Kinetic Theory Of Gases

Kinetic theory bridges the gap between the invisible chaos of molecules and the measurable quantities you already know — pressure, temperature, internal energy. For NEET, this chapter is a reliable source of formula-based questions on molecular speeds, degrees of freedom, specific heat ratios and mean free path, and it also supplies the physical justification for the gas laws you use in thermodynamics.

The Molecular Picture and Its Assumptions

An ideal gas, viewed microscopically, is a huge collection of tiny particles flying about in every direction, colliding with each other and with the walls of the container. Kinetic theory takes this picture seriously and asks: if molecules obey Newton's laws, what large-scale behaviour follows? The answer turns out to be exactly the ideal gas equation — a striking success that convinced nineteenth-century physicists that matter really is made of molecules.

The model rests on a short list of idealisations:

1. A gas contains an enormous number of identical molecules, so statistical averages are extremely stable.
2. The molecules are point-like: their total volume is negligible compared with the volume of the container.
3. Except during collisions, molecules exert no force on one another — there is no potential energy of interaction, so all the internal energy is kinetic.
4. Molecules move in straight lines with constant velocity between collisions, obeying Newtonian mechanics.
5. Collisions (molecule–molecule and molecule–wall) are perfectly elastic and take negligible time compared to the time spent travelling freely.
6. The motion is completely random: at any instant, all directions of velocity are equally likely, and there is no bulk flow. Hence the average velocity vector is zero, even though the average speed is not.

Real gases deviate from these assumptions when molecules are packed close (finite size matters) or moving slowly (attractive forces matter) — that is, at high pressure and low temperature. This is precisely why real gases approach ideal behaviour at low pressure and high temperature, and why the van der Waals corrections exist.

Pressure of an Ideal Gas and the Meaning of Temperature

Consider a cubical box of side $L$ containing $N$ molecules each of mass $m$. Take one molecule moving towards the right wall with an $x$-component of velocity $v_x$. On elastic rebound its momentum changes from $mv_x$ to $-mv_x$, so the wall receives momentum $2mv_x$. Between successive hits on the same wall the molecule travels a distance $2L$ along $x$, taking time $2L/v_x$. So the average force this one molecule exerts is $2mv_x/(2L/v_x) = mv_x^2/L$. Summing over all molecules and dividing by the wall area $L^2$:

$$P = \frac{m}{L^3}\sum v_x^2 = \frac{Nm}{V}\overline{v_x^2}$$

Randomness demands $\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2}$, and since $\overline{v^2} = \overline{v_x^2}+\overline{v_y^2}+\overline{v_z^2}$, each equals $\frac{1}{3}\overline{v^2}$. Therefore

$$P = \frac{1}{3}\frac{Nm}{V}\overline{v^2} = \frac{1}{3}\rho, v_{rms}^2$$

where $\rho$ is the gas density and $v_{rms} = \sqrt{\overline{v^2}}$ is the root-mean-square speed. An equivalent and very useful form is $PV = \frac{2}{3}\times$ (total translational kinetic energy of the gas), i.e. $P = \frac{2}{3}E_{translationalperunit~volume}$.

Now compare with the experimental law $PV = Nk_BT$. Equating:

$$\frac{1}{2}m\overline{v^2} = \frac{3}{2}k_BT$$

Temperature is nothing but a measure of the average translational kinetic energy per molecule. Two consequences worth memorising:

• The average translational KE per molecule is $\frac{3}{2}k_BT$, independent of the gas's identity. At the same temperature, hydrogen and oxygen molecules have equal average translational KE — but hydrogen, being lighter, moves much faster.
• $v_{rms} = \sqrt{\dfrac{3k_BT}{m}} = \sqrt{\dfrac{3RT}{M}} = \sqrt{\dfrac{3P}{\rho}}$, where $M$ is molar mass. So $v_{rms} \propto \sqrt{T}$ and $v_{rms} \propto 1/\sqrt{M}$.

This framework also explains Dalton's law: in a mixture, each species independently bombards the walls, so the total pressure is the sum of partial pressures, and at equilibrium all species share the same temperature and hence the same average translational KE.

Molecular Speeds and the Maxwell Distribution

Molecules do not all move at the same speed; collisions constantly redistribute energy. Maxwell showed that at a given temperature the fraction of molecules with speeds in a small range settles into a fixed distribution curve. Key features of the graph of number of molecules versus speed:

• The curve starts at zero (no molecule is exactly at rest for long), rises to a peak, and falls off with a long tail at high speeds — it is not symmetric.
• The speed at the peak is the most probable speed $v_{mp} = \sqrt{2k_BT/m}$.
• The average (mean) speed is $\bar{v} = \sqrt{8k_BT/\pi m} \approx \sqrt{2.55,k_BT/m}$.
• The rms speed is $v_{rms} = \sqrt{3k_BT/m}$.
• Ordering: $v_{mp} < \bar{v} < v_{rms}$, with the ratio $v_{mp} : \bar{v} : v_{rms} = \sqrt{2} : \sqrt{8/\pi} : \sqrt{3} \approx 1 : 1.128 : 1.224$.
• Raising the temperature shifts the peak to the right and flattens the curve (the area under it, being the total number of molecules, stays the same).
• At the same temperature, a lighter gas has a broader curve peaking at a higher speed.

A classic application is diffusion: rates of diffusion of two gases go as $1/\sqrt{M}$ (Graham's law), which is just the mass dependence of molecular speeds.

Degrees of Freedom and the Law of Equipartition

A degree of freedom is an independent coordinate (or independent mode of energy storage) needed to specify the state of a molecule. A point mass free to move in space has 3 translational degrees of freedom. Rigid rotation and vibration add more:

• Monatomic (He, Ne, Ar, and atomic vapours): 3 translational only, so $f = 3$.
• Diatomic, rigid (O₂, N₂, H₂ at ordinary temperatures): 3 translational + 2 rotational (rotation about the bond axis carries negligible moment of inertia), so $f = 5$.
• Diatomic with vibration active (at high temperature): add 1 vibrational mode, which counts twice (kinetic + potential), giving $f = 7$.
• Polyatomic, non-linear (H₂O, CH₄, NH₃, CO₂ treated as non-linear): 3 translational + 3 rotational + vibrational modes; for a rigid polyatomic $f = 6$.

The law of equipartition of energy states that in thermal equilibrium, the total energy is shared equally among all active degrees of freedom, each contributing an average of $\frac{1}{2}k_BT$ per molecule. Note carefully that for a vibrational mode this means $\frac{1}{2}k_BT$ of kinetic plus $\frac{1}{2}k_BT$ of potential energy, i.e. a full $k_BT$ per vibrational mode.

Hence the internal energy of one mole of an ideal gas with $f$ active degrees of freedom is

$$U = \frac{f}{2}RT \qquad\text{(per molecule: } \tfrac{f}{2}k_BT)$$

Remember: only the translational part, $\frac{3}{2}k_BT$ per molecule, is related to temperature through pressure and $v_{rms}$. Rotational and vibrational energy contribute to internal energy but not to the pressure formula.

Specific Heat Capacities of Gases

Since internal energy depends only on temperature for an ideal gas, at constant volume all heat supplied goes into $U$:

$$C_V = \frac{dU}{dT} = \frac{f}{2}R, \qquad C_P = C_V + R = \left(\frac{f}{2}+1\right)R, \qquad \gamma = \frac{C_P}{C_V} = 1 + \frac{2}{f}$$

These predictions match experiment well for monatomic and most diatomic gases at room temperature. Equipartition also explains the Dulong–Petit result for solids: each atom in a crystal vibrates in three directions, giving $3 \times k_BT = 3k_BT$ per atom, so molar specific heat $\approx 3R \approx 25$ J mol⁻¹ K⁻¹. For water, treating each of its atoms similarly gives roughly $9R$ per mole, close to the observed value.

The failure of equipartition at low temperatures — where measured specific heats drop far below these values, and vanish as $T \to 0$ — cannot be explained classically. It is a quantum effect: rotational and vibrational modes are "frozen out" because their energy levels are too widely spaced to be excited by the available thermal energy. This is a standard conceptual question.

For a mixture of gases, the effective $C_V$ is obtained by adding internal energies: $C_{V,mix} = (n_1C_{V1} + n_2C_{V2})/(n_1+n_2)$, and similarly for $C_P$; then $\gamma_{mix} = C_{P,mix}/C_{V,mix}$.

Mean Free Path

A molecule does not travel far before hitting another one. The mean free path $\lambda$ is the average distance travelled between successive collisions. Model a molecule of diameter $d$ moving through a gas of number density $n$: it will collide with any molecule whose centre lies within a cylinder of radius $d$ around its path, so in travelling distance $L$ it sweeps a volume $\pi d^2 L$ and suffers $n\pi d^2 L$ collisions. This gives $\lambda = L/(n\pi d^2 L) = 1/(n\pi d^2)$. Accounting for the fact that the other molecules are moving too introduces a factor $\sqrt{2}$:

$$\lambda = \frac{1}{\sqrt{2},\pi d^2 n} = \frac{k_BT}{\sqrt{2},\pi d^2 P}$$

using $n = P/k_BT$ from the ideal gas equation. Two trends are worth remembering: at fixed temperature $\lambda \propto 1/n \propto 1/P$, so halving the pressure doubles the mean free path — exactly why vacuum pumps are used to lengthen $\lambda$ inside devices like cathode ray tubes and particle accelerators. Dividing $\lambda$ by $v_{rms}$ gives the average time between collisions (the relaxation time), whose reciprocal is the collision frequency — the number of collisions a molecule suffers per second.

Common Mistakes and Exam Traps

• Assuming all gases have the same $v_{rms}$ at a given temperature. Only the average translational kinetic energy, $\tfrac{3}{2}k_BT$, is common to every gas at a shared temperature — lighter molecules must move faster to carry that same energy, since $v_{rms} \propto 1/\sqrt{M}$.
• Mixing up $k_B$ with $R$, or molecular mass with molar mass. $\tfrac{1}{2}m\overline{v^2} = \tfrac{3}{2}k_BT$ is a per-molecule statement using Boltzmann's constant and molecular mass $m$; the per-mole version $U = \tfrac{f}{2}RT$ uses the gas constant $R = N_Ak_B$ and molar mass $M = N_Am$ — swapping one pair for the other is a very common unit slip.
• Letting rotational or vibrational energy leak into the pressure formula. Pressure is fixed entirely by translational kinetic energy, $P = \tfrac{2}{3}\times$(translational KE per unit volume); a diatomic gas's larger total internal energy from rotation plays no role in this relation.
• Treating $C_V = \tfrac{f}{2}R$ as exact at every temperature. It matches experiment well near room temperature, but real gases show $C_V$ rising in steps as temperature increases (vibrational modes switching on) and falling towards the monatomic value as rotation itself freezes out at very low temperatures — a result classical equipartition cannot explain.