The mean free path of a molecule of diameter 5 × 10^{-10} { m} at the temperature 41^{} {C} and pressure 1.38 × 10^5 { Pa}, is given as \_\_\_\_ m . (Given k_B = 1.38 × 10^{-23} { J/K} ). [28 Jan 2026 Shift 2]

Correct Answer is : 22×10−82 {2} × 10^{-8}22×10−8

Correct Answer is : 22×10−82 {2} × 10^{-8}22×10−8

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Thermodynamics Part 5

Section: Physics

Question : 19 of 86

Marks: +1, -0

5 \times 10^{-10} \text{ m}

41^{\circ} \text{C}

1.38 \times 10^5 \text{ Pa}

\_\_\_\_

k_B = 1.38 \times 10^{-23} \text{ J/K}

Solution:

Concept:

Mean free path is the average distance a molecule travels between collisions.

It is given by

\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}

Explanation:

Convert temperature to Kelvin:

T = 41 + 273 = 314\ \text{K}

Given:

d = 5 \times 10^{-10}\ \text{m}

P = 1.38 \times 10^{5}\ \text{Pa}

k_B = 1.38 \times 10^{-23}\ \text{J/K}

Substitute into the formula:

\lambda = \frac{(1.38 \times 10^{-23}) \times 314}{\sqrt{2} \times \pi \times (5 \times 10^{-10})^2 \times (1.38 \times 10^5)}

Cancel

1.38

from numerator and denominator:

\lambda = \frac{10^{-23} \times 314}{\sqrt{2} \times \pi \times 25 \times 10^{-20} \times 10^5}

Simplify powers of 10:

10^{-23} \div (10^{-20} \times 10^5) = 10^{-23} \div 10^{-15} = 10^{-8}

\lambda = \frac{314}{\sqrt{2} \times \pi \times 25 \times 10^{-8}}

Use

\pi \approx 3.14

\lambda = \frac{314}{\sqrt{2} \times 3.14 \times 25 \times 10^{-8}} = \frac{4}{\sqrt{2}} \times 10^{-8} = 2\sqrt{2} \times 10^{-8}\ \text{m}

Answer:

2\sqrt{2} \times 10^{-8}\ \text{m}

, which corresponds to option C.

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