Question: Oxygen (O₂) gas at 273 K and 1 atm is confined to a cubical container 10 cm on a side. Calculate${\mathbf{\Delta U}}_{\mathbf{g}}\mathbf{/}{\mathbf{K}}_{\mathbf{avg}}$ , where${\mathbf{\Delta U}}_{\mathbf{g}}$ is the change in the gravitational potential energy of an oxygen molecule falling the height of the box and ${\mathbf{K}}_{\mathbf{avg}}$ is the molecule’s average translational kinetic energy.

The law of conservation of energy states that the sum of the mechanical energy and the internal energy remains conserved for an isolated system of gas. The mechanical energy is the sum of the potential energy and the average kinetic energy of all the molecules of the gas.

The potential energy of a gas is given as-

$\mathbf{U}\mathbf{=}\mathbf{mgh}\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\left(\mathbf{1}\right)$

Here,m is the mass of the molecule of gas, g is the acceleration due to gravity and h is the height from which molecule is falling.

The average kinetic energy of the gas is given as-

${\mathbf{K}}_{\mathbf{avg}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mV}}_{\mathbf{rms}}^{\mathbf{2}}\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\left(\mathbf{2}\right)$

Here, ${\mathbf{V}}_{\mathbf{rms}}^{}$ is the root-mean-square velocity of the molecules of the gas. It is given as-

${\mathbf{V}}_{\mathbf{rms}}^{}\mathbf{=}\sqrt{\frac{\mathbf{3}\mathbf{RT}}{\mathbf{M}}}\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\left(\mathbf{3}\right)$

1. The molar mass of the Oxygen molecule is M = 0.032 Kg/ mol
2. Temperature of gas is T = 273 K

The side of container or the height is $h=10\text{cm}=0.10\text{m}$

Using the equations (2) and (3), we get-

$K_{avg}=\frac{1}{2}m\times \frac{3RT}{M}\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\left(4\right)$

Now, let us assume that the gravitational potential energy at the base is zero.

So, the equation for change in gravitational potential energy will become

$$\begin{gathered} \Delta U_g=mgh-0 \\ =mgh \end{gathered}$$

Now, taking the ratio of gravitational potential energy to average kinetic energy, we get

$$\begin{gathered} \frac{\Delta U_g}{K_{avg}}=\frac{mgh}{\frac{1}{2}m\times \frac{3RT}{M}} \\ =\frac{2Mgh}{3RT} \end{gathered}$$

For the given values, the above equation becomes-

$$\begin{gathered} \frac{\Delta U_g}{K_{avg}}=\frac{2\times 0.032Kg/mol\times 9.8m/s^2\times 0.1m}{3\times 8.31J/Kmol\times 273K} \\ =9.21\times {10}^{-6} \end{gathered}$$

The ratio of change in gravitational potential energy to average kinetic energy is $=9.21\times {10}^{-6}$.