The mean free path of molecules of a certain gas at STP is 1500 ~d , where d is the diameter of the gas molecules. While maintaining the standard pressure, the mean free path of the molecules at 373 ~K is approximately:

Heat and Thermodynamics PYQ — JEE Main 2023 · 13 Apr · Shift 2 · Q54 with Solution

View written solutionFree

Correct answer: C

For an ideal gas,

\lambda = \frac{1}{\sqrt{2}\,\pi d^2 n}

where $d$ is molecular diameter and $n$ is number density.

Using the ideal gas law, number density is

n = \frac{P}{kT}

So,

\lambda \propto \frac{1}{n} \propto \frac{T}{P}

Pressure is maintained constant, so

\lambda \propto T

At STP, temperature is

T_1 = 273\,\text{K}

and mean free path is

\lambda_1 = 1500d

At the new temperature,

T_2 = 373\,\text{K}

Thus,

\frac{\lambda_2}{\lambda_1} = \frac{T_2}{T_1} = \frac{373}{273}

So,

\lambda_2 = 1500d \cdot \frac{373}{273}

\lambda_2 = 1500d \times 1.3663 \approx 2049d

The closest option is:

\boxed{2049d}

So the correct option is C.