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 Class 11 Physics, Ch. 12, p. 252Degrees of Freedom
8 MCQs2 revision cards9-step worked example
Source: NCERT Kinetic TheoryPYQ coverage: NEET 2026Official key: NTA-verifiedLast updated: 24 Sep 2026
Degrees of Freedom, explained for NEET
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 252, derives this directly from the equipartition theorem. The law of equipartition of energy (page 251) 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.
Can you answer these Degrees of Freedom 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?
Show answer and why every option is right or wrong
Answer: D. A monoatomic molecule has only translational motion in three independent directions (x, y, z), giving f = 3 (NCERT Class 11 Physics Chapter 12, page 252).
Why A is wrong: A is wrong because 2 would correspond to rotational DoF of a diatomic molecule, not the total DoF of a monoatomic gas.
Why B is wrong: B is wrong because f = 6 applies to a non-linear polyatomic molecule (3 translational + 3 rotational), not monoatomic.
Why C is wrong: C is wrong because f = 5 applies to a rigid diatomic molecule (3 translational + 2 rotational), not monoatomic.
MCQ 2Easy RecallPractice
At room temperature, the number of degrees of freedom of a rigid diatomic molecule such as O₂ is:
Show answer and why every option is right or wrong
Answer: A. A is correct. A diatomic molecule has 3 translational degrees of freedom and 2 rotational ones, about the two axes perpendicular to the bond. Rotation about the bond axis itself carries negligible energy because the moment of inertia about that axis is tiny. At room temperature vibration is not excited, so the total is 3 + 2 = 5.
Why B is wrong: B is wrong because 3 counts only the translational degrees of freedom. That is the count for a monatomic gas, which has no rotational energy to speak of.
Why C is wrong: C is wrong because 6 is the count for a NON-LINEAR molecule, which can rotate about three independent axes. A linear molecule has only two effective rotational axes.
Why D is wrong: D is wrong because 7 adds the two vibrational modes (kinetic and potential energy of the bond). Those are excited only at high temperature, not for a rigid molecule at room temperature.
MCQ 3Direct ApplicationPractice
The ratio of specific heats γ = C_p/C_v for a monoatomic ideal gas is:
Show answer and why every option is right or wrong
Answer: C. For monoatomic gas, f = 3. γ = 1 + 2/f = 1 + 2/3 = 5/3 ≈ 1.67 (NCERT Class 11 Physics Chapter 12, page 252).
Why A is wrong: A is wrong because γ = 1.33 = 4/3 corresponds to a non-linear polyatomic gas (f = 6), not monoatomic.
Why B is wrong: B is wrong because γ = 1.40 = 7/5 corresponds to a rigid diatomic gas (f = 5), not monoatomic.
Why D is wrong: D is wrong because γ = 1.29 = 9/7 would correspond to f = 7 (diatomic with vibration), not monoatomic.
MCQ 4Easy RecallPractice
A rigid diatomic molecule does NOT rotate about its bond axis because:
Show answer and why every option is right or wrong
Answer: C. The atoms in a diatomic molecule lie on the bond axis, so the moment of inertia about that axis is negligibly small. Classically the rotational energy in that mode is negligible; quantum-mechanically, the energy spacing is too large to excite at ordinary temperatures (NCERT Class 11 Physics Chapter 12, page 252).
Why A is wrong: A is wrong because the bond axis has the smallest moment of inertia (nearly zero), not the largest — the two perpendicular axes carry the significant moments.
Why B is wrong: B is wrong because quantum mechanics does not forbid rotation of diatomic molecules generally — it forbids excitation about the bond axis specifically because the energy gap is too large at ordinary T.
Why D is wrong: D is wrong because the issue is the negligible moment of inertia (hence negligible stored energy), not that rotational energy exceeds translational.
MCQ 5Easy RecallPractice
How many degrees of freedom does a rigid non-linear triatomic molecule (e.g. H₂O) have?
Show answer and why every option is right or wrong
Answer: B. A non-linear molecule has 3 translational + 3 rotational = 6 DoF in the rigid approximation. All three rotational axes have non-negligible moments of inertia because the atoms do not all lie on a single line (NCERT Class 11 Physics Chapter 12, page 252).
Why A is wrong: A is wrong because f = 3 accounts for translation only; a non-linear polyatomic molecule also has 3 rotational DoF.
Why C is wrong: C is wrong because f = 5 applies to linear molecules (diatomic or linear triatomic), which have only 2 rotational DoF. A non-linear molecule has 3 rotational DoF.
Why D is wrong: D is wrong because f = 9 would include vibrational modes (3 translational + 3 rotational + 3 vibrational for a non-linear triatomic), but the rigid approximation excludes vibration.
MCQ 6Direct ApplicationPractice
CO₂ is a linear triatomic molecule. In the rigid approximation, its C_v is:
Show answer and why every option is right or wrong
Answer: A. CO₂ is linear, so it has 3 translational + 2 rotational = 5 DoF (same as rigid diatomic). C_v = (5/2)R (NCERT Class 11 Physics Chapter 12, page 252).
Why B is wrong: B is wrong because (3/2)R corresponds to f = 3 (monoatomic), not a linear triatomic.
Why C is wrong: C is wrong because C_v = 3R requires f = 6, which applies to non-linear triatomic molecules (like H₂O), not linear ones like CO₂.
Why D is wrong: D is wrong because (7/2)R requires f = 7 (diatomic with vibration active). CO₂ in the rigid model has f = 5.
MCQ 7Concept TrapPractice
When vibrational modes of a diatomic gas become active at high temperature, the degrees of freedom increase from 5 to:
Show answer and why every option is right or wrong
Answer: B. Each vibrational mode contributes 2 DoF (one kinetic, one potential). A diatomic molecule has one vibrational mode, adding 2 to the rigid count of 5, giving f = 7 (NCERT Class 11 Physics Chapter 12, page 252).
Why A is wrong: A is wrong because each vibrational mode adds 2 DoF (kinetic + potential), not 1. Adding just 1 to get f = 6 misses the potential energy contribution.
Why C is wrong: C is wrong because f = 8 would require 3 extra DoF beyond the rigid 5, which has no physical basis for a single vibrational mode.
Why D is wrong: D is wrong because f = 9 would require 4 extra DoF, which does not correspond to any physical mechanism for a diatomic molecule.
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?
Show answer and why every option is right or wrong
Answer: D. γ = 1 + 2/f. For f = 3: γ = 1 + 2/3 = 5/3 ≈ 1.67. For f = 5: γ = 1 + 2/5 = 7/5 = 1.40. Since 7/5 < 5/3, γ decreases. More DoF means more ways to store energy internally without raising temperature, reducing γ (NCERT Class 11 Physics Chapter 12, page 252).
Why A is wrong: A is wrong because 7/5 (1.40) is less than 5/3 (1.67), so γ decreases, not increases. The formula γ = 1 + 2/f shows γ falls as f rises.
Why B is wrong: B is wrong because γ = 4/3 corresponds to f = 6 (non-linear polyatomic), not f = 5. For f = 5, γ = 7/5.
Why C is wrong: C is wrong because γ explicitly depends on f through γ = 1 + 2/f. Changing f from 3 to 5 necessarily changes γ.
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Degrees of Freedom: quick recall before you leave
How do you solve a Degrees of Freedom question? A 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 252).
- 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).
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What to remember before solving Degrees of Freedom questions
Which Degrees of Freedom formulas do you need for NEET?
1 formula — click to collapse
Cv from degrees of freedom
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
More in Kinetic Theory: 2 exam traps and mistakes · 4 formulas · 4 question patterns from its other lessons.
Degrees of Freedom questions from past NEET papers
1 question from NEET 2026. Answers verified against NTA official keys. — click to collapse
NEET 2026
14
23
32
41
NTA Answer: Option 1(final)
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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