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. 252Equipartition Energy
8 MCQs3 revision cards9-step worked example
Source: NCERT Kinetic TheoryPYQ coverage: NEET 2020Official key: NTA-verifiedLast updated: 24 Sep 2026
Equipartition Energy, explained for NEET
The law of equipartition of energy says: for a system in thermal equilibrium, each quadratic degree of freedom (DoF) contributes ½kT of energy per molecule, or ½RT per mole. This is the bridge between molecular motion and measurable heat capacities — and the place where NEET questions test whether you actually know the DoF count for each gas type.
The core rule (NCERT Class 11 Physics Chapter 12, page 253): A molecule with f degrees of freedom has average energy (f/2)kT per molecule, or (f/2)RT per mole. The molar specific heat at constant volume follows directly: Cᵥ = (f/2)R.
DoF count — the exam-critical table:
| Gas type | Translational | Rotational | f (rigid) | Cᵥ | Cₚ = Cᵥ + R | γ = Cₚ/Cᵥ |
|---|---|---|---|---|---|---|
| Monoatomic (He, Ne, Ar) | 3 | 0 | 3 | 3R/2 | 5R/2 | 5/3 ≈ 1.67 |
| Diatomic rigid (N₂, O₂) | 3 | 2 | 5 | 5R/2 | 7R/2 | 7/5 = 1.40 |
| Polyatomic rigid (CO₂, H₂O) | 3 | 3 | 6 | 3R | 4R | 4/3 ≈ 1.33 |
(NCERT Class 11 Physics Chapter 12, page 252)
Where aspirants lose marks: Confusing per-molecule (kT) with per-mole (RT) quantities. When a question says "average kinetic energy of a molecule," the answer uses k. When it says "of one mole," the answer uses R. Mixing these flips the numerical answer by a factor of Avogadro's number.
A second common error: applying the wrong DoF count. A diatomic molecule at moderate temperatures has f = 5 (3 translational + 2 rotational). Vibrational modes contribute only at high temperatures — unless the problem explicitly states "including vibrational modes," use the rigid-molecule f values.
The average translational KE per molecule is always (3/2)kT regardless of gas type — only translational DoFs contribute to temperature. The total energy per molecule depends on f and equals (f/2)kT.
Can you answer these Equipartition Energy 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
According to the law of equipartition of energy, each quadratic degree of freedom of a molecule in thermal equilibrium contributes an average energy of:
Show answer and why every option is right or wrong
Answer: D. The law of equipartition assigns ½kT of energy per quadratic degree of freedom per molecule (NCERT Class 11 Physics Chapter 12, page 253).
Why A is wrong: A gives kT, which is twice the correct per-DoF contribution. This would be the contribution of two degrees of freedom, not one.
Why B is wrong: B gives 2kT — no standard equipartition result corresponds to this value per degree of freedom.
Why C is wrong: C gives (3/2)kT, which is the total translational KE per molecule (3 translational DoFs × ½kT), not the per-DoF value.
MCQ 2Easy RecallPractice
A rigid diatomic molecule has how many degrees of freedom at moderate temperatures?
Show answer and why every option is right or wrong
Answer: C. A rigid diatomic molecule has 3 translational + 2 rotational = 5 degrees of freedom. Vibrational modes are frozen out at moderate temperatures (NCERT Class 11 Physics Chapter 12, page 253).
Why A is wrong: A gives 3, which is the DoF count for a monoatomic gas. A diatomic molecule has rotational modes in addition to translational.
Why B is wrong: B gives 6, which is the DoF count for a rigid polyatomic molecule (3 translational + 3 rotational), not diatomic.
Why D is wrong: D gives 7, which would apply to a diatomic molecule with vibrational modes included (f = 5 + 2). At moderate temperatures, vibrational DoFs are not active unless stated.
MCQ 3Easy RecallPractice
The ratio of specific heats γ = Cₚ/Cᵥ for a monoatomic ideal gas is:
Show answer and why every option is right or wrong
Answer: B. For monoatomic gas, f = 3, so Cᵥ = 3R/2 and Cₚ = 5R/2. γ = Cₚ/Cᵥ = 5/3 (NCERT Class 11 Physics Chapter 12, page 252).
Why A is wrong: A gives 7/5, which is γ for a rigid diatomic gas (f = 5), not monoatomic.
Why C is wrong: C gives 4/3, which is γ for a rigid polyatomic gas (f = 6), not monoatomic.
Why D is wrong: D gives 9/7, which would correspond to f = 7 (diatomic with vibration). Not applicable to monoatomic gases.
MCQ 4Direct ApplicationPractice
The average translational kinetic energy of one molecule of an ideal gas at temperature T is (3/2)kT. What is the total translational KE of one mole of this gas?
Show answer and why every option is right or wrong
Answer: A. Total translational KE per mole = Nₐ × (3/2)kT = (3/2)NₐkT = (3/2)RT, since R = Nₐk (NCERT Class 11 Physics Chapter 12, page 253).
Why B is wrong: B gives the per-molecule value, not per-mole. Scaling up to one mole requires multiplying by Avogadro's number.
Why C is wrong: C introduces a spurious T² factor. Energy scales linearly with T, not quadratically.
Why D is wrong: D gives 3RT, which would correspond to f = 6 total DoFs — the total internal energy of a polyatomic gas, not just translational KE of any gas.
MCQ 5Direct ApplicationPractice
For a rigid diatomic ideal gas, the molar specific heat at constant volume Cᵥ is:
Show answer and why every option is right or wrong
Answer: C. Rigid diatomic has f = 5. Cᵥ = (f/2)R = (5/2)R (NCERT Class 11 Physics Chapter 12, page 252).
Why A is wrong: A gives (3/2)R, which is Cᵥ for a monoatomic gas (f = 3). A diatomic molecule has 2 additional rotational DoFs.
Why B is wrong: B gives 3R, which is Cᵥ for a rigid polyatomic molecule (f = 6), not diatomic.
Why D is wrong: D gives (7/2)R, which is Cₚ (not Cᵥ) for a rigid diatomic gas. The question asks for Cᵥ, not Cₚ.
MCQ 6Direct ApplicationPractice
Two moles of a monoatomic ideal gas are at temperature 300 K. What is the total internal energy of the gas? (R = 8.314 J mol⁻¹ K⁻¹)
Show answer and why every option is right or wrong
Answer: D. Monoatomic gas: U = n × (f/2)RT = 2 × (3/2) × 8.314 × 300 = 2 × 3 × 8.314 × 300 / 2 = 7483 J (NCERT Class 11 Physics Chapter 12, page 252).
Why A is wrong: A gives 3741 J, which is the energy for 1 mole, not 2. The factor of n = 2 was missed.
Why B is wrong: B gives 2494 J, which corresponds to 1 mole with f = 2. Both the mole count and DoF count are wrong.
Why C is wrong: C gives 4988 J, which would correspond to using f = 2 (two DoFs) instead of f = 3 for monoatomic gas — or equivalently using Cᵥ = R instead of (3/2)R.
MCQ 7CalculationPractice
An ideal gas has Cₚ = (7/2)R. What is the number of degrees of freedom of each molecule of this gas?
Show answer and why every option is right or wrong
Answer: A. Cₚ = Cᵥ + R = (f/2)R + R = ((f + 2)/2)R. Given Cₚ = (7/2)R → f + 2 = 7 → f = 5 (NCERT Class 11 Physics Chapter 12, page 252).
Why B is wrong: B gives f = 3, which yields Cₚ = (5/2)R, not (7/2)R. This corresponds to a monoatomic gas.
Why C is wrong: C gives f = 6, which yields Cₚ = 4R = (8/2)R, not (7/2)R. This corresponds to a rigid polyatomic gas.
Why D is wrong: D gives f = 7, which yields Cₚ = (9/2)R. This confuses the f value with the numerator in Cₚ — the (f + 2) relation is the step students skip.
MCQ 8CalculationPractice
A container holds a mixture of 1 mole of helium (monoatomic) and 1 mole of nitrogen (diatomic, rigid) at temperature T. The total internal energy of the mixture is:
Show answer and why every option is right or wrong
Answer: B. He (f = 3): U₁ = 1 × (3/2)RT. N₂ (f = 5): U₂ = 1 × (5/2)RT. Total = (3/2 + 5/2)RT = 4RT (NCERT Class 11 Physics Chapter 12, pages 253–252).
Why A is wrong: A gives 3RT, which would be correct only if both gases were monoatomic ((3/2 + 3/2)RT). Nitrogen is diatomic with f = 5, not 3.
Why C is wrong: C gives 5RT, which would require f = 5 for both gases — treating helium as diatomic. Helium is monoatomic (f = 3).
Why D is wrong: D gives (5/2)RT, which is the internal energy of 1 mole of diatomic gas alone. The helium contribution is missing.
Free NEET study resources
Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.
Equipartition Energy: quick recall before you leave
How do you solve a Equipartition Energy question? A worked example
- 1
Given
• Average translational KE at T₁ = 27 °C = 300 K is E.• Required KE at T₂ = 2E.
- 2
Required
Temperature T₂ at which KE doubles.
- 3
Concept
Average translational KE per molecule = (3/2)kT. It is directly proportional to absolute temperature T.
- 4
Formula
⟨KE⟩ = (3/2)kT → ⟨KE⟩ ∝ T
- 5
Substitution
E₁/E₂ = T₁/T₂
E/(2E) = 300/T₂ - 6
Calculation
T₂ = 2 × 300 = 600 K = 327 °C
Note: The factor 2 in "2E" is an exact multiplier (problem-defined ratio). The integers 3 and 2 in (3/2)kT are exact counting numbers. None of these limit significant figures. - 7
Final answer
T₂ = 600 K (327 °C). Doubling absolute temperature doubles the average translational KE.
- 8
Common trap
Forgetting to convert °C to K before applying the proportionality. If you use T₁ = 27 instead of 300, you get T₂ = 54 °C — completely wrong. Always work in kelvin when using gas laws or KE formulas.
A second error: confusing this with v_rms scaling. KE ∝ T (linear), but v_rms ∝ √T. Doubling T doubles KE but only multiplies v_rms by √2. - 9
Similar NEET-style question
"The average translational KE of an oxygen molecule at temperature T is the same as that of a helium atom at temperature T. True or false? Justify using the equipartition result." (Answer: True — (3/2)kT depends only on T, not on molecular mass or type.)
---
What to remember before solving Equipartition Energy questions
Monoatomic: C_v = (3/2)R, C_p = (5/2)R, γ = 5/3. Diatomic (rigid): C_v = (5/2)R, C_p = (7/2)R, γ = 7/5. Polyatomic (rigid): C_v = 3R, C_p = 4R, γ = 4/3.
-- NCERT Class 11 Physics, Ch. 12, p. 254Which Equipartition Energy formulas do you need for NEET?
2 formulas — click to collapse
Average translational KE per molecule
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
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 · 3 formulas · 3 question patterns from its other lessons.
Equipartition Energy questions from past NEET papers
1 question from NEET 2020. Answers verified against NTA official keys. — click to collapse
NEET 2020
17/2 k_BT
21/2 k_BT
33/2 k_BT
45/2 k_BT
NTA Answer: Option 3(final)
How does NEET ask about Equipartition Energy?
1 recurring pattern 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
Sources
NCERT refs: Class 11 Physics Chapter 12, p.253 | Class 11 Physics Chapter 12, p.252
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.
Test yourself on this topic with real past-paper questions:
Practice this topic →