From what height must an oxygen molecule fall in a vacuum so that its kinetic energy at the bottom equals the average energy of an oxygen molecule at $300K$?

The average translational kinetic energy of a molecule with mass and velocity $V$is The average translational kinetic energy of a molecule is affected by its temperature, hence it is related to the temperature$T$ per molecule in the manner

${ϵ}_{\mathrm{avg}}=\frac{3}{2}{k}_{\mathrm{B}}T$

Where $k_B$is Boltzmann's constant and in SI unit its value is

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

This average energy is equal to the kinetic energy of the oxygen molecule. Plug the values for $k_B$and $T$into equation to get the kinetic energy for oxygen molecule

$\begin{array}{r}K=\frac{3}{2}{k}_{\mathrm{B}}T\\ =\frac{3}{2}1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\left(300\mathrm{K}\right)\\ =6.2\times {10}^{-21}\mathrm{J}.\end{array}$

From the conservation law of energy, the potential energy $U$of the molecule converts to the kinetic energy, so we can get the height $h$where the oxygen molecule fall by

$\begin{array}{c}U=K\\ Mgh=K\\ h=\frac{K}{Mg}.\end{array}$

The molecular mass of oxygen is $m=16u$. But oxygen is a diatomic gas, so the molecular mass of one molecule is $m=32$. Converting this to $kg$, we get the mass of one molecule of oxygen b $M=32\mathrm{u}\times \frac{1.66\times {10}^{-27}\mathrm{kg}}{1\mathrm{u}}=53\times {10}^{-27}\mathrm{kg}$

$h=\frac{K}{Mg}$

$=\frac{6.2\times {10}^{-21}\mathrm{J}}{\left(53\times {10}^{-27}\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}$

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