The mean free path of a molecule of diameter 5 × 10⁻¹⁰ m at the temperature 41° C and pressure 1.38 × 10⁵ Pa, is given as ________ m. (Given kB = 1.38 × 10⁻²³ J/K). (a) 2√(2) × 10⁻¹⁰ (b) 10√(2) × 10⁻⁸ (c) 2√(2) × 10⁻⁸ (d) 2 × 10⁻⁸

Answer: (c). λ = (kBT)/(√(2) π d² P), with T = 41 + 273 = 314 K. kBT/P = 1.38×10⁻²³×3141.38×10⁵ = 3.14×10⁻²⁶ m³ λ = 3.14×10⁻²⁶√(2)×3.14×25×10⁻²⁰ = 4×10⁻⁸√(2) λ = 2√(2)×10⁻⁸ m So the answer is (c).

The mean free path of a molecule of diameter 5 × 10⁻¹⁰ m at the… | JEE Main 2026 (28 Jan, Shift 2)

Q2JEE Main 2026 · 28 JAN · SHIFT 2Medium

Mean Free Path & Real Gases

The mean free path of a molecule of diameter $5 \times 10^{-10}$ m at the temperature $41^\circ$ C and pressure $1.38 \times 10^{5}$ Pa, is given as ________ m. (Given $k_B = 1.38 \times 10^{-23}$ J/K).

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Answer: (c)

$\lambda = \dfrac{k_BT}{\sqrt{2}\,\pi d^2 P}$ , with $T = 41 + 273 = 314$ K.
$\dfrac{k_BT}{P} = \dfrac{1.38\times10^{-23}\times314}{1.38\times10^{5}} = 3.14\times10^{-26}\text{ m}^3$
$\lambda = \dfrac{3.14\times10^{-26}}{\sqrt{2}\times3.14\times25\times10^{-20}} = \dfrac{4\times10^{-8}}{\sqrt{2}}$
$\lambda = 2\sqrt{2}\times10^{-8}$ m
So the answer is (c).

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Mean Free Path & Real Gases question from JEE Main 2026 (28 Jan, Shift 2)

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