Question

If pressure of a gas is doubled at constant temperature, mean free path becomes:

• A. Doubled
• B. Halved
• C. Same
• D. Four times

Hint:

Mean free path is inversely proportional to the number density. Doubling the pressure at constant temperature doubles the number density, which naturally cuts the distance between collisions in half.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The mean free path (\(\lambda\)) is the average distance a molecule travels between two successive collisions with other molecules.
It depends on the number density of the gas (number of molecules per unit volume) and the diameter of the molecules.
When pressure increases at a constant temperature, according to Boyle's law, the volume decreases, meaning the number density increases.
In a more crowded environment, molecules collide more frequently, which naturally reduces the distance they can travel freely before a collision occurs.
Therefore, mean free path is inversely related to pressure under isothermal conditions.
Step 2: Key Formula or Approach:
1. Standard Formula: \(\lambda = \frac{kT}{\sqrt{2}\pi d^2 P}\).
2. Proportionality: At constant temperature \(T\), \(\lambda \propto \frac{1}{P}\).
Step 3: Detailed Explanation:
From the kinetic theory expression for mean free path, \(\lambda\) is proportional to \(T/P\).
In this problem, it is stated that the temperature remains constant.
Let the initial pressure be \(P_1\) and the initial mean free path be \(\lambda_1\).
The pressure is doubled, so \(P_2 = 2P_1\).
Using the inverse proportionality:
\[ \frac{\lambda_2}{\lambda_1} = \frac{P_1}{P_2} = \frac{P_1}{2P_1} = \frac{1}{2} \]
Thus, \(\lambda_2 = \lambda_1 / 2\).
The new mean free path is exactly half of its original value because doubling the pressure makes the gas twice as dense.
Step 4: Final Answer:
The mean free path becomes halved.