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In a cubical vessel $(1 m \times 1 m \times 1 m)$ the gas molecule of diameter $1.7 \times 10^{- 8} c m$ are at a temperature of 300k and a pressure of $10^{- 4} m m$ of mercury. The mean free path of the gas molecule is

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1 metre

4 metre

2.42 metre

1cm

answer is C.

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Detailed Solution

$d = 1.7 \times 10^{- 8} c m$

$= 1.7 \times 10^{- 10} m$

$T = 300k, p = 10^{- 4} m m, n = \frac{N}{V} = \frac{P}{K_{B} T}$

$λ = \frac{1}{\sqrt{2} π . n d^{2}}$

$λ = 2.42 m$

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