# Hydrostatic pressure in a gas

I have a physical picture of hydrostatic pressure ($dp/dh = \rho g$) in liquids, interpreting it as the weight of the water column.

Now in a gas, the molecules are much further apart than in a liquid. Still the same hydrostatic pressure exists in gases (e.g. air) as well.

How am I to imagine the "weight of an air column" if molecules are not exactly lying on each other? At the moment I am imagining all those air molecules bouncing off the ground and other surfaces (elastically).

Does the hydrostatic pressure result from some time-average of such bounces? Why is the hydrostatic pressure the same in all directions then?

When a gas is too dilute, you cannot use hydrodynamics anymore and you are going to have to rely solely on statistical mechanics. This sets the difference between the continuum and molecular régimes.

Hydrodynamics in this case is an ‘equilibrium’ state where you assume the gas has a homogenous density because it has had time to thermalise. Hence all particles everywhere in the gas have the same velocity/ same rate of impinging on the wall, hence the same pressure exerting on all surface (which I reckon is referred to as Pascal’s principle). If you were to kick the gas with a piston from one side, then the ‘disturbance’ of density would travel across the gas at the speed of sound, and reflect off the other end in the case of a finite volume. Thermalisation, I.e. the transferral of the kick energy into equal momentum of all particles, will also happen in a timescale governed by the speed of sound in the gas.

The Knudsen number quantifies whether you can use the hydrodynamic approach. It compares the mean free path of the particles, I.e. the distance they cover before they hit another particle, to the physical lengthscale of the system. If the latter is much former than the latter, then you are in the molecular regime, governed by statistical mechanics. To give you an idea, an Ultra High Vacuum chamber (pressures of $10^{-11}$ mbar) can have mean free paths of tens of kilometres, albeit being about a litre in volume.

In the statistical mechanics case then yes, you rely on time averages. Which to be fair also give you the real answer in the hydrodynamic case, since a continuum fluid is none other than a mass of single particles. But hydrodynamics is a self-consistent simpler picture so one uses that if one can.

If you wish to resolve timescales that are smaller than the average time, then you have to rely on statistical mechanics also for the time evolution of pressure oscillations etc.

• Let me add that this explanation applies for a gas in a closed box without external forces. For the atmosphere, gravity has a position dependent force which means the density is not homogenous. Density is affected by this but also by wind and temperature. Sep 2 '18 at 20:36
• Thanks. If I understand correctly, in the hydrodynamic limit the gas behaves more or less like a fluid as far as pressure, etc are concerned. Sep 2 '18 at 20:55
• Yeah, well ‘fluid’ is defined in many ways but it encompasses liquids and non-dilute gases. Sep 2 '18 at 23:04
• Let me also add that Pascal’s principle is only applicable to hydrostatics. If hydrodynamics you’ll have to deal with stuff like the Venturi effect. Sep 6 '18 at 19:40

You say: "I have a physical picture of hydrostatic pressure (dp/dh=ρg) in liquids, interpreting it as the weight of the water column."

Calculationwise that's a shortcut, seeing as that it only 'holds water' in a gravity field. We basically live in a huge column of gas that reaches up to the top of the atmosphere. At that scale the shortcut also works for gases, but determining hydrostatic pressure for pressurized gas vessels by means of the acceleration that caused it, is not possible.