Cylinder submerged in salt water (Ideal gas law, pressure)

[TheKShaugh]

Homework Statement

5. A large cylinder with a diameter of 3.00 m and a height of 3.50 m is closed at the upper end and open at the lower end. It is lowered from air into sea water with the air initially at 20.0°C and then to a depth of 75.0 m. At this depth the water temperature is 4.0°C, and the cylinder is in thermal equilibrium with the water.

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

(b) To what minimum pressure must the air in the cylinder be raised to expel the water that entered? (ie. If you were to pump enough air into the cylinder to displace all the water out the bottom, what would the pressure be in the cylinder?)

Homework Equations

[tex]\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}[/tex]

[tex]P = \rho g h[/tex]

The Attempt at a Solution

In my attempt I used the following for each variable:

[tex]P_1 = 101.3Kpa \\ V_1 = \frac{\pi}{4}d^2 \cdot h \\ T_1 = 20 C \\ P_2 = \rho g (75m) \\ T_2 = 4 C \\ V_2 = ?[/tex]

My reasoning for using the pressure at a depth of 75 meters for P2 is that in order for there to be any space in the container the pressure exerted by the air and the pressure exerted by the water have to be equal (in equilibrium). For part b) I found the height of the column of water, rho g h'd it and added the pressure at 75m to get the total pressure needed. Does that all make sense?

Thanks.

 

[BvU]

It does.
Any reason for your doubt ? What did you find for V2 ? Does it look reasonable ?
If still in doubt, what would be a way to check ?

 

[ehild]

In my attempt I used the following for each variable:

[tex]P_1 = 101.3Kpa \\ V_1 = \frac{\pi}{4}d^2 \cdot h \\ T_1 = 20 C \\ P_2 = \rho g (75m) \\ T_2 = 4 C \\ V_2 = ?[/tex]

My reasoning for using the pressure at a depth of 75 meters for P2 is that in order for there to be any space in the container the pressure exerted by the air and the pressure exerted by the water have to be equal (in equilibrium). For part b) I found the height of the column of water, rho g h'd it and added the pressure at 75m to get the total pressure needed. Does that all make sense?

Thanks.

Do not forget that the variable T in the Ideal Gas Law is the absolute temperature, in Kelvins.
P2 is not only the pressure of the water column, but the atmospheric pressure should be added.

ehild

 

[Chestermiller]

Does the 75 m correspond to the top of the cylinder or the bottom of the cylinder?

Chet

 

[HalfLostAndSearching]

Your approach is correct. In order for the air in the cylinder to displace the water, the pressure exerted by the air must be equal to or greater than the pressure exerted by the water at 75 meters depth. This means that the pressure at the bottom of the cylinder (P2) must be at least equal to the pressure exerted by the water (rho g h) plus the atmospheric pressure at the surface (101.3 kPa).

Your calculation for the volume of the cylinder is also correct. However, in order to find the pressure at 75 meters depth (rho g h), you need to know the density of sea water at that depth. Once you have the pressure at 75 meters depth, you can add the atmospheric pressure to find the minimum pressure required for the air to displace the water.

Overall, your approach is correct and your reasoning is sound. Keep in mind that the ideal gas law may not be applicable in this scenario since the air in the cylinder is not at a constant temperature. However, it can still provide a good estimate for the pressure needed to displace the water.