How high does sea water rise in the bell?

[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

 

[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)

 

[M98Ranger]

, so I am not sure what I am doing wrong.

Hi,

To answer your first question, the height to which the sea water will rise in the bell can be calculated using the formula P = P_atm + p_seawater * g * h, where P is the pressure at the bottom of the bell, P_atm is the atmospheric pressure, p_seawater is the density of sea water, g is the acceleration due to gravity, and h is the height of the water column in the bell. Rearranging this formula, we get h = (P - P_atm) / (p_seawater * g). Plugging in the values given in the problem, we get h = (1.0214e5 - 1.013e5) / (1.025 * 9.8) = 0.81 m. So the sea water will rise to a height of 0.81 m in the bell.

For part b), to calculate the minimum pressure required to expel the water that entered the bell, we can use the same formula, but this time we will use the density of air (p_air) instead of sea water. We also need to consider the change in volume of the air due to the change in pressure. The final volume of the air in the bell will be the initial volume (V_initial) multiplied by the ratio of the final pressure (P_final) to the initial pressure (P_initial). So the final volume can be written as V_final = V_initial * (P_final / P_initial). Plugging this into the formula, we get P_final = (P_atm + p_seawater * g * h) * (V_initial / V_final). Since the volume of the air in the bell remains constant, V_initial = V_final and we can simplify the equation to P_final = P_atm + p_seawater * g * h. Plugging in the values, we get P_final = 1.013e5 + 1.025 * 9.8 * 0.81 = 1.0205e5 Pa. So the air in the bell must be increased to a minimum pressure of 1.0205e5 Pa in order to expel the water that entered.

For your second question, to find the number of moles of air in the bell, we can use the ideal gas law, PV = nRT