A gas has an average speed of 10 m/s and a collision frequency of 10 $s^{- 1}$ . What is its mean free path?

1 m

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2 m

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3 m

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4 m

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Solution

The correct option is A $1 m$
The collision frequency $(f)$ is the ratio of rms velocity $(v_{r m s})$ to mean free path $(λ)$ .
Here, collision frequency $f = 10 s^{- 1}$ and $v_{a v g} = 10 m / s$
So, $v_{r m s} = \sqrt{v_{a v g}^{2}} = \sqrt{10^{2}} = 10 m / s$
Thus, mean free path $λ = \frac{v_{r m s}}{f} = \frac{10}{10} = 1 m$

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A gas has an average speed of 10 m/s and a collision frequency of 10 s−1. What is its mean free path?

The correct option is A 1mThe collision frequency (f) is the ratio of rms velocity (vrms) to mean free path (λ).Here, collision frequency f=10s−1 and vavg=10m/sSo, vrms=√v2avg=√102=10m/sThus, mean free path λ=vrmsf=1010=1m

A gas has an average speed of 10 m/s and a collision frequency of 10 $s^{- 1}$ . What is its mean free path?