Question:
If the mean free path of a nitrogen molecule in a vessel containing nitrogen at a pressure of 2.1 atm and a temperature of $27^{\circ}C$ is $\lambda$, then its mean free path at a pressure of 1.65 atm and a temperature of $57^{\circ}C$ is:
If the mean free path of a nitrogen molecule in a vessel containing nitrogen at a pressure of 2.1 atm and a temperature of $27^{\circ}C$ is $\lambda$, then its mean free path at a pressure of 1.65 atm and a temperature of $57^{\circ}C$ is:
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Mean free path $\lambda \propto T/P$.
Updated On: Jun 6, 2026
- 1.4 $\lambda$
- 2.1 $\lambda$
- 2.8 $\lambda$
- 3.5 $\lambda$
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The Correct Option is A
Solution and Explanation
Step 1: Dependence.
For one gas, the mean free path follows $\lambda \propto \dfrac{T}{P}$.
Step 2: Ratio.
$\dfrac{\lambda_2}{\lambda_1} = \dfrac{T_2}{T_1}\cdot\dfrac{P_1}{P_2}$.
Step 3: Values.
$T_1 = 300$ K, $T_2 = 330$ K, $P_1 = 2.1$ atm, $P_2 = 1.65$ atm. $\dfrac{\lambda_2}{\lambda_1} = \dfrac{330}{300}\times\dfrac{2.1}{1.65} = 1.1\times1.2727 \approx 1.4$.
Step 4: Result.
$\lambda_2 \approx 1.4\,\lambda$. \[ \boxed{1.4\,\lambda} \]
For one gas, the mean free path follows $\lambda \propto \dfrac{T}{P}$.
Step 2: Ratio.
$\dfrac{\lambda_2}{\lambda_1} = \dfrac{T_2}{T_1}\cdot\dfrac{P_1}{P_2}$.
Step 3: Values.
$T_1 = 300$ K, $T_2 = 330$ K, $P_1 = 2.1$ atm, $P_2 = 1.65$ atm. $\dfrac{\lambda_2}{\lambda_1} = \dfrac{330}{300}\times\dfrac{2.1}{1.65} = 1.1\times1.2727 \approx 1.4$.
Step 4: Result.
$\lambda_2 \approx 1.4\,\lambda$. \[ \boxed{1.4\,\lambda} \]
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