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This question is a bit general but hopefully someone can point me in the right direction.

If you have a metal container, welded shut that's fitted into a hole that is say 10 meters wide and 30 meters deep.

This has the volume of about 21,205

If you pump pure oxygen into the tank until all the water was pushed out of that cylinder, what would be the pressure of the oxygen in the tank. How much oxygen would be in the tank?

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  • $\begingroup$ If there was a hole at the bottom the water would leak out under gravity. If no hole where do you expect the water to go? $\endgroup$ – Solar Mike Jun 4 '19 at 3:52
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If the hole that the water drains out from is a hole with a check valve located at the height H from the top of the tank the pressure of the gas is equal to the pressure of water at the depth H. And the bottom part below that level will be water only.

However if you want to drain the entire tank and replace the water with the gas you should have the hole on the bottom of the tank, otherwise there is always water remaining below the discharge level. In that case the pressure of the gas will be 3 atmosphere. Therefore you can use ideal gas low to calculate the weight of the oxygen:

$$ PV=nRT $$ -P is pressure

-V is volume, liters

-n is number of gas molecules

-R is the ideal gas constant= 0.08206 L atm mol–1 K–1

-T is temperature in kelvin.

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