Physics Help Forum Tank and hydrostatic pressure

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 May 20th 2017, 01:58 PM #1 Junior Member   Join Date: May 2017 Posts: 2 Tank and hydrostatic pressure Hey, Let's imagine I have a tank, filled with air at atmospheric pressure. This tank have a closing/opening valve. I dive this tank at 100m under water and opened ce valve, my filling speed will be sqrt( 2*g*h) with h=100m. Now, if i want to dive the tank (with a volume X) continuously at a velocity Vd. How can i calculate, how much time it will take to fill it ? Asa water enter the tank, the pressure won't be Patm anymore.. Thanks !
 May 20th 2017, 02:47 PM #2 Senior Member   Join Date: Aug 2010 Posts: 369 As the water goes into the tank, is the air venting or being compressed in the tank?
 May 20th 2017, 02:54 PM #3 Junior Member   Join Date: May 2017 Posts: 2 I guess the air goes away in bubble like in real life
 May 21st 2017, 06:38 AM #4 Senior Member   Join Date: Aug 2010 Posts: 369 I have no idea what "goes away in a bubble" means. If the air is not vented then it will form a bubble at the top of the tank but that is not "going away"! The pressure of the air in the tank will be given by the "ideal gas law", PV= nRT. As the tank is filled with water, the V, the volume of the air in the tank, will be the initial volume of the tank minus the new volume of water. "n" is the number of molecules of air in the tank, "R" is the "gas constant", and "T" is the temperature of air in the tank which you can take to be constant here so "nRT" is a single constant, K. You have PV= K so that P= K/V. When P is air pressure, V is the volume of the tank. You can calculate K from that. As V decreases P must increase by P= K/V. If the tank is vented so that, by "bubble away", you mean that the air bubbles up out of the tank to the top of the water, then the pressure of the water in the tank will just be the pressure of the water at the bottom of the dive.

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