Find the depth of a tube submerged in water that is half filled with air

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Homework Help Overview

The problem involves a tube submerged in water that is half filled with air, where the air behaves as an ideal gas. The context includes calculating the depth of the tube in a freshwater lake, given specific parameters such as atmospheric pressure and the length of the tube.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the pressure of the compressed air in the tube and atmospheric pressure, questioning how the volume change affects pressure. There are attempts to connect the ideal gas law to the problem, and some participants express uncertainty about the necessary variables.

Discussion Status

The discussion is active, with participants exploring the implications of the ideal gas law and the relationship between air pressure and water pressure. Guidance has been offered regarding the pressure at the air/water interface and the need to consider pressure at the same height in a fluid.

Contextual Notes

Participants note the lack of specific information regarding the volume and number of moles of air, which is acknowledged as a constraint in their reasoning.

KEØM
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Homework Statement



A tube of length L = 25 m that is open at one end contains air at atmospheric pressure. This is done in Denver so atmospheric pressure P = .667 X (1.10 X 105 Pa). The tube is thrust vertically into a freshwater lake until water rises halfway up in the pipe. Find the depth of the tube in the water. Note: air can be treated as an ideal gas.

In the attachment there is a picture on problem #5. The picture will really help clarify the problem.

Homework Equations


P = P(initial) + [tex]\rho[/tex]gd
[tex]\Sigma[/tex]F = ma
P=F/A

The Attempt at a Solution


I know the pressure of the air inside the tube and I also know that the forces exerted by the air and the water are the same but I am just not sure how to tie all of these things together to find the depth.
 

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Hint: What's the pressure of the compressed air in the tube?
 
Isn't it equal to the atmospheric pressure?
 
KEØM said:
Isn't it equal to the atmospheric pressure?
Not after being compressed. (How did its volume change?)
 
By putting it in the water the volume was halved so the pressure is the atmospheric pressure divided by 2?
 
KEØM said:
By putting it in the water the volume was halved so the pressure is the atmospheric pressure divided by 2?
No. Use the hint that the air can be treated as an ideal gas. (What's the ideal gas law?) You can assume the temperature is constant.
 
But I don't know the volume or the number of moles either.
 
KEØM said:
But I don't know the volume or the number of moles either.
You don't care about the actual volume, only that it went from V to V/2. The number of moles is constant.
 
Ok so I solved for pressure now can I use this in the formula P = P(initial) + [tex]\rho[/tex]gd but I am not sure if that will work.
 
  • #10
Make use of the fact that the pressure must be the same at the same height in a fluid. What must the water pressure be right at the air/water interface in the tube?
 

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