Thermodynamics problem, with a pulley

[Urmi Roy]

Homework Statement

The diagram is attached.
The description of the setup is in the second attachment.

I do have solutions to the problem, but I don't understand parts of it.
Questions:

a) What is the initial pressure in the liquid?
b) Please show that the magnitude of the tension in the wire is equal to the magnitude of of the spring force on piston B for all positions in process 1-2.
c) What is the change in internal energy of the fluid from process 1-2?

Homework Equations

First law of thermodynamics E2-E1=Q-W
The ideal gas law PV=mRT

The Attempt at a Solution

a) According to the free-body diagram on the third attachment, (and the fact that in the initial state the spring force Fs is 0) the pressure of the liquid pressure should be equal to PB.

However in our solutions, it says initial pressure of the liquid is PA. I'm not sure why this is true.

b) When the rope going around the pulley is pushed up on one side because the piston on side A goes up, I don't see how/ why the tension in the spring should change.

In our solutions, it says it changes as 2*σ*g*x, where x=upward displacement of piston on side A.

c) The pulley does work in pulling the piston in side A up. However in our solutions, is says we can treat the two gases , copper blocks and liquid as a system with no work/ heat exchange with the surroundings. I don't understand why they don't take into account the work by the pulley.

 

[TSny]

The Attempt at a Solution

a) According to the free-body diagram on the third attachment, (and the fact that in the initial state the spring force Fs is 0) the pressure of the liquid pressure should be equal to PB.

However in our solutions, it says initial pressure of the liquid is PA. I'm not sure why this is true.

In the free-body diagram for piston B, you forgot to include the force from the pin. It will be easier to consider the free-body diagram for piston A.

b) When the rope going around the pulley is pushed up on one side because the piston on side A goes up, I don't see how/ why the tension in the spring should change.

In our solutions, it says it changes as 2*σ*g*x, where x=upward displacement of piston on side A.

When Piston A goes up, Piston B goes down. So, the spring is compressed. Remember Hooke’s law to get the force of the spring.

c) The pulley does work in pulling the piston in side A up. However in our solutions, is says we can treat the two gases , copper blocks and liquid as a system with no work/ heat exchange with the surroundings. I don't understand why they don't take into account the work by the pulley.

The pulley doesn't do any work. However the tension in the wire does work on piston A. Likewise the spring does work on piston B. What is the total work done by both of these forces?

 

[Urmi Roy]

Thanks for the help!

In the free-body diagram for piston B, you forgot to include the force from the pin. It will be easier to consider the free-body diagram for piston A.

That's only for the initial instant (the pin) ...The free body diagram should be good for all instants after that. I think my question is more about the hydrostatics...Is the pressure at the top surface of the fluid (just below the piston), denoted by PB equal to the liquid pressure?

When Piston A goes up, Piston B goes down. So, the spring is compressed. Remember Hooke’s law to get the force of the spring.

I understand about the spring, but if the rope on one side of the pulley goes up, the other side goes down by the same length...I don't see how the tension should change.

The pulley doesn't do any work. However the tension in the wire does work on piston A. Likewise the spring does work on piston B. What is the total work done by both of these forces?

Since they both move by the same amount, and incidentally k=2*σ*g, the work done by both are equal but of opposite signs?

 

[TSny]

I think my question is more about the hydrostatics...Is the pressure at the top surface of the fluid (just below the piston), denoted by PB equal to the liquid pressure?

Before the pin is pulled, the liquid is at rest and the pressure in the liquid is uniform throughout with the liquid pressure in A equal to the liquid pressure in B. However, as I see it, after the pin is removed the liquid passes through the very narrow capillary tube from B to A. This requires the pressure of the liquid in B to be greater than the liquid pressure in A while the fluid is moving through the tube.

I understand about the spring, but if the rope on one side of the pulley goes up, the other side goes down by the same length...I don't see how the tension should change.

If the rope on the right side of the pulley goes up an amount Δx, then the rope on the left side goes down Δx. There will now be more weight of rope on the left than on the right. This will create tension in the wire connecting the rope to piston A.

Since they both move by the same amount, and incidentally k=2*σ*g, the work done by both are equal but of opposite signs?

Yes, that's right.

 

[Chestermiller]

Thanks for the help!That's only for the initial instant (the pin) ...The free body diagram should be good for all instants after that. I think my question is more about the hydrostatics...Is the pressure at the top surface of the fluid (just below the piston), denoted by PB equal to the liquid pressure?

The free body diagram also includes the force of the spring. The force at the top surface of the fluid is equal to PB*A minus the spring force.

I understand about the spring, but if the rope on one side of the pulley goes up, the other side goes down by the same length...I don't see how the tension should change.

Do a free body diagram on the rope. If the left side goes down, you need to apply tension on the right side to hold the rope in place. Don't forget that the rope has mass and weight.

As TSny suggested, you should be looking at piston A, and doing a free body diagram on that piston. What are the forces acting on piston A in the final equilibrium state?

The combination of the two pistons and the chamber assembly forms a closed system. In terms of the distance that the left side of the rope moves down, how much work is done on the closed system?

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