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.
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?
1 answer
0 answers