Solving Pressure with Gases in a Diving Bell

[Antepolleo]

Here's the problem:

4. A diving bell in the shape of a cylinder with a height of 2.10 m is closed at the upper end and open at the lower end. The bell is lowered from air into sea water ( p = 1.025 g/cm3). The air in the bell is initially at 16.0°C. The bell is lowered to a depth (measured to the bottom of the bell) of 47.0 fathoms or 86.0 m. At this depth the water temperature is 4.0°C, and the bell is in thermal equilibrium with the water.

(a) How high does sea water rise in the bell?

(b) To what minimum pressure must the air in the bell be raised to expel the water that entered?




My question is, how are you supposed to figure this out without knowing the diameter of the bell?

 

[Antepolleo]

Hmm.. I tried an approach where I let the number of moles of the gas be constant, but it got me nowhere.

 

[Janitor]

My approach would be this...

P(z) = gpz + P_0

where P(z) is the pressure as a function of depth z below sea level,
g is acceleration of gravity, p (rho) is density of water, P_0 is pressure at sea level (i.e. one atmosphere).

You have been given p and z, and you can look up g and P_0, so you can solve for P(z).

Then if we make the assumption that none of the air in the cylinder dissolves in the water as the cylinder is lowered (number of moles is constant, as you say), we should have

P(z)V(z)/T(z) = P_0 V_0/T_0

where V(z) is the volume of air in the cylinder at depth z, T(z) is the absolute temperature at depth z, V_0 is the volume of the entire cylinder, and T_0 is the absolute temperature at sea level.

You can solve this for V(z), since you know all the other quantities.

The height to which the water rises is h = L [1-V(z)/V_0] where L is the length of the cylinder.