Posted by Phil on .
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?
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 P2-P1 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.