Question: \"A diving bell is a 3-m tall cylinder closed at the upper end but ope

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

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

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

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

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

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

THANKS!!

Explanation / Answer

I agree with your answer of 2.73 (ignoring the slightly greater density of seawater 1025 kg/m^3) To remove all of the water from the bell the (absolute) pressure in the bell must equal the pressure of the atmosphere + the pressure of the seawater above the bottom of the bell So P = 1.01 * 10E5 + 1025 * 9.8 * 100 = 1.106 * 10E6 N/m^2 If they mean "gauge" pressure then P = 1.005 * 10E6 N/m^2

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