Consider two boxes containing ideal gases A and B such that their temperatures, pressures and number densities are same. The molecular size of A is half of that of B and mass of molecule A is four times that of B. If the collision frequency in gas B is $$32 \times 10^{18}/s$$ then collision frequency in gas A is _____________/s.

Collision frequency

$$z=\frac{\overline{v}}{\lambda}$$

where

$$λ=\frac{1}{\sqrt{\ 2}n\pi\ d^2}$$

So

$$z=\overline{v}(\sqrt{2}n\pi d^2)$$

Since both gases have same number density,

$$z\propto\overline{v}d^2$$

Also mean molecular speed for ideal gas is

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

So

$$z\propto\frac{d^2}{\sqrt{m}}$$

Now compare A and B:

Given molecular diameter of A is half of B,

$$d_A=\frac{d_B}{2}$$

and mass of molecule A is four times B,

$$m_A=4m_B$$

Hence

$$\frac{z_A}{z_B}=\frac{d_A^2/\sqrt{m_A}}{d_B^2/\sqrt{m_B}}$$

Substitute:

$$=\frac{\left(\frac{d_B}{2}\right)^2}{d_B^2}\cdot\frac{\sqrt{m_B}}{\sqrt{4m_B}}$$

$$=\frac{1}{4}\cdot\frac{1}{2}=\frac{1}{8}$$

Therefore

$$z_A=\frac{1}{8}z_B$$

Given

$$z_B=32\times10^{18}s^{-1}$$

so

$$z_A=\frac{1}{8}(32\times10^{18})$$

$$=4\times10^{18}s^{-1}$$