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23 Nov 2019

Question: "A diving bell is a 3-m tall cylinder closed at the upperend but open at the lower end. The temperature of the air in thebell is 20 degrees C. The bell is lowered into the ocean until itslower end is 100m deep. The temperature at that depth is 10 degreesC.

a. How high does the water rise in the bell after enough time haspassed for the air inside to reach thermal equilibrium?

I solved this by doing pV=nRT for the air in the bell both for whenit's on land and for when it's in the ocean. Then I found the ratioof the Vs in each situation and used this to get that the waterrises 2.73m up into the bell.

b. A compressed-air hose from the surface is used to expel all thewater from the bell. What minimum air pressure is required to dothis?

I'm stuck and really confused. Won't the air add moles of gas aswell as pressure? Also, whenever the system in thermal equilibrium,Pin = Pout. So won't Pin just be 10.9 atm, the pressure at thatdepth of water?

The Cramster textbook solution basically says "P2 = P1 +(waterdensity)(g)(waterheight). Looks like something about equalforces, but I don't get it.

THANKS!!

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Tod Thiel
Tod ThielLv2
28 Jul 2019

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