For New Years my oldest and I have been blowing stuff up. Including underwater. Big fun.
Got her thinking:
Scenario. A sphere filled with water. Sealed. Pump some out. Say a quarter. No air goes in.
Question: what happens inside? Nothing runs in to fill the missing volume. Does the remaining water become less dense? Does it settle at the bottom? How does it behave?
Variation: Does it matter if it’s in a gravity field or in zero gee?
Is it a solid sphere, or flexible? If solid, some water vaporizes int the lowered pressure. If flexible, the sphere collapses.
You can look up cavitation for what would happen. At first there would be negative pressure on the sphere, IIRC lower then total vacuum, till some vapor formed then it would revert to positive absolute pressure.
In a rigid sphere, it’s like the process of vacuum evaporation.
Water would evaporate rapidly into the empty space, if you pumped the water out quickly it would boil. The liquid would not become less dense, the space would fill with water vapor until it reached equilibrium vapor pressure.
Under gravity, the liquid water would sit at the bottom with the less dense vapor at the top. Without gravity I guess surface tension effects would determine what happened, perhaps the vapor would form bubbles throughout the liquid? Perhaps the surface properties of the material the sphere is made from would be significant.
It creates a vacuum and settles to the bottom.
Most of the water will settle to the bottom, with a little almost-vacuum at the top. It’ll only be an almost-vacuum, because some water will evaporate. The amount of water vapor you’ll get in the almost-empty part is the same as the amount you’d get if there was air there, and depends only on the temperature.
If it’s in zero g, then the location of the almost-vacuum bubble will be random, instead of being at the top, and the water surface will become a concave spherical surface, to minimize the surface area.
This is the way many storage tanks (cylindrical ) fail. Vessels which maybe ordinarily strong under pressure fail quite easily under vacuum. There was a myth busters episode on this.
I think this answer is the most completely correct. There would be lower than total vacuum, that is, negative absolute pressure, until some point left the liquid phase, and then the system would rapidly shift to water vapor filling the excess space. The point of this is that liquids including water do have tensile strength. Now, they have very poor notch strength, so that if some point opens up, the opening spreads very easily. But unlike what many assume, the tensile strength per se is IIRC in the hundreds of PSI. In practice, though, any microscopic suspended particle or bubble is enough of a weak spot to nucleate a tear.
At least that’s what I think I learned long ago.
Curiously, part of this came from watching a Mythbusters on Hulu. They had to build large hollow steel sphere - they were building a giant Newton’s Cradle - and that, combined with dropping small explosives into the hot tub - got her thinking.
So in the short term the water would drain and pressure would grow on the rigid steel frame until water began vaporizing and eventually a pressure balance is achieved? Do I have that right?
Keep in mind, the vapor pressure of water depends on the temperature, so it may not be that great. At, e.g., 15°C it will only equal 0.0168 atmospheres. Therefore it may never balance the outside pressure which might be 1 atm or more. The steel would have to bear the net forces.
The relevant “pressure balance” is only inside the sphere, between the liquid water and the gaseous water vapor. You can loosely think of it this way. Gas and liquid molecules are always moving with at range of different speeds/energies. At the surface of the liquid, some faster liquid molecules are constantly escaping into gas; and some slower gas molecules are constantly condensing into liquid. Equilibrium vapor pressure is achieved when there are enough molecules in the gaseous state that these two tendencies exactly balance out.
It’s a bit counterintuitive, but as Chronos pointed out, the amount of water vapor required to reach this equilibrium vapor pressure is no greater than the amount that would be required if air is present, i.e. it’s the same amount of water vapor that would be present in saturated air at 100% humidity. At room temperature at sea level that’s only about 2% of atmospheric pressure.
So the equilibrium is between the liquid water and the gaseous water, but that implies about 98% vacuum. Pressure will not equalize between the inside and outside of the sphere unless the sphere deforms or ruptures.
Or, perhaps more likely, leaks.
Hmm. All things being equal, would a semi-vacuum bubble in a zero-g water tank with wetted walls tend to remain in the bulk liquid, or would it move to a side wall and form a ‘blister’ there? I suppose it depends on the composition of the tank wall. If the wall is strongly hydrophilic, can it ‘repel’ a bubble at a distance, like a buoyancy force away from the walls caused by intermolecular forces in the water?
One of the few physics answers where the “in a gravity free environment” is the more advanced answer. 
Here’s what water in a flask inside a bell jar w/ a vacuum pump looks like.
(I have always loved watching water boil at room temperature.)