How the ceaseless random motion of molecules explains pressure, temperature, specific heats and the transport of gases — the molecular picture behind the gas laws
🔬 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. 🔉⇢
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.
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.
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.
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.
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.
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 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 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 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.
The complete chapter in reading order. Each section is narratable; the concept tabs add interactive 3D and worked examples on top of this text.
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.
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.
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. 🔉⇢
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.
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. 🔉⇢
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.
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 →
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.
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 →
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.
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. 🔉⇢
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.
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. 🔉⇢
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.
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 →
🔬 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. 🔉⇢
Source: NCERT Class XI Ch 12 Kinetic Theory
🔬 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. 🔉⇢
Source: NCERT Class XI Ch 12 Kinetic Theory
🔬 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. 🔉⇢
Source: NCERT Class XI Ch 12 Kinetic Theory
Source: NCERT Class XI Ch 12 Kinetic Theory
Source: NCERT Class XI Ch 12 Kinetic Theory
Source: NCERT Class XI Ch 12 Kinetic Theory
Source: NCERT Class XI Ch 12 Kinetic Theory
Source: NCERT XI §12.3, Example 12.1 (derived)
Source: NCERT XI §12.3, Example 12.2 (derived)
Source: NCERT XI §12.4, Exercise 12.6 (derived)
Source: NCERT XI §12.3, Example 12.4 (derived)
Source: NCERT XI §12.3, Exercise 12.5 (derived)
Source: NCERT XI §12.3 (density form, derived)
Source: NCERT XI §12.4 (derived)
Source: NCERT XI §12.4 (derived)
Source: NCERT XI §12.4 (derived)
Source: NCERT XI §12.4.2 (derived)
Source: NCERT XI §12.4.2, Example 12.5 (derived)
Source: NCERT XI §12.4.2 (derived)
Source: NCERT XI §12.4.2, Exercise 12.9 (derived)
Source: JEE-style (authored); Maxwell distribution
Source: NCERT XI §12.4.2 (derived)
Source: NCERT XI §12.5 (derived)
Source: NCERT XI §12.5 (derived)
Source: JEE-style (authored); NCERT XI §12.5
Source: NCERT XI §12.5 (derived)
Source: NCERT XI §12.6.1 (derived)
Source: NCERT XI §12.6.2 (derived)
Source: NCERT XI §12.6, Example 12.8 (derived)
Source: JEE-style (authored); NCERT XI §12.6
Source: NCERT XI §12.7 (derived)
Source: JEE-style (authored); NCERT XI §12.7
These worked examples are taught in full alongside their interactive scene:
| Quantity | Formula | What it means / when to use | Source |
|---|---|---|---|
| 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) |
| Quantity | Formula | What it means / when to use | Source |
|---|---|---|---|
| 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) |
| Quantity | Formula | What it means / when to use | Source |
|---|---|---|---|
| Law of equipartition 🔉⇢ | $\langle E\rangle=\tfrac{1}{2}k_BT$ per degree of freedom | In 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) |
| Quantity | Formula | What it means / when to use | Source |
|---|---|---|---|
| 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) |
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.
An ideal gas is expanding such that $PT^{3}$ = constant. The coefficient of volume expansion of the gas is :
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 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 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$C. Assuming the volume constant?
if the rms speed of oxygen molecules at 0$^\circ$C is 160 m/s, find the rms speed of hydrogen molecules at 0$^\circ$C.
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 :
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 :
$\left(P+\frac{a}{V^{2}}\right)(V-b)=R T$ 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:
The average kinetic energy of a molecule of the gas is
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 :
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 :-
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 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:
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 :
The root mean square velocity of molecules of gas is
An ideal gas is expanding such that $PT^2 = $ constant. The coefficient of volume expansion of the gas is
$C_v$ and $C_p$ denote the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then
A real gas behaves like an ideal gas if its
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
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 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 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?
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)
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?
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 ____
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]
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 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 :
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 :
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 :
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)?
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:
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.
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 _________.
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 ________.
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 :
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 :
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 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 :
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 vessel contains 16g of hydrogen and 128g of oxygen at standard temperature and pressure. The volume of the vessel in $cm^{3}$ is :
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 :
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 $\mathrm{k}=1.38 \times 10^{-23} \mathrm{JK}^{-1}$]
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 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 :
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 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 $\mathrm{R}=8.32 \mathrm{~J} \mathrm{~mol}^{-1} \,\mathrm{k}^{-1}$.
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 :
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 :
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 :
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)
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?
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:
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:
The ratio of speed of sound in hydrogen gas to the speed of sound in oxygen gas at the same temperature is:
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:
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 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,
The temperature at which the kinetic energy of oxygen molecules becomes double than its value at $27^{\circ} \mathrm{C}$ is
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 $\mathrm{k}_{\mathrm{B}}=1.4 \times 10^{-23} \mathrm{JK}^{-1}$ )
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 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 $\mathrm{Ar}=39.9$ )
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
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 $\_\_\_\_$ .
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:
Distribution — advanced: 16 · easy: 39 · hard: 18 · medium: 32. Every question carries a source trace; each ends in an SME-verify solution.
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MIT OCW / Walter Lewin, 3Blue1Brown, Khan, Veritasium. Each video is broken into clips — click a clip to play just that segment.
👁 Observe: How the governing relation is set up and applied to a worked number.
📚 Teaches: Tier-1 educational channel (EG Chemistry - IIT JEE & NEET); found via yt-dlp search 'मैक्सवेल वेग वितरण Maxwell speed distribution Hindi physics', oEmbed-verified live.
👁 Observe: How the governing relation is set up and applied to a worked number.
📚 Teaches: Tier-1 educational channel (Peekaboo Kidz); found via yt-dlp search 'molecular nature of matter atomic hypothesis physics', oEmbed-verified live.
👁 Observe: How the governing relation is set up and applied to a worked number.
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👁 Observe: How the governing relation is set up and applied to a worked number.
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👁 Observe: How the governing relation is set up and applied to a worked number.
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👁 Observe: How the governing relation is set up and applied to a worked number.
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👁 Observe: How the governing relation is set up and applied to a worked number.
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👁 Observe: How the governing relation is set up and applied to a worked number.
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👁 Observe: How the governing relation is set up and applied to a worked number.
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👁 Observe: How the governing relation is set up and applied to a worked number.
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👁 Observe: How the governing relation is set up and applied to a worked number.
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👁 Observe: How the governing relation is set up and applied to a worked number.
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👁 Observe: How the governing relation is set up and applied to a worked number.
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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.
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.
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.
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.
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 $C_v=\tfrac{f}{2}R$.
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.
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$^{-3}$.
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.
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 $\langle v\rangle^2$.
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 $\gamma_{mix}$.
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.
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 $\gamma_V=\tfrac{1}{V}\tfrac{dV}{dT}$.
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 $-20^\circ\text{C}$.
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?
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 $n$.
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?
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.
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.
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 $k$.
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 $1.40$.
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.
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$^{-1}$.)
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.)
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 $f=1$.
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 $\gamma=5/3$.
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.
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.
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.
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$^{-1}$.)
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.
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$^{-1}$.
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.
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.
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.
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.
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$^{-1}$.)
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 $\gamma$. ($R=8.31$.)
NCERT XI §12.5, §12.6 (derived)
| Chapter-mock score | Percentile band | Projected AIR band |
|---|---|---|
| 99.5+ percentile | 99.5+ | $\lt 1500$ |
| 99.0-99.5 percentile | 99.0-99.5 | $1500-4000$ |
| 98.0-99.0 percentile | 98.0-99.0 | $4000-9000$ |
| 95.0-98.0 percentile | 95.0-98.0 | $9000-25000$ |
| 90.0-95.0 percentile | 90.0-95.0 | $25000-55000$ |
| 80.0-90.0 percentile | 80.0-90.0 | $55000-120000$ |
| $\lt 80$ percentile | $\lt 80$ | $\gt 120000$ |
JoSAA 2023-24 closing-rank trends (indicative)
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