Explanation

Collision frequency of a gas molecule is given by

$$ Z \propto n \sigma \bar{v} $$

where $n$ is number density, $\sigma$ is collision cross-section and $\bar{v}$ is mean speed of molecules.

Since both gases have same temperature, pressure and number density,

$$ n_A = n_B $$

Collision cross-section $\sigma$ is proportional to square of molecular diameter. Given that molecular size of $A$ is half of that of $B$,

$$ \sigma_A = \left(\frac{1}{2}\right)^2 \sigma_B = \frac{1}{4}\sigma_B $$

Mean speed of molecules is given by

$$ \bar{v} \propto \frac{1}{\sqrt{m}} $$

Mass of molecule $A$ is four times that of $B$,

$$ \bar{v}_A = \frac{1}{\sqrt{4}} \bar{v}_B = \frac{1}{2}\bar{v}_B $$

Now, ratio of collision frequencies,

$$ \frac{Z_A}{Z_B} = \frac{\sigma_A \bar{v}_A}{\sigma_B \bar{v}_B} $$

$$ \frac{Z_A}{Z_B} = \frac{\frac{1}{4} \times \frac{1}{2}}{1} $$

$$ \frac{Z_A}{Z_B} = \frac{1}{8} $$

Given $Z_B = 32 \times 10^8\ \text{/s}$,

$$ Z_A = \frac{1}{8} \times 32 \times 10^8 $$

$$ Z_A = 4 \times 10^8\ \text{/s} $$

Hence, the collision frequency in gas $A$ is

$4 \times 10^8\ \text{/s}$

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