PhysicsCBSE Class 11 · Kinetic Theory

Degrees of Freedom, Specific Heats and Mean Free Path Class 11

Updated 17 October 2026 · Fact-checked

The law of equipartition assigns ½k_BT of mean energy to each quadratic degree of freedom, so a molecule with f degrees of freedom has energy (f/2)k_BT and a mole has U = (f/2)RT. Hence C_v = (f/2)R, C_p = (f/2 + 1)R and γ = 1 + 2/f (5/3 for monatomic, 7/5 for rigid diatomic). The mean free path is λ = 1/(√2 π d² n).

Understand Degrees of Freedom, Specific Heats and Mean Free Path

Degrees of freedom. The number $f$ of independent ways a molecule can store energy: a monatomic gas has $f = 3$ (translation); a rigid diatomic molecule has $f = 5$ (3 translational + 2 rotational); a rigid non-linear polyatomic molecule has $f = 6$ . At high enough temperatures, vibrational modes add 2 each (kinetic and potential), but they are frozen at ordinary temperatures in the simple model used here.

Law of equipartition of energy. In thermal equilibrium at temperature $T$ , each degree of freedom (each quadratic term in the energy) contributes on average $\frac{1}{2} k_{B} T$ per molecule, i.e. $\frac{1}{2} R T$ per mole. So
$ $U = \frac{f}{2} n R T, C_{v} = \frac{f}{2} R, C_{p} = C_{v} + R = (\frac{f}{2} + 1) R, γ = \frac{C _{p}}{C _{v}} = 1 + \frac{2}{f} .$ $
Monatomic: $C_{v} = \frac{3}{2} R$ , $C_{p} = \frac{5}{2} R$ , $γ = \frac{5}{3}$ . Rigid diatomic: $\frac{5}{2} R$ , $\frac{7}{2} R$ , $γ = \frac{7}{5}$ . Rigid non-linear polyatomic: $3 R$ , $4 R$ , $γ = \frac{4}{3}$ . At constant pressure, the fraction of heat that does work is $R / C_{p}$ .

Mixtures. For gases of mole numbers $n_{1}, n_{2}$ with $C_{v 1}, C_{v 2}$ : $C_{v} = \frac{n _{1} C _{v 1} + n _{2} C _{v 2}}{n _{1} + n _{2}}$ and $C_{p} = C_{v} + R$ .

Solids. In a solid each atom vibrates in three dimensions, with 6 quadratic terms, so $U = 3 k_{B} T$ per atom and the molar heat capacity is $3 R \approx 24.9$ J mol⁻¹K⁻¹ (the law of Dulong and Petit), giving $c = 3 R / M$ for specific heat.

Mean free path. The average distance a molecule travels between collisions is
$ $λ = \frac{1}{2 π d ^{2} n},$ $
where $d$ is the molecular diameter and $n$ the number density; $λ \propto T / P$ at fixed $d$ . The collision frequency is about $v / λ$ . At low pressure $λ$ can exceed the size of the vessel.

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

Internal energy: $U = \frac{f}{2} n R T$; f degrees of freedom.
Molar heat capacities: $C_{v} = \frac{f}{2} R, C_{p} = (\frac{f}{2} + 1) R$; Ideal gas.
Ratio: $γ = 1 + \frac{2}{f}$; 5/3, 7/5, 4/3.
Mixture: $C_{v} = \frac{n _{1} C _{v 1} + n _{2} C _{v 2}}{n _{1} + n _{2}}$; Weighted by moles.
Solid (Dulong–Petit): $C = 3 R \approx 24.9 J mol^{- 1} K^{- 1}$; Atoms vibrating in 3D.
Mean free path: $λ = \frac{1}{2 π d ^{2} n}$; Collision cross-section.

How to solve Degrees of Freedom, Specific Heats and Mean Free Path questions

Identify the molecule's degrees of freedom first; all the heat capacities follow from f.

1Decide f from the molecule type: 3 (monatomic), 5 (rigid diatomic), 6 (rigid non-linear polyatomic).
2Write C_v = (f/2)R and C_p = C_v + R; get γ = C_p/C_v.
3For mixtures, take the mole-weighted average of C_v.
4For heat transfers, use Q = nC_vΔT (constant volume) or nC_pΔT (constant pressure).
5For mean free path, find n = P/k_BT, then use λ = 1/(√2πd²n).
6Compute collision frequency as v/λ.

Degrees of Freedom, Specific Heats 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.

Warming air in a sealed tank

Suppose a sealed rigid tank holds 3.0 mol of air (treat as a rigid diatomic gas) which is heated by 20 K.

Find the heat required (R = 8.314 J/mol K).

Try it first, then show the solutionSolution

Rigid volume, so Q = nC_vΔT with C_v = 5R/2.
Q = 3.0 × 2.5 × 8.314 × 20 = 1247 J.

Answer: About 1.25 × 10³ J.

Memory hook: Constant volume uses C_v.

Cooling a copper block

Suppose a 0.500 kg copper block (M = 63.5 g/mol) is treated as a Dulong–Petit solid.

Estimate its specific heat.

Try it first, then show the solutionSolution

C = 3R = 24.94 J/mol K; c = 24.94/0.0635 = 393 J/kg K.

Answer: About 393 J kg⁻¹ K⁻¹.

Memory hook: Atoms of any solid carry about 3R each mole.

Tricks to remember Degrees of Freedom, Specific Heats and Mean Free Path

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

f decides all
C_v = f/2 R, γ = 1 + 2/f.
Use it when: Heat capacities.
Three, five, six
Monatomic 3, rigid diatomic 5, non-linear 6.
Use it when: Degrees of freedom.
Mix by moles
C_v of a mixture is a mole-weighted average.
Use it when: Mixture questions.
Low pressure, long path
λ ∝ T/P.
Use it when: Vacuum questions.

Degrees of Freedom, Specific Heats and Mean Free Path: solved examples

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

Example 1: Internal energy of a gas · 2 marks · Very short answerBoard level

Find the internal energy of 2 mol of a monatomic ideal gas at 300 K (R = 8.314 J/mol K).

Try it first, then show the solutionSolution

U = (f/2)nRT with f = 3. (1)
U = 1.5 × 2 × 8.314 × 300 = 7.48 × 10³ J. (1)

Answer: 7.48 × 10³ J.

Example 2: Water vapour as a rigid gas · 3 marks · Short answerBoard level

Treating water vapour as a rigid non-linear triatomic molecule, find the number of degrees of freedom, C_v, C_p and γ.

Try it first, then show the solutionSolution

f = 3 translational + 3 rotational = 6. (1)
C_v = 3R = 24.9 J/mol K; C_p = 4R = 33.3 J/mol K. (1)
γ = 4/3 = 1.33. (1)

Answer: f = 6; C_v = 3R; C_p = 4R; γ = 4/3.

Example 3: Mixture of gases · 3 marks · Short answerHarder board level

A mixture contains 2 mol of a monatomic gas and 3 mol of a rigid diatomic gas. Find C_v, C_p and γ of the mixture.

Try it first, then show the solutionSolution

C_v = (2 × 1.5R + 3 × 2.5R)/5 = 10.5R/5 = 2.1R. (1)
C_p = C_v + R = 3.1R. (1)
γ = 3.1/2.1 ≈ 1.48. (1)

Answer: C_v = 2.1R; C_p = 3.1R; γ ≈ 1.48.

Example 4: Heating a mixture · 4 marks · Long answerHarder board level

The 5 mol mixture above is heated by 40 K. Find the heat required (a) at constant volume and (b) at constant pressure, and the work done in case (b) (R = 8.314 J/mol K).

Try it first, then show the solutionSolution

(a) Q_v = nC_vΔT = 5 × 2.1 × 8.314 × 40 ≈ 3.49 × 10³ J. (1)
(b) Q_p = nC_pΔT = 5 × 3.1 × 8.314 × 40 ≈ 5.15 × 10³ J. (1)
W = nRΔT = 5 × 8.314 × 40 = 1.66 × 10³ J. (1)
Check: Q_p − Q_v = 5.15 − 3.49 = 1.66 kJ, equal to the work. (1)

Answer: (a) 3.49 kJ; (b) 5.15 kJ; W = 1.66 kJ.

Example 5: Mean free path and collision rate · 4 marks · Long answerHOTS / competency

A gas of molecular diameter 2.0 × 10⁻¹⁰ m and molar mass 4.0 × 10⁻³ kg/mol is at 300 K and 1.0 × 10⁵ Pa (k_B = 1.38 × 10⁻²³ J/K). Find (a) the number density, (b) the mean free path, (c) the collision frequency using the rms speed, and (d) the pressure at which the mean free path is 0.1 m.

Try it first, then show the solutionSolution

(a) n = P/k_BT = 10⁵/(1.38 × 10⁻²³ × 300) = 2.4 × 10²⁵ m⁻³. (1)
(b) λ = 1/(√2π(2 × 10⁻¹⁰)² × 2.4 × 10²⁵) = 2.3 × 10⁻⁷ m. (1)
(c) v_rms = √(3RT/M) = 1368 m/s; collision frequency = v/λ ≈ 5.9 × 10⁹ s⁻¹. (1)
(d) λ ∝ 1/P: P = 10⁵ × 2.33 × 10⁻⁷/0.1 ≈ 0.23 Pa. (1)

Answer: (a) 2.4 × 10²⁵ m⁻³; (b) 2.3 × 10⁻⁷ m; (c) 5.9 × 10⁹ s⁻¹; (d) about 0.23 Pa.

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

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

Mistake: Using γ = 5/3 for a diatomic gas.
Remembering only one value.
Do this: Diatomic (rigid) γ = 7/5; monatomic 5/3; non-linear 4/3.
Usually costs: The answer mark.
Mistake: Using C_p in Q = nC_vΔT.
Ignoring the process.
Do this: Constant volume: C_v; constant pressure: C_p.
Usually costs: Method mark.
Mistake: Averaging C_v of a mixture without mole weights.
Treating the gases equally.
Do this: Weight by the number of moles.
Usually costs: 1 mark.
Mistake: Using n as the number of moles in the mean free path formula.
Mixing up n.
Do this: n is the number per unit volume.
Usually costs: The answer mark.

Exam tips

State the value of f and the reason (translation, rotation) first.
Show C_p = C_v + R when needed.
For mixtures, show the mole-weighted average line.
Include the unit of λ (metres).
Remember vibrational modes are ignored unless the question says otherwise.

Practice questions on Degrees of Freedom, Specific Heats 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 · MCQHarder board level
When heat is supplied to a monatomic ideal gas at constant pressure, the fraction of the heat converted into work is
(a) 2/5 (b) 2/7 (c) 3/5 (d) 1/4
Hint
W/Q = R/C_p.
Answer
(a) 2/5: C_p = 5R/2.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): A gas of more complex molecules has a value of γ closer to 1.
Reason (R): γ = 1 + 2/f, and f increases with molecular complexity.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Smaller 2/f.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
Find the total translational kinetic energy of one mole of an ideal gas at 300 K (R = 8.314 J/mol K).
Hint
(3/2)RT.
Answer
K = 1.5 × 8.314 × 300 ≈ 3.74 × 10³ J.
Q4 · 3 marks · Short answerHarder board level
Use the equipartition law to show that the molar heat capacity of a solid is 3R, and use it to estimate the specific heat of iron (M = 56 g/mol).
Hint
Each atom vibrates in 3 directions with kinetic and potential energy.
Answer
Each atom has 6 quadratic terms, so energy 3k_BT per atom and U = 3RT per mole, C = 3R = 24.9 J/mol K. c = 24.9/0.056 ≈ 445 J/kg K.
Q5 · 4 marks · Long answerHOTS / competency
A dust particle of mass 1.0 × 10⁻¹⁵ kg is suspended in air at 300 K and behaves like a gas molecule (k_B = 1.38 × 10⁻²³ J/K). (a) Find its mean translational kinetic energy and rms speed. (b) Compare its speed with that of a nitrogen molecule (M = 0.028 kg/mol) at 300 K. (c) Why is the dust particle not seen moving at this speed?
Hint
K = (3/2)k_BT, same for all particles.
Answer
(a) K = 6.2 × 10⁻²¹ J; v = √(2K/m) = 3.5 × 10⁻³ m/s. (b) N₂: √(3RT/M) = 517 m/s, about 1.5 × 10⁵ times faster. (c) It is struck by air molecules an enormous number of times per second, so its path is a tiny random zigzag with a very small net displacement (Brownian motion).

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.Equipartition: energy per degree of freedom?Show answer
½k_BT per molecule.
2.f for rigid diatomic?Show answer
5.
3.γ in terms of f?Show answer
1 + 2/f.
4.C for a solid?Show answer
3R.
5.Mean free path?Show answer
1/(√2πd²n).
6.C_p − C_v?Show answer
R.

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

Why does a diatomic gas have 5 degrees of freedom?

Three for translation and two for rotation about axes perpendicular to the molecular axis; rotation about the axis itself has negligible moment of inertia.

What are frozen degrees of freedom?

Modes, such as vibration at room temperature, that cannot be excited at the thermal energy available.

Why does λ increase when pressure falls?

There are fewer molecules per unit volume, so collisions are rarer.

Is γ the same for all gases?

No, it depends on the number of degrees of freedom of the molecules.

What is the Dulong–Petit law?

The molar heat capacity of a simple solid is about 3R.