Posted by Phil on Thursday, January 8, 2009 at 8:06am.
A cylindrical diving bell 5 m in diameter and 10 m tall with an open bottom is submerged to a depth of 150 m in the ocean. The temperature of the air at the surface is 20 degrees Celsius, and the air temp 150 m down is 2 degrees Celsius. The density of seawater is 1025 kg/m^3. How high does the sea water rise in the bell when the bell is submerged?

Physics  drwls, Thursday, January 8, 2009 at 9:17am
The number of moles of air in the bell remains the same. Use the reationship
PV/T = constant. Let 1 represent sea level condtions and 2 the condistions at 150 m depth.
P1*V1/T1 = P2*V2/T2
Calculate P2P1 from the depth.
P2  P1 = (density)*g*100 m = 1.506*10^6 N/m^2
P1 = 1 atm = 1.015*10^5 N/m^2
P2/P1 = 14.8
T2/T1 = 275/293
Solve for V2/V1 . That will equal L2/L1, the ratio of the lengths of the air column in the bell. L1 = 10 m. The distance the water rises is L1  L2.
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