Question

The mean free path and the average speed of oxygen molecules at 300 K and 1 atm are 3×10−7 m and 600 m/s respectively. Find the frequency of its collisions.

Answer 1

2 × 10^9 s^-1

Answer 2

The frequency of collisions ($f$) is defined as the number of collisions a molecule experiences per unit time. It is directly related to the average speed of the molecules ($\bar{v}$) and the mean free path ($\lambda$).

The relationship is given by: $f = \frac{\bar{v}}{\lambda}$ where:

• $f$ is the collision frequency (in $s^{-1}$)
• $\bar{v}$ is the average speed (in m/s)
• $\lambda$ is the mean free path (in m)

Given values:

• Mean free path ($\lambda$) = $3 \times 10^{-7}$ m
• Average speed ($\bar{v}$) = $600$ m/s

Substitute the given values into the formula: $f = \frac{600 \text{ m/s}}{3 \times 10^{-7} \text{ m}} = \frac{600}{3} \times 10^{7} \text{ s}^{-1} = 200 \times 10^{7} \text{ s}^{-1} = 2 \times 10^{2} \times 10^{7} \text{ s}^{-1} = 2 \times 10^{9} \text{ s}^{-1}$

The frequency of its collisions is $2 \times 10^9$ collisions per second.