Question

Mean free path for an ideal gas is observed to be 20 μm while the average speed of molecules of gas is observed to be 600 m/s, then the frequency of collision is near?

• A. \( 4 \times 10^7 \)
• B. \( 1.2 \times 10^7 \)
• C. \( 3 \times 10^7 \)
• D. \( 2 \times 10^7 \)

Hint:

The frequency of collision in a gas is given by the formula \( \nu = \frac{v}{\lambda} \), where \( v \) is the average speed of the molecules and \( \lambda \) is the mean free path. This relation shows that the more frequent the collisions, the shorter the mean free path or higher the speed of the molecules.

Answer 1

The Correct Option is B

The frequency of collision \( \nu \) is related to the mean free path \( \lambda \) and the average speed \( v \) of gas molecules by the formula: \[ \nu = \frac{v}{\lambda} \] Where: - \( \lambda = 20 \, \mu m = 20 \times 10^{-6} \, \text{m} \), - \( v = 600 \, \text{m/s} \). Substitute these values into the formula: \[ \nu = \frac{600}{20 \times 10^{-6}} = \frac{600}{2 \times 10^{-5}} = 1.2 \times 10^7 \, \text{collisions per second}. \] Thus, the frequency of collisions is \( 1.2 \times 10^7 \).