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

The Correct Option is C

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:

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

Answer 2

Step 1: Recall the formula for collision frequency
The collision frequency \( Z \) for a gas is proportional to the number density \( n \) and the average speed of the molecules \( \bar{v} \): \[ Z \propto n \bar{v}. \] Since the gas is in a closed chamber, its volume is constant, hence the number density \( n \) remains constant. Therefore, \[ Z \propto \bar{v}. \]

Step 2: Express mean speed in terms of temperature
The average (mean) speed of gas molecules is proportional to the square root of temperature: \[ \bar{v} \propto \sqrt{T}. \] Hence, \[ Z \propto \sqrt{T}. \] Therefore, for two temperatures \(T_1\) and \(T_2\), \[ \frac{Z_2}{Z_1} = \sqrt{\frac{T_2}{T_1}}. \]

Step 3: Substitute the given temperatures
Given: \( T_1 = 27^\circ C = 300\,K \), \( T_2 = 127^\circ C = 400\,K. \) \[ \frac{Z_2}{Z_1} = \sqrt{\frac{400}{300}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}. \] Hence, \[ Z_2 = \frac{2}{\sqrt{3}} Z. \]

Step 4: Final conclusion
When the temperature increases from 27°C to 127°C, the collision frequency increases by a factor of \( \frac{2}{\sqrt{3}} \).

Final answer

\( \frac{2}{\sqrt{3}} Z \)