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 Class 11 Physics, Ch. 12, p. 255Mean Free Path
8 MCQs2 revision cards9-step worked example
Source: NCERT Kinetic TheoryPYQ coverage: NEET 2020, 2026Official key: NTA-verifiedLast updated: 24 Sep 2026
Mean Free Path, explained for NEET
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 12, page 253).
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.
Can you answer these Mean Free Path 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:
Show answer and why every option is right or wrong
Answer: D. Mean free path is defined as the average distance a molecule travels between two successive collisions, as stated in NCERT Class 11 Physics, Chapter 12, page 253.
Why A is wrong: A describes the distance to the wall, which is not the definition of mean free path. Mean free path concerns molecule-molecule collisions, not molecule-wall distance.
Why B is wrong: B has no physical meaning in kinetic theory. Mean free path is not referenced to the container centre.
Why C is wrong: C describes intermolecular spacing at rest, not the path traversed between collisions. These are distinct concepts — spacing is a static quantity, mean free path is a kinetic one.
MCQ 2Easy RecallPractice
In the expression λ = 1/(√2 · π · n · d²), the quantity n represents:
Show answer and why every option is right or wrong
Answer: C. In the mean free path formula, n is the number density — the number of molecules per unit volume (SI unit: m⁻³), as defined in NCERT Class 11 Physics, Chapter 12, page 253.
Why A is wrong: A confuses number density with molar concentration. The formula uses molecule count per volume, not mole count per volume. Using moles would require an extra Avogadro factor.
Why B is wrong: B describes the total molecule count N, not the density n = N/V. Using total count without dividing by volume gives a dimensionally incorrect result.
Why D is wrong: D is a fixed constant (6.022 × 10²³ mol⁻¹), not a variable. Number density n varies with the state of the gas.
MCQ 3Easy RecallPractice
The SI unit of mean free path is:
Show answer and why every option is right or wrong
Answer: D. Mean free path is a distance (average path length between collisions), so its SI unit is the metre (m).
Why A is wrong: A (m⁻¹) is the unit of number density per length or wavenumber, not a length. Mean free path is a distance.
Why B is wrong: B (m²) is an area. The d² in the denominator has area dimensions, but after all factors combine, λ has dimensions of length.
Why C is wrong: C is incorrect — λ has dimensions of length [L]. A dimensionless quantity would require all length dimensions to cancel, which they do not in this formula.
MCQ 4Direct ApplicationPractice
If the number density of gas molecules is doubled while the molecular diameter remains unchanged, the mean free path becomes:
Show answer and why every option is right or wrong
Answer: A. λ = 1/(√2 · π · n · d²). Since λ ∝ 1/n, doubling n halves λ. Direct substitution with d held constant.
Why B is wrong: B reverses the proportionality. λ is inversely proportional to n, so increasing n decreases λ — it does not increase it.
Why C is wrong: C would require n to quadruple (or d to double). Doubling n alone gives a factor of 1/2, not 1/4.
Why D is wrong: D ignores the n-dependence entirely. λ depends on number density; changing n at fixed d changes λ.
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:
Show answer and why every option is right or wrong
Answer: B. λ ∝ 1/d². Doubling d gives λ_new = λ/(2²) = λ/4. The d-squared dependence means diameter changes have a quadratic effect on mean free path.
Why A is wrong: A treats d as linear (λ ∝ 1/d), which ignores the squared dependence. The cross-sectional collision area scales as d², so λ ∝ 1/d².
Why C is wrong: C implies λ increases when d increases. Since λ ∝ 1/d², increasing d always decreases λ.
Why D is wrong: D implies λ increases by a factor of 4. The relationship is inverse-square, so doubling d decreases λ by 4, not increases it.
MCQ 6Direct ApplicationPractice
For an ideal gas at constant temperature, the pressure is tripled. The new mean free path is:
Show answer and why every option is right or wrong
Answer: B. At constant T, n = P/(kT), so n ∝ P. Tripling P triples n. Since λ ∝ 1/n, the new mean free path is λ/3.
Why A is wrong: A reverses the relationship. Increasing pressure increases number density, which decreases mean free path — not increases it.
Why C is wrong: C applies a squared factor (1/9), which would be correct if d tripled instead of P. Pressure enters linearly through n ∝ P at fixed T.
Why D is wrong: D applies a square-root factor, confusing this with formulas where √T appears (such as rms speed). Pressure enters the mean free path linearly via number density.
MCQ 7Concept TrapPractice
For an ideal gas held at constant pressure, the temperature is increased from T to 4T. The mean free path:
Show answer and why every option is right or wrong
Answer: A. At constant P, n = P/(kT), so n ∝ 1/T. Increasing T by a factor of 4 reduces n by a factor of 4. Since λ ∝ 1/n, λ increases by a factor of 4. The new mean free path is 4λ.
Why B is wrong: B incorrectly assumes λ is independent of temperature. While T does not appear explicitly in the formula, it affects n through the ideal gas relation at constant pressure.
Why C is wrong: C applies a square-root scaling (√4 = 2), confusing this with the rms speed formula where v ∝ √T. Mean free path at constant P scales linearly with T, not as √T.
Why D is wrong: D reverses the direction. Increasing temperature at constant pressure decreases number density, which increases mean free path — not decreases it.
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:
Show answer and why every option is right or wrong
Answer: C. n = P/(kT). New n = (P/2)/(k · 2T) = P/(4kT) = n₁/4. Since λ ∝ 1/n, λ₂ = 4λ₁. Both changes reduce number density — halving P halves n, doubling T halves n again — giving a combined factor of 4.
Why A is wrong: A assumes the two changes cancel. Halving P halves n (increases λ by 2), and doubling T halves n again (increases λ by another 2). The effects multiply, not cancel.
Why B is wrong: B accounts for only one of the two changes (either halving P or doubling T alone gives factor 2). Both changes must be applied together: (1/2) × (1/2) = 1/4 for n, so λ increases by 4.
Why D is wrong: D reverses the direction of both changes. Both halving pressure and doubling temperature reduce number density, which increases mean free path.
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Mean Free Path: quick recall before you leave
How do you solve a Mean Free Path question? A 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.
What to remember before solving Mean Free Path questions
Which Mean Free Path formulas do you need for NEET?
1 formula — click to collapse
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
More in Kinetic Theory: 2 exam traps and mistakes · 4 formulas · 3 question patterns from its other lessons.
Mean Free Path questions from past NEET papers
2 questions from NEET 2020, 2026. Answers verified against NTA official keys. — click to collapse
NEET 2026
1n_A = n_B
2n_A = 2n_B
3n_A = (1/4) n_B
4n_A = (1/2) n_B
NTA Answer: Option 4(final)
NEET 2020
The mean free path for a gas, with molecular diameter d and number density n can be expressed as :
11/(√2 n²π²d²)
21/(√2 nπd)
31/(√2 nπd²)
41/(√2 n²πd²)
NTA Answer: Option 3(final)
How does NEET ask about Mean Free Path?
1 recurring pattern from past papers — click to collapse
Mean free path scaling with n, d. lambda = 1/(sqrt(2)*pi*n*d^2).
InterpretationEasy
Common distractors
ignores d squared
Treats d linearly
Sources
NCERT refs: Class 11 Physics Chapter 12, p.253
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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