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 Quote by Doc Al Sure. (Better to say that the air is pushed in through the hole by the atmosphere, not pulled in, of course.)
<laughing> Touche, you got me there!

Quote by Doc Al
 q_goest said: So as the balloon sinks, if it were to be compressed, there would need to be air or something getting into the pipe to take up the decrease in volume of the balloon.
Why is that? (This is the point that I'd like to see physically justified.)
Ok, good question. Note the assumptions we’ve made:
- The container is sealed so that nothing can get in or out.
- Container is infinitely rigid, therefore internal volume can’t change.
- Water is essentially incompressible. Let’s neglect water's bulk modulus for now and add that in later.
- The balloon’s pressure is in equilibrium with the water pressure. If the balloon’s pressure was more or less than the water, it wouldn’t be in static equilibrium and the balloon would have to expand or shrink respectively until it was in equilibrium. By equilibrium, I mean that the pressure on the inside of the balloon is essentially the same as the pressure on the outside.*

Now, let’s assume that as the balloon goes down, it shrinks in volume.
1. Since the container is sealed, neither air nor water can get into the container to make up for the volume left by the shrunken balloon.
2. Since the container is rigid, the container’s volume can’t decrease to make up for the volume.
3. Since the water is incompressible, the water can’t expand to make up for this decrease in volume.
4. All we’re left with is the knowledge that the water outside the balloon must be equal in pressure to the gas inside the balloon. So if nothing in the system can make up for the shrinking balloon volume, and if the pressure inside is the same outside the balloon, then we’re left with the assumption that the balloon pressure hasn’t changed but the water pressure did!

#4 leads us to the realization that the pressure all along the column must change as the balloon sinks. And we also know that when the local water pressure drops below the saturation point, it must boil, and the lowest pressure is at the upper most surface. So when the upper most surface reaches saturation, it must boil which creates water vapor above the column.

 Quote by Doc Al Quite counterintuitive, if I understand what you're saying. Are you saying that if I pull the balloon down 1000 meters (a difference in hydrostatic pressure of about 100 atmospheres) in that closed container, that somehow that huge water pressure at the bottom is negated and that the balloon is not compressed?
You’ve left out some assumptions. What should the pressure be at the top of the column when the balloon is at the top of the column?

For example, if the pressure at the top of the column was 15 psia, then the balloon would have to have 15 psia in it to be in equilibrium. Assuming roughly 2 feet per psi for water, then 30 feet down the pressure would be 30 psia. Now if the balloon sinks 30 feet, the pressure at that level has gone to 15 psia and the top of the column has dropped to 0 psia and begins to boil. Until it did that, the balloon couldn’t shrink. But once it passes that point, the water surface begins to boil producing water vapor at a much lower density than the liquid water. Now you have something that can expand and make up for the difference in the incompressibility of the water and rigidity of the container. Now as the balloon sinks, the decrease in balloon volume can be taken up by the increase in vapor space above the water. So in this case, when the balloon is 1000 feet down, it must have shrunken to the size of a pea.

The above is simplified by assuming the bulk modulus of water is zero. But since there is some finite bulk modulus, the water actually decreases in density slightly as pressure drops. Checking a database for water shows that the density decrease between 0.4 psia (aprox. boiling at 70 F) and 15 psia is only about 0.003%. If this is taken into account, we find that the balloon shrinks very slightly with a corresponding increase in pressure and the water expands very slightly with a corresponding decrease in pressure.

* Note that in real life, there is some minor difficulty with this assumption. Rho*g*h of air inside the ballon is not equal to rho*g*h of water outside. The difference has to be made up by internal stresses of the balloon, but that’s inconsequential. One could for example, instead of using a balloon, use a small cylinder containing air with a piston. In fact, it may be easier to visuallize using a cylinder and piston arrangement instead of a balloon.