Kerala KEAM 5 June 2024 Solved Paper

The mean free path ($\lambda$) of a gas molecule is given by the following relation:$$\lambda = \frac{1}{\sqrt{2} \, n \sigma}$$where:- $n$ is the number of molecules per unit volume,- $\sigma$ is the effective collision cross-section, which depends on the molecular diameter ($d$).The effective cross-section is proportional to the square of the molecular diameter:$$\sigma \propto d^2$$Since both gases have the same number of molecules per unit volume, the ratio of their mean free paths will be inversely proportional to the ratio of their molecular diameters squared:$$\frac{\lambda_1}{\lambda_2} = \left( \frac{d_2}{d_1} \right)^2$$Given that the molecular diameters are in the ratio 1 : 3, we have:$$\frac{\lambda_1}{\lambda_2} = \left( \frac{3}{1} \right)^2 = 9$$Thus, the ratio of their mean free paths is:$$\boxed{9 : 1}$$ Quick Tip: The mean free path is inversely proportional to the square of the molecular diameter. Therefore, if the molecular diameters of the gases are in the ratio 1 : 3, their mean free paths will be in the ratio 9 : 1.