Mean Free Path — Question 7

Question 7

Mean free path can greatly exceed the typical molecular spacing

For identical dilute hard-sphere molecules of diameter $d$ and number density $\nu$, use $\lambda=1/(\sqrt2\pi d^2\nu)$. Let $\nu=10^{24}\,\mathrm{m^{-3}}$ and $d=.300$ nm. Define a characteristic spacing $s=\nu^{-1/3}$; this is a volume-per-particle scale, not a fixed lattice spacing in the gas.

1. Calculate $s$, $\lambda$ and their ratio.

2. Derive $\lambda/s$ in terms of $s/d$. If density rises by a factor of eight at fixed diameter, find the changes in spacing, mean free path and their ratio.

Question 7 - Solution

Mean free path can greatly exceed the typical molecular spacing

(a) Method and calculation

$$s=10^{-8}\,\mathrm m=10.0\,\mathrm{nm},\quad\lambda=\frac1{\sqrt2\pi(.300\times10^{-9})^2(10^{24})}\approx2.50088\,\mu\mathrm m.$$$$\lambda/s\approx250.088.$$

(b) Interpretation and further calculation

$$\nu=s^{-3}\ \Rightarrow\ \frac\lambda s=\frac1{\sqrt2\pi}\left(\frac sd\right)^2.$$$$\nu\to8\nu:\quad s\to s/2,\quad\lambda\to\lambda/8,\quad\lambda/s\to(\lambda/s)/4.$$A moving molecule usually passes between many neighbors because its collision cross section is small.

Physical check

Mean free path is not simply the distance to the nearest particle.