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. 8Equipartition Energy
8 MCQs2 revision cards9-step worked example
Source: NCERT Kinetic TheoryPYQ coverage: NEET 2020, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026
Lesson
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 8): 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 9)
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.
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
According to the law of equipartition of energy, each quadratic degree of freedom of a molecule in thermal equilibrium contributes an average energy of:
MCQ 2Easy RecallPractice
A rigid diatomic molecule has how many degrees of freedom at moderate temperatures?
MCQ 3Easy RecallPractice
The ratio of specific heats γ = Cₚ/Cᵥ for a monoatomic ideal gas is:
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?
MCQ 5Direct ApplicationPractice
For a rigid diatomic ideal gas, the molar specific heat at constant volume Cᵥ is:
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)
MCQ 7CalculationPractice
An ideal gas has Cₚ = (7/2)R. What is the number of degrees of freedom of each molecule of this gas?
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:
Quick recall before you leave
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.) ---
Before solving, remember these
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, p. 9Formulas
5 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
Ideal gas equation
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
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
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
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.
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
10.156 kg
20.125 kg
30.144 kg
40.116 kg
NTA Answer: Option 4(final)
NEET 2024Revised key
1P3 > P2 > P1
2P1 > P3 > P2
3P2 > P1 > P3
4P1 > P2 > P3
NTA Answer: Option 4(revised_final)
NEET 2023
13295°C
23097 K
3223 K
4669°C
NTA Answer: Option 1(final)
NEET 2022
15.6 m3
25.6 × 106 m3
35.6 × 103 m3
45.6 × 10–3 m3
NTA Answer: Option 1(final)
NEET 2020
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
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)*pi*n*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)
Sources
NCERT refs: Class 11 Physics Chapter 12, p.8 | Class 11 Physics Chapter 12, p.9
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