Question

Show that the mean free path for the molecules of an ideal gas at temperature $T$ and pressure $P$ is $\lambda =\frac{k_{B}T}{\sqrt{2}\pi d^{2}P}$ where $d$ is the molecular diameter .

Answer 1

N= volume of cylinder of length vt and area of cross-section

$\pi d^2\times numberofmoleculesperunitvolume$

or,

$N=\pi d^{2}nvt$

Now, mean free path of a gas molecule,

$\lambda =\frac{distancecovered}{Numberofcollisions}$

$\lambda =\frac{vt}{\pi d^{2}nvt}$

The expression for mean free path has been obtained by assuming that all the gas molecules except the one under consideration, are at rest. In fact, all the gas molecules are in random motion, their velocities being governed by Maxwell's law of distribition of velocities. On taking the motion of all the gas molecules into consideration, the mean free path comes out to be:

$\lambda =\frac{1}{\sqrt{2}\pi d^{2}n}$

In the equation, substituting for n, we have

$n=\frac{\rho }{m}=\frac{P}{K_{B}T}$

hence

$P$ is $\lambda =\frac{k_{B}T}{\sqrt{2}\pi d^{2}P}$ where $d$