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 collision frequency (\( u \)) for gas molecules, defined as the number of collisions per unit time, is determined by the mean free path (\( \lambda \)) and the average speed (\( v \)) using the equation: \[ u = \frac{v}{\lambda} \] Given: - \( \lambda = 20 \, \mu m = 20 \times 10^{-6} \, \text{m} \) - \( v = 600 \, \text{m/s} \) Substituting these values yields: \[ u = \frac{600}{20 \times 10^{-6}} = \frac{600}{2 \times 10^{-5}} = 1.2 \times 10^7 \, \text{collisions per second} \] Therefore, the calculated collision frequency is \( 1.2 \times 10^7 \) per second.