What is the ratio of the mean free paths of the molecules of two gases $A$ and $B$ having the molecular diameters $2 \mathrm{~A}^0$ and $3 \mathrm{~A}^0$ respectively under the identical conditions of pressure, temperature and volume?

The mean free path $\lambda$ of a gas molecule is given by the formula $\lambda = \dfrac{k_B T}{\sqrt{2} \pi d^2 P}$, where $k_B$ is the Boltzmann constant, $T$ is the temperature, $d$ is the molecular diameter, and $P$ is the pressure.

Under identical conditions of pressure, temperature, and volume, the variables $k_B$, $T$, and $P$ are constant for both gases $A$ and $B$.

Therefore, the mean free path is inversely proportional to the square of the molecular diameter: $\lambda \propto \dfrac{1}{d^2}$.

Given the diameters $d_A = 2 \text{ \AA}$ and $d_B = 3 \text{ \AA}$, the ratio of the mean free paths is:

$\dfrac{\lambda_A}{\lambda_B} = \dfrac{d_B^2}{d_A^2} = \left( \dfrac{3}{2} \right)^2 = \dfrac{9}{4}$.

Thus, $\dfrac{\lambda_A}{\lambda_B} = 9:4$.

Answer: $\dfrac{\lambda_A}{\lambda_B}=9: 4$