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I was reading about the gaseous state when this question struck my mind: What made us assume that, at every point inside the container, a gas exerts equal pressure? When one brings a barometer, is it true it measures the same pressure at every point inside? Is this applicable to both ideal and real gases?

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  • $\begingroup$ Where's your research? Since it's a closed container, what might change the pressure in parts of it? $\endgroup$ Nov 17, 2021 at 21:22
  • $\begingroup$ @Robbie Goodwin For interest - see my comments on gravity, speed of sound and oscillations. $\endgroup$ Nov 19, 2021 at 23:09
  • $\begingroup$ @RussellMcMahon Thanks and don't you think a container 5,000 metres tall makes sense only in English syntax, not in real physics. How is anything 5,000 metres tall usefully a "container" rather than a (micro-) environment? I don't think there are rules there, and I do think it would be odd, if not perverse, to think of a space station, a submarine or even an aeroplane as a "container" rather than a "separate environment." $\endgroup$ Nov 19, 2021 at 23:18
  • $\begingroup$ @Robbie Goodwin the concept of a 5000 metre container (scale height related) was to make the point that for more usual sizes the effects are relatively small and will usually be insignificant. Similarly, speed of sound pressure wave traveling time will usually be of low significance. $\endgroup$ Nov 20, 2021 at 0:32
  • $\begingroup$ @RussellMcMahon Yes and isn't that saying that a 5,000 metre container is outside the realms of normal consideration? $\endgroup$ Nov 20, 2021 at 18:32

8 Answers 8

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An imbalance of pressure would itself cause an internal flow in the gas. So if the gas has reached equilibrium the pressure must be the same everywhere.

The above is for a gas in ordinary circumstances, without any applied field such as a gravitational field. If there is such a field then the gas flows until the pressure gradient provides a force which just balances the effects of the field.

To calculate these effects more fully one can use the concept of chemical potential and the second law of thermodynamics.

There remains the fact that thermodynamic quantities such as pressure also undergo fluctuations. The above comments about uniformity apply to the time-averaged pressure at any point.

Generalization to fluids

The arguments above apply to fluids more generally, not just to gases (and therefore it is not restricted to ideal gas). As long as the fluid can flow then any pressure gradient will cause a flow so when a fluid reaches equilibrium in a closed container the pressure must be uniform.

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    $\begingroup$ In closed container type scenarios where piston and other devices are also there can we say that gas particles r so fast moving that gravity itself doesnt have any such big contribution to pressure gradient so ig we can say that in any compartment (in some problems the container is divided into some parts for those ) of the container average pressure is same which is calculated macroscopically? $\endgroup$
    – Orion_Pax
    Nov 14, 2021 at 12:09
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    $\begingroup$ @Orion_Pax In a gravitational field $g$ the pressure varies with height as $dp/dz = -\rho g$ where $\rho$ is the density. Since density is low for a gas in ordinary circumstances, this effect is negligible in many ordinary situations. When tackling example problems in physics, it is normally understood that we are treating the case where $g=0$ unless it is explicitly stated otherwise. $\endgroup$ Nov 14, 2021 at 12:49
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    $\begingroup$ Or a heat source, which would cause convection. $\endgroup$
    – Spencer
    Nov 14, 2021 at 20:22
  • $\begingroup$ @Andrew_Steane i understood ur point but that is valid when the molecules r mot moving along the plane perpedicular to z axis or something isnt in that derivation of dp/dz=-pg ? And this pressure balance is only possible when the molecules r coming and going at same rate at any point net molecules (+ forward, -backward) =0 right ? So only in ideal gas we can confirmly say that average pressure is constant not in case of" real gas" right ? $\endgroup$
    – Orion_Pax
    Nov 16, 2021 at 16:08
  • $\begingroup$ @Orion_Pax I added a paragraph to my answer; it is not restricted to ideal gas. $\endgroup$ Nov 16, 2021 at 16:20
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It depends on the resolution of your measuring device.

A gas contains on the order of $10^{22}$ molecules zipping about. The pressure on a wall of the container is due to the tiny force applied by these molecules when they collide with the wall. If you could take a snapshot of each of the walls at an instant in time there would be a certain number of molecules colliding with each wall. However, this number would be different from wall to wall. If you could measure small enough pressures, there would be a difference. I don't know if we have devices that can measure pressures so sensitively.

On a macro scale, the differences from wall to wall are imperceptible. It suffices to say the pressure is constant at every point on the wall of the container.

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    $\begingroup$ "A gas contains on the order of 10^22 molecules zipping about" per certain amount of volume or mass, I presume. :) Otherwise, spot on. $\endgroup$ Nov 15, 2021 at 12:45
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I was reading gaseous state when this question struck my mind what made us to assume that at every point inside the container gas exert equal pressure ?

The equilibrium pressure of a gas, just like the equilibrium temperature of a gas, is a macroscopic property applicable to the collection of gas molecules inside the container, not a microscopic property applicable to individual gas molecules at every point inside the container.

Considering the walls of the container, while the impact forces of individual molecules on the walls of the container will vary, it is the average of the impact forces of a collection of molecules that determines the macroscopic property of pressure. Similarly it is the average kinetic energy of the molecules that determines the macroscopic property of the temperature of a gas, not the kinetic energies of the individual molecules which will vary above and below the average.

Hope this helps.

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  • $\begingroup$ Does this mean the pressure measuring devices measures by itself this average net force on the walls of container ? So it stays nearly constant ? $\endgroup$
    – Orion_Pax
    Nov 14, 2021 at 11:50
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    $\begingroup$ @Orion_Pax Essentially yes. The pressure measured is the sum of the forces of all the molecules striking the wall divided by the area of the wall . So it measures the average linear momentum of the moving molecules of a gas, even though the instantaneous force at a given location can change form nanosecond to nanosecond since the speeds of individual molecules vary about the average. The lower the pressure the more sophisticated and precise the devices need to be to measure them. $\endgroup$
    – Bob D
    Nov 14, 2021 at 17:51
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    $\begingroup$ See the section Types of Vacuum Measuring Instruments in this link. solarmfg.com/wp-content/uploads/2016/02/… $\endgroup$
    – Bob D
    Nov 14, 2021 at 17:51
  • $\begingroup$ Thanks understood now that part $\endgroup$
    – Orion_Pax
    Nov 16, 2021 at 16:13
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    $\begingroup$ @Orion_Pax That assumption still holds. However the magnitude of the pressure for a real gas is generally less than an ideal gas due to the compressibility factor. $\endgroup$
    – Bob D
    Nov 16, 2021 at 16:51
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Consider a region of the fluid, and let $S$ be the surface of that region. If you take $\int_Sp\vec n$, where $\vec n$ is the normal vector, this is the net force that the pressure of the outside fluid is exerting on the mass of the fluid inside the region. If we take this force to be the only force acting on the fluid, then for the fluid inside the region to not be accelerating, this force must be zero. For this to be zero for every region, the pressure must be equal everywhere.

For fluids in a gravitational field, the force on the fluid is the pressure force plus its weight, so the pressure force must be equal in magnitude, and opposite in direction, to its weight. This pressure force is called "buoyancy", and there must be a gradient in the pressure to cause buoyancy to be equal to weight. However, for small differences in height, this change in pressure is small, and can be disregarded for many purposes.

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  • $\begingroup$ Nice one....... $\endgroup$
    – Orion_Pax
    Nov 16, 2021 at 16:35
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  1. The average kinetic energy of a gas molecule (temperature) is the same everywhere in the container, because when molecules of different energy collide, they are statistically likely to distribute the energy more evenly after the collision. Temperature evens out. The average speed of the molecules is also therefore the same everywhere.

  2. The average velocity of gas molecules everywhere is zero, assuming there are no currents. (in reality, although currents can indeed affect the relative pressure at different points, you need really strong currents to make a difference. Currents caused by convection are too small).

  3. The average density of molecules is the same everywhere. If you imagine any dividing plane, then given (1) and (2), if there were more molecules on one side, then there would be net flow to the other side.

  4. The average pressure at any point on the container's surface is proportional to the speed and density of gas molecules there, because it's the total rate of momentum transfer to the container wall, which is proportional to the molecule speed times the collision rate, and the collision rate is proportional to the speed and density, which we've already determined are the same everywhere.

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  • $\begingroup$ This answer assumes that the system has had time to reach equilibrium. In reality, things like gravity will affect a system even after it would have otherwise reached equilibrium. $\endgroup$
    – dotancohen
    Nov 15, 2021 at 11:04
  • $\begingroup$ @dotancohen how? $\endgroup$
    – Orion_Pax
    Nov 16, 2021 at 16:16
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    $\begingroup$ @Orion_Pax Pressure It is less close to being the same everywhere in those cases, but it's certainly true that pressure equalizes much more quickly than temperature, and you can have a temperature gradient and expect uniform pressure. The process of this equalization is acoustics. Again with point 4 above, pressure is momentum transfer per unit area, proportional to molecule speed^2 * density = temperature * density (for constant composition). If temperature * density is higher in one area, momentum is transferred out, i.e., it pushes its surroundings out and expands, reducing its density. $\endgroup$ Nov 17, 2021 at 3:12
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    $\begingroup$ It's not instantaneous. The moving piston creates a localized pressure change that propagates as a sound wave, which disperses into a uniform pressure change as it bounces around. Imagine that the cylinder is full of balloons squished together. The compressing side of the cylinder adds energy to the molecules in some, increasing P. They expand, transferring net momentum/energy to their neighbors, etc. Without the balloons, the situation is mostly the same. The molecules just bounce off each other instead of the balloon walls. There is diffusion, but diffusion is much slower than sound. $\endgroup$ Nov 17, 2021 at 13:37
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    $\begingroup$ Yes, that's right. And in order to properly make those assumptions you need to have a good idea about when they become invalid and how much inaccuracy they lead to. It's sometimes difficult. In a combustion engine, for example, you can't always ignore acoustic effects. $\endgroup$ Nov 17, 2021 at 18:24
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As with so many things, it's a matter of how much detail you need in your model to answer questions about the phenomenon you're studying.

Local changes in pressure propagate at the speed of sound in the medium. Assuming the pressure is equal everywhere is a simplified model, but good enough to answer questions about anything that occurs on time scales orders of magnitude longer than it takes a pressure wave to propagate through the vessel. The difference might matter for modeling an explosion, but not for pumping up a bicycle tire.

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'Gas' is a continuum (i.e. modeled as 'there are no discrete gas particles' there are gas 'parcels'). That continuum is going to be in hydrostatic equilibrium. The density, and therefore pressure, does not change significantly for gas. The result is a constant pressure in any axis.

If you measure 'every point inside the container', its seems like you are interested in a scale below the scale of the continuum itself, at which point the conceptual model will breakdown.

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  • $\begingroup$ Hmm makes sense thanks $\endgroup$
    – Orion_Pax
    Nov 17, 2021 at 15:25
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Addition to other answers:

The assumption is NOT true for a container that is large enough along the direction of the local gravity vector for differences due to gravitational force on the molecules to matter.
eg if you make a container that is about 5000 metres tall (!!!) and with rigid walls then the pressure at the top will be about half an atmosphere (about 50 kPa) less than that at the bottom. Most containers are "rather less tall" than this and the differences are usually able to be neglected.

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  • $\begingroup$ Hmm understood so in most thermodynamic processes when P is changing is it true that at every stage we r assuming P to be constant for whole gas ? I mean if the process is happening how one can reason the prssure to be uniform during the transition state inside the constainer? $\endgroup$
    – Orion_Pax
    Nov 16, 2021 at 16:20
  • $\begingroup$ @Orion_Pax. It's a matter of timescale. Pressure changes travel at the speed of sound. This varies with pressure, but in a reaction chamber in the order of a few hundred mm on a side a pressure change will cross the chamber in ~~= 2 millisecond or less. | As above a vertical pressure gradient due to gravity will occur in the steady state $\endgroup$ Nov 17, 2021 at 6:03
  • $\begingroup$ I see so horizontal pressure gradient due to piston moving/heat steady state is achieved in the whole chamber at any moment in the transition state? $\endgroup$
    – Orion_Pax
    Nov 17, 2021 at 7:50
  • $\begingroup$ @Russell_McMahon $\endgroup$
    – Orion_Pax
    Nov 19, 2021 at 13:21
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    $\begingroup$ @Orion_Pax as above - subject to speed of sound under given pressure and temperatures and any possible oscillations. All these are liable to be second order and minimal in effect in most cases. In extreme applications this may not be the case. $\endgroup$ Nov 19, 2021 at 23:07

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