# Mean potential energy per particle of Ideal gas in uniform gravitational field (high temp. limit)

I'm trying to show that for an ideal gas in a uniform gravitational field $$\vec{g} = (0, 0, -g)$$, confined to a box of height $$L$$ and base area $$A$$, the mean potential energy per particle is $$\langle v (\vec{r_i}) \rangle \approx \frac{mgL}{2},$$ provided that $$k_B T \gg mgL$$.

Thus far I've found that the distribution for the height of a particle is

$$P_z(z_i) = \frac{\beta mg}{1 - \exp(-\beta mgL)} \exp(-\beta mgz_i),$$

and so, the mean potential energy is given by

$$\langle v (\vec{r_i}) \rangle = \langle mgz_i \rangle = \frac{1}{\beta} - \frac{mgL}{\exp(\beta mgL) - 1}.$$

Now, that the mean potential energy should be $$mgL/2$$ when $$k_BT \gg mgL$$ seems very intuitive to me, however, I'm having trouble demonstrating this using the expression I've found. Any help would be greatly appreciated!

If the thermal energy of a particle is much larger than the change in potential energy when it moves from bottom to top (that's what $$k_BT>>mgL$$ means) then it doesn't care about gravity. So the density of the gas is uniform. So the center of mass is halfway up.
• Yes, that is clear to me, and is what I meant by this result appearing very intuitive. What I'd like to do is to demonstrate that the expression for the potential energy $\langle v( \vec{r_i} ) rangle\$ reduces to this. Commented Feb 7, 2021 at 7:56