Mean Free Path

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

Mean free path is the average distance a gas molecule travels between two successive collisions. The formula is:

λ = 1 / (√2 · π · n · d²)

where n is the number density (molecules per unit volume) and d is the molecular diameter (NCERT Class 11 Physics, Chapter 13, page 10).

The high-frequency confusion on this topic is treating molecular diameter d as a linear factor. It enters as d² — so doubling the molecular diameter reduces the mean free path by a factor of four, not two. This d-squared dependence is the single detail NEET questions probe most often.

What λ depends on — and what it does not. From the formula, λ depends on number density and molecular diameter. It does not directly depend on temperature or pressure as independent variables. However, for an ideal gas at fixed pressure, increasing temperature reduces n (since n = P/kT), which increases λ. At fixed temperature, increasing pressure increases n, which decreases λ. NEET questions test whether you can trace these indirect dependencies through the ideal gas relation PV = NkT.

Key proportionalities to lock in:

λ ∝ 1/n (inverse with number density)
λ ∝ 1/d² (inverse-square with molecular diameter)
At constant P: λ ∝ T (via n = P/kT)
At constant T: λ ∝ 1/P (via n = P/kT)

Watch out: When a question says "the number density is doubled," that is a direct substitution — λ halves. When it says "the pressure is doubled at constant temperature," you must first recognise that n doubles (since n ∝ P at fixed T), and then λ halves. Same result, different reasoning path. NEET distractors exploit aspirants who skip the intermediate step.

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 mean free path of a gas molecule is the average distance between:

MCQ 2Easy RecallPractice

In the expression λ = 1/(√2 · π · n · d²), the quantity *n* represents:

MCQ 3Easy RecallPractice

The SI unit of mean free path is:

MCQ 4Direct ApplicationPractice

If the number density of gas molecules is doubled while the molecular diameter remains unchanged, the mean free path becomes:

MCQ 5Direct ApplicationPractice

If the molecular diameter of a gas is doubled while the number density is kept the same, the mean free path changes by a factor of:

MCQ 6Direct ApplicationPractice

For an ideal gas at constant temperature, the pressure is tripled. The new mean free path is:

MCQ 7Concept TrapPractice

For an ideal gas held at constant pressure, the temperature is increased from T to 4T. The mean free path:

MCQ 8CalculationPractice

A gas has mean free path λ₁ at pressure P and temperature T. If the pressure is halved and the temperature is doubled simultaneously, the new mean free path λ₂ is:

Quick recall before you leave

Worked Example

1
Given
An ideal gas at temperature T has molecular diameter d and number density n, giving mean free path λ. The gas is compressed isothermally until the number density becomes 3n. Find the new mean free path.
2
Required
New mean free path λ₂ in terms of λ.
3
Concept
Mean free path formula: λ = 1/(√2 · π · n · d²). At constant temperature with the same gas, d is unchanged. Only n changes.
4
Formula
λ = 1/(√2 · π · n · d²)
5
Substitution
λ₁ = 1/(√2 · π · n · d²) λ₂ = 1/(√2 · π · 3n · d²)
6
Calculation
λ₂ / λ₁ = [1/(√2 · π · 3n · d²)] / [1/(√2 · π · n · d²)] = n / (3n) = 1/3 Note: The factor 3 in "3n" is an exact counting multiplier and does not affect significant figures.
7
Final answer
λ₂ = λ/3 The mean free path reduces to one-third of its original value.
8
Common trap
The d-squared trap: if the question had changed diameter instead of number density, a factor-of-2 change in d would give a factor-of-4 change in λ (not 2). Always check whether the question varies n or d — the scaling laws differ (linear vs. quadratic).
9
Similar NEET-style question
"If both the molecular diameter and number density of a gas are doubled, by what factor does the mean free path change?" Answer: λ_new = 1/(√2 · π · 2n · (2d)²) = 1/(√2 · π · 2n · 4d²) = λ/8. The factor is 1/8.

Before solving, remember these

Formula

Mean free path

λ = 1/(√2 π n d²), where n is number density and d is molecular diameter. Average distance between consecutive collisions.

-- NCERT, p. 10

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.10

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

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Kinetic Theory Assumptions Rms Speed Gas

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Mean Free Path

Lesson

Practice MCQs

Quick recall before you leave

λ = 1/(√2 · π · n · d²)

PV = NkT → n = N/V = P/(kT)

Worked Example

Before solving, remember these

Formulas

Valid when

Valid when

Valid when

Valid when

Valid when

Exam Traps & Common Mistakes

When it triggers

How to avoid

Correction

Past Year Questions

How NEET usually asks this

Common distractors

Common distractors

Common distractors

Common distractors

Sources

Free NEET study resources