I always hear pressure defined as the force exerted by particles on the walls of the container they're being held in. This makes sense since the mathematical definition of pressure is $ p = \frac{F}{A} $. So, can pressure exist without walls to exert force on?

My understanding of pressure motivates me to think that even without a container, particles of gas in a vacuum could create pressure since their collisions with each other result in forces being exerted on areas (the areas being the surfaces of the particles being collided with).

On the other hand, a liquid in a vacuum could not exert pressure since on a microscopic level, the particles of liquid can't collide due to the cumulative strength of the bonds holding them together.

  • 4
    $\begingroup$ "particles of gas in a vacuum" - if there's particles in it, it's not vacuum! Also, which "container" do you think the earth's atmosphere is in? $\endgroup$ – ACuriousMind Oct 4 '15 at 0:01
  • 5
    $\begingroup$ an example would be a self gravitating ball of gas, such as a star. $\endgroup$ – user83548 Oct 4 '15 at 0:07
  • $\begingroup$ Ah, yes you're correct @ACuriousMind. And thank you for the better example @brucesmitherson! $\endgroup$ – user94491 Oct 4 '15 at 0:13
  • 1
    $\begingroup$ Yes, and you're almost right about how to define pressure at a point in a gas; rather than using the collisions of one particle with another, one uses an an imaginary plane at point. $\endgroup$ – Mozibur Ullah Oct 4 '15 at 0:13
  • $\begingroup$ "an example would be a self gravitating ball of gas, such as a star." Another example of how an (enormous) pressure can exist even without containing walls is a 'dynamic' situation, such as the laser-driven implosion of a NIF capsule. In this case very large pressures are supported by the inertia of the mass of the capsule itself. Of course such a situation can only exist for a short time before the capsule flies apart. $\endgroup$ – user93237 Oct 4 '15 at 1:11

The pressure of a gas is defined as the force the gas would exert upon a surface or container. However, there is no need for a container for pressure to exist. For instance, the air you're breathing right now (unless you're in an airplane or submarine) has pressure due to the column of atmosphere above you. Stars are balls of gas (plasma, actually) that are pressurized by gravity; no containers to be seen.

| cite | improve this answer | |
  • $\begingroup$ Excellent answer, Mr. Griscom! Thank you very much. $\endgroup$ – user94491 Oct 4 '15 at 2:56

An addition to Daniel's answer:

It regularly happens that a certain definition isn't applicable when talking about liquids and gases. For example, almost all definitions in thermodynamics are defined for closed systems, which is rarely applicable in the real world.

To overcome this, there is the notion of a control volume. For this, we basically take a slice out of our medium (be it gas, liquid or anything else). We put this slice in a special box - so special, that it behaves exactly like the medium that originally surrounded our control volume. We can now forget about the rest of the medium, and just focus on what it does to the walls (boundary) of our box, and what our control volume does to the walls.

Related to your question, we can now take a little slice out of the atmosphere, and put it in our box. We can now forget about the rest of the atmosphere, i.e., put the box in a vacuum. Intuitively, you would think that the box should blow up like a balloon when put in a vacuum. However, the box behaves like there is an atmosphere, and will retain its shape - it will create a force on its walls. Now, we see pressure arise clear as the day: there's a force on the walls of the box, and since the walls have an area, you get p=F/A.

| cite | improve this answer | |
  • $\begingroup$ How is the flux through the imaginary walls handled when computing p? $\endgroup$ – Emil Oct 5 '17 at 16:33
  • $\begingroup$ @Emil we define the box such that the flux across the imaginary walls is exactly zero; that's what I meant by a box that behaves exactly as the surrounding volume. Any flux across the box walls would displace the surrounding volume, which by our definition cannot happen. $\endgroup$ – Sanchises Oct 5 '17 at 17:00