Question

If the mean free path of a gas molecule at 27°C is $ 10 \times 10^{-7} \, \text{m} $, then the mean free path at 87°C is:

• A. \( 10 \times 10^{-6} \, \text{m} \)
• B. \( 12 \times 10^{-7} \, \text{m} \)
• C. \( 15 \times 10^{-7} \, \text{m} \)
• D. \( 8 \times 10^{-7} \, \text{m} \)

Hint:

For temperature-dependent properties like mean free path, use the formula \( \lambda_2 = \lambda_1 \times \frac{T_1}{T_2} \) to find the new value based on the change in temperature.

Answer 1

The Correct Option is B

The mean free path \( \lambda \) of a gas molecule is inversely proportional to the temperature. Therefore, we can use the formula: \[ \lambda_2 = \lambda_1 \times \frac{T_1}{T_2} \] where \( \lambda_1 \) and \( \lambda_2 \) are the mean free paths at temperatures \( T_1 \) and \( T_2 \), respectively. Given that \( \lambda_1 = 10 \times 10^{-7} \, \text{m} \) at \( T_1 = 27^\circ \text{C} \) and we need to find \( \lambda_2 \) at \( T_2 = 87^\circ \text{C} \), we first convert the temperatures to Kelvin: \[ T_1 = 27 + 273 = 300 \, \text{K}, \quad T_2 = 87 + 273 = 360 \, \text{K} \] Now, applying the formula: \[ \lambda_2 = 10 \times 10^{-7} \times \frac{300}{360} = 12 \times 10^{-7} \, \text{m} \]
Thus, the mean free path at 87°C is \( 12 \times 10^{-7 \, \text{m}} \).