Kinetic Theory of an Ideal Gas

The molecular picture

A given amount of gas is a collection of an enormous number of molecules — of the order of Avogadro's number — in incessant random motion. 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 $2~\text{\AA}$. The interaction between molecules is therefore negligible, and between collisions they move freely in straight lines obeying Newton's first law.

The molecules do, however, collide — occasionally with one another, and continually with the walls of the container — and in each collision their velocities change. These collisions are taken to be elastic: total kinetic energy is conserved, and so is total momentum. The static appearance of an enclosed gas is misleading; the equilibrium is a dynamic one, in which only the average properties stay constant. It is on this picture that the entire derivation of pressure rests.

Assumptions of kinetic theory

Maxwell (1860) showed that the observed behaviour of a gas follows from a small set of assumptions about its molecules. These idealisations are what define an ideal gas — one whose molecules are point masses with no intermolecular force between them. Scan them as a checklist; NEET statement-based questions are built directly from this list.

Derivation of the pressure of an ideal gas

Consider a gas enclosed in a cube of side $l$, with the axes parallel to the sides of the cube. Take one molecule of mass $m$ with velocity $(v_x, v_y, v_z)$ that strikes the wall parallel to the $yz$-plane, of area $A = l^2$.

Because the collision is elastic, the molecule rebounds with the same speed; its $y$ and $z$ components are unchanged but the $x$-component reverses sign. The velocity after collision is $(-v_x, v_y, v_z)$, so the change in momentum of the molecule is $-mv_x - (mv_x) = -2mv_x$. By conservation of momentum, the momentum imparted to the wall is $2mv_x$.

To find the force we count how many molecules strike the wall in a small time $\Delta t$. A molecule with $x$-component $v_x$ reaches the wall if it lies within a distance $v_x \Delta t$ of it — that is, within the volume $A\, v_x \Delta t$. On the average only half of these are moving toward the wall, so the number that hit, for molecules of number density $n$, is $\tfrac{1}{2} n A\, v_x \Delta t$. The total momentum transferred to the wall in time $\Delta t$ is

$$Q = (2m v_x)\left(\tfrac{1}{2}\, n A\, v_x \Delta t\right).$$

The force on the wall is the rate of momentum transfer $Q/\Delta t$, and pressure is force per unit area:

$$P = \frac{Q}{A\,\Delta t} = n m\, v_x^2.$$

This is the contribution of one group of molecules sharing the same $v_x$. Real molecules have a distribution of velocities, so we replace $v_x^2$ by its average $\overline{v_x^2}$ and sum over all groups: $P = n m\, \overline{v_x^2}$. Now the gas is isotropic — no direction is special — so by symmetry $\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2}$. Since $\overline{v^2} = \overline{v_x^2} + \overline{v_y^2} + \overline{v_z^2}$, each component equals one-third of the whole:

$$\overline{v_x^2} = \tfrac{1}{3}\,\overline{v^2}.$$

Substituting gives the central result of the section:

$$\boxed{\,P = \tfrac{1}{3}\, n m\, \overline{v^2}\,}$$

where $\overline{v^2}$ is the mean of the squared speed. Two remarks NCERT makes are worth keeping. First, the cube was only a convenience: for any vessel shape one chooses a small planar area and the same steps follow, and neither $A$ nor $\Delta t$ survives into the result — by Pascal's law the pressure is the same everywhere in the gas. Second, intermolecular collisions were ignored, but in a steady state they do not change the velocity distribution, so they do not affect the calculated pressure.

The forms of the pressure equation

The single result $P = \tfrac{1}{3} n m \overline{v^2}$ wears several disguises in NEET problems. Recognising each instantly is half the battle.

Kinetic interpretation of temperature

Rewrite the pressure result for the whole sample. With $N = nV$ molecules,

$$PV = \tfrac{1}{3}\, n V\, m\, \overline{v^2} = \tfrac{2}{3}\, N \left(\tfrac{1}{2}\, m\, \overline{v^2}\right).$$

The quantity in the bracket, $\tfrac{1}{2} m \overline{v^2}$, is the average translational kinetic energy of a molecule. Since the internal energy $E$ of an ideal gas is purely kinetic, $E = N \times \tfrac{1}{2} m \overline{v^2}$, and the relation collapses to a compact bridge between the macroscopic and the molecular:

$$PV = \tfrac{2}{3}\, E.$$

Now combine this with the ideal gas equation $PV = N k_B T$. Equating the two expressions for $PV$ gives $\tfrac{2}{3} E = N k_B T$, that is $E = \tfrac{3}{2} N k_B T$, and per molecule

$$\boxed{\,\frac{E}{N} = \tfrac{1}{2}\, m\, \overline{v^2} = \tfrac{3}{2}\, k_B T\,}$$

This is the kinetic interpretation of temperature: the average translational kinetic energy of a molecule is directly proportional to the absolute temperature, and to nothing else. It is independent of pressure, volume and the nature of the ideal gas. Temperature — a macroscopic thermodynamic variable — is revealed as a direct measure of molecular kinetic energy, the two domains stitched together by the Boltzmann constant $k_B = 1.38\times 10^{-23}~\text{J K}^{-1}$.

Internal energy and the recovery of PV = nRT

For a monatomic ideal gas the molecules have only translational motion, so the internal energy is exactly the total translational kinetic energy. For $N$ molecules, $E = \tfrac{3}{2} N k_B T$; for $n$ moles, using $N = nN_A$ and $R = N_A k_B$,

$$U = \tfrac{3}{2}\, n R T \qquad \text{(monatomic ideal gas)}.$$

This is why the internal energy of an ideal gas depends only on temperature, not on pressure or volume — a fact that underpins the molar specific heat $C_v = \tfrac{3}{2}R$ for a monatomic gas, taken further in specific heat from kinetic theory. The same kinetic picture also regenerates the gas law it started from: substituting $\tfrac{1}{2}m\overline{v^2} = \tfrac{3}{2}k_B T$ back into $PV = \tfrac{2}{3} N\left(\tfrac{1}{2}m\overline{v^2}\right)$ gives $PV = \tfrac{2}{3}N \cdot \tfrac{3}{2}k_B T = N k_B T = n R T$.

The internal consistency is the point. Starting from molecular collisions, kinetic theory predicts a pressure; combined with the empirical ideal gas equation it interprets temperature; and that interpretation, fed back in, reproduces $PV = nRT$ exactly. Avogadro's hypothesis falls out too: at fixed $P$, $V$ and $T$, the relation $PV = N k_B T$ forces $N$ to be the same for all gases.

Average KE, rms speed and molecular mass

The energy relation immediately yields the typical molecular speed. Since $\tfrac{1}{2}m\overline{v^2} = \tfrac{3}{2}k_B T$, the root-mean-square speed is

$$v_{\text{rms}} = \sqrt{\overline{v^2}} = \sqrt{\frac{3 k_B T}{m}} = \sqrt{\frac{3 R T}{M}},$$

where $M$ is the molar mass. At a fixed temperature the average kinetic energy is identical for every gas, but a lighter molecule must move faster to carry that same energy: $v_{\text{rms}}$ is inversely proportional to $\sqrt{m}$. For nitrogen at $T = 300~\text{K}$ this gives $v_{\text{rms}} \approx 516~\text{m s}^{-1}$, comparable to the speed of sound in air. Hydrogen, being far lighter than oxygen, moves much faster at the same temperature — the reason lighter gases dominate the upper atmosphere. The speed distribution and the three speed measures ($v_{\text{rms}}$, average, most probable) are taken up in molecular speeds.