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:

• A. 1.4 $\lambda$
• B. 2.1 $\lambda$
• C. 2.8 $\lambda$
• D. 3.5 $\lambda$

Hint:

Mean free path $\lambda \propto T/P$.

Answer 1

The Correct Option is A

Step 1: What mean free path depends on.
The mean free path of a gas molecule is \[ \lambda = \frac{k_B T}{\sqrt{2}\,\pi d^2 P}, \] so for the same gas it depends on temperature and pressure as $\lambda \propto \dfrac{T}{P}$.
Step 2: Write the ratio.
\[ \frac{\lambda_2}{\lambda_1} = \frac{T_2}{T_1}\times\frac{P_1}{P_2}. \]
Step 3: Change temperatures to kelvin.
$T_1 = 27^{\circ}\text{C} = 300$ K and $T_2 = 57^{\circ}\text{C} = 330$ K.
Step 4: Put in the values.
With $P_1 = 2.1$ atm and $P_2 = 1.65$ atm, \[ \frac{\lambda_2}{\lambda_1} = \frac{330}{300}\times\frac{2.1}{1.65} = 1.1 \times 1.2727 \approx 1.4. \]
Step 5: New mean free path.
$\lambda_2 \approx 1.4\,\lambda$.
Step 6: Conclusion.
The mean free path becomes about $1.4\lambda$. \[ \boxed{1.4\,\lambda} \]