Question:
Two gases having molecular diameters \( D_1 \) and \( D_2 \), and mean free paths \( \lambda_1 \) and \( \lambda_2 \), respectively, are trapped separately in identical containers. If \( D_2 = 2D_1 \), then \( \lambda_1 / \lambda_2 = \) ..........
Two gases having molecular diameters \( D_1 \) and \( D_2 \), and mean free paths \( \lambda_1 \) and \( \lambda_2 \), respectively, are trapped separately in identical containers. If \( D_2 = 2D_1 \), then \( \lambda_1 / \lambda_2 = \) ..........
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The mean free path is inversely proportional to the square of the molecular diameter.
Updated On: Dec 12, 2025
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Correct Answer: 4
Solution and Explanation
Step 1: Use the relationship between mean free path and molecular diameter.
The mean free path \( \lambda \) is inversely proportional to the square of the molecular diameter \( D \). Mathematically, this is expressed as: \[ \lambda \propto \frac{1}{D^2} \]
Step 2: Relate the mean free paths of the two gases.
Given that \( D_2 = 2D_1 \), we can write the ratio of the mean free paths: \[ \frac{\lambda_1}{\lambda_2} = \left( \frac{D_2}{D_1} \right)^2 = \left( \frac{2}{1} \right)^2 = 4 \]
Step 3: Conclusion.
Thus, \( \lambda_1 / \lambda_2 = 4 \).
The mean free path \( \lambda \) is inversely proportional to the square of the molecular diameter \( D \). Mathematically, this is expressed as: \[ \lambda \propto \frac{1}{D^2} \]
Step 2: Relate the mean free paths of the two gases.
Given that \( D_2 = 2D_1 \), we can write the ratio of the mean free paths: \[ \frac{\lambda_1}{\lambda_2} = \left( \frac{D_2}{D_1} \right)^2 = \left( \frac{2}{1} \right)^2 = 4 \]
Step 3: Conclusion.
Thus, \( \lambda_1 / \lambda_2 = 4 \).
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