PhysicsISC Class 11 · Kinetic Theory of Gases

Degrees of Freedom, Equipartition of Energy and Mean Free Path

Updated 17 October 2026 · Fact-checked

By the law of equipartition, each degree of freedom of a molecule contributes ½kT to its mean energy. A gas with f degrees of freedom has U = (f/2)nRT, C_v = (f/2)R, C_p = (f/2 + 1)R and γ = 1 + 2/f. The mean free path is λ = 1/(√2 π d² n).

Understand Degrees of Freedom, Equipartition and Mean Free Path

Degrees of freedom. The number of independent ways in which a molecule can have energy. A monatomic molecule has f = 3 (translation). A rigid diatomic molecule has f = 5 (3 translational + 2 rotational). A non-rigid diatomic molecule at high temperature has 2 more (vibration: kinetic and potential), f = 7. A nonlinear polyatomic molecule has f = 6 if rigid.

Law of equipartition. In thermal equilibrium at temperature T, the average energy associated with each degree of freedom is ½kT per molecule (½RT per mole). Therefore the internal energy of n mol of an ideal gas is U = (f/2)nRT.

Specific heats. C_v = (dU/dT)/n = (f/2)R. Mayer's relation gives C_p = C_v + R = (f/2 + 1)R, and γ = C_p/C_v = 1 + 2/f. For monatomic gas: C_v = 3R/2, C_p = 5R/2, γ = 5/3 = 1.67. For rigid diatomic gas: C_v = 5R/2, C_p = 7R/2, γ = 7/5 = 1.4. For a mixture, the heat capacities per mole of the mixture are weighted averages by the number of moles.

Mean free path. The average distance travelled by a molecule between two successive collisions. For molecules of diameter d and number density n, λ = 1/(√2 π d² n). Since n = p/kT, λ = kT/(√2 π d² p): λ is inversely proportional to p at constant T and increases with T at constant p. The mean time between collisions is λ divided by the average speed.

Avogadro's number. The same volume of any gas at the same T and p contains the same number of molecules; N_A = 6.02 × 10²³ mol⁻¹.

Degrees of Freedom, Equipartition and Mean Free Path: formulas and key facts

Equipartition: $E p er d e g r eeo f f r ee d o m = ½ k T ∣ U = (f /2) n R T$; f = degrees of freedom.
Specific heats: $C_{v} = (f /2) R ∣ C_{p} = (f /2 + 1) R ∣ γ = 1 + 2/ f$; Mayer: C_p − C_v = R.
Mean free path: $λ = 1/ (\sqrt2 π d^{2} n) ∣ λ = k T / (\sqrt2 π d^{2} p)$; d = molecular diameter; n = number density.
Mixture: $C_{v} (mi x) = (n_{1} C_{v} 1 + n_{2} C_{v} 2) / (n_{1} + n_{2})$; Same for C_p.

Degrees of Freedom, Equipartition and Mean Free Path in real life

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.

Heat needed to warm air in a sealed room

Suppose a rigid, sealed cabin contains 3 mol of a diatomic gas (f = 5, R = 8.31 J mol⁻¹ K⁻¹), heated by 20 K at constant volume.

Find the heat needed, and the heat needed if the gas is heated at constant pressure instead.

Try it first, then show the solutionSolution

Constant V: Q = nC_vΔT = 3 × 2.5 × 8.31 × 20 = 1246.5 J.
Constant p: Q = nC_pΔT = 3 × 3.5 × 8.31 × 20 = 1745 J.
The difference nRΔT = 498.6 J is the work done in expansion.

Answer: 1247 J at constant volume; 1745 J at constant pressure.

Memory hook: More heat is needed at constant pressure because the gas also does work.

Tricks to remember Degrees of Freedom, Equipartition and Mean Free Path

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

3-5-7
Degrees of freedom of monatomic, rigid diatomic and non-rigid diatomic gas.
Use it when: Quick γ: 5/3, 7/5, 9/7.
Mean free path is a gap, not a size
λ is far greater than d at ordinary pressure.
Use it when: Order-of-magnitude questions.

Degrees of Freedom, Equipartition and Mean Free Path: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Degrees of freedom · 1 mark · MCQBoard level

The number of degrees of freedom of a monatomic gas molecule is:
(a) 1
(b) 2
(c) 3
(d) 5

Try it first, then show the solutionSolution

A point-like atom can move along x, y and z only.

Answer: (c) 3

Example 2: Internal energy · 3 marks · Short answerBoard level

Find the internal energy of 2 mol of a rigid diatomic ideal gas at 300 K (R = 8.31 J mol⁻¹ K⁻¹).

Try it first, then show the solutionSolution

f = 5 for a rigid diatomic molecule. (1 mark)
U = (f/2)nRT = 2.5 × 2 × 8.31 × 300. (1 mark)
= 12 465 J. (1 mark)

Answer: 1.25 × 10⁴ J

Example 3: Specific heats of a diatomic gas · 4 marks · Short answerHarder board level

Find C_v, C_p and γ for a rigid diatomic gas (R = 8.31 J mol⁻¹ K⁻¹), and the heat needed to warm 3 mol by 20 K at constant volume and at constant pressure.

Try it first, then show the solutionSolution

C_v = 5R/2 = 20.8 J mol⁻¹ K⁻¹; C_p = 7R/2 = 29.1 J mol⁻¹ K⁻¹; γ = 7/5 = 1.4. (1.5 marks)
Q_v = 3 × 20.775 × 20 = 1246.5 J. (1 mark)
Q_p = 3 × 29.085 × 20 = 1745 J. (1 mark)
Difference = nRΔT = 498.6 J. (0.5 mark)

Answer: C_v = 20.8, C_p = 29.1 J mol⁻¹ K⁻¹, γ = 1.4; Q_v = 1247 J; Q_p = 1745 J.

Example 4: Mixture of gases · 4 marks · Short answerHarder board level

A mixture contains 1 mol of a monatomic gas and 2 mol of a rigid diatomic gas. Find C_v, C_p and γ for the mixture.

Try it first, then show the solutionSolution

C_v(mix) = (1 × 3R/2 + 2 × 5R/2)/3. (1 mark)
= (1.5R + 5R)/3 = 6.5R/3 = 2.167R. (1 mark)
C_p = C_v + R = 3.167R. (1 mark)
γ = 3.167/2.167 = 1.46. (1 mark)

Answer: C_v = 2.17R; C_p = 3.17R; γ = 1.46.

Example 5: Mean free path · 5 marks · Long answerHOTS / competency

Nitrogen molecules have a diameter of 3.7 × 10⁻¹⁰ m. At 300 K and 1.01 × 10⁵ Pa, find the number density, the mean free path, and the mean time between collisions for an rms speed of 517 m s⁻¹. What happens to λ if the pressure is halved? (k = 1.38 × 10⁻²³ J K⁻¹)

Try it first, then show the solutionSolution

n = p/kT = 1.01 × 10⁵/(1.38 × 10⁻²³ × 300) = 2.44 × 10²⁵ m⁻³. (1 mark)
πd² = π(3.7 × 10⁻¹⁰)² = 4.30 × 10⁻¹⁹ m². (1 mark)
λ = 1/(√2 × 4.30 × 10⁻¹⁹ × 2.44 × 10²⁵) = 1/(1.484 × 10⁷) = 6.7 × 10⁻⁸ m. (1 mark)
τ = λ/v = 6.7 × 10⁻⁸/517 = 1.3 × 10⁻¹⁰ s. (1 mark)
λ ∝ 1/p, so halving the pressure doubles λ to 1.3 × 10⁻⁷ m. (1 mark)

Answer: n = 2.4 × 10²⁵ m⁻³; λ ≈ 6.7 × 10⁻⁸ m; τ ≈ 1.3 × 10⁻¹⁰ s; λ doubles.

Where students lose marks in Degrees of Freedom, Equipartition and Mean Free Path

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

Mistake: Counting the wrong f
Using f = 3 for oxygen at room temperature.
Do this: Rigid diatomic molecules have f = 5 (3 translational + 2 rotational).
Usually costs: 2 marks.
Mistake: Mixing up C_v and C_p
Using C_p to calculate the internal energy change.
Do this: ΔU = nC_vΔT for any process; Q = nC_pΔT only at constant pressure.
Usually costs: 1 to 2 marks.
Mistake: Mean free path units
Writing λ in the same unit as d without conversion.
Do this: Keep all lengths in metres; λ = 1/(√2πd²n).
Usually costs: 1 mark.

Exam tips

State f first, then compute U, C_v, C_p and γ from it.
For mixtures, average the molar heat capacities by moles.
Use n = p/kT to get the number density.
Note that equipartition holds only at temperatures where the modes are excited.

Practice questions on Degrees of Freedom, Equipartition and Mean Free Path

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 · MCQBoard level
The value of γ for a monatomic ideal gas is:
(a) 1.67
(b) 1.4
(c) 1.33
(d) 1.29
Hint
γ = 1 + 2/f with f = 3.
Answer
(a) 1.67
Q2 · 2 marks · Short answerBoard level
What are C_v and the degrees of freedom of a rigid diatomic gas? (R = 8.31 J mol⁻¹ K⁻¹)
Hint
Count translations and rotations.
Answer
f = 5 (3 translational + 2 rotational), C_v = 5R/2 = 20.8 J mol⁻¹ K⁻¹.
Q3 · 3 marks · Short answerHarder board level
Find γ for a gas molecule with 7 degrees of freedom, and the molar internal energy at 400 K (R = 8.31 J mol⁻¹ K⁻¹).
Hint
γ = 1 + 2/f; U = (f/2)RT.
Answer
γ = 1 + 2/7 = 9/7 = 1.29. U = 3.5 × 8.31 × 400 = 11 634 J mol⁻¹.
Q4 · 3 marks · Short answerHarder board level
Find the total translational kinetic energy of 1 g of helium (M = 4 g mol⁻¹) at 400 K (R = 8.31 J mol⁻¹ K⁻¹).
Hint
n = m/M; K = (3/2)nRT.
Answer
n = 0.25 mol. K = 1.5 × 0.25 × 8.31 × 400 = 1247 J. (The same as the internal energy for a monatomic gas.)
Q5 · 3 marks · Short answerHarder board level
The mean free path of molecules in a gas is 6.7 × 10⁻⁸ m at 1.01 × 10⁵ Pa and 300 K. Find it at the same temperature and a pressure of 1.01 × 10⁻² Pa, and compare it with a vessel 1 m wide.
Hint
λ ∝ 1/p at constant T.
Answer
The pressure falls 10⁷ times, so λ = 6.7 × 10⁻⁸ × 10⁷ = 0.67 m, comparable with the vessel size: collisions with the walls then dominate.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose 3 mol of argon (monatomic, R = 8.31 J mol⁻¹ K⁻¹) is warmed by 100 K.
(i) Find the heat needed at constant volume. (1 mark)
(ii) Find the heat needed at constant pressure. (1 mark)
(iii) Find the work done by the gas in the constant-pressure case. (1 mark)
(iv) Check the first law and find γ. (1 mark)
Hint
C_v = 3R/2; C_p = 5R/2.
Answer
(i) 3 × 1.5 × 8.31 × 100 = 3740 J. (ii) 3 × 2.5 × 8.31 × 100 = 6233 J. (iii) W = nRΔT = 3 × 8.31 × 100 = 2493 J. (iv) ΔU + W = 3740 + 2493 = 6233 J = Q ✓; γ = 5/3 = 1.67.

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.State the law of equipartition of energy.Show answer
Each degree of freedom has an average energy of ½kT per molecule.
2.γ for a rigid diatomic gas?Show answer
7/5 = 1.4.
3.Formula for the mean free path?Show answer
1/(√2 π d² n).
4.How does λ change with pressure?Show answer
λ ∝ 1/p at constant T.

Degrees of Freedom, Equipartition and Mean Free Path: frequently asked questions

Why does oxygen show f = 5 and not 7 at room temperature?

The vibrational modes are not excited at room temperature; they contribute only at high temperatures, when f approaches 7.

Why is C_p always greater than C_v?

At constant pressure the gas does work in expanding, so extra heat nR ΔT is required.

Is the mean free path the same as the molecular diameter?

No. At normal pressure it is hundreds of times larger than the diameter of a molecule.