JEE Physics 2023 Question with Solution

Given: Mean free path at STP is $1500d$ at $273 \, \text{K}$ and pressure is kept constant.

Find: Mean free path at $373 \, \text{K}$.

The mean free path is given by

$$\lambda = \frac{RT}{\sqrt{2}\pi d^2 N_A P}$$

So, at constant pressure, $\lambda \propto T$.

Hence,

$$\frac{\lambda}{\lambda_{\text{STP}}} = \frac{T}{T_{\text{STP}}}$$

Substituting the given values,

$$\frac{\lambda}{1500d} = \frac{373}{273}$$

Therefore,

$$\lambda = 1500d \cdot \frac{373}{273} \approx 2049d$$

Thus, the mean free path at $373 \, \text{K}$ is approximately $2049d$. The correct option is D.

The solution marks option C, but the worked value is $2049d$, which matches option D.

Given: Pressure remains standard, and initial mean free path is $1500d$ at $273 \, \text{K}$.

Find: New mean free path at $373 \, \text{K}$.

Since all quantities except temperature are unchanged in

$$\lambda = \frac{RT}{\sqrt{2}\pi d^2 N_A P}$$

the ratio method is sufficient:

$$\lambda_2 = \lambda_1 \frac{T_2}{T_1}$$

Now put

$$\lambda_1 = 1500d, \quad T_2 = 373, \quad T_1 = 273$$

So,

$$\lambda_2 = 1500d \times \frac{373}{273}$$

Evaluating,

$$\frac{373}{273} \approx 1.366$$

Thus,

$$\lambda_2 \approx 1500d \times 1.366 \approx 2049d$$

Therefore, the required answer is $2049d$.