In a certain particle accelerator, protons travel around a circular path of diameter $\mathbf{23}\mathbf{.}\mathbf{0}\mathit{m}\mathbf{}$in an evacuated chamber, whose residual gas is at$\mathbf{295}\mathit{K}\mathbf{}$and$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}\mathit{t}\mathit{o}\mathit{r}\mathit{r}\mathbf{}$pressure.

1. Calculate the number of gas molecules per cubic centimeter at this pressure.
2. What is the mean free path of the gas molecules if the molecular diameter is$\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{}\mathit{c}\mathit{m}$?

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

$d=23.0\text{m}$

$T=295\text{\hspace{0.17em}K}$

The expression for the mean free path is given by,

$\lambda =\frac{1}{\sqrt{2}\pi {\text{d}}^2\left(\frac{\text{N}}{\text{V}}\right)}$

Here$\lambda$ 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$1.33\times {10}^{-4}\text{Pa}$ for P , $1.38\times {10}^{-23}J/K$for K and $295\text{\hspace{0.17em}K}$for T into the above equation,

$\frac{N}{V}=\frac{1.33\times {10}^{-4}}{1.38\times {10}^{-23}\times 295}$

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

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

The mean free path is given by

$\lambda =\frac{1}{\sqrt{2}\pi d^2\left(\frac{N}{V}\right)}$

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

$\begin{array}{c}\lambda =\frac{1}{\sqrt{2}\left(3.142\right){(2.0\times {10}^{-10})}^2(3.267\times {10}^{16})}\\ =172.24\\ \approx 172\text{\hspace{0.17em}m}\end{array}$

Therefore, the mean free path is$172\text{\hspace{0.17em}m}.$