# Pressure of thermodynamical systems and gravity

1. Aug 26, 2014

### atat1tata

In almost every textbook I have seen, pressure is said to be one of the most important state variables of a thermodynamical system. But if the system is three-dimensional and on planet Earth, it is not constant! This problem, however, is neglected in every reference I consulted.

It seems to me that it is to be addressed, since there is also potential energy (that makes the energy of the particles dependent on height) to consider.

The only solution I see is to assign a thermodynamical system for each infinitesimal height of the volume of the actual system. But considering, for example, a chemical reaction occurring in a bucket of water, how can it be described in terms of Gibbs free energy, that is useful only when the pressure is constant?

2. Aug 26, 2014

### SteamKing

Staff Emeritus
Well, pressure is a state variable, which suggests that it, you know, varies. Why do you think that pressure would be constant?

3. Aug 26, 2014

### atat1tata

Because in all the textbooks that I know of, they speak of "the pressure of a gas", or the "pressure of a bucket of water", and they do not consider $\rho g h$

4. Aug 26, 2014

### Useful nucleus

The thermodynamic framework as presented in textbooks usually consider what is called a simple system which is by definition, one that is homogeneous and is not exposed to fields. Taking into account the gravitational field is possible. Instead of considering the pressure (P) one can consider σ which is the stress tensor. In that case, the forces need not be isotropic as you described.

EDIT: I had a solid in mind when I wrote this. I think the answer given by WannabeNewton below is more relevant.

Last edited: Aug 26, 2014
5. Aug 26, 2014

### SteamKing

Staff Emeritus
$\rho g h$ of what? The atmosphere? The water in the bucket?

Gasses can be confined within closed vessels, where the pressure inside the vessel is independent of the pressure outside.

It's very much unclear what you are having a problem with.

6. Aug 26, 2014

### atat1tata

Of the water in the bucket (or of the gas inside the vessel, although this can be easily negligible). I'm saying that the pressure inside a "thermodynamic system" is actually non-uniform, but calculations are done as if it were.

7. Aug 26, 2014

### WannabeNewton

If you immerse an ideal gas in a gravitational field you simply have to wait for the gas to equilibriate and you can assign it a pressure in the usual way. The gas will come to equilibrium in such a way that the potential energy due to the gravitational field is taken into account in the pressure i.e. the pressure will be a function of the elevation. Nowhere in thermodynamics is pressure required to be uniform in space. In order to have a pressure you simply need a system that is in equilibrium and an ideal gas in a gravitational field will certainly come to equilibrium eventually. The only uniformity across the gas is that of the temperature $T$ which is a result of equilibrium$^{\dagger}$.

This is a very easy calculation to perform. Clearly by symmetry the non-uniformity in the gas distribution, and the pressure, will only vary along $z$ so consider an infinitesimal slab of thickness $dz$ of the ideal gas. Let the slab have cross sectional area $A$. The number of gas particles in the slab is $N = \frac{PA dz}{k_B T}$ from the equation of state. The gas particles exert a net force $F(z + dz) - F(z) = -Nmg = -mgA\frac{P}{k_B T}dz$ so $dP = \frac{dF}{A} = -mg\frac{P}{k_B T}dz$ or $\frac{dP}{P} = -\frac{mg}{k_B T}dz$. This tells you how the pressure varies with elevation.

$^{\dagger}$ Actually, as an aside if you're interested, that the temperature is uniform across a gas in a gravitational field in equilibrium is only true in Newtonian gravity. In general relativity even the temperature of the gas varies with elevation in equilibrium essentially because $T \sim E$ and the energy $E$ itself gravitates. See Ehrenfest-Tolman effect.

8. Aug 26, 2014

### SteamKing

Staff Emeritus
You're still not clear. What calculations?

The pressure in a bucket of water is not constant; it varies with the depth of the water, as the formula ρgh implies. At the free surface, the pressure is equal to atmospheric and increases as the depth of the water.

9. Aug 26, 2014

### Staff: Mentor

All of the textbooks I know of pay attention to the gravitational pressure gradient when it matters (examples: hydro plant, open pumping system) and ignore it when it doesn't (examples: closed pumping system, pressurized air system). So that's the general answer to your query: they ignore it when it doesn't matter.

Do you have any actual examples we can discuss, because I think either your memory or understanding is failing you. In particular, one common bucket of water problem is "how far does water shoot sideways out of this hole in this bucket?" and obviously you need pressure due to gravity to solve it.

10. Aug 26, 2014

### Staff: Mentor

In the case of chemical reaction in a bucket (where you are using the Gibbs free energy to determine the equilibrium constant), the effects of hydrostatic pressure variations are typically negligible compared to the atmospheric pressure. Check it out for yourself to see how much the pressure varies compared to 1 atm. But, if you really want to, you can take the pressure variation into account.

Chet