PhysicsCBSE Class 11 · Chapter 12

Kinetic Theory Class 11: Notes, RMS Speed and Solved Numericals

Updated 14 October 2026 · Fact-checked

Kinetic Theory is Chapter 12 of CBSE Class 11 Physics. It covers the ideal gas equation and work done in compressing a gas, the assumptions of kinetic theory, pressure of an ideal gas, kinetic interpretation of temperature and rms speed, degrees of freedom, the law of equipartition (statement), specific heats of gases, and mean free path.

What Kinetic Theory is about

A gas is made of an enormous number of tiny molecules rushing about at hundreds of metres per second and colliding with one another and with the walls. Kinetic theory explains the large-scale behaviour of a gas (pressure, temperature, specific heat) from this picture.

You start with the ideal gas equation $P V = μ R T = N k_{B} T$ , which sums up Boyle's, Charles' and Avogadro's laws, and Dalton's law of partial pressures. Then, with a few simple assumptions, you derive the pressure of a gas from molecules bouncing off the walls: $P = \frac{1}{3} nm \overline{v^{2}}$ . Comparing with $P V = N k_{B} T$ gives the key result that the average kinetic energy of a molecule is $\frac{3}{2} k_{B} T$ : temperature is a measure of molecular motion. This gives the rms speed $\frac{3 R T}{M}$ .

The law of equipartition of energy (statement only) gives each degree of freedom $\frac{1}{2} k_{B} T$ , which predicts the specific heats of monatomic, diatomic and polyatomic gases and of solids. Finally, the mean free path, the average distance between collisions, explains why smells spread slowly even though molecules are fast. This chapter explains the results of Thermodynamics, and the ideas of thermal energy return in Class 12 (thermal speeds of electrons in a conductor).

Weightage: This chapter is part of Unit IX, Behaviour of Perfect Gases and Kinetic Theory of Gases. The syllabus groups Unit IX with Unit VII, Properties of Bulk Matter and Unit VIII, Thermodynamics: together they carry 20 of the 70 theory marks in the school annual exam.Source: CBSE Physics (042) syllabus 2026-27, Class XI course structure, Unit IX Behaviour of Perfect Gases and Kinetic Theory of Gases

Key concepts

Molecular nature of matter and the gas laws

NCERT section 12.2 and 12.3

Matter is made of molecules. In gases they are far apart (about ten times their size) and interact only during collisions, which is why gases are easy to compress.

At low pressure and high temperature, real gases obey:

Boyle's law: at constant T, $P V$ = constant.
Charles' law: at constant P, $V \propto T$ .
Avogadro's hypothesis: equal volumes of all gases at the same T and P contain the same number of molecules. One mole contains $N_{A} = 6.02 \times 1 0^{23}$ particles and occupies 22.4 L at STP.

Together they give the ideal gas equation:
$P V = μ R T = N k_{B} T$
where μ is the number of moles, R = 8.314 J mol $^{- 1}$ K $^{- 1}$ , N the number of molecules and $k_{B} = \frac{R}{N _{A}} = 1.38 \times 1 0^{- 23}$ J K $^{- 1}$ (Boltzmann constant).

Dalton's law of partial pressures: in a mixture of ideal gases, the total pressure is the sum of the pressures each gas would exert alone: $P = P_{1} + P_{2} + \dots$

A gas that obeys PV = μRT at all pressures and temperatures is an ideal gas; real gases come close at low pressure and high temperature.

Worked example

How many gas molecules are there per cubic metre at STP (1.013 × 10 $^{5}$ Pa, 273 K)? ( $k_{B} = 1.38 \times 1 0^{- 23}$ J K $^{- 1}$ ) [1 mark]

Try it first, then show the solutionSolution

$n = \frac{N}{V} = \frac{P}{k _{B} T} = \frac{1.013 \times 1 0 ^{5}}{1.38 \times 1 0 ^{- 23} \times 273}$
$n \approx 2.69 \times 1 0^{25} m^{- 3}$

Answer: About $2.7 \times 1 0^{25}$ molecules per m $^{3}$

Work done in compressing a gas

NCERT section Syllabus topic (uses 11.8.2)

The syllabus lists the work done in compressing a gas alongside the equation of state. For a gas compressed from $V_{1}$ to $V_{2}$ , the work done on the gas is $- \int_{V_{1}}^{V_{2}} P d V$ .

At constant pressure: $W_{o n} = P (V_{1} - V_{2})$ .
Isothermally (slowly, in contact with a heat bath), using $P = \frac{μ R T}{V}$ :

$W_{o n} = μ R T ln \frac{V _{1}}{V _{2}}$
This energy leaves the gas as heat, since its internal energy (which depends only on T) does not change.

On a P-V graph, the work is the area under the curve between $V_{2}$ and $V_{1}$ .

Worked example

One mole of an ideal gas at 300 K is compressed isothermally from 20 L to 10 L. How much work is done on it? (R = 8.314 J mol $^{- 1}$ K $^{- 1}$ , ln 2 = 0.693) [2 marks]

Try it first, then show the solutionSolution

$W_{o n} = μ R T ln \frac{V _{1}}{V _{2}} = 1 \times 8.314 \times 300 \times ln 2$
$W_{o n} \approx 1729 J$ (and 1729 J of heat is given out)

Answer: About 1.73 kJ

Assumptions of the kinetic theory of an ideal gas

NCERT section 12.4

1. A gas consists of a very large number of identical molecules that are like tiny hard spheres; their own volume is negligible compared with the volume of the gas.
2. The molecules are in constant random motion in all directions, with a range of speeds.
3. There are no forces between molecules except during collisions; between collisions they move in straight lines at constant speed.
4. Collisions with each other and with the walls are perfectly elastic, and the time of a collision is negligible compared with the time between collisions.
5. In equilibrium, the molecular density is uniform and no direction is preferred.
6. Newton's laws apply to the molecules.

Real gases depart from ideal behaviour mainly because assumptions 1 and 3 fail at high pressure and low temperature.

Worked example

Which two assumptions of kinetic theory fail for a real gas at high pressure? [1 mark]

Try it first, then show the solutionSolution

At high pressure molecules are crowded, so their own volume is no longer negligible, and they are close enough for intermolecular forces to matter.

Answer: Negligible molecular volume, and no intermolecular forces

Pressure of an ideal gas

NCERT section 12.4.1

Consider a cube of side l with N molecules, each of mass m. A molecule with velocity component $v_{x}$ hits a wall perpendicular to x and bounces back elastically.

Momentum given to the wall per collision: $2 m v_{x}$ .
In time Δt, the molecules with velocity $v_{x}$ that reach the wall are those within $v_{x} Δ t$ of it: $\frac{1}{2} n A v_{x} Δ t$ of them (half move towards the wall), where n is the number per unit volume.
Momentum given in Δt: $nm A v_{x}^{2} Δ t$ , so pressure $P = nm \overline{v_{x}^{2}}$ .
Since motion is random, $\overline{v_{x}^{2}} = \overline{v_{y}^{2}} = \overline{v_{z}^{2}} = \frac{1}{3} \overline{v^{2}}$ .

$P = \frac{1}{3} nm \overline{v^{2}} = \frac{1}{3} ρ \overline{v^{2}}$

where ρ = nm is the density of the gas. Also $P V = \frac{2}{3} E$ , where E is the total translational kinetic energy of the molecules.

Worked example

A gas has density 1.3 kg m $^{- 3}$ at a pressure of $1.0 \times 1 0^{5}$ Pa. Find the rms speed of its molecules. [2 marks]

Try it first, then show the solutionSolution

$P = \frac{1}{3} ρ v_{r m s}^{2} \Rightarrow v_{r m s} = \frac{3 P}{ρ}$
$v_{r m s} = \frac{3 \times 1 0 ^{5}}{1.3} \approx 480 m s^{- 1}$

Answer: About 480 m s $^{- 1}$

Kinetic interpretation of temperature and rms speed

NCERT section 12.4.2

From $P V = \frac{1}{3} N m \overline{v^{2}}$ and $P V = N k_{B} T$ :
$\frac{1}{2} m \overline{v^{2}} = \frac{3}{2} k_{B} T$

The average translational kinetic energy of a molecule depends only on the absolute temperature, not on the type of gas, its pressure or its volume. Temperature is a measure of the average kinetic energy of the molecules. At the same temperature, molecules of all gases have the same average KE; lighter molecules move faster.

Root mean square speed:
$v_{r m s} = \overline{v^{2}} = \frac{3 k _{B} T}{m} = \frac{3 R T}{M}$
where M is the molar mass (in kg mol $^{- 1}$ ). So $v_{r m s} \propto \frac{T}{M}$ .

For a mixture at the same T, $\frac{v _{r m s, 1}}{v _{r m s, 2}} = \frac{M _{2}}{M _{1}}$ ; this is the basis of separating uranium isotopes by diffusion.

Worked example

Find the rms speed of oxygen molecules (M = 0.032 kg mol $^{- 1}$ ) at 300 K. (R = 8.314 J mol $^{- 1}$ K $^{- 1}$ ) [2 marks]

Try it first, then show the solutionSolution

$v_{r m s} = \frac{3 R T}{M} = \frac{3 \times 8.314 \times 300}{0.032}$
$v_{r m s} = 2.34 \times 1 0^{5} \approx 484 m s^{- 1}$

Answer: About 484 m s $^{- 1}$

Degrees of freedom and the law of equipartition of energy

NCERT section 12.5

The degrees of freedom of a molecule are the independent ways in which it can move (and store energy).

Monatomic (He, Ar): 3 translational.
Diatomic (O $_{2}$ , N $_{2}$ ), rigid: 3 translational + 2 rotational = 5. (Rotation about the line joining the atoms does not count.) At high temperatures vibration adds 2 more (kinetic and potential), giving 7.
Polyatomic (non-linear): 3 translational + 3 rotational, plus vibrational modes f.

Law of equipartition of energy (statement): in thermal equilibrium at absolute temperature T, the total energy is shared equally among all energy modes; each translational and rotational degree of freedom has an average energy of $\frac{1}{2} k_{B} T$ , and each vibrational mode $k_{B} T$ (kinetic plus potential).

The syllabus requires only the statement, not a proof.

Worked example

Find the average energy of a molecule of a rigid diatomic gas at temperature T. [1 mark]

Try it first, then show the solutionSolution

5 degrees of freedom (3 translational + 2 rotational), each $\frac{1}{2} k_{B} T$
Average energy $= \frac{5}{2} k_{B} T$

Answer: $\frac{5}{2} k_{B} T$

Specific heats of gases and solids from equipartition

NCERT section 12.6

For one mole, the internal energy U is $N_{A}$ times the average energy per molecule, and $C_{v} = \frac{d U}{d T}$ , $C_{p} = C_{v} + R$ , $γ = \frac{C _{p}}{C _{v}}$ .

Gas	U per mole	$C_{v}$	$C_{p}$	γ
Monatomic	$\frac{3}{2} R T$	$\frac{3}{2} R$	$\frac{5}{2} R$	$\frac{5}{3} \approx 1.67$
Diatomic (rigid)	$\frac{5}{2} R T$	$\frac{5}{2} R$	$\frac{7}{2} R$	$\frac{7}{5} = 1.40$
Diatomic (vibrating)	$\frac{7}{2} R T$	$\frac{7}{2} R$	$\frac{9}{2} R$	$\frac{9}{7} \approx 1.29$
Polyatomic (f vibrational modes)	$(3 + f) R T$	$(3 + f) R$	$(4 + f) R$	$\frac{4 + f}{3 + f}$

Solids: each atom vibrates in 3 dimensions with energy $3 k_{B} T$ , so $U = 3 R T$ per mole and $C = 3 R$ (Dulong and Petit's law), which matches many solids at room temperature.

The predicted values agree well with experiment for monatomic and most diatomic gases; the differences at low temperature need quantum theory.

Worked example

One mole of helium is mixed with one mole of nitrogen. Find $C_{v}$ , $C_{p}$ and γ for the mixture. [2 marks]

Try it first, then show the solutionSolution

$C_{v} = \frac{1 \times \frac{3}{2} R + 1 \times \frac{5}{2} R}{2} = 2 R$
$C_{p} = C_{v} + R = 3 R$ , so $γ = \frac{3}{2} = 1.5$

Answer: $C_{v} = 2 R$ , $C_{p} = 3 R$ , γ = 1.5

Mean free path

NCERT section 12.7

Molecules of a gas travel fast (hundreds of m s $^{- 1}$ ) but keep colliding, so they zigzag. The mean free path l is the average distance a molecule travels between two successive collisions.

A molecule of diameter d sweeps out a cylinder of cross-section $π d^{2}$ ; any molecule whose centre lies within it is hit. Allowing for the motion of the other molecules:
$l = \frac{1}{2 nπ d ^{2}}$
where n is the number of molecules per unit volume.

l is larger for a less dense gas (low pressure) and for smaller molecules.
For air at STP, l is about $1 0^{- 7}$ m, a few hundred molecular diameters. That is why the smell of perfume takes time to spread across a room even though the molecules move fast.

Worked example

Estimate the mean free path of nitrogen molecules at STP. (d = $3 \times 1 0^{- 10}$ m, n = $2.69 \times 1 0^{25}$ m $^{- 3}$ ) [2 marks]

Try it first, then show the solutionSolution

$l = \frac{1}{2 \times 2.69 \times 1 0 ^{25} \times π \times ( 3 \times 1 0 ^{- 10} ) ^{2}}$
$l \approx 9.3 \times 1 0^{- 8} m$ (about 300 molecular diameters)

Answer: About $9 \times 1 0^{- 8}$ m

Physics in real life: Kinetic Theory

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Pressure law

Suppose a car tyre is filled to an absolute pressure of 2.0 atm at 27 °C, and after a long drive the air in it is at 47 °C. Its volume hardly changes.

What is the new absolute pressure?

Try it first, then show the solutionSolution

Constant V: $\frac{P _{2}}{P _{1}} = \frac{T _{2}}{T _{1}} = \frac{320}{300}$
$P_{2} = 2.0 \times \frac{320}{300} \approx 2.13 a t m$

Answer: About 2.13 atm

Memory hook: Hotter molecules hit the walls harder and more often.

Mean free path

Imagine someone opens a bottle of perfume at the front of a classroom. Perfume molecules move at a few hundred metres per second, yet the smell takes many seconds to reach the back.

Suppose a molecule travels about $1 0^{- 7}$ m between collisions. Explain the delay.

Try it first, then show the solutionSolution

Each molecule collides billions of times per second and changes direction each time
Its path is a random zigzag, so its net progress across the room is very slow (diffusion), helped mostly by air currents

Answer: Frequent collisions over a tiny mean free path make the molecules wander instead of travelling straight.

Memory hook: Fast molecules, slow smell: blame the collisions.

Number of molecules and energy

Suppose a bedroom has 60 m $^{3}$ of air at 20 °C and 1 atm.

Estimate the number of air molecules and their total translational kinetic energy.

Try it first, then show the solutionSolution

$N = \frac{P V}{k _{B} T} = \frac{1.013 \times 1 0 ^{5} \times 60}{1.38 \times 1 0 ^{- 23} \times 293} \approx 1.5 \times 1 0^{27}$
Total KE $= \frac{3}{2} P V = 1.5 \times 1.013 \times 1 0^{5} \times 60 \approx 9.1 \times 1 0^{6} J$

Answer: About $1.5 \times 1 0^{27}$ molecules; about 9 MJ

Memory hook: Total KE of a gas = 1.5 × PV.

Topics in this chapter

Kinetic Theory formula sheet

Ideal gas equation: $P V = μ R T = N k_{B} T, k_{B} = \frac{R}{N _{A}} = 1.38 \times 1 0^{- 23} J K^{- 1}$
Dalton's law: $P = P_{1} + P_{2} + \dots = (μ_{1} + μ_{2} + \dots) \frac{R T}{V}$
Isothermal compression work (on the gas): $W = μ R T ln \frac{V _{1}}{V _{2}}$
Pressure of an ideal gas: $P = \frac{1}{3} nm \overline{v^{2}} = \frac{1}{3} ρ \overline{v^{2}}, P V = \frac{2}{3} E$
Average kinetic energy: $\frac{1}{2} m \overline{v^{2}} = \frac{3}{2} k_{B} T$
RMS speed: $v_{r m s} = \frac{3 k _{B} T}{m} = \frac{3 R T}{M} = \frac{3 P}{ρ}$
Equipartition: $energy per degree of freedom = \frac{1}{2} k_{B} T$
Specific heats: $C_{v} = \frac{f}{2} R, C_{p} = (\frac{f}{2} + 1) R, γ = 1 + \frac{2}{f}$; f = number of degrees of freedom (3 monatomic, 5 rigid diatomic).
Mean free path: $l = \frac{1}{2 nπ d ^{2}}$

How the school exam asks Kinetic Theory

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Ideal gas equation	Number of molecules in a volume; pressure of a mixture (Dalton); changes in P, V, T; bubble rising in a lake.	PV = Nk_BT for molecules, PV = μRT for moles; always kelvin; absolute pressure includes the atmosphere.
Derivation of gas pressure	3-5 marks: state assumptions and derive $P = \frac{1}{3} nm \overline{v^{2}}$ ; deduce KE = $\frac{3}{2} k_{B} T$ .	Momentum change per collision, number of collisions per second, isotropy ( $\overline{v_{x}^{2}} = \overline{v^{2}} /3$ ).
RMS speed and temperature	v_rms of a gas at T; ratio for two gases; temperature at which a speed is reached; average KE.	$v_{r m s} \propto T / M$ ; use M in kg mol $^{- 1}$ .
Equipartition and specific heats	Degrees of freedom; $C_{v}$ , $C_{p}$ and γ for different gases and mixtures; Dulong-Petit.	Count f; $C_{v} = f R /2$ ; for mixtures take mole-weighted averages of $C_{v}$ .
Mean free path	Formula and dependence on n and d; estimate at STP or low pressure; why smells spread slowly.	n = P/(k_BT); $l \propto 1/ (n d^{2})$ .

Tricks to remember Kinetic Theory

Short memory aids. Say each one aloud once, then use it on a question the same day.

3/2 kT
Average translational KE per molecule = $\frac{3}{2} k_{B} T$ : three directions, half kT each.
Use it when: KE and temperature questions.
Root 3RT over M
$v_{r m s} = 3 R T / M$ ; lighter gas, faster molecules.
Use it when: RMS speed numericals.
f = 3, 5, 7
Monatomic 3, rigid diatomic 5, vibrating diatomic 7; then $C_{v} = \frac{f}{2} R$ and $γ = 1 + \frac{2}{f}$ .
Use it when: Specific heats and γ.
One third rho v squared
$P = \frac{1}{3} ρ \overline{v^{2}}$ : the 1/3 comes from three equal directions.
Use it when: Pressure derivation.
Mean free path: fewer and smaller means farther
$l \propto \frac{1}{n d ^{2}}$ : lower density or smaller molecules give longer free paths.
Use it when: Mean free path reasoning.

Kinetic Theory: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: How rms speed depends on temperature · 1 mark · MCQExam level

The rms speed of the molecules of a gas is proportional to:
(a) T
(b) $T^{2}$
(c) $T$
(d) $1/ T$

Try it first, then show the solutionSolution

$v_{r m s} = 3 R T / M$ .

Answer: (c) $T$

Example 2: Same temperature, same KE · 1 mark · Assertion-reasonExam level

Assertion (A): At the same temperature, hydrogen and oxygen molecules have the same average translational kinetic energy.
Reason (R): The average translational kinetic energy of a gas molecule depends only on the absolute temperature.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

$\frac{1}{2} m \overline{v^{2}} = \frac{3}{2} k_{B} T$ has no m or gas-type on the right.

Answer: (a)

Example 3: Hydrogen and oxygen speeds · 2 marks · Very short answerHarder exam level

Find the rms speed of hydrogen molecules at 300 K and compare it with that of oxygen at the same temperature. (R = 8.314 J mol $^{- 1}$ K $^{- 1}$ ) [2 marks]

Try it first, then show the solutionSolution

$v_{H_{2}} = \frac{3 \times 8.314 \times 300}{0.002} \approx 1934 m s^{- 1}$
Oxygen is 16 times heavier, so it is 4 times slower: about 484 m s $^{- 1}$

Answer: About 1.93 km s $^{- 1}$ ; four times oxygen's

Example 4: Rising air bubble · 3 marks · Short answerHarder exam level

An air bubble of volume 2.0 cm $^{3}$ rises from the bottom of a pond 20 m deep, where the temperature is 10 °C, to the surface at 27 °C. Find its volume at the surface. (Atmospheric pressure $1.013 \times 1 0^{5}$ Pa, ρ = 1000 kg m $^{- 3}$ , g = 9.8 m s $^{- 2}$ ) [3 marks]

Try it first, then show the solutionSolution

Pressure at the bottom $= 1.013 \times 1 0^{5} + 1000 \times 9.8 \times 20 \approx 2.97 \times 1 0^{5} P a$
$V_{2} = V_{1} \frac{P _{1}}{P _{2}} \cdot \frac{T _{2}}{T _{1}} = 2.0 \times \frac{2.97 \times 1 0 ^{5}}{1.013 \times 1 0 ^{5}} \times \frac{300}{283}$
$V_{2} \approx 6.2 c m^{3}$

Answer: About 6.2 cm $^{3}$

Example 5: Gas mixture · 2 marks · Very short answerHarder exam level

A 2 L vessel contains 0.1 mol of oxygen and 0.2 mol of nitrogen at 300 K. Find the partial pressure of oxygen and the total pressure. (R = 8.314 J mol $^{- 1}$ K $^{- 1}$ ) [2 marks]

Try it first, then show the solutionSolution

$P_{O_{2}} = \frac{0.1 \times 8.314 \times 300}{2 \times 1 0 ^{- 3}} \approx 1.25 \times 1 0^{5} P a$
Total (Dalton): $\frac{0.3 \times 8.314 \times 300}{2 \times 1 0 ^{- 3}} \approx 3.74 \times 1 0^{5} P a$

Answer: $1.25 \times 1 0^{5}$ Pa; $3.74 \times 1 0^{5}$ Pa

Example 6: Kinetic theory in full · 5 marks · Long answerHarder exam level

(a) Derive $P = \frac{1}{3} ρ \overline{v^{2}}$ for an ideal gas. (b) Show that $P V = \frac{2}{3} E$ , where E is the total translational KE. (c) Find the rms speed of nitrogen molecules at 27 °C. (d) By what factor does it change if the absolute temperature is made four times? (M = 0.028 kg mol $^{- 1}$ , R = 8.314 J mol $^{- 1}$ K $^{- 1}$ ) [5 marks]

Try it first, then show the solutionSolution

(a) Momentum per collision $2 m v_{x}$ ; collisions on area A in Δt: $\frac{1}{2} n A v_{x} Δ t$ ; so $P = nm \overline{v_{x}^{2}} = \frac{1}{3} nm \overline{v^{2}} = \frac{1}{3} ρ \overline{v^{2}}$
(b) $P V = \frac{1}{3} N m \overline{v^{2}} = \frac{2}{3} (N \cdot \frac{1}{2} m \overline{v^{2}}) = \frac{2}{3} E$
(c) $v_{r m s} = \frac{3 \times 8.314 \times 300}{0.028}$
$v_{r m s} \approx 517 m s^{- 1}$
(d) $v_{r m s} \propto T$ : it doubles, to about 1034 m s $^{- 1}$

Answer: About 517 m s $^{- 1}$ ; doubles

Example 7: Equal speeds · 2 marks · Very short answerHOTS / competency

At what temperature will the rms speed of oxygen molecules equal that of hydrogen molecules at 300 K? Would the average kinetic energies then be equal? [2 marks]

Try it first, then show the solutionSolution

$\frac{T _{O_{2}}}{32} = \frac{300}{2} \Rightarrow T_{O_{2}} = 4800 K$
No: average KE depends only on T, so oxygen at 4800 K has 16 times the average KE of hydrogen at 300 K.

Answer: 4800 K; KEs are not equal (16 : 1)

Example 8: Mean free path in a vacuum · 3 marks · Short answerHOTS / competency

A vacuum pump lowers the pressure in a chamber to $1 0^{- 3}$ Pa at 300 K. Estimate the mean free path of air molecules (d = $3 \times 1 0^{- 10}$ m), and say why this matters in devices such as old television tubes. [3 marks]

Try it first, then show the solutionSolution

$n = \frac{P}{k _{B} T} \approx 2.4 \times 1 0^{17} m^{- 3}$
$l = \frac{1}{2 nπ d ^{2}} \approx 10 m$
This is much longer than the tube, so electrons (and molecules) cross it without hitting gas molecules: that is why such devices need a good vacuum.

Answer: About 10 m; particles cross the tube without collisions

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Air in a classroom

Suppose a classroom contains 150 m $^{3}$ of air at 27 °C and 1.013 × 10 $^{5}$ Pa. Treat air as an ideal diatomic gas of molar mass 0.029 kg mol $^{- 1}$ . Take $k_{B} = 1.38 \times 1 0^{- 23}$ J K $^{- 1}$ , R = 8.314 J mol $^{- 1}$ K $^{- 1}$ .

(i) · 1 mark
Find the average translational KE of one air molecule.
Answer
Answer: About $6.2 \times 1 0^{- 21}$ J
$\frac{3}{2} \times 1.38 \times 1 0^{- 23} \times 300$
(ii) · 1 mark
Estimate the number of molecules in the room.
Answer
Answer: About $3.7 \times 1 0^{27}$
$P V / k_{B} T$
(iii) · 1 mark
Find the rms speed of the molecules.
Answer
Answer: About 508 m s $^{- 1}$
$3 \times 8.314 \times 300/0.029$
(iv) · 1 mark
How many degrees of freedom does an air molecule have at room temperature, and what is γ?
Answer
Answer: 5; γ = 1.4
Rigid diatomic: 3 translational + 2 rotational.

Specific heats in the lab

Suppose a student measures the molar specific heat at constant volume of three gases at room temperature and gets about 12.5, 20.8 and 25 J mol $^{- 1}$ K $^{- 1}$ . Take R = 8.31 J mol $^{- 1}$ K $^{- 1}$ .

(i) · 1 mark
Which value fits a monatomic gas such as argon?
Answer
Answer: 12.5
$\frac{3}{2} R \approx 12.5$
(ii) · 1 mark
Which value fits a rigid diatomic gas such as nitrogen?
Answer
Answer: 20.8
$\frac{5}{2} R \approx 20.8$
(iii) · 1 mark
How many degrees of freedom does the third gas effectively have?
Answer
Answer: About 6
$C_{v} = \frac{f}{2} R \Rightarrow f = \frac{2 \times 25}{8.31} \approx 6$ (a non-linear polyatomic molecule: 3 translational + 3 rotational)
(iv) · 1 mark
Find γ for the third gas.
Answer
Answer: About 1.33
$1 + \frac{2}{6}$

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Find the ratio of the rms speeds of hydrogen and oxygen molecules at the same temperature. [2 marks] (2 marks)

Step	Marks
$v_{r m s} \propto \frac{1}{M}$ at the same T	1
$\frac{v _{H_{2}}}{v _{O_{2}}} = \frac{32}{2} = 4$ , so 4 : 1	1

Starting from $P = \frac{1}{3} nm \overline{v^{2}}$ , show that the average kinetic energy of a molecule of an ideal gas is $\frac{3}{2} k_{B} T$ , and write the rms speed. [3 marks] (3 marks)

Step	Marks
Multiply by V: $P V = \frac{1}{3} N m \overline{v^{2}}$ (N = nV)	1
Compare with $P V = N k_{B} T$ : $\frac{1}{3} m \overline{v^{2}} = k_{B} T \Rightarrow \frac{1}{2} m \overline{v^{2}} = \frac{3}{2} k_{B} T$	1
$v_{r m s} = \frac{3 k _{B} T}{m} = \frac{3 R T}{M}$	1

(a) State any three assumptions of the kinetic theory of gases. (b) Derive an expression for the pressure exerted by an ideal gas. (c) Find the rms speed of helium atoms at 300 K. (M = 0.004 kg mol $^{- 1}$ , R = 8.314 J mol $^{- 1}$ K $^{- 1}$ ) [5 marks] (5 marks)

Step	Marks
(a) Negligible molecular volume; no forces except during collisions; perfectly elastic collisions; random motion (any three)	1
(b) Momentum change per collision with a wall: $2 m v_{x}$ ; number hitting area A in Δt: $\frac{1}{2} n A v_{x} Δ t$	1
Force/area: $P = nm \overline{v_{x}^{2}}$	1
Isotropy: $\overline{v_{x}^{2}} = \frac{1}{3} \overline{v^{2}}$ , so $P = \frac{1}{3} nm \overline{v^{2}}$	1
(c) $v_{r m s} = \frac{3 \times 8.314 \times 300}{0.004} \approx 1.37 \times 1 0^{3} m s^{- 1}$	1

Where students lose marks in Kinetic Theory

The slips examiners see most often, and what to write instead.

Mistake: Using molar mass in g mol $^{- 1}$
Copying the number from the periodic table.
Wrong working
$v_{r m s} = 3 \times 8.314 \times 300/32$
Right working
$M = 0.032$ kg mol $^{- 1}$
Do this: Convert M to kg mol $^{- 1}$ in SI formulas.
Usually costs: Whole numerical.
Mistake: Using °C in gas formulas
Habit.
Do this: T must be in kelvin in PV = μRT, $\frac{3}{2} k_{B} T$ and $v_{r m s}$ .
Usually costs: Whole numerical.
Mistake: Saying heavier molecules have more kinetic energy at the same temperature
Thinking KE grows with mass.
Do this: At the same T all gases have the same average KE per molecule; heavier ones just move slower.
Usually costs: 1 mark.
Mistake: Counting 6 degrees of freedom for a diatomic molecule
Including rotation about its own axis.
Do this: A rigid diatomic molecule has 3 translational + 2 rotational = 5.
Usually costs: 1 mark.
Mistake: Writing a proof of equipartition
Over-preparing.
Do this: The 2026-27 syllabus asks for the statement only; spend the time on applications to specific heats.
Usually costs: Time.

NCERT exercise guide: Kinetic Theory

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercises 12.1-12.6 (ideal gas behaviour)

What it tests: Molecular volume fraction, molar volume, PV/T plots, oxygen cylinders, a rising air bubble, and counting molecules in a room.

How to approach it: PV = μRT or Nk_BT with absolute pressure and kelvin; for changes, $\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$ .

Same kind as Exercise 12.6

Estimate the number of air molecules in 1 litre of air at 27 °C and 1 atm. ( $k_{B} = 1.38 \times 1 0^{- 23}$ J K $^{- 1}$ )

Show the solutionSolution

$N = \frac{P V}{k _{B} T} = \frac{1.013 \times 1 0 ^{5} \times 1 0 ^{- 3}}{1.38 \times 1 0 ^{- 23} \times 300}$
$N \approx 2.4 \times 1 0^{22}$

Answer: About $2.4 \times 1 0^{22}$

Exercises 12.7-12.9 (energy and speed)

What it tests: Average thermal energy of helium atoms, comparing gases in three vessels, and the temperature at which argon matches helium's rms speed.

How to approach it: Average KE = $\frac{3}{2} k_{B} T$ for any gas; $v_{r m s} \propto T / M$ .

Same kind as Exercise 12.9

At what temperature is the rms speed of nitrogen molecules (M = 28 g mol $^{- 1}$ ) equal to that of helium atoms (M = 4 g mol $^{- 1}$ ) at 300 K?

Show the solutionSolution

Equal $v_{r m s}$ means equal $\frac{T}{M}$ : $\frac{T}{28} = \frac{300}{4}$
T = 2100 K

Answer: 2100 K

Exercise 12.10 (mean free path)

What it tests: Mean free path and collision frequency of a gas molecule.

How to approach it: n = P/(k_BT); $l = 1/ (2 nπ d^{2})$ ; collision frequency ≈ $v_{r m s} / l$ .

Same kind as Exercise 12.10

Estimate the mean free path and the collision frequency of helium atoms (d = $2 \times 1 0^{- 10}$ m, $v_{r m s} \approx 1370$ m s $^{- 1}$ ) at 1 atm and 300 K.

Show the solutionSolution

$n = \frac{P}{k _{B} T} = \frac{1.013 \times 1 0 ^{5}}{1.38 \times 1 0 ^{- 23} \times 300} \approx 2.45 \times 1 0^{25} m^{- 3}$
$l = \frac{1}{2 \times 2.45 \times 1 0 ^{25} \times π \times 4 \times 1 0 ^{- 20}} \approx 2.3 \times 1 0^{- 7} m$
Collision frequency $\approx \frac{v _{r m s}}{l} = \frac{1370}{2.3 \times 1 0 ^{- 7}} \approx 6 \times 1 0^{9} s^{- 1}$

Answer: About $2.3 \times 1 0^{- 7}$ m; about $6 \times 1 0^{9}$ collisions per second

Practice questions on Kinetic Theory

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
The number of degrees of freedom of a rigid diatomic molecule is:
(a) 3
(b) 5
(c) 6
(d) 7
Hint
3 translational + 2 rotational.
Answer
(b) 5
Q2 · 1 mark · MCQExam level
The molar specific heat at constant volume of a monatomic ideal gas is:
(a) $\frac{1}{2} R$
(b) $\frac{3}{2} R$
(c) $\frac{5}{2} R$
(d) 3R
Hint
Three translational degrees of freedom.
Answer
(b) $\frac{3}{2} R$
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): Real gases behave most like ideal gases at low pressure and high temperature.
Reason (R): Under these conditions, molecular volume and intermolecular forces become negligible.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Molecules far apart and fast.
Answer
(a)
Q4 · 1 mark · Very short answerExam level
State the law of equipartition of energy.
Hint
½ k_BT per mode.
Answer
In thermal equilibrium, energy is shared equally among all degrees of freedom: each translational and rotational degree of freedom has average energy $\frac{1}{2} k_{B} T$ , and each vibrational mode $k_{B} T$ .
Q5 · 1 mark · Very short answerHarder exam level
Find the average translational kinetic energy of a gas molecule at 127 °C. ( $k_{B} = 1.38 \times 1 0^{- 23}$ J K $^{- 1}$ )
Hint
$\frac{3}{2} k_{B} T$ .
Answer
$\frac{3}{2} \times 1.38 \times 1 0^{- 23} \times 400 \approx 8.3 \times 1 0^{- 21}$ J.
Q6 · 2 marks · Very short answerHarder exam level
Find $C_{p}$ , $C_{v}$ and γ for a diatomic gas whose molecules also vibrate (one vibrational mode).
Hint
Vibration adds $k_{B} T$ per molecule.
Answer
Energy per molecule $= \frac{5}{2} k_{B} T + k_{B} T = \frac{7}{2} k_{B} T$ , so $C_{v} = \frac{7}{2} R$ , $C_{p} = \frac{9}{2} R$ , γ = 9/7 ≈ 1.29.
Q7 · 2 marks · Very short answerHarder exam level
The pressure of a gas is doubled at constant temperature. What happens to (a) the rms speed of its molecules, (b) the mean free path?
Hint
$v_{r m s}$ depends on T only; $l \propto 1/ n$ .
Answer
(a) Unchanged (T is the same). (b) n doubles (n = P/k_BT), so the mean free path halves.
Q8 · 2 marks · Very short answerHarder exam level
Calculate the rms speed of nitrogen molecules at 0 °C. (M = 0.028 kg mol $^{- 1}$ , R = 8.314 J mol $^{- 1}$ K $^{- 1}$ )
Hint
T = 273 K.
Answer
$v_{r m s} = \frac{3 \times 8.314 \times 273}{0.028} \approx 493$ m s $^{- 1}$ .
Q9 · 3 marks · Short answerHarder exam level
A cylinder holds 20 L of oxygen at 27 °C and an absolute pressure of 10 atm. How many moles does it contain, and what mass? (1 atm = $1.013 \times 1 0^{5}$ Pa, R = 8.314 J mol $^{- 1}$ K $^{- 1}$ )
Hint
μ = PV/RT.
Answer
$μ = \frac{1.013 \times 1 0 ^{6} \times 0.020}{8.314 \times 300} \approx 8.1$ mol; mass $= 8.1 \times 32 \approx 260$ g.
Q10 · 3 marks · Short answerHOTS / competency
Explain why the molar specific heat of hydrogen is about $\frac{3}{2} R$ at very low temperatures, rises to $\frac{5}{2} R$ near room temperature, and increases further at very high temperatures.
Hint
Which modes are active?
Answer
At very low temperatures only translation is active (3 degrees of freedom: $\frac{3}{2} R$ ). Near room temperature rotation also takes part (5: $\frac{5}{2} R$ ). At very high temperatures vibration also gets energy, adding R and giving about $\frac{7}{2} R$ . Classical equipartition cannot explain why modes 'switch on'; quantum theory does.
Q11 · 3 marks · Short answerHOTS / competency
Two vessels hold equal volumes of helium and argon at the same temperature and pressure. Compare (a) the number of atoms, (b) the average KE per atom, (c) the rms speeds. (M: He 4, Ar 40 g mol $^{- 1}$ )
Hint
Avogadro; KE depends on T; $v \propto 1/ M$ .
Answer
(a) Equal (Avogadro's hypothesis: same P, V, T). (b) Equal ( $\frac{3}{2} k_{B} T$ ). (c) $\frac{v _{H e}}{v _{A r}} = \frac{40}{4} = 10 \approx 3.16$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Write the ideal gas equation in terms of the number of molecules.Show answer
$P V = N k_{B} T$ .
2.State Dalton's law of partial pressures.Show answer
The total pressure of a mixture of ideal gases is the sum of their partial pressures.
3.Write the expression for the pressure of an ideal gas.Show answer
$P = \frac{1}{3} nm \overline{v^{2}}$ .
4.What is the average kinetic energy of a gas molecule at temperature T?Show answer
$\frac{3}{2} k_{B} T$ .
5.Write the formula for rms speed.Show answer
$v_{r m s} = 3 R T / M$ .
6.State the law of equipartition of energy.Show answer
In equilibrium, each translational or rotational degree of freedom has average energy $\frac{1}{2} k_{B} T$ (each vibrational mode $k_{B} T$ ).
7.What is γ for a monatomic gas?Show answer
5/3.
8.Write the formula for mean free path.Show answer
$l = 1/ (2 nπ d^{2})$ .

Best order to study it

Read the molecular picture of matter; learn the gas laws and the ideal gas equation in both forms.
Learn Dalton's law and the work done in compressing a gas.
Learn the assumptions of kinetic theory.
Derive the pressure of an ideal gas.
Deduce average KE = $\frac{3}{2} k_{B} T$ and the rms speed.
Learn degrees of freedom and the equipartition statement; work out $C_{v}$ , $C_{p}$ , γ.
Finish with the mean free path and Avogadro's number.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you write PV = μRT and PV = Nk_BT and convert between them?
Can you find the pressure of a gas mixture?
Can you list the assumptions of kinetic theory?
Can you derive $P = \frac{1}{3} nm \overline{v^{2}}$ ?
Can you show that average KE is proportional to T?
Can you compute $v_{r m s}$ and compare it for two gases?
Can you count degrees of freedom for monatomic, diatomic and polyatomic gases?
Can you find $C_{v}$ , $C_{p}$ and γ, including for mixtures?
Can you explain Dulong and Petit's law?
Can you estimate a mean free path?

Kinetic Theory: frequently asked questions

What was removed from this chapter for 2026-27?

The section on the specific heat capacity of water was removed. The law of equipartition of energy is needed as a statement only.

Is the derivation of the pressure of a gas important?

Yes. It is the main derivation of the chapter and a natural long-answer question in a school exam. Learn the steps: momentum per collision, collisions per second, the factor 1/2 for direction, and \overline{v_x^2} = \overline{v^2}/3.

Why is γ smaller for diatomic gases than for monatomic gases?

Diatomic molecules can also rotate, so they store extra energy in rotation. C_v is larger (5R/2 instead of 3R/2) while C_p - C_v stays R, so the ratio C_p/C_v is smaller.

Do all molecules in a gas have the same speed?

No. They have a spread of speeds that keeps changing through collisions. The rms speed is a typical value defined so that \frac{1}{2}mv_{rms}^2 equals the average kinetic energy.

How does this chapter connect to Thermodynamics and Class 12?

It explains why U of an ideal gas depends only on T, and gives the values of C_v and γ used in adiabatic processes. In Class 12, thermal speeds of free electrons in a conductor use the same idea.

Based on: CBSE Physics (042) syllabus 2026-27 (Classes XI-XII); NCERT Physics Part II, Textbook for Class XI, Chapter 12 (Reprint 2026-27)