Question

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

• A. 1500d
• B. 750d
• C. 2049d
• D. 1098d

Answer 1

The Correct Option is C

To solve this problem, we need to find the mean free path of gas molecules at 373K while keeping the pressure constant, given that the mean free path at STP (Standard Temperature and Pressure) is 1500d, where d is the diameter of the gas molecules.

The mean free path \lambda of a gas is given by the formula:

\lambda = \frac{k \cdot T}{\sqrt{2} \cdot \pi \cdot d^2 \cdot P}

where:

• k is the Boltzmann constant,
• T is the absolute temperature,
• d is the diameter of the molecules, and
• P is the pressure.

At STP, the conditions are 273K for temperature and 1 atm for pressure, leading to the mean free path formula:

\lambda_{\text{STP}} = \frac{k \cdot 273}{\sqrt{2} \cdot \pi \cdot d^2 \cdot 1}

Given \lambda_{\text{STP}} = 1500d, we can express it as:

\frac{k \cdot 273}{\sqrt{2} \cdot \pi \cdot d^2} = 1500d

Now, at 373K with the same pressure, the mean free path \lambda_{\text{373K}} becomes:

\lambda_{\text{373K}} = \frac{k \cdot 373}{\sqrt{2} \cdot \pi \cdot d^2 \cdot 1}

The ratio of the mean free path at 373K to that at STP is:

\frac{\lambda_{\text{373K}}}{\lambda_{\text{STP}}} = \frac{373}{273}

Simplifying this ratio:

\frac{\lambda_{\text{373K}}}{\lambda_{\text{STP}}} = \frac{373}{273} \approx 1.366

Then the mean free path at 373K is:

\lambda_{\text{373K}} = \lambda_{\text{STP}} \times 1.366 = 1500d \times 1.366 \approx 2049d

Thus, the mean free path of the molecules at 373K is approximately 2049d, which matches the given correct answer option.