Q142 / 160 · TS EAMCET 2021 (05 Aug Shift 1) · Physics

The mean free path for a gas at temperature $300 K$ and pressure $600$ torr is $10^{- 7} m .$ The mean free path of the gas at a temperature $400 K$ and pressure $200$ torr will be

Solution

The mean free path in terms of pressure,

λ = \frac{k T}{\sqrt{2} d^{2} P}

,
here

k

is the Boltzmann constant, now make the proper relation between pressure and temperature of the gas,

\frac{λ_{1}}{λ_{2}} = \frac{\frac{k T_{1}}{\sqrt{2} d^{2} P_{1}}}{\frac{k T_{2}}{\sqrt{2} d^{2} P_{2}}} = \frac{\frac{T_{1}}{P_{1}}}{\frac{T_{2}}{P_{2}}} = \frac{T_{1} P_{2}}{P_{1} T_{2}} = \frac{300 \times 600}{400 \times 200} \Rightarrow \frac{10^{- 7}}{λ_{2}} = \frac{300 \times 600}{400 \times 200} \Rightarrow λ_{2} = \frac{10^{- 7} \times 8}{18} = 4.44 \times 10^{- 8} m

Hence, the correct answer is option: 'B'.