Kinetic TheoryNEET Physics · Class 11 · NCERT Chapter 12

Introduction

Kinetic theory explains the macroscopic behaviour of a gas (its pressure, temperature, energy) in terms of the motion of its molecules. It is the bridge between thermodynamics, which only deals with bulk quantities, and the actual microscopic picture: countless tiny molecules zipping around and colliding.

For NEET 2027 you can expect 1 to 2 questions from Kinetic Theory. The repeated favourites are: the formula $P=\frac{1}{3}\rho v_{\text{rms}}^2$, average translational KE equals $\frac{3}{2}{k}_{B}T$, the ratio of v_p, v_avg, v_rms, equipartition (with f as the number of degrees of freedom), and Cv from f. Lock those down and you have most of the marks.

Postulates of kinetic theory

The kinetic theory of an ideal gas rests on the following assumptions:

1. A gas consists of a very large number of tiny molecules in random, ceaseless motion.
2. The size of a molecule is negligible compared to the average distance between molecules.
3. The molecules exert no force on each other except during collisions (no intermolecular forces).
4. All collisions (molecule with molecule, molecule with wall) are perfectly elastic.
5. The motion is isotropic: on average, equal numbers of molecules move in any direction.
6. The duration of a collision is negligible compared to the time between collisions.
7. Newton's laws of motion describe the molecular motion.

Real gases follow these assumptions only approximately. They are most accurate at low pressures and high temperatures, where molecules are far apart and their volumes and forces matter least.

Ideal gas equation

The ideal gas law combines Boyle's, Charles's and Avogadro's laws into one equation:

Symbols:

• $P$ in pascals, $V$ in cubic metres, $T$ in kelvins.
• $n$ is number of moles, $N$ is total number of molecules, $N_A=6.022\times 10^{23}$ per mole.
• $R=8.314$ J per mol per K. $k_B=1.38\times 10^{-23}$ J per K.

At STP (standard temperature 273 K and standard pressure 101325 Pa), one mole of any ideal gas occupies 22.4 L.

Pick the unknown. Enter the other three quantities. Use SI units (Pa, m³, mol, K).

Pressure of an ideal gas

Imagine N molecules of mass m in a cubical box of side L. Pick the x-direction. A molecule with x-velocity $v_x$ hits the right wall, bounces off elastically with x-velocity $-v_x$, and the wall absorbs a momentum change of $2m{v}_x$ per collision. The time between successive hits on the right wall (round trip) is $2L/v_x$. So the time-averaged force on the wall from this one molecule is $(2m{v}_x)/(2L/v_x)=m{v}_x^2/L$.

Sum over all N molecules and divide by area $L^2$:

Because motion is isotropic, $⟨v_x^2⟩=⟨v_y^2⟩=⟨v_z^2⟩=\frac{1}{3}⟨v^2⟩$. Defining $v_{\text{rms}}^2=⟨v^2⟩$, we get the central result:

Each molecule slamming the wall transfers momentum. Add up all the molecules in a unit volume and the pressure comes out as one third of rho times v_rms squared.

Kinetic interpretation of temperature

Multiply both sides of the pressure equation by V:

Compare with the ideal gas law $PV=N{k}_{B}T$. We get the famous result:

Temperature is just the average translational kinetic energy of the molecules, scaled by $\frac{2}{3k_B}$. The same formula holds for any ideal gas, regardless of molecular mass. So at the same T, all gases have the same average translational KE per molecule. Heavier molecules just move slower.

Temperature is a direct measure of the average translational kinetic energy of the gas molecules. The same formula works for any ideal gas, no matter the molecular mass.

RMS, average and most probable speed

From $\frac{1}{2}m{v}_{\text{rms}}^2=\frac{3}{2}{k}_{B}T$, we get the RMS speed:

Two more averages defined from the Maxwell distribution:

The fixed ratio between them is a NEET classic:

Always remember $v_{\text{rms}}$ is the largest, $v_p$ is the smallest. They all scale with $\sqrt{T/M}$.

Pick a gas, set a temperature, and compare the three speed measures. Notice that v_rms is always the largest.

Selected: N₂ (M = 28 g/mol)

Maxwell-Boltzmann speed distribution

Molecules do not all move at the same speed. Maxwell worked out the actual distribution:

Key features:

• The curve starts at zero (no molecules with v = 0), rises to a peak at v_p, then decays exponentially.
• Higher T flattens the curve and shifts the peak to higher v.
• The total area under the curve is 1 (probability normalisation).
• Fraction of fast molecules (v above some threshold) grows fast with T.

Maxwell-Boltzmann distribution: most molecules cluster near v_p with a long tail of fast ones. Heat the gas, and the curve flattens and shifts to the right.

Equipartition of energy

The equipartition theorem says: in thermal equilibrium, every quadratic degree of freedom carries the same average energy of $\frac{1}{2}{k}_{B}T$ per molecule (or $\frac{1}{2}RT$ per mole).

Degrees of freedom (f) for common gases at room temperature:

• Monoatomic (He, Ne, Ar): 3 translational only. f = 3.
• Diatomic (H₂, O₂, N₂): 3 translational + 2 rotational. f = 5.
• Polyatomic non-linear (H₂O, NH₃, CH₄): 3 translational + 3 rotational. f = 6.
• Linear polyatomic (CO₂): 3 translational + 2 rotational. f = 5 at moderate T.

Vibration is frozen out at room T for most diatomics; it only kicks in at high temperatures. NEET problems will usually quote f or a temperature where vibration matters explicitly.

Total internal energy per molecule:

Each quadratic degree of freedom carries half k_B T of energy on average. Pick a gas and a temperature.

Diatomic (room T), examples: H₂, O₂, N₂. f = 5.

Specific heats from kinetic theory

From $U=(f/2)nRT$ and the definitions of Cv and Cp:

Standard values to memorise:

• Monoatomic (f = 3): Cv = (3/2) R, Cp = (5/2) R, γ = 5/3 ≈ 1.667.
• Diatomic (f = 5): Cv = (5/2) R, Cp = (7/2) R, γ = 7/5 = 1.4.
• Polyatomic (f = 6): Cv = 3 R, Cp = 4 R, γ = 4/3 ≈ 1.333.

Mean free path

On average, how far does a molecule travel between two successive collisions? Imagine a molecule of diameter d moving through a gas of stationary scatterers. It sweeps out a cylinder of cross-section $\pi d^2$ per unit length. With number density $n$, the average distance between collisions is:

The factor $\sqrt{2}$ accounts for the relative motion between target and projectile (both move, not just one). In terms of pressure and temperature, using $n=P/(k_{B}T)$:

For air at STP (d ≈ 0.37 nm, n ≈ 2.5 × 10²⁵ /m³), lambda is about 70 nm. Lower the pressure and lambda grows rapidly. In a good vacuum it can reach metres.

Mean free path lambda is the average distance a molecule travels between collisions. It depends on the molecular size and how packed the gas is.

Real gases vs ideal gas

The ideal gas model fails at:

• High pressures: molecular volumes become a significant fraction of the container.
• Low temperatures: intermolecular attractions slow molecules and reduce pressure.
• Near the liquefaction point, where the gas is about to condense.

For NEET problems, treat the gas as ideal unless the question explicitly asks about real gas behaviour or van der Waals corrections.

Worked NEET problems

Summary cheat sheet

• Ideal gas: $PV=nRT=N{k}_{B}T$, $R=8.314$ J/(mol·K).
• Pressure: $P=\frac{1}{3}\rho v_{\text{rms}}^2$.
• Temperature: $⟨K{E}_{\text{trans}}⟩=\frac{3}{2}{k}_{B}T$ per molecule.
• Speeds: $v_p:v_{\text{avg}}:v_{\text{rms}}=\sqrt{2}:\sqrt{8/\pi }:\sqrt{3}$.
• Equipartition: half k_B T per quadratic degree of freedom per molecule.
• Internal energy: $U=(f/2)nRT$.
• Specific heats: $C_V=(f/2)R$, $C_P=(f+2)/2\cdot R$, $\gamma =1+2/f$.
• γ values: mono 5/3, di 7/5, poly 4/3.
• Mean free path: $\lambda =1/(\sqrt{2}\pi d^{2}n)$.
• STP: 273 K and 101325 Pa, one mole occupies 22.4 L.

Next: try the interactive widgets for ideal gas law, RMS speed, Maxwell distribution and mean free path, or work through the 32 NEET PYQs with full solutions. To time yourself, take the free 10-question mock test.