The mean free path for a gas at temperature 3 | vedclass

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(D) The mean free path $\lambda$ of a gas molecule is given by the formula: $\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}$.From this relation, we can se

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$4 	imes 10^{-7} \ m$

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(D) The mean free path $\lambda$ of a gas molecule is given by the formula: $\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}$.From this relation, we can see that $\lambda \propto \frac{T}{P}$.Therefore, the ratio of the mean free paths at two different states is given by: $\frac{\lambda_2}{\lambda_1} = \frac{T_2}{T_1} 	imes \frac{P_1}{P_2}$.Given values are: $T_1 = 300 \ K$, $P_1 = 600 \ 	ext{torr}$, $\lambda_1 = 10^{-7} \ m$, $T_2 = 400 \ K$, $P_2 = 200 \ 	ext{torr}$.Substituting these values into the ratio formula:$\lambda_2 = \lambda_1 	imes \frac{T_2}{T_1} 	imes \frac{P_1}{P_2} = 10^{-7} 	imes \frac{400}{300} 	imes \frac{600}{200}$.$\lambda_2 = 10^{-7} 	imes \frac{4}{3} 	imes 3 = 10^{-7} 	imes 4 = 4 	imes 10^{-7} \ m$.

Temperature	Pressure thermometer $A$	Pressure thermometer $B$
Triple-point of water	$1.250 \times 10^{5} \; Pa$	$0.200 \times 10^{5} \; Pa$
Normal melting point of sulphur	$1.797 \times 10^{5} \; Pa$	$0.287 \times 10^{5} \; Pa$

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

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