# In a gravitational field, will the temperature of an ideal gas will be lower at higher altitude?

I know there are similar questions, but I have some arguments which seem to explain that temperature of an ideal gas in a gravitational field will be lower at higher altitudes. I am assuming that:

1.Molecules of the gas are point sized.

2.They interact only during a collision.

3.All collisions are elastic.

4.Time of collision is negligibly small.

During an elastic collision of 2 equal masses, say A & B, velocities of A and B are just exchanged. If A and B are molecules of the same gas, then they would be indistinguishable in their appearance. Even if they exchanged their velocities, using their indistinguishability and their zero size, one could say that particles passed right through each other because in reality there is no label A or B on molecules. One can't distinguish whether they collided elastically or passed through each other unaffected.

In a lump of ideal gas in a box made up of rigid walls, it is like every molecule is moving freely as if it is alone in the box. So treating each molecule as isolated, I could assert that it will slow down when moving upwards (against the gravitational field) & this applies to all the molecules in that box. Hence, the average kinetic energy at higher altitudes will be lesser than those of the lower altitudes & so the temperature will be lesser at higher altitudes. So am I wrong in concluding this? Or are the assumptions that I made too impractical?

• You seem to be forgetting that in any region where temperature is measured, for every molecule traveling upward and slowing down a certain amount, there is, statistically, one traveling downward and speeding up by the same amount. – D. Ennis Dec 5 '17 at 15:04
• @D.Ennis We are measuring average speed, not accelerations. So, in a region, I look for speeds only, no matter what their change is. – Yaman Sanghavi Dec 5 '17 at 16:24
• I'm afraid that I don't follow your objection to my comment. – D. Ennis Dec 5 '17 at 19:05
• @D.Ennis What I meant was If I map temperature as a function of altitude, molecules at a higher altitude will be slower than those at lower altitudes. Any molecule that is at some height will be slower compared to itself if it were at the bottom. – Yaman Sanghavi Dec 6 '17 at 6:51
• @D.Ennis Ah, I think the problem might be because I am comparing a molecule with itself at another instant of time. What do you say? – Yaman Sanghavi Dec 6 '17 at 6:52

Yes, the temperature of a gas (either real or ideal) is lower at the top of a container. A simple example is Earth's atmosphere, which is colder at the top than at the bottom.

There's an obvious caveat: if the gas is in thermal equilibrium with its container, e.g. because the container is quite small or the pressure is quite high, then of course there can't be a temperature gradient.

The reason temperature varies with elevation is not quite what you came up with. I think your argument is valid within your definition of an ideal gas: the particles that climb high up in your box clearly have less kinetic energy than they did on the ground, so their temperature is reduced. (It helps your illustration to imagine that the box is 10 miles high and 10 feet wide.) But in the usual definitions, ideal gas molecules are small hard spheres, so their collisions induce scattering, which scrambles their speeds and directions. I think the effect of this scattering would be that the kinetic energy "lost" by a single particle as a result of rising up in the box would be refunded by collisions with other particles being scattered from below or above.

To show that the temperature of a gas decreases with increasing elevation, I reproduce a simplified version of the argument in Exercise 2042 from Doron Cohen's online stat-mech notes.

Consider a box of fixed volume in a gravitational field, containing an ideal gas that is not at thermal equilibrium. An arbitrarily thin "layer" of gas at the bottom of the box has pressure $P_0$, number density $n_0$, and temperature $T_0$, with $P_0=n_0T_0$ (eliding the gas constant for convenience). As some gas rises (or falls), it does adiabatic work on its surroundings, which changes the energy content of the rising (or falling) gas. Therefore, at every "layer" at height $h$ we have $$P(h)=Cn(h)^\gamma$$ where the constant is $C=T_0n_0^{1-\gamma}$ and $\gamma$ is another constant, the so-called adiabatic index.

The pressure must vary continuously up the box, which leads to the condition $$dP(h)=-mgn(h)\space dh$$ where $m$ is the mass of the particle and $g$ is the local acceleration due to gravity.

We have $P(h)=Cn(h)^\gamma$ from above, so $dP(h)=C\gamma n(h)^{\gamma-1} dn$. Integrating the continuity condition above with this definition of $dP$ brings us (eventually) to a final result,

$$n(h)=n_0\left(1-\frac{\gamma-1}\gamma \cdot \frac{mgh}{T_0}\right)$$

and since temperature is $T=P/n$, the height-dependent temperature is

$$T(h)=P(h)/n(h)=Cn(h)^{\gamma-1}=T_0-\left(\frac{\gamma-1}\gamma\right)mgh.$$

So we see that the pressure gradient induced by gravity produces a temperature gradient: the higher you go in the box, the colder the gas gets.

• The reason why the atmosphere is colder higher up is that the system is heated by absorption of solar energy at the bottom, And it is radiating at the top. – Pieter Dec 6 '17 at 9:46

Your thinking is sound. Thermal equilibrium for an isolated (no energy in/out at boundary) gas in a gravitational field means thermal energy + gravitational energy = constant per unit of gas mater (mol, kg, etc). So, indeed, the temperature of the bottom of the column will be higher than the top of the column. To visualize this using the billiard ball gas model, consider that at any given instant every billiard ball moving up is transferring kinetic energy (temperature) to gravitational potential energy, slowing down, and every ball moving down is speeding up (hotter) thus in equilibrium there is a thermal gradient.