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 $1.2 \times 10^7$.