If the collision frequency of hydrogen molecules in a closed chamber at $$27°C$$ is $$Z$$, then the collision frequency of the same system at $$127°C$$ is :

Collision frequency of H$$_2$$ at 27°C is Z. Find collision frequency at 127°C.

The collision frequency $$Z \propto n^2 \sigma^2 \sqrt{T}$$ where $$n$$ is number density and $$\sigma$$ is collision cross-section. In a closed chamber at constant volume, $$n$$ is constant and thus $$Z \propto \sqrt{T}$$.

For $$T_1 = 27 + 273 = 300$$ K and $$T_2 = 127 + 273 = 400$$ K,

$$\frac{Z_2}{Z_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{400}{300}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}},$$ so $$Z_2 = \frac{2}{\sqrt{3}} Z.$$

The correct answer is Option B: $$\frac{2}{\sqrt{3}}Z$$.