Question

If the mean free path of a gas molecule at $27^{\circ}\text{C}$ is $10\times10^{-7}\ \text{m}$, its mean free path at $87^{\circ}\text{C}$ is:

• A. $12\times10^{-7}\ m$
• B. $8\times10^{-7}\ m$
• C. $6\times10^{-7}\ m$
• D. $10\times10^{-7}\ m$
• E. $14\times10^{-7}\ m$

Hint:

Always convert temperature to Kelvin for gas law calculations.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The mean free path (\( \lambda \)) is the average distance a molecule travels between successive collisions with other molecules. It depends on the temperature, pressure, and size of the molecules. We need to find how it changes with temperature.
Step 2: Key Formula or Approach:
The formula for the mean free path is: \[ \lambda = \frac{1}{\sqrt{2}\pi d^2 n} \] where \( d \) is the molecular diameter and \( n \) is the number density (number of molecules per unit volume, N/V). The number density \( n \) depends on pressure (P) and temperature (T) according to the ideal gas law, \( PV = Nk_BT \), which gives \( n = N/V = P/(k_BT) \). Substituting this into the formula for \( \lambda \): \[ \lambda = \frac{k_B T}{\sqrt{2}\pi d^2 P} \] This formula shows the dependencies: - If the pressure (P) is kept constant, the mean free path is directly proportional to the absolute temperature (T): \( \lambda \propto T \). - If the volume (V) is kept constant, then the pressure is proportional to T (\( P \propto T \)), making \( \lambda \) independent of temperature. Since the problem does not specify the conditions, the standard assumption in such cases is that the pressure is constant (e.g., the gas is in a container with a movable piston open to the atmosphere). We will proceed with the assumption of constant pressure. Step 3: Detailed Explanation:
Assuming constant pressure, we have the relationship \( \frac{\lambda_1}{T_1} = \frac{\lambda_2}{T_2} \). First, convert the temperatures from Celsius to Kelvin (the absolute temperature scale). - \( T_1 = 27 °\text{C} + 273.15 \approx 300 \text{ K} \) - \( T_2 = 87 °\text{C} + 273.15 \approx 360 \text{ K} \) We are given: - \( \lambda_1 = 10 \times 10^{-7} \text{ m} \) at \( T_1 = 300 \text{ K} \) We need to find \( \lambda_2 \) at \( T_2 = 360 \text{ K} \). Using the ratio: \[ \lambda_2 = \lambda_1 \left(\frac{T_2}{T_1}\right) \] \[ \lambda_2 = (10 \times 10^{-7} \text{ m}) \times \left(\frac{360 \text{ K}}{300 \text{ K}}\right) \] \[ \lambda_2 = (10 \times 10^{-7}) \times (1.2) \] \[ \lambda_2 = 12 \times 10^{-7} \text{ m} \] Step 4: Final Answer:
The mean free path at 87 °C is \( 12 \times 10^{-7} \) m.