Question

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:

• A. $\frac{\sqrt{3}}{2} Z$
• B. $\frac{4}{3} Z$
• C. $\frac{2}{\sqrt{3}} Z$
• D. $\frac{3}{4} Z$

Answer 1

Answer: (C)

$\frac{2}{\sqrt{3}} Z$

Answer 2

Assuming the mean free path remains constant, the collision frequency $f$ is proportional to the square root of temperature ($\sqrt{T}$):

$f \propto \sqrt{T}$

Given:

$T_1 = 27^\circ C = 300 \, \text{K}, \quad T_2 = 127^\circ C = 400 \, \text{K}$

The ratio of collision frequencies is:

$\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{400}{300}} = \sqrt{\frac{4}{3}}$

Therefore:

$f_2 = \sqrt{\frac{4}{3}} \cdot f_1 = \frac{2}{\sqrt{3}} f_1$