Mean Free Path — Question 5

Question 5

Counting binary collisions requires a factor of one-half

A uniform gas contains $N=2.00\times10^{20}$ molecules. Its mean speed is 400 m/s and mean free path 100 nm. Use per-molecule collision rate $z=\overline v/\lambda$ as the model rate.

1. Find the mean free time and total number of distinct binary collisions per second.

2. If the gas amount doubles at fixed volume and temperature, find how $\lambda$, $z$ and the distinct total collision rate change.

Question 5 - Solution

Counting binary collisions requires a factor of one-half

(a) Method and calculation

$$\tau=\lambda/\overline v=10^{-7}/400=2.50\times10^{-10}\,\mathrm s,\quad z=4.00\times10^9\,\mathrm{s^{-1}}.$$$$\dot N_{\mathrm{collisions}}=\tfrac12Nz=4.00\times10^{29}\,\mathrm{s^{-1}}.$$

(b) Interpretation and further calculation

$$\nu\to2\nu:\quad\lambda\to\lambda/2,\quad z\to2z,\quad N\to2N.$$$$\dot N_{\mathrm{collisions}}\to4\dot N_{\mathrm{collisions}}=1.60\times10^{30}\,\mathrm{s^{-1}}.$$Each binary encounter is counted for both partners in $Nz$, so division by two avoids duplication.

Physical check

The total collision rate scales as density squared at fixed volume and thermal speed.