Question:
If pressure of a gas is doubled at constant temperature, mean free path becomes:
If pressure of a gas is doubled at constant temperature, mean free path becomes:
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Mean free path is inversely proportional to the number density. Doubling the pressure at constant temperature doubles the number density, which naturally cuts the distance between collisions in half.
Updated On: Jun 3, 2026
- Doubled
- Halved
- Same
- Four times
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The Correct Option is B
Solution and Explanation
Concept:
The mean free path (\( \lambda \)) is the average distance traveled by a gas molecule between two successive collisions. It is determined by the size of the molecules and the number of molecules per unit volume (density).
• Standard formula: \( \lambda = \frac{kT}{\sqrt{2}\pi d^2 P} \)
• \( k \): Boltzmann constant.
• \( T \): Absolute temperature.
• \( d \): Diameter of the gas molecule.
• \( P \): Pressure of the gas.
• Proportionality: At constant temperature (\( T \)), \( \lambda \propto \frac{1}{P} \).
Step 1: Establishing the relationship at constant temperature.
According to the kinetic theory of gases, the mean free path is inversely proportional to the pressure when temperature remains constant: \[ \lambda_1 P_1 = \lambda_2 P_2 \]
Step 2: Analyzing the pressure change.
The problem states that the pressure is doubled: \[ P_2 = 2P_1 \] Substituting this into the proportionality: \[ \lambda_2 = \lambda_1 \times \frac{P_1}{2P_1} = \frac{\lambda_1}{2} \]
Step 3: Conclusion.
Since the new mean free path \( \lambda_2 \) is half of the original \( \lambda_1 \), the mean free path becomes halved.
• Standard formula: \( \lambda = \frac{kT}{\sqrt{2}\pi d^2 P} \)
• \( k \): Boltzmann constant.
• \( T \): Absolute temperature.
• \( d \): Diameter of the gas molecule.
• \( P \): Pressure of the gas.
• Proportionality: At constant temperature (\( T \)), \( \lambda \propto \frac{1}{P} \).
Step 1: Establishing the relationship at constant temperature.
According to the kinetic theory of gases, the mean free path is inversely proportional to the pressure when temperature remains constant: \[ \lambda_1 P_1 = \lambda_2 P_2 \]
Step 2: Analyzing the pressure change.
The problem states that the pressure is doubled: \[ P_2 = 2P_1 \] Substituting this into the proportionality: \[ \lambda_2 = \lambda_1 \times \frac{P_1}{2P_1} = \frac{\lambda_1}{2} \]
Step 3: Conclusion.
Since the new mean free path \( \lambda_2 \) is half of the original \( \lambda_1 \), the mean free path becomes halved.
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