Question 186: Thermodynamics & KTG

Question

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

Accepted Answer

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

Answer

$$5 	imes 10^8 / \mathrm{s}$$

Answer

$9 	imes 10^5 / \mathrm{s}$

Answer

$2 	imes 10^{10} / \mathrm{s}$

Answer

$2 	imes 10^9 / \mathrm{s}$

Thermodynamics & KTG

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.

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