Question

At about what pressure would the mean free path of an air molecule at room temperature equal $10cm$, the size of a typical laboratory apparatus?

Short answer

The mean free path pressure is $P=0.146\mathrm{N}\cdot {\mathrm{m}}^{-2}$

Step-by-step solution

Step: 1 To find pressure of mean free path:

The equation as

$\begin{array}{r}PV=NKT\end{array}$

$V=\pi (2\gamma {)}^2\ell$

By equating,

$P=\frac{UT}{4\pi {\gamma }^2\lambda }$

Subtituting the values 

subtituting the values of

$T=100\mathrm{dr},\gamma =0.01,P=0.082\mathrm{N}/{\mathrm{m}}^2$as

$\begin{array}{r}P=\frac{1.381\times {10}^{-23}\times 100}{4\pi \times {\left(2\times {10}^{-10}\right)}^2\times 0.01}\\ \end{array}$

$P=0.274N{m}^2.$