Kinetic Theory

JEE Main level, single correct, medium

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Question 5 of 168

The mean free path of a molecule of diameter 5×10−10 m5 \times 10^{-10} \mathrm{~m} at the temperature 41∘C41^{\circ} \mathrm{C} and pressure 1.38×105 Pa1.38 \times 10^{5} \mathrm{~Pa}, is given as ____\_\_\_\_ m.
(Given kB=1.38×10−23 J/Kk_{B}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}).

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Answer: (B) 22×10−82 \sqrt{2} \times 10^{-8}

Mean free path is calculated using λ=kBT2πd2P\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}
Given: d=5×10−10d = 5 \times 10^{-10} m, T=41°C=314T = 41°C = 314 K, P=1.38×105P = 1.38 \times 10^5 Pa, kB=1.38×10−23k_B = 1.38 \times 10^{-23} J/K
λ=1.38×10−23×3142×π×(5×10−10)2×1.38×105\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}
=1.38×10−23×3142×π×25×10−20×1.38×105= \frac{1.38 \times 10^{-23} \times 314}{\sqrt{2} \times \pi \times 25 \times 10^{-20} \times 1.38 \times 10^5}
Numerator: 433.32×10−23433.32 \times 10^{-23}
Denominator: 1.414×3.14159×34.5×10−15=153.18×10−151.414 \times 3.14159 \times 34.5 \times 10^{-15} = 153.18 \times 10^{-15}
λ=433.32×10−23153.18×10−15=2.83×10−8\lambda = \frac{433.32 \times 10^{-23}}{153.18 \times 10^{-15}} = 2.83 \times 10^{-8} m = 22×10−82\sqrt{2} \times 10^{-8} m

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