Mean Free Path — Question 2

Question 2

Heating changes collision rates differently at fixed volume and fixed pressure

For identical dilute hard-sphere molecules of diameter $d$ and number density $\nu$, use $\lambda=1/(\sqrt2\pi d^2\nu)$. Initially the gas has mean free path $\lambda_0$ and mean molecular speed $\overline v_0$. Its temperature doubles without changing diameter. Adopt collision rate $z=\overline v/\lambda$ and $\overline v\propto\sqrt T$. Compare fixed amount at fixed volume with fixed amount at fixed pressure.

1. Find the ratios of mean free path and collision rate for each process.

2. For initial $\lambda_0=100$ nm and $\overline v_0=500$ m/s, find the new mean free times $\tau=1/z$.

Question 2 - Solution

Heating changes collision rates differently at fixed volume and fixed pressure

(a) Method and calculation

At fixed volume, number density is unchanged:$$\lambda_V/\lambda_0=1,\quad z_V/z_0=\sqrt2.$$At fixed pressure, $\nu=p/(k_BT)$ halves:$$\lambda_p/\lambda_0=2,\quad z_p/z_0=\sqrt2/2.$$

(b) Interpretation and further calculation

$$\tau_0=\lambda_0/\overline v_0=10^{-7}/500=2.00\times10^{-10}\,\mathrm s.$$$$\tau_V=\tau_0/\sqrt2=1.41421\times10^{-10}\,\mathrm s,\quad\tau_p=\sqrt2\tau_0=2.82843\times10^{-10}\,\mathrm s.$$Faster molecules collide less frequently in the fixed-pressure case because dilution increases their flight distance more strongly.

Physical check

A temperature dependence for mean free path is meaningful only with a specified density or pressure constraint.