# How high does sea water rise in the bell?

1. Nov 25, 2003

### vmind

Hi,

I would appreciate help with the following problems:

1.) A diving bell in the shape of a cylinder with a height of 1.00 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/cm^3). The air in

the bell is initially at 22.0 degrees C. The bell is lowered to

a depth (measured to the bottom of the bell) of 46.0 fathoms or

84.1 m. At this depth the water temperature is 4.0 degrees C,

and the air in 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 increased

for the water that entered to be expelled?

Here is what I have done so far for this question:

a). P= P_atm + p_seawater * g * h = 1.013e5 + 1.025 * 9.8 * 84.1 =

= 1.0214e5 Pa.

PV=nRT

(1.0214e5)*V= nR(22.0)

I am not sure how to find the number of moles.

2.) A cylinder that has a 40.0 cm radius and 50.0 cm deep is filled with air at 20.0 degrees and 1.00 atm. A 24.0 kg piston is now lowered into the cylinder, compressing the air trapped inside. Finally, a 77.0 kg mans stands on the piston, further compressing the air, which remains at 20 degrees C.

a). How far down (delta h in mm) does the piston move when the man steps on it?

b.) To what temperature should the gas be heated to raise the piston and the man back to h_initial?

Here is what I have so far for this question:

W_piston + W_man = (24.0+77.0)* g= 989.8 N

B_air = p_air * V_displaced air * g

Total weight = B_air

(1.2kg/m^3)* V_displace air = (24.0+77.0)

V= 8.42 m^3

V_cylinder =0.400^2*pi*0.500= 2.513e-1 m^3

For some reason my V_displaced air is more than V_cylinder

2. Nov 25, 2003

### Chi Meson

to start with,You don't need to know the number of moles because this will remain constant.

AND: the PV = nRT equation is for use with the Kelvin temperature scale.
Try it again (0 C = 273 K)