Degrees of Freedom

8 MCQs9-step worked example

Source: NCERT Kinetic TheoryPYQ coverage: NEET 2020, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

Degrees of freedom (DoF) is the number of independent ways a gas molecule can store energy. Each quadratic term in the molecule's total energy expression — whether translational, rotational, or vibrational — counts as one degree of freedom. The equipartition theorem assigns each DoF an average energy of ½kT per molecule, or equivalently ½RT per mole.

Counting DoF by molecule type:

Monoatomic (He, Ne, Ar): 3 translational DoF only. No rotational or vibrational modes contribute at ordinary temperatures. So f = 3.
Diatomic rigid (O₂, N₂ at moderate T): 3 translational + 2 rotational = 5. Rotation about the bond axis contributes negligibly (moment of inertia is near zero for that axis). So f = 5.
Diatomic with vibration (high T): 3 translational + 2 rotational + 2 vibrational (one kinetic + one potential) = 7.
Polyatomic rigid (CO₂ linear rigid: f = 5; H₂O non-linear rigid: f = 6, i.e. 3 translational + 3 rotational).

From DoF to specific heat — this is where NEET questions live. Once you know f, the molar specific heat at constant volume is:

C_v = (f/2)R

and C_p = C_v + R, giving γ = C_p/C_v = 1 + 2/f.

NCERT Class 11 Physics Chapter 12 (Part 2), page 9, derives this directly from the equipartition theorem. The law of equipartition of energy (page 8) is the foundational statement: each quadratic DoF contributes ½kT.

Watch out: The most common confusion is miscounting DoF — particularly forgetting that a linear triatomic molecule has only 2 rotational DoF (like a diatomic), not 3. A non-linear molecule has 3 rotational DoF. This single miscount shifts C_v, C_p, and γ simultaneously.

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

A monoatomic ideal gas has how many degrees of freedom?

MCQ 2Direct ApplicationPractice

For a rigid diatomic gas molecule, the molar specific heat at constant volume C_v is:

MCQ 3Direct ApplicationPractice

The ratio of specific heats γ = C_p/C_v for a monoatomic ideal gas is:

MCQ 4Easy RecallPractice

A rigid diatomic molecule does NOT rotate about its bond axis because:

MCQ 5Easy RecallPractice

How many degrees of freedom does a rigid non-linear triatomic molecule (e.g. H₂O) have?

MCQ 6Direct ApplicationPractice

CO₂ is a linear triatomic molecule. In the rigid approximation, its C_v is:

MCQ 7Concept TrapPractice

When vibrational modes of a diatomic gas become active at high temperature, the degrees of freedom increase from 5 to:

MCQ 8CalculationPractice

If the degrees of freedom of a gas molecule increase from 3 to 5 (at constant temperature), how does γ = C_p/C_v change?

Worked Example

1
Given
- Gas A: monoatomic, rigid → f_A = 3 - Gas B: diatomic, rigid → f_B = 5
2
Required
The ratio γ_A / γ_B.
3
Concept
The ratio of specific heats depends only on degrees of freedom: γ = 1 + 2/f. This follows from the equipartition theorem assigning ½R per mole per DoF to C_v, and C_p = C_v + R (NCERT Class 11 Physics Chapter 12, page 9).
4
Formula
γ = 1 + 2/f
5
Substitution
γ_A = 1 + 2/3 = 5/3 γ_B = 1 + 2/5 = 7/5
6
Calculation
γ_A / γ_B = (5/3) / (7/5) = (5/3) × (5/7) = 25/21 Note: all numbers here (3, 5, 2, 7) are exact counting integers representing degrees of freedom or arising from the formula structure. They do not limit significant figures.
7
Final answer
γ_A / γ_B = 25/21 ≈ 1.19
8
Common trap
A common confusion: students sometimes assign f = 2 to monoatomic (thinking "mono = one, but it moves in 2D") or f = 3 to diatomic (thinking rotation about the bond axis counts). The correct counts are f = 3 (monoatomic: 3 translational) and f = 5 (rigid diatomic: 3 translational + 2 rotational). Miscounting either shifts the entire ratio.
9
Similar NEET-style question
"The ratio C_p/C_v for a gas whose molecules have 6 degrees of freedom is: (A) 4/3 (B) 5/3 (C) 7/5 (D) 9/7." Answer: γ = 1 + 2/6 = 4/3 → option (A). ---

Before solving, remember these

Law

Law of equipartition of energy

Each quadratic degree of freedom contributes (½)kT to the average energy per molecule. Monoatomic: 3 translational DoF → (3/2)kT. Diatomic: 3 trans + 2 rot → (5/2)kT. Polyatomic: 6 DoF → 3kT.

-- NCERT, p. 8

Formulas

5 formulas — click to collapse

Average translational KE per molecule

⟨ K E ⟩ = \frac{3}{2} k T

Microscopic interpretation of temperature: T is direct measure of average translational kinetic energy.

Symbol	Quantity	SI Unit
k	Boltzmann constant	J/K
T	absolute temperature	K

Valid when

Translational degrees of freedom only
Ideal gas

Cv from degrees of freedom

C_{v} = \frac{f}{2} R

Each quadratic DoF contributes (1/2)R to molar Cv. Mono: f=3, Cv=3R/2; di-rigid: f=5, Cv=5R/2; poly-rigid: f=6, Cv=3R.

Symbol	Quantity	SI Unit
Cv	molar specific heat	J/mol/K
f	degrees of freedom	-
R	gas constant	J/mol/K

Valid when

Equipartition holds (temperature high enough)
Quadratic energy modes

Ideal gas equation

P V = n R T = N k T

Fundamental equation of state of ideal gas relating pressure, volume, temperature.

Symbol	Quantity	SI Unit
P	pressure	Pa
V	volume	m^3
n	moles	mol
R	8.314	J/mol/K
N	molecule count	-
k	Boltzmann 1.38e-23	J/K
T	temp	K

Valid when

Gas obeys ideal gas approximation (low pressure, high temperature relative to phase transitions)

Mean free path of gas molecule

λ = \frac{1}{2 π n d ^{2}}

Average distance between successive molecular collisions.

Symbol	Quantity	SI Unit
lambda	mean free path	m
n	number density	1/m^3
d	molecular diameter	m

Valid when

Hard-sphere model
Equilibrium gas

RMS speed of gas molecules

v_{rms} = \frac{3 R T}{M} = \frac{3 k T}{m}

Root-mean-square molecular speed; depends on T and molar mass M (or molecular mass m).

Symbol	Quantity	SI Unit
R	gas constant	J/mol/K
T	temp	K
M	molar mass	kg/mol
k	Boltzmann	J/K
m	molecular mass	kg

Valid when

Ideal gas
Maxwell-Boltzmann distribution

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

2 items — click to collapse

Trap

v_rms vs T: linear vs sqrt scaling

Category: Similar Terms

Student treats v_rms ∝ T instead of √T. Doubling T does NOT double v_rms; it multiplies by √2.

When it triggers

Question asks for new v_rms after T change.

How to avoid

v_rms = √(3RT/M). v_rms ∝ √T. To double v_rms, T must quadruple.

Mistake

Scales v_rms linearly with T instead of √T.

Root cause: formula misuse

Correction

v_rms = √(3RT/M), so v_rms ∝ √T. To double v_rms, T must quadruple (factor of 4). Common error: assume doubling T doubles v_rms.

Past Year Questions

6 questions from NEET 2020, 2022, 2023, 2024, 2025. Answers verified against NTA official keys. — click to collapse

NEET 2025

An oxygen cylinder of volume 30 litre has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressures at temperature 27°C. The mass of the oxygen withdrawn from the cylinder is nearly equal to: 100 [Given, R= Jmol–1K–1, and molecular mass of O2 = 32, 1 atm pressure = 1.01 × 105 N/m] 12

10.156 kg

20.125 kg

30.144 kg

40.116 kg

NTA Answer: Option 4(final)

NEET 2024Revised key

The following graph represents the T-V curves of an ideal gas (where T is the temperature and V the volume) at three pressures P1, P2 and P3 compared with those of Charles’s law represented as dotted lines. Then the correct relation is:

1P3 > P2 > P1

2P1 > P3 > P2

3P2 > P1 > P3

4P1 > P2 > P3

NTA Answer: Option 4(revised_final)

NEET 2023

The temperature of a gas is –50°C. To what temperature the gas should be heated so that the rms speed is increased by 3 times?

13295°C

23097 K

3223 K

4669°C

NTA Answer: Option 1(final)

NEET 2022

The volume occupied by the molecules contained in 4.5 kg water at STP, if the intermolecular forces vanish away is

15.6 m3

25.6 × 106 m3

35.6 × 103 m3

45.6 × 10–3 m3

NTA Answer: Option 1(final)

NEET 2020

The average thermal energy for a mono-atomic gas 16. U — is : (k is Boltzmann constant and T, absolute B ( U =k B U =T) temperature) 7 7

12 k B T 1 1

22 k B T 3 3

32 k B T 5 5

42 k B T

NTA Answer: Option 3(final)

NEET 2020

, fl d 23. The mean free path for a gas, with molecular fl n , U diameter d and number density n can be expressed — as : 1 1

12 n 2 2 d 2 1 1

22 n d 1 1

32 n d2 1 1

42 n 2 d 2

NTA Answer: Option 3(final)

How NEET usually asks this

4 recurring patterns from past papers — click to collapse

Average translational KE per molecule = (3/2)kT (monoatomic). Total via DoF.

Direct ApplicationEasy

Common distractors

uses 3 2 RT not 3 2 kT

Confuses per-mole RT with per-molecule kT

PV = nRT. Given two of (P, V, T, n) and changes, find missing.

Multi StepEasy

Common distractors

forgets temperature conversion

Mixes °C with K

Mean free path scaling with n, d. lambda = 1/(sqrt(2)pin*d^2).

InterpretationEasy

Common distractors

ignores d squared

Treats d linearly

Find new T given v_rms scales by factor k. v_rms ∝ sqrt(T), so T_new = T*k^2.

Multi StepMedium

Common distractors

uses linear scaling

Treats v_rms ∝ T not sqrt(T)

Test yourself on this topic with real past-paper questions:

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