Pressure, water and gravity

I can't see why not. At any depth the pressure above and below the piston is approximately equal so the resultant force acting is simply the weight. This assumes that the frictional resistance at the tube sides is negligible.

Is the top of the tube above the water level? If so, the weight would have to be heavier than the total volume of water. Right? At a length of 25,000 ft. and a diameter of 1 ft. the water in the tube would weigh 1,225,188.7 lbs.

The piston, if made of solid lead would be 2,200.7 ft. in length. Actually that would balance the weight of the water, so it would have to be a bit heavier.

I don't think the tube has anything to do with it since the weight of the piston immediately transmits excess pressure to the bottom of the tube forcing the water out. The only principle which matters (Archimedes) is the density of the water which will always be less than the average density of the piston.

I wasn't sure if I was thinking about this the right way. Consider; the pressure at the bottom of the tube is equal to the weight of all the water from the bottom to the top, if the piston is forcing water out the bottom, but no new water is replacing it from the top, it wouldn't be heavy enough to continue forcing water out the bottom. Or am I missing something?

Ok,my apologies, you're right.I did not read the 'leaving' an empty tube' bit !

In that case the piston would stop as the piston + tube would be equivalent to a ship with a partially vacated ballast tank.
The 'ship' would 'float' when it displaced its is own weight of water i.e. the piston would stop when the piston had displaced its own weight of water.