Kinetic Interpretation Temperature

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 single idea behind this topic: temperature is not what a thermometer reads — it is the average translational kinetic energy of a molecule, made measurable.

Here is the connection that NEET tests. Start from the pressure expression that kinetic theory derives:

PV = (1/3) N m v²_rms

Compare this with the ideal gas result PV = NkT. Equating the right-hand sides gives:

(1/2) m v²_rms = (3/2) kT

This is the kinetic interpretation of temperature (NCERT Class 11 Physics, Chapter 13, page 6). The left side is the average translational kinetic energy of one molecule. The right side says it equals (3/2)kT. Temperature T is therefore a direct, linear measure of average translational KE — nothing more.

What this means for problem-solving:

The average translational KE per molecule is (3/2)kT. It depends only on T — not on the gas species, not on molar mass, not on pressure.
At the same temperature, a helium atom and a nitrogen molecule have the same average translational KE. Their speeds differ (lighter molecules move faster), but their KE is identical.
The constant k (Boltzmann constant, 1.38 × 10⁻²³ J/K) bridges the microscopic and macroscopic worlds. Per mole, the energy is (3/2)RT.

The high-frequency confusion NEET exploits (pattern: average KE at thermal equilibrium): distractors offer (3/2)RT instead of (3/2)kT. The distinction is per-molecule (kT) versus per-mole (RT). When a stem says "average KE of a molecule," the answer uses k. When it says "total KE of one mole," the answer uses R. Mixing them up costs 5 marks (4 lost + 1 penalty).

Watch-out: at a given temperature, average translational KE is the same for all ideal gas molecules — this is species-independent. Any option that makes KE depend on molecular mass for translational motion at thermal equilibrium is wrong.

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

The average translational kinetic energy of an ideal gas molecule at temperature T is:

MCQ 2Easy RecallPractice

The Boltzmann constant k is related to the gas constant R by:

MCQ 3Easy RecallPractice

At a given temperature T, the average translational kinetic energy of molecules is:

MCQ 4Direct ApplicationPractice

The average translational kinetic energy of a molecule at 300 K is:

MCQ 5Direct ApplicationPractice

A gas mixture contains helium (molar mass 4 g/mol) and argon (molar mass 40 g/mol) at thermal equilibrium at temperature T. What is the ratio of average translational KE of a helium atom to that of an argon atom?

MCQ 6Direct ApplicationPractice

The total translational kinetic energy of all molecules in 2 moles of an ideal gas at temperature T is:

MCQ 7Concept TrapPractice

Two containers hold oxygen (O₂) and hydrogen (H₂) at the same temperature. Which statement is correct?

MCQ 8CalculationPractice

At what temperature will the average translational kinetic energy of an ideal gas molecule equal 1.0 eV?

Quick recall before you leave

Worked Example

Pattern: Average translational KE per molecule = (3/2)kT (pattern: straightforward application of the kinetic interpretation of temperature).

1
Given
A vessel contains an ideal gas at temperature T = 600 K. Find the average translational kinetic energy per molecule. Given: T = 600 K, k = 1.38 × 10⁻²³ J/K.
2
Required
Average translational KE per molecule, ⟨KE⟩.
3
Concept
The kinetic interpretation of temperature states that the average translational KE of a molecule depends only on absolute temperature, via ⟨KE⟩ = (3/2)kT (NCERT Class 11 Physics, Chapter 13, page 6). No information about the gas species is needed.
4
Formula
⟨KE⟩ = (3/2)kT
5
Substitution
⟨KE⟩ = (3/2)(1.38 × 10⁻²³ J/K)(600 K)
6
Calculation
First: kT = 1.38 × 10⁻²³ × 600 = 1.38 × 6.00 × 10⁻²¹ = 8.28 × 10⁻²¹ J. Then: (3/2) × 8.28 × 10⁻²¹ = 1.242 × 10⁻²⁰ J. Note on exact constants: the factor 3/2 is an exact mathematical constant (from 3 translational degrees of freedom). It does not limit significant figures. The given values k and T each have 3 significant figures, so the answer is reported to 3 significant figures.
7
Final answer
⟨KE⟩ = 1.24 × 10⁻²⁰ J
8
Common trap
Using R instead of k. If you mistakenly write ⟨KE⟩ = (3/2)RT = (3/2)(8.314)(600) = 7483 J, you get the energy per mole, not per molecule. The stem says "per molecule" — that means k.
9
Similar NEET-style question
At what temperature will the average translational KE of an ideal gas molecule be twice the value at 300 K? Quick solve: since ⟨KE⟩ ∝ T, doubling the KE requires doubling T. Answer: 600 K.

Before solving, remember these

Formula

Kinetic interpretation of temperature

Average translational KE per molecule: <½ m v²> = (3/2) k T. So v_rms = √(3kT/m) = √(3RT/M). Temperature is a measure of microscopic kinetic energy.

-- NCERT, p. 6

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)

Sources

NCERT refs: Class 11 Physics Chapter 13, p.6

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