Chapter 13 — Kinetic Theory of Gases

The kinetic theory of gases bridges the microscopic world (molecules in random motion) and the macroscopic world (pressure, temperature, internal energy). Starting from a handful of postulates and Newtonian mechanics, we will derive the ideal gas law, the meaning of temperature, the specific heats of gases, and the distance a molecule travels between collisions — the mean free path.

This chapter is one of the most "derivational" of the Class XI syllabus. Every equation in this chapter has a one-line physical interpretation. Master the derivations and the formulas write themselves.

Concept Map

• 13.1 Molecular nature of matter — Avogadro's hypothesis, mole, $N_A$.
• 13.2 Behaviour of gases — ideal gas equation in three equivalent forms.
• 13.3 Postulates of the kinetic theory of an ideal gas.
• 13.4 Derivation of pressure: $P = \tfrac{1}{3}\rho\,\langle v^2\rangle$.
• 13.5 Kinetic interpretation of temperature: $\langle E_k\rangle = \tfrac{3}{2}kT$.
• 13.6 RMS, mean, and most probable speeds; Maxwell–Boltzmann distribution (qualitative).
• 13.7 Law of equipartition of energy; degrees of freedom.
• 13.8 Specific heats of gases — $C_V$, $C_P$, $\gamma$ for mono-, di-, polyatomic gases.
• 13.9 Specific heats of solids and water — Dulong–Petit law.
• 13.10 Mean free path — derivation in terms of $\sigma$ and number density $n$.

13.1 Molecular Nature of Matter

Definition

Matter is made of atoms and molecules in ceaseless motion. The state of aggregation (solid, liquid, gas) is decided by the relative strength of intermolecular forces and the thermal energy.

• Solid: strong forces, fixed mean positions, vibration about lattice sites.
• Liquid: weaker forces, no fixed positions but definite volume.
• Gas: negligible mean intermolecular force compared with $kT$; molecules occupy the whole container.

Avogadro's Hypothesis

Equal volumes of all gases, at the same temperature and pressure, contain the same number of molecules.

The number of molecules in one mole of any substance is Avogadro's number

A mole is the amount of substance containing as many elementary entities as there are atoms in $12\ \text{g}$ of $^{12}\!C$. Useful relations:

Worked Example

How many molecules are there in $1.0\ \text{cm}^3$ of an ideal gas at STP ($T=273\ \text{K}$, $P=1.013\times 10^5\ \text{Pa}$)?

Using $PV = Nk_BT$:

This is Loschmidt's number — handy to memorise.

Pitfalls

• Confusing $N$ (number of molecules) with $n$ (number of moles). Always check units of the gas constant: $R$ goes with $n$, $k_B$ goes with $N$.
• Avogadro's hypothesis applies only to gases, not to liquids or solids.

13.2 Behaviour of Gases — The Ideal Gas Equation

Definition

A gas obeying the equation

at all $P$ and $T$ is called an ideal gas. Real gases obey it well at high temperature and low pressure (when molecules are far apart and intermolecular forces are negligible).

Three Equivalent Forms

Let $M$ be the molar mass, $\rho$ the mass density and $N$ the total number of molecules. Starting from $PV = nRT$:

Derivation of the third form:

The Empirical Gas Laws (re-derived)

• Boyle's law ($T$, $n$ fixed): $PV =$ const.
• Charles's law ($P$, $n$ fixed): $V/T =$ const.
• Gay-Lussac's law ($V$, $n$ fixed): $P/T =$ const.
• Avogadro's law ($P$, $T$ fixed): $V/n =$ const.

All four follow as special cases of $PV = nRT$.

Worked Example

An air bubble of volume $V_1 = 1.0\ \text{cm}^3$ at the bottom of a lake ($T_1=280\ \text{K}$, depth $40\ \text{m}$) rises to the surface ($T_2 = 300\ \text{K}$, $P_\text{atm}=1.013\times 10^5\ \text{Pa}$). Find its volume at the surface.

Pressure at the bottom: $P_1 = P_\text{atm} + \rho g h = 1.013\times 10^5 + (1000)(9.8)(40) \approx 4.93\times 10^5\ \text{Pa}$.

Using $\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}$:

Pitfalls

• Temperature must be in kelvin. Using $^\circ$C silently divides answers by absurd amounts.
• $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ when $P$ is in pascal and $V$ in m$^3$. Mixing units (atm with litres, etc.) without converting is the single most common error.
• Real gases deviate near liquefaction; Van der Waals correction $\left(P + \dfrac{a}{V^2}\right)(V-b) = nRT$ is needed there.

13.3 Kinetic Theory of an Ideal Gas — Postulates

The kinetic theory rests on the following idealised assumptions:

1. A gas consists of a very large number of identical molecules in random motion.
2. The size of a molecule is negligibly small compared to the average inter-molecular distance.
3. Molecules do not exert forces on each other except during collisions.
4. Collisions (with each other and the walls) are perfectly elastic and of negligible duration.
5. Between collisions molecules move in straight lines with constant velocity (no external forces, gravity ignored over container scales).
6. The container is large enough that boundary effects are negligible; the molecular distribution is isotropic and homogeneous.

These six postulates are sufficient to derive the entire macroscopic gas behaviour for a dilute gas.

Pitfalls

• "Negligible size" is not the same as "point mass": collisions still occur because molecules have a finite cross-section $\sigma = \pi d^2$ — needed in 13.10.
• The theory works for monatomic gases as written; for diatomic gases we additionally need the equipartition theorem (13.7).

13.4 Pressure of an Ideal Gas

Definition

Pressure is the force per unit area exerted on the walls of the container by molecular impacts.

Derivation (step-by-step)

Consider $N$ molecules of mass $m$ in a cubical box of side $L$ (volume $V = L^3$). Let molecule $i$ have velocity $\vec v_i = (v_{ix}, v_{iy}, v_{iz})$ with speed $v_i^2 = v_{ix}^2 + v_{iy}^2 + v_{iz}^2$.

Step 1 — Momentum change in one collision.

The molecule strikes the wall normal to the $x$-axis and rebounds elastically:

Step 2 — Time between two successive hits on the same wall.

The molecule must travel a distance $2L$ along $x$:

Step 3 — Force from one molecule.

Step 4 — Total force on the wall, sum over all molecules.

Step 5 — Pressure.

where $\langle v_x^2\rangle = \dfrac{1}{N}\sum v_{ix}^2$.

Step 6 — Use isotropy.

By symmetry $\langle v_x^2\rangle = \langle v_y^2\rangle = \langle v_z^2\rangle$, so

Step 7 — Final expression.

where $\rho = Nm/V$ is the gas density.

Equivalent Form

Define $v_{\rm rms} = \sqrt{\langle v^2\rangle}$:

Worked Example

Calculate $v_{\rm rms}$ for nitrogen at $T = 300\ \text{K}$ ($M = 28\ \text{g/mol}$).

Using $PV = nRT$ and $P = \tfrac{1}{3}\rho v_{\rm rms}^2$:

Pitfalls

• The factor $1/3$ comes from isotropy, not from the cubical shape — the result is shape-independent.
• $\langle v\rangle^2 \ne \langle v^2\rangle$. Always square first, then average.
• $\rho$ here is mass density, not number density.

13.5 Kinetic Interpretation of Temperature

Definition

Temperature is a measure of the average translational kinetic energy of a molecule.

Derivation

From 13.4, $PV = \tfrac{1}{3}Nm\langle v^2\rangle$. Multiply and divide by 2:

Compare with the empirical $PV = Nk_BT$:

Per mole:

Worked Example

The average translational KE of any gas molecule at $T = 300\ \text{K}$:

Independent of the gas! Hydrogen, helium, oxygen — same translational KE per molecule at the same $T$. What differs is the speed (since masses differ).

Pitfalls

• This is translational KE only. Diatomic molecules also have rotational and vibrational KE — those come from equipartition (13.7), and they multiply with extra factors of $\tfrac{1}{2}k_BT$ per degree of freedom.
• At the same $T$, $v_{\rm rms} \propto 1/\sqrt{M}$ — lighter gases move faster (basis of Graham's law of diffusion).

13.6 RMS, Mean, and Most Probable Speeds

The Three Speeds

The Maxwell–Boltzmann distribution of molecular speeds at temperature $T$ is

From it three characteristic speeds emerge:

• Most probable speed (peak of $f(v)$): $\displaystyle v_p = \sqrt{\frac{2k_BT}{m}} = \sqrt{\frac{2RT}{M}}.$
• Mean (average) speed: $\displaystyle \bar v = \sqrt{\frac{8k_BT}{\pi m}} = \sqrt{\frac{8RT}{\pi M}}.$
• Root-mean-square speed: $\displaystyle v_{\rm rms} = \sqrt{\frac{3k_BT}{m}} = \sqrt{\frac{3RT}{M}}.$

Order

For every gas at every temperature:

with numerical ratios $v_p : \bar v : v_{\rm rms} = \sqrt{2}:\sqrt{8/\pi}:\sqrt{3} \approx 1.41 : 1.60 : 1.73$.

Qualitative Shape of the Distribution

• Skewed to the right; long tail at high speeds.
• The peak shifts right and broadens as $T$ increases.
• Lighter gases have wider, flatter distributions.

Worked Example

For oxygen ($M = 32\ \text{g/mol}$) at $T = 300\ \text{K}$:

Pitfalls

• $\bar v \ne v_{\rm rms}$ — never use them interchangeably.
• All three speeds scale as $\sqrt{T/M}$, so the ratios above are universal.
• Doubling $T$ multiplies every speed by $\sqrt{2}$, not by 2.

13.7 Law of Equipartition of Energy

Statement

In thermal equilibrium, each independent quadratic term in the total energy expression of a molecule has an average energy equal to $\tfrac{1}{2}k_BT$ (per molecule), i.e. $\tfrac{1}{2}RT$ (per mole).

A "quadratic term" means an energy contribution of the form $\tfrac{1}{2}mv_i^2$, $\tfrac{1}{2}I\omega_i^2$, or $\tfrac{1}{2}kx_i^2$.

Why $\tfrac{1}{2}k_BT$?

For one translational degree of freedom along $x$,

But $\langle E_k\rangle = \tfrac{1}{2}m\langle v^2\rangle = \tfrac{3}{2}k_BT$, hence

Similarly for $v_y$ and $v_z$. The result then generalises to any quadratic term in the Hamiltonian — this is the equipartition theorem.

Degrees of Freedom (DOF)

The number of independent quadratic terms in the energy.

For a diatomic molecule, rotation about the bond axis has negligible moment of inertia, so it does not contribute — hence only 2 rotational DOF.

Average Energy per Molecule

and per mole, $U = (f/2)RT$.

Pitfalls

• Each vibrational mode contributes two quadratic terms (kinetic + potential), hence $k_BT$ — not $\tfrac{1}{2}k_BT$.
• At room temperature, vibrational modes of diatomic gases are usually "frozen out" (quantum effect) — use $f=5$ for N$_2$, O$_2$.
• Equipartition is a classical result; it fails for stiff modes at low $T$.

13.8 Specific Heats of Gases

Definition

• $C_V$ = molar heat capacity at constant volume.
• $C_P$ = molar heat capacity at constant pressure.
• $\gamma = C_P / C_V$ = adiabatic ratio.

For one mole, $U = (f/2)RT$, so

By the first law and ideal-gas property $(\partial H/\partial T)_P = (\partial U/\partial T)_V + R$ (Mayer's relation):

Values by Molecule

Worked Example

Calculate $\gamma$ for a mixture of $n_1$ moles of monoatomic and $n_2$ moles of diatomic gas (both rigid).

For $n_1 = n_2 = 1$: $C_V = 2R$, $C_P = 3R$, $\gamma = 3/2$.

Pitfalls

• Mayer's relation $C_P - C_V = R$ holds for ideal gases only.
• $\gamma$ is dimensionless; do not write "$\gamma$ J/mol K".
• For a mixture, average $C_V$ and $C_P$ weighted by moles — not $\gamma$.

13.9 Specific Heats of Solids and Water

Solids — Dulong–Petit Law

In a crystalline solid each atom oscillates about a lattice site in 3D, with kinetic + potential energy. Each atom therefore has $6$ quadratic terms, giving

This Dulong–Petit law holds at high temperature for most solids (Cu, Fe, Al, Pb). At low $T$, quantum effects (Einstein, Debye) suppress $C$.

Water (and Liquids)

Treat water as a solid with 3 atoms per molecule, each with 6 quadratic DOF:

Experimental value $\approx 75.3\ \text{J mol}^{-1}\text{K}^{-1}$ — remarkable agreement.

Pitfalls

• For solids and liquids, "$C_P$" and "$C_V$" differ very little — both are simply called the specific heat.
• The classical Dulong–Petit value fails for diamond at room $T$ because the bonds are extremely stiff — quantum freezing.

13.10 Mean Free Path

Definition

The mean free path $\lambda$ is the average distance a molecule travels between successive collisions.

Derivation

Picture a molecule as a sphere of diameter $d$ moving through a "frozen" gas with the rest of the molecules at rest. In time $t$ it sweeps out a cylinder of radius $d$ (any centre within this radius collides):

If $n = N/V$ is the number density, the number of collisions in time $t$ is

So the mean free path is

The other molecules are not at rest. A more careful calculation that uses the relative velocity (factor of $\sqrt 2$) gives

Worked Example

For nitrogen at STP, $d \approx 3.7\times 10^{-10}\ \text{m}$, $n \approx 2.7\times 10^{25}\ \text{m}^{-3}$:

A molecule has $\sim$ thousands of millions of collisions per second.

Pitfalls

• The cross-section uses diameter $d$, not radius. $\sigma = \pi d^2$.
• $\lambda \propto T/P$ — at low pressure (vacuum), $\lambda$ can become larger than the container, and "kinetic theory" gives way to "molecular flow".
• The $\sqrt{2}$ factor comes from relative motion of the molecules — often omitted at the school level but expected in JEE-advanced.

Solved Problems

Problem 1. A gas at $T_1 = 300\ \text{K}$ is heated to $T_2 = 1200\ \text{K}$ at constant volume. By what factor does $v_{\rm rms}$ change? By what factor does the pressure change?

Solution. $v_{\rm rms}\propto\sqrt T$, so $v_{\rm rms,2}/v_{\rm rms,1}=\sqrt{4}=2$. At constant $V$, $P\propto T$, so $P_2/P_1 = 4$.

Problem 2. Compute the ratio $v_{\rm rms}(\text{H}_2)/v_{\rm rms}(\text{O}_2)$ at the same temperature.

Solution. $v_{\rm rms}\propto 1/\sqrt M$. $\sqrt{32/2}=4$. So H$_2$ is 4 times faster.

Problem 3. Two moles of an ideal monoatomic gas are heated from $300\ \text{K}$ to $400\ \text{K}$ at constant pressure. Find $Q$, $\Delta U$, $W$.

Solution. $C_P=\tfrac{5}{2}R$, $C_V=\tfrac{3}{2}R$.

Problem 4. A mixture has 2 moles of He and 1 mole of N$_2$ (rigid). Find $C_V$, $C_P$, $\gamma$.

Solution.

Problem 5. Show that the average KE of a molecule of an ideal gas at $300\ \text{K}$ is $\approx 6.2\times 10^{-21}\ \text{J}$ and that this is independent of mass.

Solution. Already done in 13.5. The mass-independence is the hallmark of kinetic temperature.

Problem 6. Estimate the mean free path of oxygen at $1\ \text{atm}$ and $300\ \text{K}$, given $d=3.6\times 10^{-10}\ \text{m}$.

Solution.

Problem 7. An ideal diatomic gas (rigid rotor) is compressed adiabatically from $V$ to $V/2$. By what factor does $T$ rise?

Solution. $TV^{\gamma-1}=$ const, $\gamma=7/5$ so $\gamma-1=2/5$.

JEE / NEET Edge Cases

• Effusion (Graham's law). Rate of effusion $\propto v_{\rm rms}\propto 1/\sqrt M$. Useful in isotope separation problems.
• Real-gas correction. Near liquefaction $P$ is reduced (intermolecular attraction) and effective volume is reduced (molecular size). Van der Waals: $(P+a/V^2)(V-b)=nRT$. NEET sometimes tests sign of $a,b$.
• Mixture pressure (Dalton). $P_\text{tot}=\sum P_i$ where $P_i = n_i RT/V$.
• Speed distribution shifts. Heating shifts the Maxwell peak right and lowers it (area is conserved).
• Adiabatic exponent of a mixture. $\gamma_\text{mix}$ is not the weighted average of $\gamma$'s; use $C_V$ and $C_P$ weighted by moles.
• Equipartition at low $T$. Hydrogen at 50 K behaves like monoatomic ($\gamma=5/3$) because rotation freezes; at 300 K it is diatomic ($\gamma=7/5$); at very high $T$ vibration unfreezes ($\gamma=9/7$).
• Number of collisions per second with a wall of area $A$: $Z = \tfrac{1}{4}n\bar v A$. Useful in effusion derivations.
• $1/\sqrt 2$ factor in $\lambda$ is the most-tested derivation subtlety.

Quick Recap

• Ideal gas: $PV = nRT = Nk_BT = (\rho/M)RT$.
• Pressure from kinetic theory: $P=\tfrac{1}{3}\rho\langle v^2\rangle$.
• $\langle E_k\rangle_\text{trans} = \tfrac{3}{2}k_BT$ — independent of mass.
• $v_{\rm rms}=\sqrt{3RT/M}$, $\bar v=\sqrt{8RT/\pi M}$, $v_p=\sqrt{2RT/M}$.
• Equipartition: $\tfrac{1}{2}k_BT$ per quadratic degree of freedom.
• $C_V = (f/2)R$, $C_P-C_V = R$, $\gamma = 1 + 2/f$.
• Dulong–Petit: $C_\text{solid}\approx 3R$.
• Mean free path: $\lambda = 1/(\sqrt 2 n\pi d^2) = k_BT/(\sqrt 2 \pi d^2 P)$.

Formula Sheet