_A$ is approximately easy
Step solution + source
One mole contains $N_A=6.02\times10^{23}$ entities. It converts between the molecular and macroscopic worlds: the molar mass in grams holds $N_A$ molecules, so a single molecule's mass is $M/N_A$. 🔉⇢

Source: NCERT-derived

Q5 Avogadro's hypothesis states that equal volumes of gases at the same T and P contain easy
Step solution + source
At the same temperature and pressure, equal volumes hold equal numbers of molecules, $N/V$ being fixed by $P/kT$. This follows from $PV=NkT$: at common $P,T$ the number density $n=N/V=P/kT$ is the same for every gas. 🔉⇢

Source: NCERT-derived

Q6 The number density $n$ of a gas is defined as medium
Step solution + source
Number density $n=N/V$ counts molecules per cubic metre. It links to pressure through $P=nkT$ and to mean spacing through $d\approx n^{-1/3}$, and it drops as a gas expands. 🔉⇢

Source: NCERT-derived

Q7 If a gas of number density $n$ has molecular mass $m$, its mass density is medium
Step solution + source
Mass density is molecule mass times number per volume: $\rho=mn=mN/V$. For an ideal gas this also equals $\rho=PM/RT$ once you write $m=M/N_A$ and $n=PN_A/RT$. 🔉⇢

Source: NCERT-derived

Q8 The mean intermolecular spacing in a gas at STP is of order medium
Step solution + source
At STP $n\approx2.7\times10^{25}\,\text{m}^{-3}$, so $d\approx n^{-1/3}\approx3\times10^{-9}\,\text{m}$. Since a molecule is $\sim3\times10^{-10}\,\text{m}$, spacing is about ten diameters, leaving plenty of empty space for free motion. 🔉⇢

Source: NCERT-derived

Q9 The number of molecules in $2\,\text{mol}$ of an ideal gas is medium
Step solution + source
Molecule count is $N=\nu N_A=2\times6.02\times10^{23}=1.204\times10^{24}$. The number of molecules scales linearly with the number of moles, independent of which gas it is. 🔉⇢

Source: NCERT-derived

Q10 At STP, the volume occupied by the molecules themselves in $22.4\,\text{L}$ of gas is hard
Step solution + source
With $d\approx10\times$ the molecular size, the packing fraction is $\sim(1/10)^3=10^{-3}$. Thus molecular volume is $\lesssim0.1\%$ of $V$: $V_{mol}/V\sim10^{-3}$, justifying the ideal-gas neglect of molecular volume. 🔉⇢

Source: NCERT-derived

Q11 A cube of side $1\,\text{cm}$ of a solid contains atoms of order hard
Step solution + source
With spacing $\sim2\times10^{-10}\,\text{m}$ in a solid, atoms per edge $\approx0.01/2\times10^{-10}=5\times10^{7}$, so per cube $N\approx(5\times10^{7})^3\approx10^{23}$; order-of-magnitude $10^{22}$–$10^{23}$. 🔉⇢

Source: NCERT-derived

Q12 Estimate the number of moles in $1\,\text{L}$ of water (density $1000\,\text{kg m}^{-3}$, molar mass $18\,\text{g}$). advanced
Step solution + source
Mass is $1\,\text{L}\times1000\,\text{kg m}^{-3}=1\,\text{kg}=1000\,\text{g}$, so $\nu=1000/18\approx55.6\,\text{mol}$. Hence $N=55.6\times N_A\approx3.3\times10^{25}$ molecules of water. 🔉⇢

Source: NCERT-derived

Q13 Two vessels of equal volume at the same T and P hold $\text{H}_2$ and $\text{O}_2$. The ratio of the number of molecules $N_{H_2}:N_{O_2}$ is advanced
Step solution + source
By Avogadro's law, equal volumes at equal $T,P$ contain equal numbers of molecules: $N=PV/kT$ is independent of species, so $N_{H_2}:N_{O_2}=1:1$ even though their masses differ by 16. 🔉⇢

Source: NCERT-derived

Q14 The ideal gas equation can be written as easy
Step solution + source
The equation of state is $PV=nRT=NkT$, with $R=8.314\,\text{J mol}^{-1}\text{K}^{-1}$ and $k=R/N_A$. It combines Boyle's, Charles' and Avogadro's laws into one relation valid for a dilute (ideal) gas. 🔉⇢

Source: NCERT-derived

Q15 Boyle's law (fixed T and amount) states that easy
Step solution + source
At constant temperature and amount, pressure and volume vary inversely: $P\propto1/V$, i.e. $PV=\text{const}$. Doubling the pressure halves the volume, a $PV$-hyperbola isotherm. 🔉⇢

Source: NCERT-derived

Q16 Charles' law (fixed P and amount) states that easy
Step solution + source
At constant pressure the volume of a fixed amount of gas is proportional to absolute temperature: $V/T=\text{const}$. This only holds with $T$ measured from absolute zero in kelvin. 🔉⇢

Source: NCERT-derived

Q17 Dalton's law of partial pressures says that for a non-reacting mixture easy
Step solution + source
Because ideal-gas molecules do not interact, the total pressure of a mixture of ideal gases is the sum of partial pressures: $P=P_1+P_2+\dots=(n_1+n_2+\dots)RT/V$. 🔉⇢

Source: NCERT-derived

Q18 Boltzmann's constant $k$ equals easy
Step solution + source
Boltzmann's constant is the gas constant per molecule, $k=R/N_A=8.314/6.02\times10^{23}\approx1.38\times10^{-23}\,\text{J K}^{-1}$. It appears whenever a gas law is written per molecule, e.g. $PV=NkT$. 🔉⇢

Source: NCERT-derived

Q19 At STP ($0^\circ\text{C}$, $1\,\text{atm}$) one mole of an ideal gas occupies easy
Step solution + source
Molar volume at STP is $V_m=RT/P=8.314\times273/1.013\times10^{5}\approx0.0224\,\text{m}^3=22.4\,\text{L}$, the same for every ideal gas by Avogadro's law. 🔉⇢

Source: NCERT-derived

Q20 The gas constant $R$ has the value easy
Step solution + source
The universal gas constant is $R=8.314\,\text{J mol}^{-1}\text{K}^{-1}$ (or $0.0821\,\text{L atm mol}^{-1}\text{K}^{-1}$). It equals $N_A k$ and sets the scale of $PV=nRT$. 🔉⇢

Source: NCERT-derived

Q21 A gas at $27^\circ\text{C}$ is heated at constant pressure until its volume doubles. Its final temperature is medium
Step solution + source
By Charles' law $V/T=\text{const}$: doubling $V$ doubles absolute $T$. From $T_1=300\,\text{K}$, $T_2=600\,\text{K}=327^\circ\text{C}$. Work in kelvin, then convert back. 🔉⇢

Source: NCERT-derived

Q22 A gas at $1\,\text{atm}$ is compressed isothermally to one-third of its volume. Its new pressure is medium
Step solution + source
Isothermal means $PV=\text{const}$, so $P_2=P_1V_1/V_2=1\times3=3\,\text{atm}$. Pressure and volume trade off inversely at fixed temperature. 🔉⇢

Source: NCERT-derived

Q23 Two moles of $\text{He}$ and three moles of $\text{Ne}$ share a vessel at $T$. If He alone would exert $2P_0$, the total pressure is medium
Step solution + source
Partial pressure is proportional to moles: $P_i=n_iRT/V$. He gives $2P_0$ for 2 mol, so per mole $P_0$; Ne (3 mol) gives $3P_0$. Total $=2P_0+3P_0=5P_0$ by Dalton's law. 🔉⇢

Source: NCERT-derived

Q24 The mass density of an ideal gas is given by medium
Step solution + source
From $PV=(m/M)RT$ with $\rho=m/V$ we get $\rho=PM/RT$. Density rises with pressure and molar mass and falls with temperature, so hot gas is lighter and rises. 🔉⇢

Source: NCERT-derived

Q25 If pressure and volume are both doubled, the absolute temperature of a fixed amount of ideal gas becomes medium
Step solution + source
Since $PV/T=\text{const}$, $T_2=T_1(P_2V_2)/(P_1V_1)=T_1(2)(2)=4T_1$. The combined gas law multiplies the pressure and volume factors. 🔉⇢

Source: NCERT-derived

Q26 A real gas approaches ideal behaviour best at hard
Step solution + source
At low pressure (large spacing) and high temperature (large KE), intermolecular forces and molecular volume are negligible, so $PV/nRT\to1$. Deviations grow when the gas is dense or cold, near liquefaction. 🔉⇢

Source: NCERT-derived

Q27 For a fixed mass of gas, which quantity stays constant along an isotherm? hard
Step solution + source
An isotherm has $T=\text{const}$, and $PV=nRT$ makes $PV=\text{const}$. On a $P$–$V$ diagram this is a rectangular hyperbola; each point has the same $PV$. 🔉⇢

Source: NCERT-derived

Q28 A vessel holds $\text{O}_2$ at pressure $P$. An equal number of $\text{N}_2$ molecules is added at the same T. The new pressure is hard
Step solution + source
Pressure depends on number of molecules, not their mass: $P=NkT/V$. Doubling $N$ at fixed $T,V$ doubles pressure to $2P$, whatever the species — a consequence of Dalton's law. 🔉⇢

Source: NCERT-derived

Q29 A closed rigid vessel of $\text{O}_2$ at $300\,\text{K}$, $1\,\text{atm}$ is heated to $600\,\text{K}$. The pressure becomes medium
Step solution + source
At constant volume $P/T=\text{const}$ (Gay-Lussac): $P_2=P_1T_2/T_1=1\times600/300=2\,\text{atm}$. Rigid vessel means $V$ fixed. 🔉⇢

Source: NCERT-derived

Q30 Two moles of an ideal gas occupy $49.2\,\text{L}$ at $300\,\text{K}$. The pressure is (use $R=8.314$) advanced
Step solution + source
$P=nRT/V=2\times8.314\times300/(49.2\times10^{-3})\approx1.01\times10^{5}\,\text{Pa}$, i.e. about one atmosphere. Convert litres to $\text{m}^3$ before substituting. 🔉⇢

Source: NCERT-derived

Q31 The number density of molecules in an ideal gas at pressure $P$ and temperature $T$ is advanced
Step solution + source
From $P=nkT$, $n=P/kT$. At STP $n=1.013\times10^{5}/(1.38\times10^{-23}\times273)\approx2.7\times10^{25}\,\text{m}^{-3}$, the Loschmidt number. 🔉⇢

Source: NCERT-derived

Q32 A key assumption of the kinetic theory is that gas molecules are in easy
Step solution + source
The theory assumes molecules of a gas are in incessant random motion, colliding elastically with each other and the walls. Pressure $P=\tfrac13 n m\overline{v^2}$ arises from the momentum these collisions deliver to the walls. 🔉⇢

Source: NCERT-derived

Q33 In kinetic theory, collisions of molecules with the walls are taken to be easy
Step solution + source
Wall collisions are elastic, so kinetic energy is conserved and the wall merely reverses the normal velocity component. A molecule hitting with $v_x$ rebounds with $-v_x$, transferring momentum $2mv_x$. 🔉⇢

Source: NCERT-derived

Q34 The pressure exerted by an ideal gas is easy
Step solution + source
The central kinetic-theory result is $P=\tfrac13\,n\,m\,\overline{v^2}$ where $n=N/V$. The factor $\tfrac13$ comes from averaging over the three equivalent directions, $\overline{v_x^2}=\tfrac13\overline{v^2}$. 🔉⇢

Source: NCERT-derived

Q35 In terms of density $\rho$, the gas pressure is easy
Step solution + source
Writing $\rho=nm$, the pressure relation becomes $P=\tfrac13\rho\,\overline{v^2}$. This lets one find the mean-square speed directly from measurable $P$ and $\rho$. 🔉⇢

Source: NCERT-derived

Q36 When a molecule of mass $m$ and speed $v_x$ rebounds elastically off a wall, the momentum transferred is easy
Step solution + source
The normal velocity reverses from $+v_x$ to $-v_x$, so the molecule's momentum changes by $2mv_x$ and the wall receives $2mv_x$ per impact. Summing these impulses over many molecules gives the pressure. 🔉⇢

Source: NCERT-derived

Q37 Which is NOT an assumption of the kinetic theory of an ideal gas? easy
Step solution + source
Ideal-gas molecules are taken as free except during brief elastic collisions; forces act only during contact, not continuously. The neglected long-range attraction is exactly what distinguishes an ideal gas from a real one, where $a/V^2$ corrections appear. 🔉⇢

Source: NCERT-derived

Q38 Kinetic theory gives $PV=\dfrac{2}{3}E$, where $E$ is the medium
Step solution + source
Since $P=\tfrac13 nm\overline{v^2}$ and $E=\tfrac12 Nm\overline{v^2}$, we get $PV=\tfrac23E$. Thus pressure is a direct measure of the total translational kinetic energy per unit volume. 🔉⇢

Source: NCERT-derived

Q39 For a gas, $\dfrac{1}{2}m\,\overline{v^2}$ represents the medium
Step solution + source
$\tfrac12 m\overline{v^2}$ is the mean translational KE of one molecule; multiplied by $N$ it gives $E$. Kinetic theory shows this equals $\tfrac32kT$, tying speed to temperature. 🔉⇢

Source: NCERT-derived

Q40 If the volume is halved at constant temperature, the pressure predicted by kinetic theory medium
Step solution + source
At fixed $T$, $\overline{v^2}$ is unchanged, so $P=\tfrac13(N/V)m\overline{v^2}\propto1/V$. Halving $V$ doubles $P$, exactly Boyle's law recovered from molecular motion. 🔉⇢

Source: NCERT-derived

Q41 For a gas with $P=1.0\times10^{5}\,\text{Pa}$ and $\rho=1.25\,\text{kg m}^{-3}$, the mean-square speed $\overline{v^2}$ is medium
Step solution + source
From $P=\tfrac13\rho\overline{v^2}$, $\overline{v^2}=3P/\rho=3\times10^{5}/1.25=2.4\times10^{5}\,\text{m}^2\text{s}^{-2}$, so $v_{rms}\approx490\,\text{m s}^{-1}$. 🔉⇢

Source: NCERT-derived

Q42 Kinetic theory predicts that at fixed temperature the pressure of an ideal gas is independent of hard
Step solution + source
Since $\tfrac12 m\overline{v^2}=\tfrac32kT$, the product $m\overline{v^2}=3kT$ depends only on $T$. Hence $P=\tfrac13 n\,m\overline{v^2}=nkT$ has no leftover mass dependence — heavier molecules simply move slower. 🔉⇢

Source: NCERT-derived

Q43 At the same temperature, hydrogen and oxygen have equal hard
Step solution + source
Average KE is $\tfrac32kT$, set by temperature alone, so it is equal for $\text{H}_2$ and $\text{O}_2$ at the same $T$. Their speeds differ because $\overline{v^2}=3kT/m$ favours the lighter molecule. 🔉⇢

Source: NCERT-derived

Q44 The relation $PV=\tfrac{2}{3}E$ combined with $PV=NkT$ shows the energy per molecule is hard
Step solution + source
Equating $\tfrac23E=NkT$ gives $E=\tfrac32NkT$, so per molecule $\tfrac{E}{N}=\tfrac32kT$. This is the microscopic meaning of temperature for a monatomic gas. 🔉⇢

Source: NCERT-derived

Q45 A gas has $v_{rms}=500\,\text{m s}^{-1}$ at pressure $P$. If it is compressed isothermally to double the pressure, $v_{rms}$ becomes advanced
Step solution + source
Isothermal compression keeps $T$ and hence $v_{rms}=\sqrt{3kT/m}$ constant. Doubling pressure by halving volume changes $n$, not the molecular speed; $v_{rms}$ depends only on temperature. 🔉⇢

Source: NCERT-derived

Q46 For $2\times10^{25}$ molecules of mass $5\times10^{-26}\,\text{kg}$ in $0.02\,\text{m}^3$ with $\overline{v^2}=3\times10^{5}\,\text{m}^2\text{s}^{-2}$, the pressure is advanced
Step solution + source
$P=\tfrac13\tfrac{N}{V}m\overline{v^2}=\tfrac13\times(2\times10^{25}/0.02)\times5\times10^{-26}\times3\times10^{5}=5\times10^{6}\,\text{Pa}$. Substitute carefully and keep SI units. 🔉⇢

Source: NCERT-derived

Q47 If the absolute temperature of an ideal gas is quadrupled, its $v_{rms}$ advanced
Step solution + source
Since $v_{rms}=\sqrt{3kT/m}\propto\sqrt{T}$, a factor 4 in $T$ gives a factor $\sqrt4=2$ in speed. Speed grows only as the square root of temperature. 🔉⇢

Source: NCERT-derived

Q48 The kinetic interpretation of temperature states that the average kinetic energy of a molecule is easy
Step solution + source
Kinetic theory shows the average kinetic energy of a molecule is proportional to the absolute temperature of the gas: $\tfrac12 m\overline{v^2}=\tfrac32kT$. Temperature is thus a direct measure of molecular motion. 🔉⇢

Source: NCERT-derived

Q49 The average translational kinetic energy of a gas molecule is easy
Step solution + source
For any ideal gas, $\langle KE\rangle=\tfrac32kT$ per molecule, independent of mass or species. At $300\,\text{K}$ this is $\tfrac32\times1.38\times10^{-23}\times300\approx6.2\times10^{-21}\,\text{J}$. 🔉⇢

Source: NCERT-derived

Q50 The root-mean-square speed of gas molecules is easy
Step solution + source
Setting $\tfrac12 m\overline{v^2}=\tfrac32kT$ gives $v_{rms}=\sqrt{3kT/m}=\sqrt{3RT/M}$, using $k=R/N_A$ and $M=mN_A$. Lighter, hotter gases move faster. 🔉⇢

Source: NCERT-derived

Q51 At absolute zero, the kinetic-theory model predicts the molecular motion easy
Step solution + source
Since $\langle KE\rangle=\tfrac32kT$, as $T\to0$ the average kinetic energy tends to zero. Classically all translational motion stops; absolute zero is the temperature of vanishing molecular kinetic energy. 🔉⇢

Source: NCERT-derived

Q52 At the same temperature, the ratio of $v_{rms}$ of hydrogen to oxygen ($M=2$ vs $32$) is easy
Step solution + source
Since $v_{rms}\propto1/\sqrt{M}$ at fixed $T$, $\dfrac{v_{H_2}}{v_{O_2}}=\sqrt{32/2}=\sqrt{16}=4$. The lighter molecule is four times faster. 🔉⇢

Source: NCERT-derived

Q53 Temperature is a measure of easy
Step solution + source
Temperature reflects the mean random kinetic energy per molecule, $\tfrac32kT$, not the total energy (which also depends on $N$) nor pressure. Hotter means faster molecules on average. 🔉⇢

Source: NCERT-derived

Q54 If the temperature of a gas rises from $300\,\text{K}$ to $1200\,\text{K}$, the average KE per molecule medium
Step solution + source
Average KE is $\tfrac32kT\propto T$. Raising $T$ from 300 to $1200\,\text{K}$ (factor 4) multiplies the KE by 4. Note the speed only doubles, since $v\propto\sqrt T$. 🔉⇢

Source: NCERT-derived

Q55 The $v_{rms}$ of oxygen ($M=0.032\,\text{kg mol}^{-1}$) at $300\,\text{K}$ is about medium
Step solution + source
$v_{rms}=\sqrt{3RT/M}=\sqrt{3\times8.314\times300/0.032}\approx483\,\text{m s}^{-1}$. This is comparable to the speed of sound in the gas, as expected. 🔉⇢

Source: NCERT-derived

Q56 Two gases at the same temperature must have equal medium
Step solution + source
$\langle KE\rangle=\tfrac32kT$ depends only on $T$, so it is equal for both gases. Their speeds and densities differ because those also depend on molecular mass. 🔉⇢

Source: NCERT-derived

Q57 To double the $v_{rms}$ of a gas initially at $300\,\text{K}$, its temperature must be raised to medium
Step solution + source
Since $v_{rms}\propto\sqrt T$, doubling the speed needs a factor 4 in $T$: $T_2=4\times300=1200\,\text{K}$. Speed scales with the square root of absolute temperature. 🔉⇢

Source: NCERT-derived

Q58 The average kinetic energy per molecule at $27^\circ\text{C}$ is (use $k=1.38\times10^{-23}$) hard
Step solution + source
$\langle KE\rangle=\tfrac32kT=\tfrac32\times1.38\times10^{-23}\times300\approx6.2\times10^{-21}\,\text{J}$. Remember to convert $27^\circ\text{C}$ to $300\,\text{K}$ first. 🔉⇢

Source: NCERT-derived

Q59 A gas is heated so its $v_{rms}$ increases by $50\%$. The ratio of final to initial temperature is hard
Step solution + source
$v_{rms}\propto\sqrt T$, so $T_2/T_1=(v_2/v_1)^2=(1.5)^2=2.25$. A 50% speed rise needs a 125% temperature rise. 🔉⇢

Source: NCERT-derived

Q60 At what temperature is the $v_{rms}$ of $\text{O}_2$ equal to the $v_{rms}$ of $\text{H}_2$ at $300\,\text{K}$? ($M_O=32,M_H=2$) hard
Step solution + source
Equal $v_{rms}$ needs $T/M$ equal: $T_O/32=300/2$, so $T_O=300\times16=4800\,\text{K}$. The heavier gas needs a proportionally higher temperature to match the light gas's speed. 🔉⇢

Source: NCERT-derived

Q61 The total translational kinetic energy of $2\,\text{mol}$ of an ideal gas at $300\,\text{K}$ is advanced
Step solution + source
$E=\tfrac32nRT=\tfrac32\times2\times8.314\times300\approx7483\,\text{J}$. This is the translational part only; a diatomic gas would carry more via rotation. 🔉⇢

Source: NCERT-derived

Q62 A vessel contains a mixture of $\text{He}$ and $\text{Ar}$ at $400\,\text{K}$. The ratio of average KE per molecule He:Ar is advanced
Step solution + source
Average KE per molecule is $\tfrac32kT$, the same for both since they share one temperature. Mass affects speed, not energy: $\langle KE\rangle_{He}:\langle KE\rangle_{Ar}=1:1$. 🔉⇢

Source: NCERT-derived

Q63 If a monatomic gas is compressed so its temperature rises from $300\,\text{K}$ to $675\,\text{K}$, its $v_{rms}$ increases by a factor advanced
Step solution + source
$v_{rms}\propto\sqrt T$, so the factor is $\sqrt{675/300}=\sqrt{2.25}=1.5$. The speed grows as the square root of the temperature ratio. 🔉⇢

Source: NCERT-derived

Q64 The law of equipartition of energy states that easy
Step solution + source
In thermal equilibrium the total energy is equally distributed in all possible energy modes, each quadratic degree of freedom carrying an average $\tfrac12kT$. This is known as the law of equipartition of energy. 🔉⇢

Source: NCERT-derived

Q65 Each quadratic degree of freedom of a molecule carries an average energy of easy
Step solution + source
Equipartition assigns $\tfrac12kT$ to each quadratic term in the energy (each velocity component, each rotation, each vibrational coordinate). A monatomic gas with 3 translational modes thus has $\tfrac32kT$ per molecule. 🔉⇢

Source: NCERT-derived

Q66 A monatomic ideal gas molecule has how many degrees of freedom? easy
Step solution + source
A point-like monatomic molecule has only translational motion in 3 directions, so $f=3$. Its energy per molecule is $\tfrac{f}{2}kT=\tfrac32kT$; there are no rotational modes to excite. 🔉⇢

Source: NCERT-derived

Q67 A diatomic molecule (no vibration) has degrees of freedom easy
Step solution + source
A rigid diatomic molecule adds 2 rotational modes about the two axes perpendicular to the bond, giving $f=5$. Rotation about the bond axis has negligible moment of inertia and is not counted. 🔉⇢

Source: NCERT-derived

Q68 The internal energy of $n$ moles of an ideal gas with $f$ degrees of freedom is easy
Step solution + source
Summing $\tfrac12kT$ over $f$ modes and over $N=nN_A$ molecules gives $U=\tfrac{f}{2}nRT$. For monatomic $f=3$ ($U=\tfrac32nRT$), for diatomic $f=5$ ($U=\tfrac52nRT$). 🔉⇢

Source: NCERT-derived

Q69 For a diatomic gas at temperature $T$, the internal energy per mole is medium
Step solution + source
With $f=5$, $U=\tfrac{f}{2}RT=\tfrac52RT$ per mole. The two extra rotational modes beyond the monatomic $\tfrac32RT$ add $RT$ of energy. 🔉⇢

Source: NCERT-derived

Q70 A vibrational mode contributes to the energy medium
Step solution + source
A vibration has both kinetic and potential quadratic terms, each worth $\tfrac12kT$, so one active vibrational mode contributes $kT$. This is why high-temperature diatomic gases have larger heat capacities. 🔉⇢

Source: NCERT-derived

Q71 The ratio of internal energies of equal moles of monatomic and diatomic gases at the same T is medium
Step solution + source
$U\propto f$: monatomic $f=3$, diatomic $f=5$, so $U_{mono}:U_{di}=3:5$ at equal $n$ and $T$. More degrees of freedom store more energy. 🔉⇢

Source: NCERT-derived

Q72 At high temperature a diatomic gas whose vibration is active has degrees of freedom medium
Step solution + source
Adding one vibrational mode (2 quadratic terms) to the 5 rigid-diatomic modes gives $f=7$, so $U=\tfrac72nRT$. Vibration only 'switches on' once $kT$ exceeds the vibrational quantum. 🔉⇢

Source: NCERT-derived

Q73 The average energy associated with one translational degree of freedom at $300\,\text{K}$ is medium
Step solution + source
One quadratic mode holds $\tfrac12kT=\tfrac12\times1.38\times10^{-23}\times300\approx2.07\times10^{-21}\,\text{J}$. Three such modes give the familiar $\tfrac32kT$ translational energy. 🔉⇢

Source: NCERT-derived

Q74 For a triatomic non-linear gas molecule (rigid), the number of degrees of freedom is hard
Step solution + source
A non-linear molecule can rotate about all three axes, adding 3 rotational modes to the 3 translational, so $f=6$ and $U=3nRT$ per mole (rigid). A linear triatomic instead has only 2 rotational modes. 🔉⇢

Source: NCERT-derived

Q75 The internal energy of $3\,\text{mol}$ of a monatomic gas at $400\,\text{K}$ is hard
Step solution + source
$U=\tfrac32nRT=\tfrac32\times3\times8.314\times400\approx1.5\times10^{4}\,\text{J}$. For monatomic gases only the 3 translational modes contribute. 🔉⇢

Source: NCERT-derived

Q76 A gas mixture has $2\,\text{mol}$ monatomic and $2\,\text{mol}$ diatomic gas at $T$. Its total internal energy is advanced
Step solution + source
Monatomic: $\tfrac32\times2\,RT=3RT$; diatomic: $\tfrac52\times2\,RT=5RT$; total $U=3RT+5RT=8RT$. Add the contributions of each component using its own $f$. 🔉⇢

Source: NCERT-derived

Q77 Equipartition predicts the molar specific heat at constant volume of a rigid diatomic gas to be advanced
Step solution + source
Since $U=\tfrac{f}{2}RT$ per mole, $C_v=\dfrac{dU}{dT}=\tfrac{f}{2}R=\tfrac52R$ for $f=5$. Equipartition thus links microscopic degrees of freedom to a measurable heat capacity. 🔉⇢

Source: NCERT-derived

Q78 For an ideal gas, Mayer's relation is easy
Step solution + source
The molar heat capacities obey $C_p-C_v=R$. Physically, heating at constant pressure must also supply the work of expansion, exactly $R$ per mole per kelvin more than at constant volume. 🔉⇢

Source: NCERT-derived

Q79 The molar $C_v$ of a gas with $f$ degrees of freedom is easy
Step solution + source
From $U=\tfrac{f}{2}RT$, $C_v=dU/dT=\tfrac{f}{2}R$. Monatomic gives $\tfrac32R$, rigid diatomic $\tfrac52R$; more modes mean a larger heat capacity. 🔉⇢

Source: NCERT-derived

Q80 The adiabatic ratio $\gamma$ is defined as easy
Step solution + source
The ratio of specific heats is $\gamma=C_p/C_v=1+2/f$. It governs adiabatic processes ($PV^\gamma=\text{const}$) and the speed of sound, and it decreases as $f$ grows. 🔉⇢

Source: NCERT-derived

Q81 For a monatomic ideal gas, $\gamma$ equals easy
Step solution + source
With $f=3$, $C_v=\tfrac32R$, $C_p=\tfrac52R$, so $\gamma=C_p/C_v=5/3\approx1.67$. Equivalently $\gamma=1+2/f=1+2/3$. 🔉⇢

Source: NCERT-derived

Q82 For a rigid diatomic gas, $\gamma$ equals easy
Step solution + source
With $f=5$, $C_v=\tfrac52R$ and $C_p=\tfrac72R$, giving $\gamma=7/5=1.4$. This is the value used for air ($\text{N}_2,\text{O}_2$) at ordinary temperatures. 🔉⇢

Source: NCERT-derived

Q83 The molar $C_p$ of a monatomic ideal gas is medium
Step solution + source
$C_p=C_v+R=\tfrac32R+R=\tfrac52R$ for a monatomic gas. The extra $R$ over $C_v$ pays for expansion work at constant pressure. 🔉⇢

Source: NCERT-derived

Q84 The Dulong–Petit law gives the molar specific heat of most solids as about medium
Step solution + source
A solid atom has 3 vibrational modes, each with kinetic and potential $\tfrac12kT$, totalling $3kT$ per atom, so $C=3R\approx25\,\text{J mol}^{-1}\text{K}^{-1}$. This is the Dulong–Petit value. 🔉⇢

Source: NCERT-derived

Q85 Relation between $\gamma$ and degrees of freedom $f$ is medium
Step solution + source
Using $C_v=\tfrac{f}{2}R$ and $C_p=C_v+R$, $\gamma=C_p/C_v=(f+2)/f=1+2/f$. As $f\to\infty$, $\gamma\to1$; the stiffest gases have the largest $\gamma$. 🔉⇢

Source: NCERT-derived

Q86 Two moles of a monatomic gas need how much heat to raise its temperature by $10\,\text{K}$ at constant volume? ($R=8.314$) medium
Step solution + source
$Q=nC_v\Delta T=2\times\tfrac32R\times10=30R\approx249\,\text{J}$. At constant volume all the heat becomes internal energy, no work is done. 🔉⇢

Source: NCERT-derived

Q87 For a gas with $\gamma=1.5$, the number of degrees of freedom is medium
Step solution + source
From $\gamma=1+2/f$, $f=2/(\gamma-1)=2/0.5=4$. This inversion recovers $f$ directly from a measured heat-capacity ratio. 🔉⇢

Source: NCERT-derived

Q88 For a diatomic gas whose vibrational mode is active ($f=7$), $\gamma$ equals hard
Step solution + source
With $f=7$, $\gamma=1+2/7=9/7\approx1.29$. As vibration switches on at high temperature, $\gamma$ drops from the rigid-diatomic $1.4$ toward smaller values. 🔉⇢

Source: NCERT-derived

Q89 At constant pressure, $2\,\text{mol}$ of a diatomic gas is heated by $20\,\text{K}$. The heat supplied is ($R=8.314$) hard
Step solution + source
$Q=nC_p\Delta T=2\times\tfrac72R\times20=140R\approx1164\times2\approx2328\,\text{J}$. At constant pressure use $C_p=\tfrac72R$, which includes expansion work. 🔉⇢

Source: NCERT-derived

Q90 The internal-energy change when $2\,\text{mol}$ of a monatomic gas is heated $20\,\text{K}$ at constant pressure is hard
Step solution + source
$\Delta U=nC_v\Delta T=2\times\tfrac32R\times20=60R\approx499\,\text{J}$, regardless of the path — $\Delta U$ depends only on $T$. The extra heat at constant $P$ went into work, not $\Delta U$. 🔉⇢

Source: NCERT-derived

Q91 A mixture of $1\,\text{mol}$ monatomic and $1\,\text{mol}$ diatomic gas has an effective $C_v$ of advanced
Step solution + source
Effective $C_v=\dfrac{n_1C_{v1}+n_2C_{v2}}{n_1+n_2}=\dfrac{\tfrac32R+\tfrac52R}{2}=2R$. Mole-weighted averaging of heat capacities gives the mixture value. 🔉⇢

Source: NCERT-derived

Q92 For the same mixture ($1\,\text{mol}$ mono + $1\,\text{mol}$ di), the effective $\gamma$ is advanced
Step solution + source
Effective $C_v=2R$, and $C_p=C_v+R=3R$, so $\gamma=C_p/C_v=3R/2R=1.5$. Compute the mixture's $C_v$ and $C_p$ separately, then take the ratio. 🔉⇢

Source: NCERT-derived

Q93 The mean free path is easy
Step solution + source
The average distance between two successive collisions is called the mean free path $\lambda$. Between collisions the molecule moves in a straight line; $\lambda$ measures how far, on average, it gets. 🔉⇢

Source: NCERT-derived

Q94 The mean free path is given by easy
Step solution + source
For molecules of diameter $d$ and number density $n$, $\lambda=\dfrac{1}{\sqrt2\,\pi d^2 n}$. The $\sqrt2$ accounts for the relative motion of the other molecules. 🔉⇢

Source: NCERT-derived

Q95 The mean free path depends on number density $n$ as easy
Step solution + source
Since $\lambda=1/(\sqrt2\pi d^2 n)$, doubling the number density halves the mean free path: $\lambda\propto1/n$. A denser gas means more frequent collisions and shorter free flights. 🔉⇢

Source: NCERT-derived

Q96 At constant temperature, increasing the pressure of a gas makes the mean free path easy
Step solution + source
At fixed $T$, $n=P/kT\propto P$, and $\lambda\propto1/n\propto1/P$. Higher pressure packs molecules closer, shortening the mean free path. 🔉⇢

Source: NCERT-derived

Q97 In terms of $P$ and $T$, the mean free path varies as easy
Step solution + source
Using $n=P/kT$, $\lambda=\dfrac{kT}{\sqrt2\pi d^2 P}\propto T/P$. Heating at fixed pressure lengthens the free path; compressing shortens it. 🔉⇢

Source: NCERT-derived

Q98 The collision frequency $\nu$ (collisions per second) is related to $\lambda$ and mean speed $\bar v$ by medium
Step solution + source
A molecule of mean speed $\bar v$ covers one free path $\lambda$ between collisions, so it collides $\nu=\bar v/\lambda$ times per second. Faster molecules or shorter paths mean more collisions. 🔉⇢

Source: NCERT-derived

Q99 If the diameter of the molecules were doubled (same $n$), the mean free path would medium
Step solution + source
Since $\lambda\propto1/d^2$, doubling $d$ multiplies $d^2$ by 4 and divides $\lambda$ by 4. Bigger molecules present a larger collision cross-section. 🔉⇢

Source: NCERT-derived

Q100 The mean free path at STP for air is of the order medium
Step solution + source
With $d\sim3\times10^{-10}\,\text{m}$ and $n\sim2.7\times10^{25}\,\text{m}^{-3}$, $\lambda=1/(\sqrt2\pi d^2 n)\sim10^{-7}\,\text{m}$, about $1000$ molecular diameters. 🔉⇢

Source: NCERT-derived

Q101 If the number density of a gas is halved at constant temperature, the mean free path medium
Step solution + source
$\lambda\propto1/n$, so halving $n$ doubles $\lambda$. Fewer molecules per volume means each travels farther before a collision. 🔉⇢

Source: NCERT-derived

Q102 A gas has mean free path $\lambda_0$ at $300\,\text{K}$, $1\,\text{atm}$. At $600\,\text{K}$, $1\,\text{atm}$ it becomes hard
Step solution + source
At constant pressure $\lambda\propto T$ (because $n=P/kT\propto1/T$). Doubling $T$ from 300 to $600\,\text{K}$ doubles $\lambda$ to $2\lambda_0$. 🔉⇢

Source: NCERT-derived

Q103 A gas at pressure $P$ has mean free path $\lambda$. At the same temperature and pressure $4P$, the mean free path is hard
Step solution + source
At fixed $T$, $\lambda\propto1/P$, so quadrupling the pressure quarters the mean free path: $\lambda'=\lambda/4$. Denser gas, shorter flights. 🔉⇢

Source: NCERT-derived

Q104 Given $d=2\times10^{-10}\,\text{m}$ and $n=2.5\times10^{25}\,\text{m}^{-3}$, the mean free path is about advanced
Step solution + source
$\lambda=\dfrac{1}{\sqrt2\pi d^2 n}=\dfrac{1}{1.414\times3.14\times(2\times10^{-10})^2\times2.5\times10^{25}}\approx2.3\times10^{-7}\,\text{m}$. Compute $d^2$ first, then the product. 🔉⇢

Source: NCERT-derived

Q105 A molecule with $\bar v=480\,\text{m s}^{-1}$ has mean free path $1.2\times10^{-7}\,\text{m}$. Its collision frequency is advanced
Step solution + source
$\nu=\bar v/\lambda=480/(1.2\times10^{-7})=4\times10^{9}\,\text{s}^{-1}$. A molecule suffers billions of collisions each second at ordinary densities. 🔉⇢

Source: NCERT-derived

⏱️ Mock Test 30 Q · 60 min · +4 correct, −1 wrong, 0 unattempted

Rules: No calculator beyond basic arithmetic. Take $R=8.314\,\mathrm{J\,mol^{-1}K^{-1}}$, $k=1.38\times10^{-23}\,\mathrm{J\,K^{-1}}$ and $N_A=6.02\times10^{23}\,\mathrm{mol^{-1}}$. Temperatures in kelvin. Negative marking rewards accuracy over guessing.

🎬 Video Lectures clip-indexed · NPTEL / IIT / MIT

MIT OCW / Walter Lewin, 3Blue1Brown, Khan, Veritasium. Each video is broken into clips — click a clip to play just that segment.

maxwell speed distribution curve || हिंदी में maxwell boltzmann speed wise distribution Explained!! 🔉⇢
EG Chemistry - IIT JEE & NEET

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📑 Clips (4)
  • 0:00–2:32Temperature & rms speed — segment 1अजय को और अकल कब क्या था मैक्सिमम नेचुरल गर्भ क्या दिया था तो देख यार इस एक्चुअल कर्व को कैसी मनी प्लांट किया था पहले यह सांस लेते हैं अपन तो सबसे पहली बात कि यह जो एक्सेस है इस एक्सेस के ऊपर स्पीड को रखा गया है इस…temperature & rms speed
  • 2:32–5:02Temperature & rms speed — segment 2मॉलिक्यूल रहता है और वह इस तरीके से बना लेते हैं कुछ ग्रुप बना लेते हैं और उस ग्रुप का जो होता है उसको हम इस पर रखते हैं और उस ग्रुप का जो होता है उसको हम डालते हैं यह ग्रुप है यह ग्रुप का ग्रुप आफ थे टोटल नंबर ऑफ…temperature & rms speed
  • 5:02–7:33Temperature & rms speed — segment 3इस पॉइंट रिकॉर्डिंग इस पॉइंट को इस पॉइंट इस पॉइंट को इस पॉइंट टू ऑन पूरे ग्राफ में जो भी पॉइंट से लेफ्ट उसके कर्ज पंडित तुम्हारे पास एक्सट्रैक्शन हो रहा होगा और एक बात जो तय है वह यह है कि कम ऑफ ट्रैक से जोड़ने वाला है…temperature & rms speed
  • 7:33–8:49Temperature & rms speed — segment 4क्या मिल जाता है एरिया अंडर कर तो इस कर्ज के अंदर का जो एरिया है वह में टोटल नंबर ऑफ हृदय का यह चिन्ह अगर मान लो कि मैं तुमसे पूछूं कि यह बता दो कि आप मन रूपी लीटर पूछ कितने यह यहां पर नहीं दिया और बोला कि लेटेस्ट…temperature & rms speed
What Is An Atom? | The Dr. Binocs Show | Best Learning Videos For Kids | Peekaboo Kidz 🔉⇢
Peekaboo Kidz

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📑 Clips (3)
  • 0:07–2:37Molecular nature of matter — segment 1[Music] oh goodness me thief thief oh it's you little kitty who and you broke auntie's favorite flowerpot come on we need to fix it before she comes oh thanks let's see if this helps oh you're already here friends…molecular nature of matter
  • 2:37–5:09Molecular nature of matter — segment 2even though sage khanna was believed to be the first one to come up with this idea but the credit for making this concept popular and proposing the first atomic theory goes to a great philosopher democritus who…molecular nature of matter
  • 5:09–7:18Molecular nature of matter — segment 3is unfortunately misleading the real picture for now is this isn't it cool friends now go and surprise your teachers trip your time did you know that the atoms are mostly empty space yes an atom is about 99.99999 empty…molecular nature of matter
The Atomic/Molecular Nature of Matter​ 🔉⇢
Manorama Horizon

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📑 Clips (2)
  • 0:34–3:06Molecular nature of matter — segment 1Did you know that the molecular theory of matter that you started to learn about in primary school wasn't completely accepted until late into the 19th century? That is after the first Indian War of Independence. That is…molecular nature of matter
  • 3:06–5:00Molecular nature of matter — segment 2pressures and temperatures same volumes of different gases carry equal numbers of molecules. So little by little people all over the world gradually added more pieces into this puzzle. Today we have reached a point…molecular nature of matter
Atom Structure | Matter | Physics | FuseSchool 🔉⇢
FuseSchool - Global Education

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📑 Clips (2)
  • 0:05–2:35Molecular nature of matter — segment 1in the previous video you learnt about the structure of atoms you also learned that elements are atoms of different types you have probably come across atoms and elements before in chemistry physics and chemistry…molecular nature of matter
  • 2:35–4:29Molecular nature of matter — segment 2makes sense it has gained more negative charge and so the atom has gone from neutral to relatively negative a positive ion is an atom that has lost one or more electrons again this makes sense as the atom has lost some…molecular nature of matter
Kinetic Molecular Theory and the Ideal Gas Laws 🔉⇢
Professor Dave Explains

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📑 Clips (2)
  • 0:00–2:30Ideal gas equation — segment 1hey it's Professor Dave let's talk about ideal gases science Dave expains let's recall our definition of a gas as the phase of matter where atoms of a substance are in motion and fill their container if we make a couple…ideal gas equation
  • 2:30–5:02Ideal gas equation — segment 2absolute temperature scale called the Kelvin scale 1° Kelvin is the same magnitude as 1° celsi but 0 Kelvin is absolute zero the lowest temperature possible a complete absence of heat energy this helps us avoid weird…ideal gas equation
Molecular Kinetic Theory (simple derivation) - Kinetic Theory (Lesson 4) 🔉⇢
vt.physics

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📑 Clips (2)
  • 0:00–2:30Kinetic theory of pressure — segment 1there are more sophisticated methods to derive the equation for the pressure of an ideal gas but if you're looking for an easily understandable and straightforward derivation of the gas pressure equation using molecular…kinetic theory of pressure
  • 2:30–4:14Kinetic theory of pressure — segment 2times by the speed squared divided by volume this is beginning to look a lot like the kinetic theory equation that we're trying to derive but where has the factor of a third come from well pressure is a scalar quantity…kinetic theory of pressure
12th Physics | Chapter 3 | Kinetic Theory of Gases & Radiation | Lecture 2 | Maharashtra Board | 🔉⇢
JR Tutorials

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📑 Clips (6)
  • 0:01–2:32Kinetic theory of pressure — segment 1Hello everyone, how are you all, how is your studies going, everything is going well and welcome to our lovely channel, beautiful channel, in Zar Tutorial, as you know we have started the third chapter of 12th which is…kinetic theory of pressure
  • 2:32–5:03Kinetic theory of pressure — segment 2what will be the total mass, the total mass is nothing but capital M which is nothing but small M times N, okay, if someone asks me, brother, tell me the mass of the entire gas, then what will be the mass of the entire…kinetic theory of pressure
  • 5:03–7:33Kinetic theory of pressure — segment 3and it is travelling till the surface, that means it is travelling this much, that means how much is it travelling, DL, when is this, when is it travelling on the L side and when have we considered, when is this, when…kinetic theory of pressure
  • 7:33–10:06Kinetic theory of pressure — segment 4talking about x direction then everything will be in x direction but if particles change then what will be the velocity ux1 means first particle in x direction + uh2 second particle in x direction By doing this, how…kinetic theory of pressure
  • 10:06–12:38Kinetic theory of pressure — segment 5Here, how many particles are there here? Number of particles. Mass of the gas. So this will be the pressure. This is nothing but total mass/three. C in RMS squared. This is the volume and this is the velocity. This one…kinetic theory of pressure
  • 12:38–15:09Kinetic theory of pressure — segment 6relation of RMS velocity. Okay, RMS velocity in temperature, we will try to find it in terms of temperature. How will we do it? Look, we will use the same relation which we have already derived. What have we derived and…kinetic theory of pressure
8.01x - Lect 33 - Kinetic Gas Theory, Ideal Gas Law, Phase Transitions 🔉⇢
Lectures by Walter Lewin. They will make you ♥ Physics.

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📑 Clips (6)
  • 0:03–2:34Kinetic theory of pressure — segment 1liquids are incompressible gases are not incompressible when you decrease the volume of a gas by 50% that's no problem it's impossible to do that for a liquid in liquids the atoms and the molecules effectively touch…kinetic theory of pressure
  • 2:34–5:05Kinetic theory of pressure — segment 2matter always has this number of Mo molecules approximately now each of these substances have very different masses if I take for instance carbon then one mole of carbon would weigh very close to 12 gram if I take…kinetic theory of pressure
  • 5:05–7:37Kinetic theory of pressure — segment 3so a is four that's where you see your four G if you take oxygen has eight protons and eight neutrons so a is 16 but you have O2 in gas Forum so now your atomic mass number has to be doubled is 32 and so an a mole of o2…kinetic theory of pressure
  • 7:37–10:08Kinetic theory of pressure — segment 4R 1.03 excuse me you know I'm a little bit ahead of myself you know R which is 8.3 you know the temperature which is 293 and you know the pressure which is 1.03 * 10 5 and when you calculate that you find something very…kinetic theory of pressure
  • 10:09–12:40Kinetic theory of pressure — segment 5collision and it comes back in exactly the same direction so there is momentum transfer and the momentum transfer for one Collision is 2 MV because it comes in with MV in this direction it comes back with MV in that…kinetic theory of pressure
  • 12:40–15:12Kinetic theory of pressure — segment 6means is that if the oxygen molecule this is eight times more massive this one then the velocity is the square root of eight times smaller because the ratio 32 over 4 is 8 so the oxygen molecules have a lower speed so…kinetic theory of pressure
Kinetic Interpretation of Temperature | YOLO JEE Advance Physics with Vikrant Kirar 🔉⇢
Crash Up

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📑 Clips (5)
  • 0:02–2:35Temperature & rms speed — segment 1Kinetic Interpretation of Temperature. What is temperature? How can we understand that? Let us try to see it according to the Kinetic Theory of Gases. So look, in this way, suppose we had a gas which was in a cube.…temperature & rms speed
  • 2:35–5:08Temperature & rms speed — segment 2multiplied by total mass divided by mass per mole * RT. So what will this thing become? This will be the number of moles. Look, if we have 5 kg in 1 mole. It happens. For example, here we have 20 kg. Total mass is. So…temperature & rms speed
  • 5:08–7:38Temperature & rms speed — segment 3asked for the kinetic energy of one particular atom, we cannot say which value it has, because the molecular velocities all differ — but we can state the average kinetic energy…temperature & rms speed
  • 7:38–10:09Temperature & rms speed — segment 4Both are occupying the same volume. There are so many small particles. The helium people don't even know that there is good argon here too. For that also both types of molecules can move completely. How will you solve…temperature & rms speed
  • 10:09–11:57Temperature & rms speed — segment 5grams. Now it has become kilogram. Its value per kilogram per mole kg comes out to be approximately 1.37 kilometer per second. Solve it using a calculator or whatever way you want to do approximation. [sound of clearing…temperature & rms speed
MOST PROBABLE, AVERAGE, RMS SPEED OF GAS MOLECULES || KINETIC THEORY OF GASES || WITH EXAM NOTES || 🔉⇢
Pankaj Physics Gulati

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Full lecture — no clip index.

Law of Equipartition of Energy | YOLO JEE Advance Physics with Vikrant Kirar 🔉⇢
Crash Up

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📑 Clips (5)
  • 0:00–2:30Degrees of freedom & equipartition — segment 1Law of Equivalent Partition of Energy Equivalent partition means that the energy we have is divided equally into different degrees of freedom. Let us first start understanding it. See, we had already derived that our…degrees of freedom & equipartition
  • 2:30–5:06Degrees of freedom & equipartition — segment 2energy modes with n mode having average kinetic equally distributed in all possible energy modes. Note: The average value of all the possible energy modes that we have is called the number of molecules. Degrees of…degrees of freedom & equipartition
  • 5:06–7:39Degrees of freedom & equipartition — segment 3Now, a very good question to solve is kV. When do we have to find the value of r? When do we have to find the number of moles? What do we have to do? 96 grams of oxygen gas is at 27 degrees Celsius. Find something. But…degrees of freedom & equipartition
  • 7:39–10:10Degrees of freedom & equipartition — segment 418750 6.5 water, so our approximation is quite correct, our answer is done, so now let's see, we have to see the concepts, do not go into calculation right now, internal energy of gas per mole, this energy that we got…degrees of freedom & equipartition
  • 10:10–10:28Degrees of freedom & equipartition — segment 5and here also we will get water, this is also the correct answer, let's move ahead. All of you have seen the video, all of them should comment so that the comments can be distributed equally among all. Please comment,…degrees of freedom & equipartition
Degrees of freedom | Kinetic theory of gases | IIT JEE 🔉⇢
Tuition in

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📑 Clips (2)
  • 0:00–2:30Degrees of freedom & equipartition — segment 1[Music] let's visualize the concept of degrees of freedom of a gas from the thermodynamics chapter so first we have three kinds of gases the first is a monoatomic one wherein if you pick up any molecule it is just…degrees of freedom & equipartition
  • 2:30–3:03Degrees of freedom & equipartition — segment 2degrees of freedom as far as its rotation is concerned if you plate place two of these balls on the x axis it can rotate about Y axis it can rotate about z axis it also has a significant significant rotation about x…degrees of freedom & equipartition
Thermodynamics SPECIFIC HEATS - cv & cp - in 12 Minutes! 🔉⇢
Less Boring Lectures

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  • 0:00–2:31Specific heats & gamma — segment 1we define the specific heat of a substance as the amount of heat necessary to bring one kilogram of that substance up by one degree celsius different substances will require a different amount of heat to bring their…specific heats & gamma
  • 2:31–5:01Specific heats & gamma — segment 2quick parenthesis here lower kc is called specific heat and lower kc times the mass is capital c and we call it heat capacity this applies to both cv and cp the differential form of the first law of thermodynamics or…specific heats & gamma
  • 5:01–7:31Specific heats & gamma — segment 3was no heat exchanged with the surrounding the internal energy of the gas remained the same despite changes of pressure and volume this means that the internal energy of an ideal gas is only a function of its…specific heats & gamma
  • 7:32–10:02Specific heats & gamma — segment 4more on that later of course a diagram like this will not offer enough precision for us to find the specific heat for a given we would be almost guessing it by eyeballing it with almost no accuracy and that's why we…specific heats & gamma
  • 10:02–12:32Specific heats & gamma — segment 5change from a reference temperature to the temperature we're interested in and we don't even need to know what that reference temperature is if we're looking for the change in enthalpy between 240 and 270 we're really…specific heats & gamma
"Cp, Cv & Gamma Explained | Specific Heats of Different Gases Made Simple" 🔉⇢
Shridhar Jagtap

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  • 0:00–2:33Specific heats & gamma — segment 1I'm alive, everything's simple, and 20 modules, I realized that Brazil wants to stop realizing that it had lost at home with a phishing goal instead of a chip in Rome. For defending that thesis, he was the only one to…specific heats & gamma
  • 2:33–5:13Specific heats & gamma — segment 250 point. Don't forget the process, you should always save from criticism. A lot of kids is building a hip hop hub that lives in the favela, rather than Minister Peter Solberg and the vehicle, rather than sitting rich…specific heats & gamma
  • 5:13–7:46Specific heats & gamma — segment 3kind of The crisis has areas of illegal business with a bis and cut the target in the first simple child but civilians while for the agreement he saw and lived under the command of the Swan Cuba duo the old man is 30…specific heats & gamma
  • 7:46–8:14Specific heats & gamma — segment 436 goals as in the book a decade ago and 11 trucks while the year he graduated in cash and help us camozzi psv rafael lucas a long time because of this independent pointed out instead ofspecific heats & gamma
Derivation of CP - CV = R Class 11 Physics Term 2 Important Topics 2022, Mayer's Formula 🔉⇢
Mandeep Education Academy

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  • 0:00–2:30Specific heats & gamma — segment 1Hello hello friends welcome back to mid june k tube channel so in the list of important derivations of class 11th physics time two today we have a very important mode on religious share topic and this is called molar…specific heats & gamma
  • 2:30–5:01Specific heats & gamma — segment 2derivation is first of all, we put the gas in a cylinder, this is a diagram, inside this cylinder there is thermodynamics clear graduation, children, it is ground and fixed in this calendar, and how is the oil in it,…specific heats & gamma
  • 5:01–7:23Specific heats & gamma — segment 3we cannot stop it from expanding because to stop it the pressure will have to be increased, the blood pressure is content, so you will see in this, let's assume the heat given is D cute, the work is DW, that is Magan…specific heats & gamma
Mean Free Path 🔉⇢
myEdu Learn

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  • 0:01–2:35Mean free path — segment 1mean free path the mean free path is the average distance traveled by a moving molecule between collisions imagine gas leaking out of a pipe it would take a while for the gas to diffuse and spread into the environment…mean free path
  • 2:35–3:37Mean free path — segment 2square the average distance between two successive collisions called the mean free path I is equals to average V tau is equals to one by n PI D square in this derivation we imagined the other molecules to be at rest but…mean free path
Class 11th – Mean Free Path | Kinetic Theory of Gases | Tutorials Point 🔉⇢
TutorialsPoint

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  • 0:00–2:30Mean free path — segment 1mean free path when the molecules are in kinetic condition they are moving in the gas then it is likely that they may strike other molecules when do they strike and when do they do not strike if a molecule is moving in…mean free path
  • 2:30–3:23Mean free path — segment 2here then VT what does it give length so what is the formula of length of the path that is 1 upon n PI d square and this is length of which path when there is a second collision so that is the length between two…mean free path
12th Physics | Chapter 3 | Kinetic Theory of Gases & Radiation | Lecture 1 | Maharashtra Board | 🔉⇢
JR Tutorials

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  • 0:01–2:32Kinetic theory of pressure — segment 1हेलो गैस कैसे हैं आप लोगों के संजीव पांडे और स्वागत है आपका अपने प्यारे से चैनल में खूबसूरत से चैनल में जर ट्यूटोरियल में ओके सो जैसा की आपको पता है सी हैव कंप्लीटेड डी चैप्टर नंबर सेकंड हम लोग मैकेनिकल प्रॉपर्टीज ऑफ…kinetic theory of pressure
  • 2:32–5:02Kinetic theory of pressure — segment 2विजुलाइज डेट बॉडी वेयर हम देख के ही उसे बॉडी का बिहेवियर बता सकते हैं उसकी प्रॉपर्टी बता सकते हैं ओके उसकी वेलोसिटी क्या है बता सकते हैं ओके वो कितना वजन है ये सारा चीज बता सकते हैं बट जब हम लोग गैस की बात करते हैं तो…kinetic theory of pressure
  • 5:02–7:32Kinetic theory of pressure — segment 3करेगा उसे गैस को हम लोग क्या बोलेंगे उसे गैस को बोलेंगे हम लोग आइडियल गैस आइडियल का एक्चुअल मतलब क्या होता है की जो गैस के बीच में जो इंटरमॉलिक्युलर स्पेस है ना जो इंटरमॉलिक्युलर फोर्स है मतलब जो इंटरेक्शन है बिटवीन…kinetic theory of pressure
  • 7:32–10:03Kinetic theory of pressure — segment 4रहेगा जो शॉप रहेगा वो जनरली क्या रहेगा उनका जो शॉप रहेगा वो जनरली क्या रहेगा स्फेरिकल रहेगा ओके आईडेंटिकल शॉप आईडेंटिकल मतलब सारे मॉलेक्युलिस अगर मैन लो ऑक्सीजन की बात कर रहा हूं तो ऑक्सीजन के सारे मॉलेक्युलिस का जो शॉप…kinetic theory of pressure
  • 10:03–12:33Kinetic theory of pressure — segment 5डेट इसे नथिंग बट अन फ्री पथ डेट इसे नथिंग बट अन फ्री पथ व्हाट इसे फ्री पथ फ्री पथ इस डी डिस्टेंस बिटवीन तू सक्सेसिव कोलाइजन ओके मैन लो की यहां पे कंटेनर है मैन लो अब पार्टिकल यहां से चालू हुआ और यहां तक गया यहां से यहां…kinetic theory of pressure
  • 12:33–15:05Kinetic theory of pressure — segment 6डायरेक्शन में भी वायर डायरेक्शन में और स डायरेक्शन में वो तीनों डायरेक्शन में ट्रेवल कर सकता है तो इट इसे वेरी डिफिकल्ट तू कैलकुलेट और तू विजुलाइज डी वेलोसिटी ओके इन वैन अन सिंगल डायरेक्शन तो इसी लिए हम लोग आरएमएस…kinetic theory of pressure
Kinetic Theory Of Gases In One Shot | Class 12 Physics Radiation | Maharashtra HSC Board | MHT CET 🔉⇢
PW Maharashtra

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (PW Maharashtra); found via yt-dlp search 'गैसों का अणुगति सिद्धांत kinetic theory of gases Hindi physics NCERT', oEmbed-verified live.

📑 Clips (6)
  • 0:00–2:33Kinetic theory of pressure — segment 1नमस्कार विद्यार्थी मित्र न आय होप यू आर डूइंग वेल विदाउट वेस्टिंग टाइम सी आर गोइंग तू स्टार्ट आर न्यू चैप्टर डेट इस व्हाट आई नीड थ्योरी ऑफ गैसेस ठीक है आपका परंतु अपन आ बाकी जो चैप्टर buchit ले लिया है सेमीकंडक्टर…kinetic theory of pressure
  • 2:33–5:18Kinetic theory of pressure — segment 2दिस इसे क्लियर आते डी आइडियल गैस बिहेव करते मंजे कई टेंशन मत दे इंटरमॉलिक्युलर अट्रैक्शंस यत ना हिरन अट्रैक्शन इन डी साइंस इंटरेक्शन नहीं रहना आई होप दिस इस क्लियर फाइन डिस्टेंस वेलोसिटी अवर कांस्टेंट वेलोसिटी मैन…kinetic theory of pressure
  • 5:18–7:52Kinetic theory of pressure — segment 3रेट ऑफ चेंज ऑफ मोमेंटम मास * वेलोसिटी करता है वेलोसिटी पकड़ वेलोसिटी पकड़ यह साइट लाइक वॉलपेपर विच से तीन कॉम्पोनेंट होना vxier पहला पॉइंट क्लियर करता है 2 मिनट आता है इथे कोलाइड होता है या वो अल्लाह परत जाता है वो…kinetic theory of pressure
  • 7:52–10:27Kinetic theory of pressure — segment 4है तुम सब माइंस एमवीसी करंट वेलोसिटी डायरेक्शन चेंज होता है मोमेंटम जोड़ा चेंज ट्रांसफर तो मोमेंटम कितनी चेंज अलग हेयर कलर तो है यहां फेस वॉश देता है एक परत जाता है मुझे ठस एल डिस्टेंस में कंसीडर करता है एम x1² / एल…kinetic theory of pressure
  • 10:27–12:59Kinetic theory of pressure — segment 5एवरेज टोटल एवरेज वेलोसिटी अगर मिलता है एवरेज ऑफ स्क्वायर ऑफ एक्स कॉम्पोनेंट ऑफ डी वेलोसिटी करंट की स्क्वायर है तो स्क्वायर शुद्ध मॉलिक्यूल नंबर ऑफ मॉलेक्युलिस मुझे हाफ ब्रैकेट बराबर कार्बन हैव ब्रैकेट * वेलोसिटी मुझे…kinetic theory of pressure
  • 12:59–15:36Kinetic theory of pressure — segment 6नॉट डी नंबर ऑफ मॉल ठीक है अंडर रूट ऑफ थ्री आरटी / एम नॉट वैन जस्ट आई सी आर एम एस डायरेक्टली प्रोपोर्शनल टेंपरेचर बस इतना माला एनर्जी फ्री रहता है ताना काइनेटिक एनर्जी है ना मैं काइनेटिक एनर्जी का फॉर्मूला क्या है 1/2…kinetic theory of pressure
आदर्श गैंस समीकरण ! Ideal gas equation ! Basic chemistry 🔉⇢
Shiv Coaching Classes India

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Shiv Coaching Classes India); found via yt-dlp search 'आदर्श गैस समीकरण ideal gas equation Hindi physics class 11', oEmbed-verified live.

📑 Clips (4)
  • 0:00–2:33Ideal gas equation — segment 1लुट है रेस्टोरेंट कैसे आप सभी का खून आप सभी का आपके अपने चैनल सी कोचिंग क्लासेस पर स्वागत है आज एक महत्वपूर्ण टॉपिक पर चेक करेंगे आदर्श गैस धीमी कर दी कि आदर्श के समीकरण 11th क्लास में है न इसके बाद आप ग्रेजुएशन लेवल…ideal gas equation
  • 2:33–5:04Ideal gas equation — segment 2तो वृद्धि होगी समझ गए कि इस पर संबंधित गैस पर था और बताया था कि निश्चित मात्रा में गैस की निश्चित मात्रा का आयतन सब्सक्राइब कर सकते हैं इसको ले सकते हैं इसके लिए आपको लिखना है गैस से संबंधित बताया था क्विक गैस का दाम…ideal gas equation
  • 5:04–7:34Ideal gas equation — segment 3आनुपातिक व स्थिरांक है लुट यहां तक तो आप समझ गए होंगे कहां कहां से आ रहा है तो मैंने क्या बताएं आपको बालकनी हम की जो आयतन है वह प्रसव के बीच कम है इसके बाद चार्ल्स कर ने बताया चार्ल्स ने क्या बताया था जो वॉल्यूम है…ideal gas equation
  • 7:34–8:05Ideal gas equation — segment 4सालों अच्छे से तैयार कर लोगे आदर्श गैस समीकरण के चलते बेरोजगारी दीजिए मिलते हैं अगले विडियो में वीडियो पसंद आए तो लाइक कर देना और साथ में इसको नोट परचेस कर है तो हंड्रेड रुपीज़ में कांटेक्ट नंबर 96 8529 सिंह 049 यह मेरा…ideal gas equation
class 11 physics , आदर्श गैस समीकरण, Equation of Ideal Gas 🔉⇢
ps study guru

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (ps study guru); found via yt-dlp search 'आदर्श गैस समीकरण ideal gas equation Hindi physics class 11', oEmbed-verified live.

📑 Clips (3)
  • 0:01–2:33Ideal gas equation — segment 1[संगीत] मैं पंकज कुमार वर्मा स्वागत करता हूं आप सबका मोस्ट इंपॉर्टेंट क्वेश्चन करवाने वालों मैंने आपको इसकी कौन-कौन से गुण होते हैं यह बताया था आज हम आदर्श गैस समीकरण को सिद्ध करने वाले तो चलिए इतने कितनी सरल तरीके से…ideal gas equation
  • 2:33–5:06Ideal gas equation — segment 2लिए क्या करेंगे अपन लीजिए दूध तीनों चीज लिखिए वायर चार्ज इनके नियम के अनुसार होता क्या है उतने यहां पर हम लिख देंगे तो देखो बैल की नियम के अनुसार बाय के नियम के अनुसार अब वायरल के नियम के अनुसार क्या होता है हमें पता है…ideal gas equation
  • 5:06–6:51Ideal gas equation — segment 3गैस नियतांक ही क्योंकि हम गैस समीकरण निकल रहे हैं इसलिए हमने यहां पर गैस नियतांक रख दिया तो कभी-कभी तो आपसे यही पूछा जाता है की पीवी = एनआरटी सिद्ध कीजिए ये भी गैस समीकरण है और या फिर इसे सैप लिख सकते हैं देखो यदि गैस…ideal gas equation
Relation between kinetic energy and temperature of a molecule | KTG | Deduction of the kinetic ex... 🔉⇢
Tutor Talk

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Tutor Talk); found via yt-dlp search 'ताप की गतिज व्याख्या rms speed Hindi physics class 11', oEmbed-verified live.

📑 Clips (4)
  • 0:00–2:34Temperature & rms speed — segment 1हेलो बच्चों, इस वीडियो में हम बात करेंगे ताप की अनुगिक व्याख्या के बारे में जो क्लास 11 भौतिक विज्ञान का एक महत्वपूर्ण विषय है और इस सूत्र का निगमन करेंगे। देखो आदर्श गैस के दाब का सूत्र का निगमन हम कर चुके हैं। उसकी…temperature & rms speed
  • 2:34–5:04Temperature & rms speed — segment 2pv इक्वल टू होता है nr टी ठीक है ये वाला सूत्र आपने याद रखना है आदर्श गैस समीकरण का सूत्र आपको याद होगा अभी p का मतलब होता है दाब v का मतलब आयतन n का मतलब मोलों की संख्या r मतलब स्थिरांक और और t मतलब तापमान। अब ये तो था…temperature & rms speed
  • 5:04–7:37Temperature & rms speed — segment 3हम यह वैल्यू यहां फिल कर पाएं। देखो इस स्टेप को बहुत ध्यान से समझना पड़ेगा। देखो r/ t / n है। ये वैल्यू तो यहां फिल हो ही नहीं सकती। कुछ मैचिंग जरूर हो रही है कि mb² mb² मैच हो रहा है यहां पर। मैं आपको समझाता हूं कैसे…temperature & rms speed
  • 7:37–7:53Temperature & rms speed — segment 4तो यहां से यह क्लियर होता है कि जो ताप होता है वह गति के वर्ग के समानुपाती होता है। तो यहां पर यह वीडियो कंप्लीट होती है। अगर आपको वीडियो अच्छी लगी हो तो वीडियो को लाइक और चैनल को सब्सक्राइब करें। थैंक्स फॉरtemperature & rms speed
Class 11 Physics अणुगति सिद्धान्त | ताप की गतिज व्याख्या | Class 11 Physics KINETIC THEORY 🔉⇢
Doubtnut Classes

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Doubtnut Classes); found via yt-dlp search 'ताप की गतिज व्याख्या rms speed Hindi physics class 11', oEmbed-verified live.

📑 Clips (6)
  • 0:08–2:38Temperature & rms speed — segment 1हेलो एवरीवन कैसे हो आप सब मैं हूं आपकी मैमिला क्रांति और आज मैं अपने सभी प्यारे बच्चों के लिए लेकर के आई हूं अणु गति सिद्धांत का लेक्चर नंबर कौन सा है भाई ये लेक्चर नंबर सिक्स है और आज हम इस लेक्चर नंबर सिक्स में करेंगे…temperature & rms speed
  • 2:38–6:17Temperature & rms speed — segment 2लिया गया है एकांक आयतन में अणुओं की संख्या है यहां पर जो है हम n की जगह में n / v n / v mv1 ये भी हम क्या कर सकते हैं रख सकते हैं ये जो v के वर्ग के ऊपर हम ये बार का चिन्ह लगाते हैं ये क्या है v के वर्ग का औसत चाल है…temperature & rms speed
  • 6:17–8:47Temperature & rms speed — segment 3हां भाई तो यहां पर हां तो हम क्या देख रहे हैं यह जो समीकरण हम देख रहे हैं ठीक है यहां पर यह गतिज ऊर्जा है ठीक है इसके बाद हम यहां पर क्या कर सकते हैं ये जो है इसको हम इधर भी ट्रांसफर कर सकते हैं जब हम यहां पर आयतन को…temperature & rms speed
  • 8:48–11:20Temperature & rms speed — segment 4आप लोग देख लो यहां पर यह क्या है गतिज ऊर्जा है और यह क्या है एक नियतांक है इसको हम फिर से नियतांक लेंगे क्योंकि आ क्या है रिडब नियतांक तो वो भी फिक्स रहता है यहां पर य एकांक आयतन में अणुओं की संख्या कैपिटल n इसका मान भी…temperature & rms speed
  • 11:20–13:55Temperature & rms speed — segment 5जो है ना मैं अगर बोल के नहीं भी बताऊंगी ना तो आपको देख कर के समझ में आना चाहिए ठीक है देख कर के समझ में आना चाहिए ये क्या है ये वाला हिस्सा क्या है ये क्या है गतिज ऊर्जा है किसके समानुपाती है तो ताप के समानुपाती है ये…temperature & rms speed
  • 13:55–16:27Temperature & rms speed — segment 6rt2 ठीक है तो हमको क्या करना था ताप से हम हम लोगों को व्याख्या करनी थी क्या करना है हमको ताप से व्याख्या करना है इसलिए हम लोगों को करना क्या था इस इक्वेशन में हमको ताप को घुसाना था हमको क्या करना था इस इक्वेशन में इस…temperature & rms speed
Statistical Physics | Theory of Equipartition of Energy | BSc 2nd Year Physics MGSU @FastXBSc 🔉⇢
Exam Suchna

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Exam Suchna); found via yt-dlp search 'ऊर्जा समविभाजन स्वतंत्रता की कोटि equipartition Hindi physics', oEmbed-verified live.

📑 Clips (3)
  • 0:00–2:33Degrees of freedom & equipartition — segment 1नमस्कार साथियों आज हम पढ़ने वाले हैं ऊर्जा सम विभाजन सिद्धांत पहले इसकी डेफिनेशन देखते हैं प्रत्येक स्वतंत्रता कोटि के लिए स्वतंत्रता कोटि को f से डिनोट करते हैं औसत ऊर्जा का मान 1/2 kb2 होता है जहां केब बोल्ट मान…degrees of freedom & equipartition
  • 2:33–5:05Degrees of freedom & equipartition — segment 2की प्रायिकता प्रायिकता को हम प से डिनोट करेंगे यह क्या होगी q1 से q ए तक पव से पए तक डी कव से डी क ए तक पव से डी पए तक इक्वल टू ये क्या हो जाएगी सी e की पावर माइनस बीटा डी कव से डी क ए तक पव से डीप ए तक आगे बढ़ते हैं…degrees of freedom & equipartition
  • 5:05–6:39Degrees of freedom & equipartition — segment 3की वैल्यू क्या आई रूट बीटा प ब बीटा पडी एक ये कैसे आया x स् इक्व टू में क्या है बीटा प स्क्वा है तो x के लिए हम क्या मान सकते हैं रूट बीटा प तो डी प के लिए वैल्यू क्या आई 1 ब रूट बी d एक इसको इक्वेशन हम फोर मानते हैं…degrees of freedom & equipartition
L-4, स्वतंत्रता की कोटियाँ (D O F) | अध्याय-13, अणुगतिक सिद्धांत (K T G) Class 11th Physics 🔉⇢
Learn and Share

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Learn and Share); found via yt-dlp search 'ऊर्जा समविभाजन स्वतंत्रता की कोटि equipartition Hindi physics', oEmbed-verified live.

📑 Clips (6)
  • 0:00–2:32Degrees of freedom & equipartition — segment 1लुट लो हुआ था हैं तो प्यारी बच्चों गुड इवनिंग एवरीवन फर्स्ट आफ ऑल सबसे बहुत बड़ा सॉरी कि बहुत ज्यादा लेट हो गया मेरी क्लास ठीक है तो आई हॉप शमा कर देंगे दुनिया को तो बच्चा बन बात कर रहे हैं क्लास इलेवेंथ है जिसके बारे…degrees of freedom & equipartition
  • 2:32–5:07Degrees of freedom & equipartition — segment 2बच्चा है जो टॉपिक है वह बहुत ही प्यारा है क्योंकि आज का हमारा टॉपिक है स्वतंत्रता की कोटियां इन ए डिग्री आफ फ्रीडम देखो आया वह सब बच्चों ने डिग्री आफ फ्रीडम ऑलरेडी कि मिट्टी में भी पढ़ा होगा ठीक है तो आज का वही टॉपिक…degrees of freedom & equipartition
  • 5:07–7:37Degrees of freedom & equipartition — segment 3के व्यक्तियों की संख्या उसी को बोलते हैं ठीक है डेफिनेशन तो यह एक डेफिनेशन आफ और भी बोल सकते हैं कितनी दिशाओं में कर सकता है ठीक है तो ऐसा किसी द्वारा है कि किसी कारण द्वारा कि जितनी दिशाओं में कि गति की जाती है गति की…degrees of freedom & equipartition
  • 7:37–10:09Degrees of freedom & equipartition — segment 4गति मोशन पोस्टर मतलब रोटेशनल मोशन कर सकता है घुर्णन गति कर सकता भी यह कंडोम आलू अच्छे अति कर सकता है अपनी एक्सप्रेस इस घूम सकता है ठीक है यह तो फिर ऐसे से घूम सकता है तो यह उसकी * करती थी यह कंपनी गति भी कर सकता है सुबह…degrees of freedom & equipartition
  • 10:09–12:43Degrees of freedom & equipartition — segment 5बताया कि जितने प्रकार की गतियां कर सकता है एक कण में पिछले तीन तरीके की हत्या कर सकते हैं बच्चों पहले होता है स्थानांतरण करती अब अपनी देंगे इधर घाघरा है ठीक है दूसरा होता घूर्णन गति लक्ष अपने अक्ष पर पिंपल्स उसके रहमान…degrees of freedom & equipartition
  • 12:43–15:14Degrees of freedom & equipartition — segment 6मिल करके धो करके ठीक है इस तरीके से अलग-अलग निकायों के बारे में बात करूं तो वो यह बात करें तो अब हम क्या करेंगे अलग-अलग नियुक्त नंबर पर उनके साथ थे यह समझना भी बच्चों कि हम अभी सिर्फ इस थानांतर्गत के बारे में बात कर रहे…degrees of freedom & equipartition
L-5, माध्य मुक्त पथ | Mean free path | अध्याय-13, अणुगतिक सिध्दांत | Kinetic Theory | 11th Physics 🔉⇢
Learn and Share

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Learn and Share); found via yt-dlp search 'माध्य मुक्त पथ mean free path Hindi physics class 11', oEmbed-verified live.

📑 Clips (6)
  • 0:02–2:32Mean free path — segment 1सो प्यारे बच्चों नमस्कार आप की तबीयत फिर से आप सबका स्वागत करते हैं आपके अपने इस चैनल लर्न एंड शेयर की यूट्यूब प्लेटफार्म पे तो सुबह प्रातः होने वाली क्लास आज दिल है हो करके इवनिंग में हो गई है तो उसे सबके लिए आप सब कुछ…mean free path
  • 2:32–5:02Mean free path — segment 2तो चले स्टार्ट करें आज की क्लास रेडी आप लोग सब बच्चे बहुत कम दिख रहे हैं मेरे को आज बच्चे यार मॉर्निंग में तो बहुत सारे आते हैं अभी सब सो रहे हो क्या चलिए स्टार्ट करें फिर ओके तो आज हमें पढ़ना है उससे पहले मैं आपको…mean free path
  • 5:02–7:34Mean free path — segment 3अक्षरों में गति कर सकता है एक्स में भी कर सकता है ए में भी जा सकता है और स में भी जा सकता है ठीक है पहले पढ़ा था क्योंकि उसे स्थानांतरण गति करने के लिए बनाया कंपन और गुणन गति करना उसके लिए संभव नहीं है तो हम क्या करेंगे…mean free path
  • 7:34–10:14Mean free path — segment 4उसके लिए ऊर्जा कर सकता था उसे केस में इसके पास गतिज ऊर्जा हो जाएगी 1/2 किमी तो मैन लो अगर आपका अनु जो है अक्ष में गति कर रहा है सिर्फ एक्स अक्ष में गति कर रहा है और कहीं नहीं जा रहा तो उसके पास जो एनर्जी होगी वो हाफ के…mean free path
  • 10:14–12:48Mean free path — segment 5फॉर्मूला 3 तो 3 * 3 = 9 - 2 करोगे तो सात ए जाएगी तो उसके लिए हमारे पास जो है द्वीप परमाणु अनु के लिए जो है जैसे हमारे पास फाइव बाय तू केबीडी था वैसे तीन अनु के लिए सात केबीटी / 7 केबीटी हो सकता है ठीक है इस तरीके से…mean free path
  • 12:48–15:26Mean free path — segment 6यूनिवर्सल गैस कांस्टेंट है ठीक गैस नहीं है तो आप इस तरीके से भी इसको लिख सकते हो चलिए अब बच्चों बात करते हैं एक टॉपिक जो हम बार-बार पढ़ते हुए हैं पिछले चैप्टर में पढ़ा है उससे भी विचलित चैप्टर में पढ़ा वो पढ़ना पड़ेगा…mean free path
L-5, माध्य मुक्त पथ (mean free path) | अध्याय-13, अणुगतिक सिद्धांत (K T G) Class 11th Physics 🔉⇢
Learn and Share

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Learn and Share); found via yt-dlp search 'माध्य मुक्त पथ mean free path Hindi physics class 11', oEmbed-verified live.

📑 Clips (6)
  • 0:00–2:35Mean free path — segment 1कि अ है तो प्यारे बच्चों नमस्कार गुड इवनिंग एवरीवन आपकी जीतू भैया फिर से हैं आप लोगों के लिए एक और नया वीडियो लेकर के आपके अपने चैनल एंड शेयर पर और बात कर रहे हैं अपन क्लास इलेवेंथ फिजिक्स के बारे में जाकर चंद्रमा…mean free path
  • 2:35–5:05Mean free path — segment 2है क्योंकि सबको गुड इवनिंग गुड इवनिंग दिव्या जिस पेन अशोक चलो सबको नहीं है तो बच्चों लास्ट क्लास में जो टॉपिक तो बहुत ही प्यारा रहा था आप सबको पता है अपन बात करें थे ऊर्जा कसम में टॉपिक आज हम आगे कंटिन्यू करेंगे कि वह…mean free path
  • 5:05–7:35Mean free path — segment 3है यहां बोल सकते जैसे किसी बच्चे ने बोला नंबर ऑफ़ बोंस मतलब बंदरों की संख्या कितने बंद बन रहे हैं तो बंदूक की संख्या भी बोल सकते हो हां बिल्कुल ठीक है दोनों ही करेक्टर अच्छा चलो ठीक है सबको अच्छी बात है चलो तो बच्चों का…mean free path
  • 7:35–10:12Mean free path — segment 4परमाणु सिर्फ एक ठीक है एक परमाण्विक कणो अगर मेरे पास आ हैं तो एक परमाण्विक कणो के लिए जो हमने फोर व्हील ड्राइव करके निकाला था जब हमने बात करी थी किसी भी कंटेनर किसी भी पात्र की दीवारों पर लगने वाले लाभ के बारे में तो…mean free path
  • 10:12–12:43Mean free path — segment 5कि वह लड्डू के लिए यह तो आपको पता है कि मैंने किसी थोड़ा कर दिया था आवोगाद्रो संख्या से ऐड और बट्टू के बीच यौन है और आपको पता है कि इसके वीक में लिखा था आर्य ने इस सफल हो चुका है भी लास्ट क्लास में जो 5 मिनट मैंने बताया…mean free path
  • 12:43–15:16Mean free path — segment 6बारे में बात नहीं चल रही है मैं अनमोल पलों के बारे में बात कर रहा हूं प्रियांशु ने सही लिखा है लेकिन उसने एक मॉल में लिखा अनमोल के लिए पूछ रहा आपसे तो गया वापस दिल्ली की ओर कौन एंड अपडेटेड ओं अच्छा ठीक है यह चैनल पर…mean free path
Gas laws on the basis of kinetic theory in Hindi | H-3 | Thermodynamics 🔉⇢
Sacademy

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Sacademy); found via yt-dlp search 'गैस का दाब अणुगति सिद्धांत pressure of gas kinetic theory Hindi', oEmbed-verified live.

📑 Clips (6)
  • 0:00–2:35Kinetic theory of pressure — segment 1है दोस्तों एक प्रॉब्लम इस YouTube चैनल प्रोडक्ट है उस व्यक्ति की क्या है हिंदी वर्जन का यह माप थन लगते हैं जिसमें हम कब्जे के सिद्धांत के आधार पर गैस नियमों के बारे में जानकारी प्राप्त करेंगे तो कि हमारे क्षेत्र में…kinetic theory of pressure
  • 2:35–5:08Kinetic theory of pressure — segment 2किसी भी गैस के लिए निश्चित आइटम में इनवेस्ट रहता है अतः मन पर क्यों एवं शुक्र और बुध कांटेक्ट है एंड वेरी कॉन्फिडेंट हैं ऐड उबाल लें और डिस्कंटेंट है तो पीड़ित व्यक्तियों ने टांग तरफ बराबर दिए डा यही बॉईल का नियम है…kinetic theory of pressure
  • 5:08–7:41Kinetic theory of pressure — segment 3कैंसर हो जाएगा 24 कंट्रोल एंड पॉइंट्स पे अब एंड तक है के लिए तक है यदि कम और नियत हो तो इस डिफिकल्ट टो t&amp;c कि यदि आप इंजीनियर को अर्थात यह विधि गैस का दाम यह छोड़ तो वीएस डिफिकल्ट रोटी और यह हमारा चार्ल्स का नियम है…kinetic theory of pressure
  • 7:41–10:12Kinetic theory of pressure — segment 4समान होती है ए मॉडल किसी गैस के एक अणु का द्रव्यमान लेमन है इसमें प्रति घन मीटर हुए दौरों की संख्या एनिमल है तथा इसके अनुभवों का वर्ग माध्य मूल वैध सीमन है कि एवं यदि कोई दूसरे क्रम ले ले में जुटे गैस के एक अनुमान शैंपू…kinetic theory of pressure
  • 10:12–12:43Kinetic theory of pressure — segment 5इनटू थे झाल संबोधित आप एवं ढाबों पर समाप्त आइटम वाली गैसों में अंडों की संख्या समान होती है यही अभद्र की परिकल्पना है है नेक्स्ट रेड कलर का आंशिक डा नी पधारो तरह का आंशिक दाब का नियम है कि टेस्ट अनुसार यदि एक पात्र में…kinetic theory of pressure
  • 12:43–15:13Kinetic theory of pressure — segment 6कि यह क्रीम अभी तो पींठ गैस के संपूर्ण आइटम में अलग-अलग गैसों के ढाबे में अतः गैस का परिणाम विद अब उन्हें गैसों के अलग-अलग आज सितारों के योग के बराबर होता यही डायरेक्टर का अनफिट का नियम है कि अब इस टॉपिक हमारा है गांव…kinetic theory of pressure
Specific Heat of Gases गैसों की विशिष्ट ऊष्मा ll Cp & Cv ll Molar specific Heat मोलर विशिष्ट ऊष्मा 🔉⇢
Ajay Kumar Maurya

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Ajay Kumar Maurya); found via yt-dlp search 'गैसों की विशिष्ट ऊष्मा Cp Cv specific heat Hindi physics class 11', oEmbed-verified live.

📑 Clips (3)
  • 0:00–2:34Specific heats & gamma — segment 1कि आज का टॉपिक है स्पेसिफिक हीट आफ गेसचर्स गैसों की विशिष्ट ऊष्मा इस टॉपिक आफ डिस्कशन किया जाएगा तो स्पेसिफिक हीट आफ गेस्ट यह गैसों की विशिष्ट ऊष्मा यह डिपेंड करती है कि हम कौन सा थर्मोडायनेमिक या उस मार्ग अतिक्रमण कर…specific heats & gamma
  • 2:34–5:05Specific heats & gamma — segment 2बेसिक यूनिट यहां से यह का योनि पर ग्राम पर पर डिग्री सेल्सियस होता है कि लो कैलोरी पर केजी पर डिग्री सेल्सियस ऐसा ही सिस्टम होगा इसको चाहिए हम जून में भी लिख सकते हैं जून पर केजी पर डिग्री सेल्सियस दोनों मात्रक यूज किए…specific heats & gamma
  • 5:05–7:34Specific heats & gamma — segment 3आफ इक्वल टू द टेंपरेचर ऑफ वन आफ डिग्री सेल्सियस वॉल्यूम किसी गैस कि स्थिर आयतन पर विशिष्ट ऊष्मा को दी गई सूचना के बराबर होता है जो तापमान तापमान डिग्री सेल्सियस पर आ की मात्रा वाले कहते हैं अब में तो टीवी पर पर डिग्री…specific heats & gamma
Specific Heat of Gases (गैसों की विशिष्ट ऊष्मा).Relation between Cv & Cp. Constant Volume & Pressure 🔉⇢
AKC TECHNICAL CLASSES

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (AKC TECHNICAL CLASSES); found via yt-dlp search 'गैसों की विशिष्ट ऊष्मा Cp Cv specific heat Hindi physics class 11', oEmbed-verified live.

📑 Clips (6)
  • 0:01–2:34Specific heats & gamma — segment 1तो दोस्तों एक बार फिर से स्वागत है आपका आपके अपने youtube1 का पार्ट थर्ड और यही पार्ट थर्ड जो है आपका लास्ट होगा चैप्टर सेकंड का क्योंकि आपके बुक में कुछ एक्स्ट्रा दिया गया है लेकिन वो सब आपके सिलेबस में नहीं है तो आइए…specific heats & gamma
  • 2:34–5:05Specific heats & gamma — segment 2के द्वारा किया गया कार्य जो होता है मान लीजिए कोई आपका सिस्टम है जब यह सिस्टम कोई कार्य करता है जब यह सिस्टम कोई कार्य करता है तो इसमें जो वर्क डन यानी डेल्टा डब्लू जो मिलता है वो हमेशा क्या होता है पॉजिटिव होता है और…specific heats & gamma
  • 5:05–7:36Specific heats & gamma — segment 3ओके आपका पिस्टन मूव कर सकता है तो इसमें अगर हम टेंपरेचर इसके टेंपरेचर को अगर बढ़ा रहे हैं इसके ट यहां से हम नीचे मान लीजिए इसके टेंपरेचर को बढ़ा रहे हैं तो क्या कह रहा है कि किसी पदार्थ की एकांक मात्रा का जो टेंपरेचर…specific heats & gamma
  • 7:36–10:08Specific heats & gamma — segment 4होता है सीवी सी क्या होता है ये होता है आपका स्पेसिफिक हीट एट कांस्टेंट वैल्यू तो देखिए जो आपका फार्मूला आता है जो आपका फार्मूला आता है सीप माइ सीब = r सीप - सीब = जो r आपका फार्मूला आता है इसको हम किसी एक को लेकर प्रूफ…specific heats & gamma
  • 10:08–12:39Specific heats & gamma — segment 5का प्रेशर कांस्टेंट प्रेशर पर जो वर्क डन होता है इंटरनल एनर्जी होती है और हीट ट्रांसफर जो होता है उन्हीं सबको लेकर हम इसमें सॉल्व करेंगे तो आप लोगों को थोड़ा कंफ्यूजन जरूर होगा लेकिन आप लोग समझने का प्रयास जरूर कीजिएगा…specific heats & gamma
  • 12:39–15:10Specific heats & gamma — segment 6समझना है और इस सिंपल लैंग्वेज में आपको हम समझाएंगे भी और पांच नंबर में आपका क्वेश्चन आता भी है एग्जाम में देखिए कैसे तो हमने क्या किया है कि मान लीजिए कोई गैस है उसको हम गर्म कर रहे हैं और वह एक से दो तक के पॉइंट का…specific heats & gamma
पदार्थ की अवस्थाएं | matter and its states | Chemistry | रासायन विज्ञान | Study vines official 🔉⇢
Study Vines official

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Study Vines official); found via yt-dlp search 'पदार्थ की आणविक प्रकृति molecular nature of matter Hindi physics', oEmbed-verified live.

📑 Clips (6)
  • 0:00–2:30Molecular nature of matter — segment 1नमस्कार दोस्तों आपका स्वागत है आपके अपने चैनल स्टडी वाइ ऑफिशियल पर तो आज की इस वीडियो में हम केमिस्ट्री का फर्स्ट टॉपिक पदार्थ की अवस्थाओं के बारे में पढ़ेंगे तो इंपॉर्टेंट टॉपिक है पहले हम पूरे टॉपिक को अच्छे से…molecular nature of matter
  • 2:30–5:01Molecular nature of matter — segment 2तो मास किसे कहते हैं क्वांटिटी ऑफ मैटर को किसी वस्तु में उपस्थित पदार्थ की मात्रा को उसका द्रव्यमान कहा जाता है तो पदार्थ के उदाहरण क्या हैं तो जो भी चीज स्थान घेरे और जिसका द्रव्यमान हो वह पदार्थ कहलाती है जैसे जल वायु…molecular nature of matter
  • 5:01–7:34Molecular nature of matter — segment 3तो आधुनिक समय में हमें पदार्थ की कितनी अवस्थाएं पता चल चुकी हैं तो पांच अवस्थाएं पता चल चुकी हैं पहले तीन तो हमने देख ही लिए ठोस द्रव और गैस तो पदार्थ की चौथी अवस्था कौन सी है तो पदार्थ की चौथी अवस्था है प्लाज्मा और…molecular nature of matter
  • 7:34–10:04Molecular nature of matter — segment 4नगण्य संपीड्यता का गुण होता है संपीड़ित क्या होती है कंप्रेसिबिलिटी यानी कि जो ठोस वस्तु होती है उसे बाहर बल लगाकर कंप्रेस नहीं किया जा सकता जैसे जो मुलायम चीज होती है उसे हम कंप्रेस कर सकते हैं लेकिन जो ठोस वस्तु होती…molecular nature of matter
  • 10:04–12:37Molecular nature of matter — segment 5का आकार निश्चित होता है क्या नहीं होता अगर आप पानी को बाल्टी में डालेंगे तो उसका आकार अलग हो जाएगा उसको बोतल में डालेंगे तो उसका आकार अलग हो जाएगा तो द्रव अवस्था में जो पदार्थ होता है इसका आकार अनिश्चित होता है परंतु…molecular nature of matter
  • 12:38–15:11Molecular nature of matter — segment 6होता है तो गैसों में अंतर आणविक स्थान बहुत अधिक होता है और इनमें उच्च संपीड़ित का गुण होता है तो इनमें हाई कंप्रेसिबिलिटी कैसे होती है क्योंकि इनमें जो मॉलिक्यूलर स्पेस है वो बहुत ज्यादा होता है तो जब मॉलिक्यूलर स्पेस…molecular nature of matter
मैटर | पदार्थों की अवस्थाएँ | What Is Matter In Hindi | Dr.Binocs Show | Educational Videos 🔉⇢
Binocs Ki Duniya

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Binocs Ki Duniya); found via yt-dlp search 'पदार्थ की आणविक प्रकृति molecular nature of matter Hindi physics', oEmbed-verified live.

📑 Clips (3)
  • 0:04–2:35Molecular nature of matter — segment 1[संगीत] और वैसे भी हमारे चारों तरफ मैटर ही तो है हान मेरे दोस्तों मुझे यकीन है आपने मैटर टर्म के बारे में जरूर सुना होगा और कैसे हर वह चीज जो हमारे जीवन का हिस्सा है जैसे की हम जो सांस लेते हैं किताबें पढ़ते हैं जब…molecular nature of matter
  • 2:35–5:05Molecular nature of matter — segment 2करते हैं सबसे ओबवियस स्टेट ऑफ मैटर से जो है सॉलिड स्टेट ये वो चीजें हैं जिनकी अपनी एक शॉप होती है और किसी भी टेंपरेचर में फ्लो नहीं करती तो सॉलिड के एग्जांपल्स हैं स्टडी टेबल में आती है लेकिन जब पिघलती है तो लिक्विड बन…molecular nature of matter
  • 5:05–6:26Molecular nature of matter — segment 3स्टिक के छोरी पर बंद दीजिए उसके बाद मीटर स्टिक को एक धागे पर ऐसे लटकाए की वो बैलेंस हो जाए अब एक बलून में हवा भरिए और दोबारा मीट्स टिक में लगाइए आप देखेंगे हवा वाले बलून का सिरा नीचे की तरफ जाएगा क्योंकि हवा की वजह से…molecular nature of matter
The ideal gas law (PV = nRT) | Intermolecular forces and properties | AP Chemistry | Khan Academy 🔉⇢
Khan Academy

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Khan Academy); found via yt-dlp search 'ideal gas equation PV=nRT derivation Khan Academy', oEmbed-verified live.

📑 Clips (3)
  • 0:00–2:30Ideal gas equation — segment 1in this video we're going to talk about ideal gases and how we can describe what's going on with them so the first question you might be wondering is what is an ideal gas and it really is a bit of a theoretical…ideal gas equation
  • 2:30–5:01Ideal gas equation — segment 2changing the temperature so i'm not changing t and n what's going to happen to the pressure well that gas is going to per square inch or per square area exert more and more force it gets harder and harder for me to…ideal gas equation
  • 5:01–6:20Ideal gas equation — segment 3or another way to say it is you could say that volume is going to be equal to some constant that's what proportionality is just talking about it's going to be equal to some constant let's call it r times all of this…ideal gas equation
Definition of an ideal gas, ideal gas law | Physical Processes | MCAT | Khan Academy 🔉⇢
khanacademymedicine

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (khanacademymedicine); found via yt-dlp search 'ideal gas equation PV=nRT derivation Khan Academy', oEmbed-verified live.

📑 Clips (3)
  • 0:01–2:33Ideal gas equation — segment 1okay in our last video we talked about gas pressure and we got on that subject by making some observations about the air that's inside of a balloon so I've got a red balloon and I'll give it a white string my balloon…ideal gas equation
  • 2:33–5:03Ideal gas equation — segment 2give us p is equal to R time and that's our constant time NT over V and so if we multiply both sides by V and do just a little bit of rearranging we're going to get a pretty important equation called the ideal gas…ideal gas equation
  • 5:03–5:44Ideal gas equation — segment 3elastic because again we need to assume that none of our kinetic energy is loss in the collisions of these particles so we make these three conditions and and while no gases are actually ideal the ideal gas equation…ideal gas equation
Worked example:RMS speed and average KE of gas molecules | Kinetic theory | Physics | Khan Academy 🔉⇢
Khan Academy India - English

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Khan Academy India - English); found via yt-dlp search 'rms speed average kinetic energy temperature Khan Academy', oEmbed-verified live.

📑 Clips (3)
  • 0:00–2:34Temperature & rms speed — segment 1let's solve a couple of questions on RMS speed and average kinetic energy of gas molecules first question says at what temperature is the RMS speed of nitrogen the same as that of Oxygen's RMS speed at 2 degree celsius…temperature & rms speed
  • 2:34–5:07Temperature & rms speed — segment 2this comes out to be equal to minus 32.375 degree celsius we need to report the answer to two significant figures so we can do that and we can write this as minus 32 degrees Celsius so this is minus 32. okay let's look…temperature & rms speed
  • 5:07–5:30Temperature & rms speed — segment 3wrong in fact the right answer is option F because it only depends on temperature and they are kept at the same temperature so the kinetic energy for oxygen and hydrogen must be the same you can try more questions from…temperature & rms speed
Law of Equipartition | Kinetic theory | Grade 11 | Physics | Khan Academy 🔉⇢
Khan Academy India - English

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Khan Academy India - English); found via yt-dlp search 'law of equipartition of energy degrees of freedom physics', oEmbed-verified live.

📑 Clips (3)
  • 0:00–2:30Degrees of freedom & equipartition — segment 1Hello there. In this video, we will be learning about the law of equipartition of energy. Now, this law of equipartition of energy states that for a system in the thermal equilibrium, the total energy is distributed…degrees of freedom & equipartition
  • 2:30–5:02Degrees of freedom & equipartition — segment 2freedom. And in the last two columns, we can also see the variation with respect to temperature. Now, let's say if the temperature is less than 70 Kelvin, then the situation would be something different. And if the…degrees of freedom & equipartition
  • 5:03–5:58Degrees of freedom & equipartition — segment 3And when the molecule is linear, then the formula becomes 3n - 5. This happens when the molecule is linear. So, let's say we talk about a non-linear polyatomic molecule. So, in that particular case, the total degrees of…degrees of freedom & equipartition
Mean free path | Kinetic theory | Grade 11 | Physics | Khan Academy 🔉⇢
Khan Academy India - English

👁 Observe: How the governing relation is set up and applied to a worked number.

📚 Teaches: Tier-1 educational channel (Khan Academy India - English); found via yt-dlp search 'mean free path of gas molecules derivation physics', oEmbed-verified live.

📑 Clips (3)
  • 0:00–2:32Mean free path — segment 1Hello there. In this video, we will be understanding about mainfree path and deriving a formula for it. So let's say in a given gas volume, there are n number of gas molecules. So what actually happens is when this…mean free path
  • 2:32–5:02Mean free path — segment 2cylinder type region. So we can say that all the molecules that would be lying within this cylinder would be the one that would be interacting with this particular molecule and thus colliding with it. Let's say for this…mean free path
  • 5:02–7:06Mean free path — segment 3say that number of collisions per unit time that would be equal to n pi d² into v delta t divided by delta t. So we can say that per unit time the number of collisions that are happening is n pi d² * v. So we can write…mean free path

💬 Doubt Solving Ask-any-doubt RAG + FAQ

🤖 Ask any doubt RAG · NCERT-grounded, cited

Answers are grounded in tier-1 material (NCERT / JEE PYQ) and cited — the AI will say so if a doubt isn't in the indexed syllabus.

Frequently-asked doubts

Why is the rms speed larger than the average speed?
Because squaring emphasises the fast molecules. When you square each speed, average, and take the square root, the fast tail of the Maxwell distribution contributes disproportionately, so $v_{rms}=\sqrt{\langle v^2\rangle}$ always exceeds the ordinary mean $\langle v\rangle$. Numerically $v_{rms}:\langle v\rangle:v_p=1.225:1.128:1$. Use $v_{rms}=\sqrt{3k_BT/m}$ whenever the quantity is tied to kinetic energy or temperature, and the mean speed $\langle v\rangle=\sqrt{8k_BT/\pi m}$ only when a problem explicitly asks for the average speed, as in mean-free-path and effusion estimates.
Does the average kinetic energy of a molecule depend on the type of gas?
No. The kinetic interpretation of temperature says the average translational kinetic energy per molecule is $\tfrac{3}{2}k_BT$ for every ideal gas — monatomic, diatomic or polyatomic — at a given temperature. It is independent of mass, pressure, volume and the chemical identity of the gas. What does depend on the gas is the rms speed (lighter molecules move faster) and the total internal energy (molecules with more degrees of freedom store more energy per molecule), but the translational part per molecule is universal.
What exactly is a degree of freedom, and why does a vibration count as two?
A degree of freedom is an independent quadratic term in the energy of a molecule. Translation contributes three ($\tfrac{1}{2}mv_x^2$ and so on), rotation adds up to two for a linear molecule and three for a nonlinear one, each a $\tfrac{1}{2}I\omega^2$ term. A vibration is special: it stores both kinetic energy ($\tfrac{1}{2}m\dot y^2$) and potential energy ($\tfrac{1}{2}ky^2$), two quadratic terms, so by equipartition it carries $2\times\tfrac{1}{2}k_BT=k_BT$ and counts as two degrees of freedom.
When can I use $PV=\mu RT$ and when does it fail?
The ideal-gas equation holds when molecules are far apart and their mutual interactions are negligible — that is, at low pressure and high temperature. It fails at high pressure (molecules crowd together, their finite size matters) and low temperature (attractions become important and the gas may liquefy). The compressibility factor $Z=PV/\mu RT$ measures the deviation; $Z=1$ is ideal. For JEE numerical problems the gas is almost always taken as ideal unless the question explicitly invokes a van der Waals correction.
Why does a gas have two specific heats while a solid effectively has one?
A gas can absorb heat while its volume is held fixed (no work done, so all heat raises internal energy, capacity $C_v$) or while its pressure is held fixed (the gas expands and does work, so more heat is needed for the same temperature rise, capacity $C_p$). Hence $C_p>C_v$ and $C_p-C_v=R$ for an ideal gas. A solid barely changes volume on heating, so its two specific heats are almost equal and one value ($\approx3R$, Dulong-Petit) suffices.
How does the mean free path depend on pressure and temperature?
The mean free path is $l=\dfrac{1}{\sqrt{2}\,n\pi d^2}$, which depends on the number density $n$ and molecular diameter $d$. At constant temperature, since $P=nk_BT$, increasing pressure raises $n$ and shortens $l$ (they are inversely proportional). At constant volume, changing temperature does not change $n$, so $l$ is unchanged — although the molecules collide more often because they move faster. Only through a change in density does $l$ change.
Is the pressure formula affected by ignoring collisions between molecules?
No, not in steady state. NCERT derives $P=\tfrac{1}{3}nm\langle v^2\rangle$ ignoring intermolecular collisions, and argues that in a steady state, for every molecule knocked out of a velocity group by a collision another is knocked into it, so the velocity distribution — and hence the pressure — is unaffected. The shape of the container and the chosen time interval also drop out of the final result, so the formula is general.
Why do light gases like hydrogen and helium escape a planet's atmosphere?
At a given temperature all gases share the same average kinetic energy, so the lightest molecules have the highest rms speed ($v_{rms}\propto1/\sqrt{m}$). When a fast molecule near the top of the atmosphere exceeds the escape velocity it leaves for good. Over geological time the low-mass gases are lost preferentially, which is why Earth's atmosphere is mostly heavy N$_2$ and O$_2$ with almost no free H$_2$, and why the low-gravity Moon retains no atmosphere at all.
What is the difference between $\langle v^2\rangle$ and $\langle v\rangle^2$?
They are not equal: the mean of the squares exceeds the square of the mean whenever the speeds vary, because squaring weights larger speeds more heavily. Only if every molecule had exactly the same speed would they coincide. This is why the rms speed $\sqrt{\langle v^2\rangle}$ is genuinely different from (and larger than) the mean speed $\langle v\rangle$, and it is a common exam trap to conflate the two.
Why is the specific heat of hydrogen temperature-dependent if equipartition says it is fixed?
Classical equipartition assigns $\tfrac{1}{2}k_BT$ to every mode and predicts a constant specific heat, but experiment shows $C_v$ of hydrogen rising in steps: only translation at very low temperature, translation plus rotation near room temperature, and vibration too only above about $1000$ K. The reason is that energy is quantised — a mode cannot absorb energy until $k_BT$ is comparable to its quantum spacing — so modes 'switch on' with temperature. This freezing-out is a classical failure that heralds quantum mechanics.
How is temperature related to molecular motion?
Temperature is a direct measure of the average translational kinetic energy of the molecules: $\tfrac{1}{2}m\langle v^2\rangle=\tfrac{3}{2}k_BT$. Absolute zero would correspond to zero translational energy. This is why temperature is independent of the amount, pressure or type of gas, and why two gases in thermal equilibrium have equal average molecular kinetic energies even if their molecules have very different masses and speeds.
Why does a gas take so long to diffuse across a room if molecules move at hundreds of m/s?
Because a molecule almost never travels in a straight line. At atmospheric density the mean free path is only about $10^{-7}$ m, so a molecule collides roughly a billion times a second and executes a random walk. Net displacement in a random walk grows only as the square root of the number of steps, so despite instantaneous speeds of hundreds of metres per second the effective spreading (diffusion) across a room takes minutes.
Why do a gas's specific heats come in a pair $C_p$ and $C_v$, and why is $C_p$ always the larger?
When you heat a gas at constant volume, no work is done, so every joule goes into raising the internal energy and hence the temperature — this defines $C_v$. When you heat it at constant pressure, the gas must also expand and push back the surroundings, doing work $R\,\Delta T$ per mole, so more heat is needed for the same temperature rise — this defines $C_p$. The extra work is exactly $R$ per mole, giving Mayer's relation $C_p-C_v=R$ for any ideal gas, which is why $C_p>C_v$ always. Their ratio $\gamma=C_p/C_v$ then depends only on the degrees of freedom of the molecule.
If all gas molecules have the same average energy, why do balloons of hydrogen and carbon dioxide behave so differently?
At a given temperature every molecule does share the same average translational energy $\tfrac{3}{2}k_BT$, but that is only part of the story. Lighter hydrogen molecules move much faster ($v_{rms}\propto1/\sqrt m$), so they effuse and diffuse out through the balloon skin far quicker, which is why a hydrogen balloon deflates sooner than a heavier carbon-dioxide one. The two gases also differ in atomicity and hence in specific heat and $\gamma$, so they respond differently to heating and to adiabatic changes. Equal energy per molecule does not mean equal behaviour.

Trap-answer taxonomy

Trap: Using mean speed where rms is required

Students plug the mean speed $\langle v\rangle$ into energy or temperature relations that actually call for $v_{rms}$, or vice versa, because both are 'average speeds'.

Fix: Tie the choice to the physics: kinetic energy and temperature use $v_{rms}=\sqrt{3k_BT/m}$; mean-free-path and effusion rates use the mean speed $\langle v\rangle=\sqrt{8k_BT/\pi m}$. Never assume they are equal.

Trap: Thinking heavier gas has more kinetic energy

It seems intuitive that a heavier molecule at the same temperature carries more kinetic energy, so students give unequal energy ratios in mixture problems.

Fix: At a common temperature every molecule has the same average translational energy $\tfrac{3}{2}k_BT$. Mass affects speed ($v_{rms}\propto1/\sqrt m$), not the translational energy per molecule.

Trap: Applying $\gamma=1.4$ to every diatomic gas

Students memorise $\gamma=7/5$ for diatomic gases and apply it even when vibrational modes are active or the molecule is non-rigid.

Fix: $\gamma=1.4$ holds only for a rigid diatomic at ordinary temperature ($f=5$). If vibration is active, $f=7$ and $\gamma\to9/7\approx1.29$. Count the active degrees of freedom first.

Trap: Forgetting a vibration is two degrees of freedom

When counting degrees of freedom, students give a vibrational mode one $\tfrac{1}{2}k_BT$ instead of the full $k_BT$.

Fix: A vibration has both kinetic and potential quadratic terms, so it contributes $2\times\tfrac{1}{2}k_BT=k_BT$ and counts as two degrees of freedom in $C_v=\tfrac{f}{2}R$.

Trap: Believing mean free path depends on temperature at fixed volume

Since heating speeds molecules up, students assume the mean free path must change with temperature.

Fix: $l=1/(\sqrt2\,n\pi d^2)$ depends only on number density, not speed. At constant volume heating leaves $n$ (hence $l$) unchanged; it only raises the collision frequency because the molecules move faster.

Trap: Confusing number density with amount

In pressure and mean-free-path formulas students substitute the number of moles or total number of molecules $N$ where the number density $n=N/V$ is required.

Fix: $P=nk_BT$ and $l\propto1/n$ use number per unit volume. Always divide by the volume; check units are m$^{-3}$.

Trap: Mixing up per-molecule and per-mole constants

Students use $R$ where $k_B$ belongs (per molecule) or $k_B$ where $R$ belongs (per mole), because $R=N_Ak_B$.

Fix: Per-molecule energies use $k_B$ ($\tfrac{3}{2}k_BT$); per-mole quantities use $R$ ($U=\tfrac{3}{2}RT$ for a mole). Match the constant to whether you are counting molecules or moles.

Trap: Assuming $\langle v^2\rangle=\langle v\rangle^2$

Students square the mean speed to get the mean-square speed, or treat the two as interchangeable in the pressure formula.

Fix: The mean of the square is always greater than the square of the mean for a spread of speeds. Keep $\langle v^2\rangle$ (which sets pressure and temperature) distinct from $\langle v\rangle^2$.

🚪 Dive Deeper Mystery room · 41 discoveries

Discovered 0 / 41

JEE Advanced archive — DISCOVER → ATTEMPT → REVEAL

A vessel contains a mixture of $2$ mol of helium (monatomic) and $3$ mol of oxygen (rigid diatomic) at temperature $T$. Find the equivalent molar specific heat at constant volume $C_v$ of the mixture and the effective adiabatic exponent $\gamma_{mix}$.

Attempt, then reveal full solution
For a mixture, $C_v^{mix}=\dfrac{\mu_1 C_{v1}+\mu_2 C_{v2}}{\mu_1+\mu_2}$. Helium: $C_{v1}=\tfrac{3}{2}R$; oxygen: $C_{v2}=\tfrac{5}{2}R$. So $C_v^{mix}=\dfrac{2(\tfrac{3}{2}R)+3(\tfrac{5}{2}R)}{5}=\dfrac{3R+7.5R}{5}=\dfrac{10.5R}{5}=2.1R$. Then $C_p^{mix}=C_v^{mix}+R=3.1R$, and $\gamma_{mix}=\dfrac{C_p}{C_v}=\dfrac{3.1}{2.1}=1.476$.

JEE-style (authored); NCERT XI §12.6

Two ideal gases have molecules of mass $m_1$ and $m_2$ with $m_2=4m_1$, held at the same temperature. Find the ratio of (i) their rms speeds and (ii) the average kinetic energy per molecule.

Attempt, then reveal full solution
(i) $v_{rms}=\sqrt{3k_BT/m}$, so $\dfrac{v_{rms,1}}{v_{rms,2}}=\sqrt{m_2/m_1}=\sqrt{4}=2$: the lighter gas is twice as fast. (ii) Average kinetic energy $\tfrac{3}{2}k_BT$ depends only on temperature, not mass, so the ratio is $1:1$. This separation of 'speed depends on mass, energy does not' is the crux of most mixture problems.

NCERT XI §12.4 (derived), Example 12.5

An ideal gas undergoes a process in which $PT^2=\text{constant}$. Find its coefficient of volume expansion $\gamma_V=\tfrac{1}{V}\tfrac{dV}{dT}$.

Attempt, then reveal full solution
From $PV=\mu RT$, $P=\mu RT/V$. Substitute into $PT^2=k$: $(\mu RT/V)T^2=k\Rightarrow V=\dfrac{\mu R}{k}T^3$. Then $\dfrac{dV}{dT}=\dfrac{3\mu R}{k}T^2=\dfrac{3V}{T}$, so $\gamma_V=\dfrac{1}{V}\dfrac{dV}{dT}=\dfrac{3}{T}$. The coefficient of volume expansion is $3/T$.

JEE Main 2008-style; NCERT XI §12.3

Estimate the temperature at which the rms speed of argon atoms ($M=39.9$ u) equals the rms speed of helium atoms ($M=4.0$ u) at $-20^\circ\text{C}$.

Attempt, then reveal full solution
Equal rms speeds require $\dfrac{3k_BT_{Ar}}{m_{Ar}}=\dfrac{3k_BT_{He}}{m_{He}}$, i.e. $\dfrac{T_{Ar}}{M_{Ar}}=\dfrac{T_{He}}{M_{He}}$. With $T_{He}=253$ K, $T_{Ar}=253\times\dfrac{39.9}{4.0}=253\times9.975\approx2523$ K. The much heavier argon needs a far higher temperature to match helium's speed.

NCERT XI §12.4, Exercise 12.9

A gas of rigid diatomic molecules is heated at constant volume so that its rms speed doubles. By what factor does (i) the absolute temperature and (ii) the internal energy of a fixed amount of the gas change?

Attempt, then reveal full solution
(i) $v_{rms}\propto\sqrt{T}$, so doubling $v_{rms}$ requires $T\to4T$: temperature increases four-fold. (ii) Internal energy $U=\mu C_vT=\mu\tfrac{5}{2}RT\propto T$, so $U$ also increases by a factor of $4$. Both scale with $T$, hence a factor of $4$.

JEE-style (authored); NCERT XI §12.4.2, §12.6

In a mixture, $2$ mol of a monatomic gas is mixed with $n$ mol of a rigid diatomic gas, and the mixture behaves like a gas with $\gamma=1.5$. Find $n$.

Attempt, then reveal full solution
$C_v^{mix}=\dfrac{2(\tfrac{3}{2}R)+n(\tfrac{5}{2}R)}{2+n}$ and $\gamma=1+\dfrac{R}{C_v^{mix}}=1.5\Rightarrow C_v^{mix}=2R$. So $\dfrac{3R+2.5nR}{2+n}=2R\Rightarrow 3+2.5n=2(2+n)=4+2n\Rightarrow0.5n=1\Rightarrow n=2$. Two moles of the diatomic gas are needed.

JEE Advanced-style (authored); NCERT XI §12.6

The mean free path of a nitrogen molecule ($d\approx3.0\times10^{-10}$ m) is $l$ at $2.0$ atm and $300$ K. If the gas is heated to $600$ K at constant volume, how does $l$ change? If instead the pressure is halved at constant temperature, how does $l$ change?

Attempt, then reveal full solution
$l=\dfrac{1}{\sqrt{2}\,n\pi d^2}$ depends only on number density $n$ (and $d$). At constant volume, heating does not change $n$, so $l$ is unchanged (though collisions become more frequent because $\langle v\rangle$ rises). Halving the pressure at fixed $T$ halves $n$ (since $P=nk_BT$), so $l$ doubles. Mean free path tracks density, not temperature at fixed volume.

NCERT XI §12.7 (derived), Exercise 12.10

A closed vessel of fixed volume contains $1$ mol of H$_2$ and $1$ mol of He in equilibrium at temperature $T$. Compare (i) the average kinetic energy per molecule, (ii) the rms speeds, and (iii) the total internal energies of the two gases.

Attempt, then reveal full solution
(i) Average KE per molecule $=\tfrac{3}{2}k_BT$ for both — equal. (ii) $v_{rms}\propto1/\sqrt{M}$; $M_{H_2}=2$, $M_{He}=4$, so $\dfrac{v_{H_2}}{v_{He}}=\sqrt{4/2}=\sqrt2\approx1.41$: hydrogen is faster. (iii) H$_2$ is diatomic ($U=\tfrac{5}{2}RT$), He monatomic ($U=\tfrac{3}{2}RT$), so $U_{H_2}:U_{He}=5:3$ despite equal translational energy — the extra rotational modes carry the difference.

JEE Advanced 2015-style; NCERT XI §12.5

Uranium hexafluoride is used to separate $^{235}$U from $^{238}$U. Taking the molecular masses as $349$ and $352$ u, estimate the percentage difference in their rms speeds at any common temperature.

Attempt, then reveal full solution
$\dfrac{v_{349}}{v_{352}}=\sqrt{\dfrac{352}{349}}=\sqrt{1.00859}\approx1.00429$. The percentage difference is $\approx0.43\%$. Because the difference is so tiny, a single diffusion stage barely enriches the mixture and the process must be cascaded thousands of times.

NCERT XI §12.4, Example 12.6

One mole of an ideal monatomic gas is taken through a process where the molar heat capacity is $C=2R$. Express this as a polytropic process $PV^k=\text{const}$ and find $k$.

Attempt, then reveal full solution
For a polytropic process $PV^k=$const, the molar heat capacity is $C=C_v+\dfrac{R}{1-k}$. Here $C_v=\tfrac{3}{2}R$ and $C=2R$, so $\dfrac{R}{1-k}=2R-\tfrac{3}{2}R=\tfrac{1}{2}R\Rightarrow1-k=2\Rightarrow k=-1$. The process is $PV^{-1}=$const, i.e. $P\propto V$.

JEE Advanced-style (authored); NCERT XI §12.6

Air is a mixture of about $4$ parts N$_2$ to $1$ part O$_2$ by mole. Treating both as rigid diatomic, estimate $\gamma$ for air and compare with the measured $1.40$.

Attempt, then reveal full solution
Both constituents are rigid diatomic with $C_v=\tfrac{5}{2}R$, so any mixture of them also has $C_v=\tfrac{5}{2}R$, giving $\gamma=\dfrac{C_p}{C_v}=\dfrac{7/2}{5/2}=1.40$ exactly. The prediction matches experiment because at ordinary temperatures the vibrational modes of N$_2$ and O$_2$ are frozen out.

JEE-style (authored); NCERT XI §12.6.2

Estimate the ratio of the molecular volume to the actual volume occupied by oxygen at STP, taking the molecular diameter as $3$ Å.

Attempt, then reveal full solution
Volume of one molecule $\approx\tfrac{4}{3}\pi r^3$ with $r=1.5\times10^{-10}$ m $=1.41\times10^{-29}$ m$^3$. Number density $n=2.7\times10^{25}$ m$^{-3}$. Molecular fraction $=n\times V_{mol}=2.7\times10^{25}\times1.41\times10^{-29}\approx3.8\times10^{-4}$. Only about $0.04\%$ of the gas volume is actually occupied by molecules — justifying the point-molecule assumption.

NCERT XI §12.3, Exercise 12.1

A cylinder of fixed volume $44.8$ litres contains helium at STP. How much heat is required to raise its temperature by $15.0^\circ$C? ($R=8.31$ J mol$^{-1}$K$^{-1}$.)

Attempt, then reveal full solution
At STP one mole occupies $22.4$ L, so the cylinder holds $2$ mol. Fixed volume $\Rightarrow$ use $C_v=\tfrac{3}{2}R$ (helium monatomic). $Q=\mu C_v\Delta T=2\times\tfrac{3}{2}R\times15.0=45R=45\times8.31\approx374$ J.

NCERT XI §12.6, Example 12.8

A flask contains argon and chlorine in a $2:1$ ratio by mass at $27^\circ$C. Find the ratio of (i) the average kinetic energy per molecule and (ii) the rms speeds. ($M_{Ar}=39.9$, $M_{Cl_2}=70.9$ u.)

Attempt, then reveal full solution
(i) Average KE per molecule $=\tfrac{3}{2}k_BT$ depends only on temperature, so the ratio is $1:1$ regardless of the mass composition. (ii) $\dfrac{v_{rms,Ar}^2}{v_{rms,Cl}^2}=\dfrac{M_{Cl}}{M_{Ar}}=\dfrac{70.9}{39.9}=1.77$, so $\dfrac{v_{rms,Ar}}{v_{rms,Cl}}=\sqrt{1.77}=1.33$. The mass ratio $2:1$ is irrelevant to both answers.

NCERT XI §12.4, Example 12.5

For a polyatomic gas with $3$ translational, $3$ rotational and $f$ vibrational modes, write $C_v$, $C_p$ and $\gamma$, and evaluate them for $f=0$ and $f=1$.

Attempt, then reveal full solution
Total quadratic modes $=3+3+2f$ (each vibration counts twice). $C_v=\tfrac{1}{2}(6+2f)R=(3+f)R$, $C_p=(4+f)R$, $\gamma=\dfrac{4+f}{3+f}$. For $f=0$: $C_v=3R$, $C_p=4R$, $\gamma=\tfrac{4}{3}\approx1.33$. For $f=1$: $C_v=4R$, $C_p=5R$, $\gamma=\tfrac{5}{4}=1.25$. Note $C_p-C_v=R$ holds throughout.

NCERT XI §12.6.3 (derived)

One mole of a monatomic ideal gas at $300$ K is compressed adiabatically and reversibly until its volume is halved. Find the final temperature and the factor by which the rms speed of its atoms increases. Take $\gamma=5/3$.

Attempt, then reveal full solution
For a reversible adiabatic process $TV^{\gamma-1}=$const, so $T_2=T_1(V_1/V_2)^{\gamma-1}=300\times2^{2/3}$. Since $2^{2/3}=1.587$, $T_2=300\times1.587\approx476$ K. The rms speed scales as $v_{rms}\propto\sqrt T$, so it increases by the factor $\sqrt{T_2/T_1}=\sqrt{1.587}=1.26$. The compression does work on the gas, raising both its temperature and the mean molecular speed even though no heat is added — a neat link between a bulk thermodynamic process and the microscopic speed distribution.

JEE Advanced-style (authored); NCERT XI §12.4, §12.6

Two identical vessels are connected by a fine tube. One holds helium ($M=4$), the other neon ($M=20$), at the same temperature and pressure. Compare the rates at which the two gases would effuse through an identical pinhole, and the time each takes to lose half its molecules.

Attempt, then reveal full solution
Effusion rate $\propto\langle v\rangle\propto1/\sqrt M$, so $\dfrac{r_{He}}{r_{Ne}}=\sqrt{\dfrac{20}{4}}=\sqrt5\approx2.24$: helium effuses about $2.24$ times faster. The half-emptying time is inversely proportional to the rate, so $\dfrac{t_{He}}{t_{Ne}}=\dfrac{1}{\sqrt5}\approx0.447$; helium empties in under half the time neon takes. This $1/\sqrt M$ dependence is exactly Graham's law of effusion, a direct consequence of the kinetic-theory mean speed and the reason lighter gases always leak away first.

NCERT XI §12.4 (derived); T. Graham

For a rigid diatomic ideal gas, verify that the difference of the molar specific heats equals $R$ and find what fraction of the internal energy is translational.

Attempt, then reveal full solution
A rigid diatomic molecule has $f=5$ ($3$ translational $+2$ rotational). $C_v=\tfrac{5}{2}R$ and $C_p=C_v+R=\tfrac{7}{2}R$, so $C_p-C_v=R$ as required by the ideal-gas relation, because heating at constant pressure must additionally supply the expansion work $R\,\Delta T$ per mole. Of the $5$ quadratic modes, $3$ are translational, so the translational share of the internal energy is $3/5=60\%$; the remaining $40\%$ is rotational. The equal splitting of $\tfrac{1}{2}RT$ per mode makes the fraction simply the ratio of mode counts.

NCERT XI §12.5, §12.6 (derived)

A sealed rigid container holds $N=3.0\times10^{22}$ molecules of an ideal gas at $27^\circ$C and pressure $1.0\times10^5$ Pa. Find the volume of the container and the total translational kinetic energy of the gas.

Attempt, then reveal full solution
Using $PV=Nk_BT$: $V=\dfrac{Nk_BT}{P}=\dfrac{3.0\times10^{22}\times1.38\times10^{-23}\times300}{1.0\times10^5}$. The numerator is $3.0\times10^{22}\times4.14\times10^{-21}=124.2$, so $V=1.24\times10^{-3}$ m$^3\approx1.24$ litre. Total translational kinetic energy $=\tfrac{3}{2}Nk_BT=\tfrac{3}{2}PV=\tfrac{3}{2}\times1.0\times10^5\times1.24\times10^{-3}\approx186$ J. Note the tidy shortcut that the translational energy of any ideal gas is always $\tfrac{3}{2}PV$.

JEE Main-style (authored); NCERT XI §12.3, §12.4.2

The rms speed of oxygen molecules ($M=32$ u) at a certain temperature is $480$ m/s. At what temperature will hydrogen molecules ($M=2$ u) have the same rms speed, and what is the oxygen temperature? ($R=8.31$ J mol$^{-1}$K$^{-1}$.)

Attempt, then reveal full solution
For oxygen, $v_{rms}=\sqrt{3RT/M}\Rightarrow T_{O}=\dfrac{M v_{rms}^2}{3R}=\dfrac{(32\times10^{-3})(480)^2}{3\times8.31}=\dfrac{32\times10^{-3}\times230400}{24.93}\approx296$ K. Equal rms speed requires equal $T/M$, so $T_H=T_O\times\dfrac{M_H}{M_O}=296\times\dfrac{2}{32}=18.5$ K. Hydrogen reaches the same speed at a far lower temperature because it is sixteen times lighter, illustrating $v_{rms}\propto\sqrt{T/M}$.

NCERT XI §12.4 (derived), Example 12.5

Five molecules have speeds $2, 3, 4, 5$ and $6$ (in units of $100$ m/s). Compute the mean speed, the rms speed and the most-probable speed for this small sample, and verify that the rms exceeds the mean.

Attempt, then reveal full solution
Mean speed $\langle v\rangle=\dfrac{2+3+4+5+6}{5}=4.0$ ($\times100$ m/s). Mean-square speed $\langle v^2\rangle=\dfrac{4+9+16+25+36}{5}=\dfrac{90}{5}=18$, so $v_{rms}=\sqrt{18}=4.243$ ($\times100$ m/s). Every value occurs once, so no single speed is more probable here, but the calculation confirms $v_{rms}=424$ m/s $>\langle v\rangle=400$ m/s. The rms is larger precisely because squaring gives extra weight to the faster molecules — the general result $v_{rms}>\langle v\rangle$ in miniature.

NCERT XI §12.4 (derived), Example 12.4

One mole of an ideal diatomic gas at $300$ K is heated at constant pressure until its volume doubles. Find the heat supplied and the fraction of it that becomes work done by the gas. Take $R=8.31$ J mol$^{-1}$K$^{-1}$.

Attempt, then reveal full solution
At constant pressure $V\propto T$, so doubling the volume doubles the temperature to $600$ K, $\Delta T=300$ K. Heat $Q=C_p\Delta T=\tfrac{7}{2}R\times300=1050R=1050\times8.31\approx8726$ J. Work done $W=R\Delta T=300R=2493$ J. The fraction that becomes work is $\dfrac{W}{Q}=\dfrac{R}{C_p}=\dfrac{2}{7}\approx0.286$, i.e. about $29\%$; the remaining $71\%$ goes into internal energy. The split $R:C_p$ is fixed by the atomicity.

JEE Main-style (authored); NCERT XI §12.6

Estimate the collision frequency of a nitrogen molecule at STP, given mean free path $l=1.0\times10^{-7}$ m and mean speed $\langle v\rangle=4.7\times10^2$ m/s.

Attempt, then reveal full solution
The collision frequency is the number of collisions per second, $\nu=\dfrac{\langle v\rangle}{l}=\dfrac{4.7\times10^2}{1.0\times10^{-7}}=4.7\times10^9$ per second — nearly five billion collisions each second. The mean time between collisions is $\tau=1/\nu\approx2.1\times10^{-10}$ s. This staggering rate, despite the tiny mean free path, is what keeps a gas in rapid local equilibrium and justifies treating collisions as effectively instantaneous.

NCERT XI §12.7 (derived), Example 12.9

A monatomic and a diatomic ideal gas have the same number of moles and the same temperature. Find the ratio of their total internal energies and the ratio of the heat needed to raise each by the same $\Delta T$ at constant volume.

Attempt, then reveal full solution
Internal energy $U=\mu C_v T$. Monatomic $C_v=\tfrac{3}{2}R$, diatomic (rigid) $C_v=\tfrac{5}{2}R$, so $\dfrac{U_{mono}}{U_{di}}=\dfrac{3/2}{5/2}=\dfrac{3}{5}$. Heat at constant volume is $Q=\mu C_v\Delta T$, so the ratio is the same, $3:5$. The diatomic gas stores more energy per molecule and needs more heat for the same temperature rise because it has two extra rotational degrees of freedom to fill.

JEE-style (authored); NCERT XI §12.5, §12.6

The pressure of an ideal gas is $P=1.0\times10^5$ Pa and its density is $\rho=1.2$ kg m$^{-3}$. Find the rms speed of its molecules directly, without needing the temperature or molar mass.

Attempt, then reveal full solution
From $P=\tfrac{1}{3}nm\langle v^2\rangle$ and recognising $nm=\rho$ (mass per unit volume), $P=\tfrac{1}{3}\rho v_{rms}^2$, so $v_{rms}=\sqrt{\dfrac{3P}{\rho}}=\sqrt{\dfrac{3\times1.0\times10^5}{1.2}}=\sqrt{2.5\times10^5}\approx500$ m/s. This elegant form shows the rms speed follows from bulk pressure and density alone — no thermometer or molecular mass required — and is the value for air at ordinary conditions.

NCERT XI §12.4.1 (derived)

Compute the mean free path of a nitrogen molecule at STP, taking the molecular diameter $d=3.0\times10^{-10}$ m and number density $n=2.7\times10^{25}$ m$^{-3}$. Comment on its size relative to the molecular diameter.

Attempt, then reveal full solution
The mean free path is $l=\dfrac{1}{\sqrt2\,n\pi d^2}$. Compute the denominator: $\pi d^2=\pi(3.0\times10^{-10})^2=2.83\times10^{-19}$ m$^2$; then $\sqrt2\,n\pi d^2=1.414\times2.7\times10^{25}\times2.83\times10^{-19}=1.08\times10^{7}$ m$^{-1}$. Hence $l=9.3\times10^{-8}$ m, roughly $10^{-7}$ m. This is about $300$ times the molecular diameter of $3\times10^{-10}$ m, confirming that a molecule travels a long way — in molecular terms — between successive collisions, which is why the point-molecule picture works so well.

NCERT XI §12.7 (derived), Example 12.9

A vessel contains $2.0$ g of hydrogen ($M=2$) and $8.0$ g of oxygen ($M=32$) at $300$ K in a volume of $10$ litre. Find the total pressure using Dalton's law. ($R=8.31$ J mol$^{-1}$K$^{-1}$.)

Attempt, then reveal full solution
Moles: hydrogen $\mu_1=2.0/2=1.0$ mol; oxygen $\mu_2=8.0/32=0.25$ mol; total $\mu=1.25$ mol. By Dalton's law the total pressure is that of all the molecules together: $P=\dfrac{\mu RT}{V}=\dfrac{1.25\times8.31\times300}{10\times10^{-3}}=\dfrac{3116}{0.010}\approx3.12\times10^{5}$ Pa. Each gas contributes its partial pressure in proportion to its mole fraction — hydrogen $80\%$, oxygen $20\%$ — because in kinetic theory every molecule drums on the walls independently with the same average energy.

NCERT XI §12.3 (derived); J. Dalton

One mole of an ideal polyatomic gas (nonlinear, $3$ translational $+3$ rotational modes, vibrations frozen) is at $400$ K. Find its internal energy and its molar specific heat ratio $\gamma$. ($R=8.31$.)

Attempt, then reveal full solution
A nonlinear polyatomic molecule with frozen vibrations has $f=6$ quadratic modes, so $C_v=\tfrac{f}{2}R=3R$ and internal energy $U=\mu C_vT=1\times3R\times400=1200R=1200\times8.31\approx9972$ J. Then $C_p=C_v+R=4R$ and $\gamma=\dfrac{C_p}{C_v}=\dfrac{4R}{3R}=\dfrac{4}{3}\approx1.33$. The larger number of degrees of freedom pushes $\gamma$ well below the diatomic $1.4$, exactly as equipartition predicts.

NCERT XI §12.5, §12.6 (derived)

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