Integrated circuits are manufactured in vacuum chambers in which the air pressure is $1.0\times {10}^{-10}\mathrm{mm}$ of $\mathrm{Hg}$. What are (a) the number density and (b) the mean free path of a molecule? Assume $T=20°\mathrm{C}$.

Integrated circuits air pressure $=1.0\times {10}^{-10}$of$Hg.$

Molecular density determines how tightly the atoms are bonded in a system. It is determined by equation (18.2) in the form of the number of atoms per cubic meter in a system.

$numberdensity=\frac{N}{V}\to \left(1\right)$

According to the ideal gas law, the volume of the container $V$, the pressure $p$exerted by the gas, the temperature $T$of the gas, and the number of moles $n$of the gas in the container are all related.

$pV=N{k}_{\mathrm{B}}T$

$\frac{N}{V}=\frac{p}{k_{\mathrm{B}}T}\to \left(2\right)$

$k_{\mathrm{B}}$is Boltzmann's constant and in SI unit its value is

$k_{\mathrm{B}}=1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}$

The pressure is given in$\mathrm{mmHg},$

Convert it into Pascal by

$p=\left(1.0\times {10}^{-10}\mathrm{mmHg}\right)\left(\frac{133\mathrm{Pa}}{1\mathrm{mmHg}}\right)=133\times {10}^{-10}\mathrm{Pa}$

The equation (18.7) in the form of a negative number converts the Celsius scale to the Kelvin scale
$\begin{array}{r}{T}_{\mathrm{K}}=T_{\mathrm{C}}+273\end{array}$

localid="1648281401508" $={20}^{\circ }\mathrm{C}+273$

$=293\mathrm{K}$

Put the values for $p,Tand{k}_B$into equation (2) to get number density number by,

$\begin{array}{r}\text{number density}=\frac{p}{k_{\mathrm{B}}T}\end{array}$

localid="1648281496189" $=\frac{133\times {10}^{-10}\mathrm{Pa}}{\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(293\mathrm{K}\right)}$

$=3.3\times {10}^{12}{\mathrm{m}}^{-3}$

Hence, the number density is$3.3\times {10}^{12}{m}^{-3}$.

Integrated circuits air pressure$=1.0\times {10}^{-10}$of$Hg.$

Due to collisions with other molecules, a molecule undergoing distance travel experiences a delay between diffusing to a different position due to these collisions.

The molecules need a certain amount of time to diffuse to a new position.

In equation (20.3), the mean free path $\lambda$is the distance the molecules travel between collisions on average.

$\lambda =\frac{1}{4\sqrt{2}\pi (N/V)r^2}\to \left(3\right)$

$r=$Radius of the molecule,

$N=$Number of molecules

Volume.
Diatomic gas of nitrogen's radius is, $r=1.0\times {10}^{-10}\mathrm{m}$

Put the values for$(N/V)$and $r$into equation (3) to get$\lambda$by,

$\begin{array}{r}\lambda =\frac{1}{4\sqrt{2}\pi (N/V)r^2}\end{array}$

localid="1648282220501" $=\frac{1}{4\sqrt{2}\pi \left(3.3\times {10}^{12}{\mathrm{m}}^{-3}\right){\left(1.0\times {10}^{-10}\mathrm{m}\right)}^2}$

$=1.7\times {10}^6\mathrm{m}$

Therefore, the mean free path of a molecule is$\lambda =1.7\times {10}^6\mathrm{m}.$