At what pressure will the mean free path in room-temperature $\left(20°\mathrm{C}\right)$nitrogen be $1.0\mathrm{m}$?

Nitrogen room temperature $={20}^{\circ }C$

Mean free path$=1.0m$

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.

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

localid="1648278027720" $\frac{N}{V}=\frac{p}{k_{\mathrm{B}}T}\to \left(2\right)$

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

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

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.

localid="1648278041399" $\lambda =\frac{1}{4\sqrt{2}\pi (N/V)r^2}\to \left(3\right)$

$r=$Radius of the molecule,

$N=$Number of molecules

$V=$Volume.

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

If $\lambda =1.0m$, then the answer to equation (3) is $N/V$, which gives us the equation for the pressure $p,$

$\lambda =\frac{1}{4\sqrt{2}\pi \left(p/k_{\mathrm{B}}T\right)r^2}$

$p=\frac{k_{\mathrm{B}}T}{4\sqrt{2}\pi \lambda r^2}\to \left(4\right)$

To convert Celsius to Kelvin, we must first convert the units between the two scales. Kelvins are the units for the Kelvin scale. It can be converted between the two scales by using equation (18.7).

$T_{\mathrm{K}}=T_{\mathrm{C}}+273\to \left(5\right)$

Substitute the value for $T_{\mathrm{C}}=20°\mathrm{C}$into equation (5) to get $T_{\mathrm{K}}$

$T_{\mathrm{K}}=T_{\mathrm{C}}+273$

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

$=293\mathrm{K}$

Put the values for $k_{\mathrm{B}},T,\lambda$and $r$into equation (4) to get$p,$

$\begin{array}{r}p=\frac{k_{\mathrm{B}}T}{4\sqrt{2}\pi \lambda r^2}\end{array}$

$=\frac{\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(293\mathrm{K}\right)}{4\sqrt{2}\pi \left(1.0\mathrm{m}\right){\left(1.0\times {10}^{-10}\mathrm{m}\right)}^2}$

$=0.023\mathrm{Pa}$

Hence, if the mean free path in room temperature $\left({20}^{\circ }C\right)$of nitrogen $(1.0m)$then the pressure is$0.023Pa$.