Assertion: Mean free path of a gas molecule v | vedclass

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Question

(A) The mean free path $\lambda$ of a gas molecule is the average distance between two successive collisions.It is given by the formula $\lambda = \fr

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If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

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(A) The mean free path $\lambda$ of a gas molecule is the average distance between two successive collisions.It is given by the formula $\lambda = \frac{k T}{\sqrt{2} \pi \sigma^2 P}$,where $k$ is the Boltzmann constant,$T$ is the temperature,$\sigma$ is the molecular diameter,and $P$ is the pressure.Since density $\rho = \frac{m}{V} = \frac{M P}{R T}$,we can express the mean free path in terms of density as $\lambda = \frac{m}{\sqrt{2} \pi \sigma^2 \rho}$,where $\rho$ is the density of the gas.From these relations,it is clear that $\lambda \propto \frac{1}{P}$ and $\lambda \propto \frac{1}{\rho}$.Therefore,the mean free path is inversely proportional to both the pressure and the density of the gas.Since the Assertion states that it varies inversely with density and the Reason states it varies inversely with pressure,both are correct.Furthermore,the relationship between pressure and density (at constant temperature) explains why the mean free path depends on both in an inverse manner,making the Reason a correct explanation for the Assertion.

$Assertion:$ Mean free path of a gas molecule varies inversely as the density of the gas.
$Reason:$ Mean free path varies inversely as the pressure of the gas.

Similar Questions

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$Assertion:$ Mean free path of a gas molecule varies inversely as the density of the gas.$Reason:$ Mean free path varies inversely as the pressure of the gas.