Question

A cylindrical diving bell, 2.80 m in diameter and 6.00 m tall with an open bottom, is submerged to a depth of 220 m in the ocean. The surface temperature is 25.0°C, and the temperature 220 m down is 5.00°C. The density of seawater is 1025 kg/m³. How high does the seawater rise in the bell when it is submerged?

Answer 1

ANSWER
For a volume of air, the relation "PV/T = constant".

let PVT and P₁V₁T₁are the values of pressure, volume and temperture on the surface and at the bottom of the cylindricla bell respectively. Then   PV / T = P₁V₁ / T₁
   V₁ = V(P/P₁)(T₁/T)
P = 1 atmosphere = 100,000 N/m²,
V = π r²h=πX(1.4)²X6 =36.9264m³
T = 25°C = (25 + 273)°K = 298°K

T₁ = 5°C = 278°K
P₁is the atmospheric pressure P plus the pressure of a column of sea water of 220 m height.
P₁ = P+ρgh = 100,000 + 1025×9.8×220 = 2309900N/m²
V₁ = π r²h₁
V₁= V (P/P₁) (T₁/T)
π r²h₁ = 36.9264 X (100,000 / 2309900)X ( 278 / 298)
3.14X1.4X1.4Xh₁ = 36.9264 X 0.043291917 X 0.932885906
h₁ = 1.491326896 / 3.14X1.4X1.4
     h₁ = 0.24 m
Therfore, the rise in sea water in the bell is h₀= h-h₁   h=6m
     h₀= 6-0.24
   h₀= 5.76m