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

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Homework Help Overview

The problem involves a large cylinder submerged in salt water, focusing on the behavior of air within the cylinder as it is lowered to a depth of 75 meters. The initial conditions include air at 20.0°C, and the water temperature at depth is 4.0°C. The discussion centers on calculating the height of the water that rises in the cylinder and determining the minimum pressure required to expel the water.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of the Ideal Gas Law and hydrostatic pressure to the problem. There is an exploration of the variables involved, including initial and final pressures, volumes, and temperatures. Questions arise regarding the interpretation of pressure at depth and the need to consider atmospheric pressure.

Discussion Status

The discussion is ongoing, with participants clarifying their reasoning and checking assumptions. Some guidance has been offered regarding the absolute temperature in the Ideal Gas Law and the need to include atmospheric pressure in calculations. There is no explicit consensus yet, as participants continue to explore different aspects of the problem.

Contextual Notes

Participants are working under the constraints of the problem statement, which specifies the dimensions of the cylinder and the conditions at the given depth. There is an emphasis on ensuring that all relevant pressures are accounted for in the calculations.

TheKShaugh
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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



\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

P = \rho g h

The Attempt at a Solution



In my attempt I used the following for each variable:

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 = ?

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.
 
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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 ?
 
TheKShaugh said:
In my attempt I used the following for each variable:

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 = ?

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

Chet
 

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