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:
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:
Updated On: Jan 13, 2026
- \( \frac{\sqrt{3}}{2} Z \)
- \( \frac{4}{3} Z \)
- \( \frac{2}{\sqrt{3}} Z \)
- \( \frac{3}{4} Z \)
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The Correct Option is C
Approach Solution - 1
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:
\[ f_2 = \sqrt{\frac{4}{3}} \cdot f_1 = \frac{2}{\sqrt{3}} f_1 \]
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Approach Solution -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
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
\( \frac{2}{\sqrt{3}} Z \)
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