91Ó°ÊÓ

$\begin{array}{c}P=1.00×{10}^{−6}\text{torr}\\ =1.33×{10}^{−4}\text{Pa}\end{array}$

$d=23.0\text{m}$

$T=295\text{‿é}$

The expression for the mean free path is given by,

$\mathrm{λ}=\frac{1}{\sqrt{2}\mathrm{π}{\text{d}}^2\left(\frac{\text{N}}{\text{V}}\right)}$

Here$\mathrm{λ}$ is the mean free path, $\frac{N}{V}$is the molecule density in $molecule/c{m}^3$.

The expression for the ideal gas equation is given by,

$pV=nRT$

Here P is the pressure, V is the volume, n is the number of moles, Ris the universal gas constant, T is the temperature.

The relation between number of molecules and Avogadro number is given by,

$n=\frac{N}{N_A}$

Here n is the number of moles, N is the number of molecules, $N_A$is Avogadro number.

The relation between universal gas constant and Boltzmann constant is given by,

$k=\frac{R}{N_A}$

Here K is the Boltzmann constant.

According to the gas law,

$pV=NkT$

$\frac{N}{V}=\frac{p}{kT}$

Substitute$1.33×{10}^{−4}\text{Pa}$ for P , $1.38×{10}^{−23}\text{ }J/K$for K and $295\text{‿é}$for T into the above equation,

$\frac{N}{V}=\frac{1.33×{10}^{−4}}{1.38×{10}^{−23}×295}$

$\begin{array}{c}\frac{N}{V}=3.267×{10}^{16}\\ ≈3.27×{10}^{16}\frac{\text{molecules}}{{\text{m}}^{\text{3}}}\\ =3.27×{10}^{10}\frac{\text{molecules}}{{\text{cm}}^{\text{3}}}\end{array}$

Therefore, the number of gas molecules per cubic centimeter is$3.27×{10}^{10}.$

The mean free path is given by

$\mathrm{λ}=\frac{1}{\sqrt{2}\mathrm{π}{d}^2\left(\frac{N}{V}\right)}$

Substitute$2.0×{10}^{−10}$ for d , $3.267×{10}^{16}\text{ }molecules/c{m}^3$for $\frac{N}{V}$into the above equation,

$\begin{array}{c}\mathrm{λ}=\frac{1}{\sqrt{2}\left(3.142\right){(2.0×{10}^{−10})}^2(3.267×{10}^{16})}\\ =172.24\\ ≈172\text{″¾}\end{array}$

Therefore, the mean free path is$172\text{″¾}.$