Answer to Question #245005 in Physics for Diane

Question

A diving bell in the shape of a cylinder with a height of hb= 1.2m is closed at the upper end and open at the lower end. The bell lowered from the air into sea water with mask density p=1.033 g/cm3. The air in the bell is initially at 24.0 Celsius. The bell is lowered to a depth (measured to the bottom of the bell) of d=140 m. At this depth the water temperature is 6.0 Celsius, and the bell is in thermal equilibrium with the water. Take g=9.7 m/s2 and 1 atm as 101300 Pa.

to what minimum pressure must the air in the bell be raised to expel the water that entered? Input your answer in kPa to one decimal please.

Answer 1

Apply combined gas law. P₀ - pressure at sea level, P₁ - pressure at 140 m.

$$\frac{P_0V_0}{T_0}=\frac{P_1V_1}{T_1},\\\space\\ \frac{P_0\pi r^2h}{T_0}=\frac{(P_0+\rho gD)V_1}{T_1}.\\\space\\ V_1=\frac{T_1P_0\pi r^2h}{T_0(P_0+\rho gD)}.$$

This is the volume of air trapped in the bell at 140 m below the water surface. The height of air in the bell is

$$a=\frac{V_1}{\pi r^2}=\frac{T_1P_0h}{T_0(P_0+\rho gD)}.$$

The height of water in the bell is

$$w=h-a=h\frac{T_0P_0+T_0\rho gD-T_1P_0}{T_0(P_0+\rho gD)}.$$

To expel the water in the bell, the pressure of air inside must 'oppose' the pressure of water at 140-m-depth minus the pressure of water inside the bell:

$$P_i=\rho g(D-w)=\\\space\\=\rho g\bigg(D-h\frac{T_0P_0+T_0\rho gD-T_1P_0}{T_0(P_0+\rho gD)}\bigg),\\\space\\ P_i=1391.5\text{ kPA}.$$