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If you were to sink a container to the bottom of a deep ocean and seal it there, then bring it up to the surface, would it retain its pressure?

The answer for a gas is obviously yes, but what about for a liquid like water which is incompressible? Once the crushing weight of the water column above is removed, does the water retain it's quality of "pressurizedness" or return to normal water? I guess a clear way to test this would be to bottle up a deep water fish and bring it up to the surface and see if it explodes.

While we're at it, what about a solid? Barring any elasticity and incidental temperature change, will a solid object break a non-sealed glass container which is exactly fitted to it and then placed in vacuum?

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I pity the fish, though. –  Hanno Fietz Dec 7 '12 at 11:37
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The problem in your post is that you don't consider the force resisting the water pressure, which is exerted by the walls of your container.

To do this, imagine a two-chamber container, with outer walls infinitely strong, but with a piston in between them which is maintained at its central position by some force that you exert. Chamber A is filled with air at sea level and then closed, chamber B is left open as you sink the container. The force you need to exert on the piston raises as you sink it, in proportion with water pressure, say at some depth it is area times $P_1$. Now close chamber B too. You can now bring your container anywhere, this won't change the balance inside.

And you can still exert the same pressure $P_1$ on the piston, and water is under pressure $P_1$. But you can also vary this pressure on the piston, to any value you like and bring the water to some other pressure level, without having any movement of the piston (as long as you remain in the incompressibility regime). In particular, releasing the force suddenly won't lead to a burst, because water is incompressible.

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Pressure is synonymous with energy - its a kind of kinetic potential energy, not unlike heat. Kinetic theory relates these two quantities, together with volume.

Let's make some assumptions. When we bottle something we expect its volume to stay about the same - the glass shouldn't flex. Also we pretend the bottom of the ocean is a comfy room temperature, so that heat is a constant too.

Now it might make more sense why pressurisedness (energy) doesn't change. Where could that energy go?

The confusion lies in what incompressible means. Compressibility relates changes to volume with changes in pressure/potential energy.

Consider a piston containing water and one containing air. Recall that energy = force * distance.

If we pressurised the air, it would compress easily at first. The force would be small and the distance large. Eventually the piston would reach our desired pressure, and the force would reach a maximum (when the air pressure balances it).

If we pressurised the water, it's volume would barely change. The force would be large and applied over a very short distance. However the same pressure (and force on the piston) would be reached.

This means that both pistons contain the same potential energy! They are just as equally compressed despite very different volumes. How come we don't see more pressurised solids in day to day life? I think this might be because they just can't compressed as easily as water or liquid. The most common application is the spring, where we call this pressure "stress". Engineers well now that not many solids behave very "elastically" - when they are suitably compressed (compression is called "strain" here), they are likely to deform/bend (energy loss) rather than return to their original shape.

See the similar question about why iron balls don't bounce.

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The potential energy of the piston with air and water will not be the same. Because this energy should be the same as the work done by the piston, which is defined as $dW=-pdV$. Since the volume change of the air is much bigger, at the same pressure, it has to experienced larger work. –  fibonatic Apr 11 at 10:21
    
@fibonatic but p is a function of V! it is certainly not a constant - like i said the air piston has a low pressure over a larger volume, where as the water experiences a rapid increase in p with V. that's why they are d's and not deltas. (i could be completely wrong of course, but i don't have time to give this further thought right now sorry) - i guess some accomodation has to be made to ensure same amount of air/water is compressed in the final state - the point is the energy per unit volume is the same in both cases –  user3125280 Apr 11 at 10:39
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Water is slightly compressible, so it will hold its pressure as long as the container does not stretch.

But since it's only slightly compressible, if the container bursts under pressure it will probably not be an explosive failure. This is because at the time of failure, unlike a gas, the water does not push for a long enough time on the failing part of the container to generate much speed. This is why pressure containers are often pressure tested with water or oil instead of air or other gasses.

If a solid is slightly compressible, it will retain pressure inside a container. In practice, if a incompressible solid is enclosed in a pressurized container, there will usually be some gas or liquid mixed in with it that will retain the pressure.

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Your last point is similar to the reason that nuclear plants have pressurizers. A pressure boundary filled with a nearly incompressible fluid will experience large pressure swings due to small disturbances. Classic example is that of a glass filled with water shattering epically when shot with a bullet, whereas it would only leave small bullet holes if it were filled with a gas. –  AlanSE Sep 28 '11 at 3:54
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