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Q & A: Water and air pressure

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Most recent answer: 10/22/2007
Q:
A pipe fifty feet long is submerged in a swimming pool 60 feet long and ten feet deep. The pipe is capped at one end and that end is stood straight up in the pool. What will happen to the water in the pipe?
- Raul Melgar (age 19)
Rock Island
A:
We're not going to do your homework for you, but here are some things to think about.

The pressure of the water at the open bottom of the pipe is the same as the pressure of the water in the swimming pool at that depth. The change in water pressure from one depth to another is just rho*g*(change in depth), where rho is the density of water (1 gram per cubic centimeter, or 1000 kg per cubic meter, using MKS units). The gravitational acceleration g is 9.81 meters/second^2, and h is measured in meters. This gives pressures in Pascals.

The water pressure at the top of the swimming pool is 1 atmosphere, due to the air pressing down on the top. You can find how far up in the tube the water can go because at some height in the tube the pressure will be zero. Above this height there will be a vacuum in the tube. Water is held up by pressure, and not "sucked up" by vacuum (the vacuum exerts no force on the water).

Now this situation won't last. At the interface between water and the space in the tube with vacuum, the water will evaporate very quickly (and may even boil!). Eventually, the portion of the tube with a vacuum in it will be full of water vapor. The pressure will be the "vapor pressure of water" at the temperature this is all at. The water will then settle down to a height below that originally calculated. If you let this settle down over a long period of time, you will have built a thermometer.

Tom J.

(published on 10/22/2007)

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