The mean free path and the average speed of oxygen molecules at 300 K and 1 atm are $3 \times 10^{-7} \mathrm{~m}$ and $600 \mathrm{~m} / \mathrm{s}$, respectively. Find the frequency of its collisions.

<p>To determine the frequency of collisions for oxygen molecules, we use the formula for frequency:</p> <p>$ \text{Frequency} = \frac{1}{T} = \frac{V_{\text{avg}}}{\lambda} $</p> <p>where $ V_{\text{avg}} $ is the average speed of the molecules and $ \lambda $ is the mean free path.</p> <p>Given:</p> <p><p>Average speed ($ V_{\text{avg}} $) = 600 m/s</p></p> <p><p>Mean free path ($ \lambda $) = $ 3 \times 10^{-7} $ m</p></p> <p>Substitute these values into the formula:</p> <p>$ \text{Frequency} = \frac{600}{3 \times 10^{-7}} = 2 \times 10^9 \, \text{s}^{-1} $</p> <p>Therefore, the frequency of collisions is $ 2 \times 10^9 \, \text{collisions per second} $.</p>