The Distribution of Molecular Speeds — Question 7

Question 7

Effusing molecules are weighted toward the faster part of the reservoir

A tiny hole in a large equilibrium reservoir produces collisionless effusion. The supplied normalized speed density of the escaping molecules is $g(v)=vf(v)/\overline v$, where $f(v)=Cv^2e^{-av^2}$ and $a=m/(2k_BT)$. Use $\int_0^\infty v^3e^{-av^2}dv=1/(2a^2)$ and $\int_0^\infty v^5e^{-av^2}dv=1/a^3$.

1. Find the mean translational energy of an escaping molecule and compare it with the reservoir mean.

2. Find the mode of the escaping speed density. Does effusion sample molecules uniformly from the reservoir speed distribution?

Question 7 - Solution

Effusing molecules are weighted toward the faster part of the reservoir

(a) Method and calculation

$$\langle v^2\rangle_{\mathrm{escape}}=\frac{\int_0^\infty v^3f(v)dv}{\int_0^\infty vf(v)dv}=\frac{1/a^3}{1/(2a^2)}=\frac2a.$$$$\langle E\rangle_{\mathrm{escape}}=\frac m2\frac2a=2k_BT=\frac43\langle E\rangle_{\mathrm{reservoir}}.$$

(b) Interpretation and further calculation

$$g(v)\propto v^3e^{-av^2},\quad\frac{d\ln g}{dv}=\frac3v-2av=0.$$$$v_{{\mathrm{mp,escape}}}=\sqrt{3/(2a)}=\sqrt{3k_BT/m}.$$The crossing flux adds a speed factor, so faster molecules have a larger chance of escaping per unit time.

Physical check

The result describes the emitted ensemble while reservoir temperature is specified; sustained unheated loss can cool the reservoir.