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(B) The mean free path $\bar{l}$ of a gas molecule is given by the formula:$\bar{l} = \frac{1}{\sqrt{2} \pi d^{2} n}$where $d$ is the molecular diamet

Vedclass · Accepted Answer

$4:1$

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(B) The mean free path $\bar{l}$ of a gas molecule is given by the formula:$\bar{l} = \frac{1}{\sqrt{2} \pi d^{2} n}$where $d$ is the molecular diameter and $n$ is the number density.Given that the gases are under identical conditions of temperature,pressure,and volume,the number density $n$ remains constant for both gases.From the formula,we observe that $\bar{l} \propto \frac{1}{d^{2}}$.Therefore,the ratio of the mean free paths is:$\frac{\bar{l}_{1}}{\bar{l}_{2}} = \left(\frac{d_{2}}{d_{1}}\right)^{2}$Given $d_{1} = 1\,\mathring{A}$ and $d_{2} = 2\,\mathring{A}$,we substitute these values:$\frac{\bar{l}_{1}}{\bar{l}_{2}} = \left(\frac{2}{1}\right)^{2} = \frac{4}{1}$Thus,the ratio of the mean free paths is $4:1$.

Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters $1\,\mathring{A}$ and $2\,\mathring{A}$. The gases may be considered under identical conditions of temperature,pressure,and volume.

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