Bulk Modulus, Expansion Coeff. and compression of a liquid in a piston

[klawlor419]

Homework Statement

A liquid is enclosed in a metal container that is provided with a piston of the same metal. The system is originally at a pressure of 1.00 atm (1.013*10^5 Pa) and at a temperature of 30.0 C. The piston is forced down until the pressure on the liquid is increased by 50.0 atm, and then clamped in this position.

Find the new temperature at which the pressure of the liquid is again 1.00 atm.

Assume that the cyclider is sufficiently strong so that its volume is not altered by the pressure, only by changes in temperature.

Compressibility: k=8.50*10^-10 [Pa^-1]
β(Liquid)=4.80*10^-4
β(Metal)=3.90*10^-5

Homework Equations

Bulk Modulus: B= Δp/(ΔV/V)
Linear Volume expansion: ΔV=β*ΔT*V
Compressibility: k=1/B

The Attempt at a Solution

So using the bulk modulus definition I found an expression that relates changes in pressure to changes in temperature. With this I found for the liquid in the system the change in temperature is equal to 8.97 C. Meaning the temperature of the liquid after compression is 38.97 C.

I understand that a change in pressure of the system leads to a change in temperature of the system; which in turn leads to a change in volume of the systems components. The way I'm seeing the problem now is now that I know the temperature of the liquid after compression I can assume the temperature of the container is the same.

The pressure is 51 atm after compression, therefore to get it back to 1 atm we must change the temperature of the system to account for a pressure change of -50 atm. But if I do this I'm literally just reversing what I did in the last step. I don't know what I'm missing. I know I'm not using the expansion coefficient of the metal in any way despite that it was given so it probably involves that.

 

[haruspex]

It's not clear to me whether the initial compression is adiabatic. If it's slow then you can take it as isothermal. Anyway, the compression phase involved either a V and P change, or a V, P and T change. The decompression involves a V, P and T change, but this time the V change is only in relation to the thermal contraction, so will be smaller. It is therefore not reversing the compression phase.
You'll need the difference in expansion rates to get the V change that results from the temperature changes.

 

[klawlor419]

It's not clear to me whether the initial compression is adiabatic. If it's slow then you can take it as isothermal. Anyway, the compression phase involved either a V and P change, or a V, P and T change. The decompression involves a V, P and T change, but this time the V change is only in relation to the thermal contraction, so will be smaller. It is therefore not reversing the compression phase.
You'll need the difference in expansion rates to get the V change that results from the temperature changes.

I don't think the compression is adiabatic, or at least it doesn't specify in the problem statement. I copied the problem statement exactly from the text so I think it involves the V, P, and T change.

With bulk modulus of the liquid and a volume change attributable to a change in temperature, the change in pressure;
Δp=B*β*ΔT
is also attributable to the change in temperature. β=β(Liquid) in this eq.

So now your saying that I need the difference in the volumetric expansion/contraction of the new system. The liquid will expand/contract more readily; is this what creates the second pressure change?

 

[Chestermiller]

Homework Statement

A liquid is enclosed in a metal container that is provided with a piston of the same metal. The system is originally at a pressure of 1.00 atm (1.013*10^5 Pa) and at a temperature of 30.0 C. The piston is forced down until the pressure on the liquid is increased by 50.0 atm, and then clamped in this position.

Find the new temperature at which the pressure of the liquid is again 1.00 atm.

Assume that the cyclider is sufficiently strong so that its volume is not altered by the pressure, only by changes in temperature.

Compressibility: k=8.50*10^-10 [Pa^-1]
β(Liquid)=4.80*10^-4
β(Metal)=3.90*10^-5

Homework Equations

Bulk Modulus: B= Δp/(ΔV/V)
Linear Volume expansion: ΔV=β*ΔT*V
Compressibility: k=1/B

The Attempt at a Solution

So using the bulk modulus definition I found an expression that relates changes in pressure to changes in temperature. With this I found for the liquid in the system the change in temperature is equal to 8.97 C. Meaning the temperature of the liquid after compression is 38.97 C.

I understand that a change in pressure of the system leads to a change in temperature of the system; which in turn leads to a change in volume of the systems components. The way I'm seeing the problem now is now that I know the temperature of the liquid after compression I can assume the temperature of the container is the same.

The pressure is 51 atm after compression, therefore to get it back to 1 atm we must change the temperature of the system to account for a pressure change of -50 atm. But if I do this I'm literally just reversing what I did in the last step. I don't know what I'm missing. I know I'm not using the expansion coefficient of the metal in any way despite that it was given so it probably involves that.

It seems to me, from the problem statement, that, during the initial compression phase, the liquid and cylinder/piston were both held at constant 30 C temperature. If the volume of the liquid initially were V₀, what would be the liquid volume V₁ after raising the pressure to 50 atm? Now, in phase 2, the temperature of the liquid and the cylinder/piston are both raised to temperature T. In this phase, the volume of the chamber will increase as a result of thermal expansion of the cylinder/piston. Starting from volume V₁, if the temperature of the cylinder/piston is raised from 30C to temperature T, what will be the new volume V₂ of the container (in terms of T)? Next, focusing on the liquid, what is its new volume V₂ if, starting at volume V₁, its temperature is raised from 30 C to temperature T, and its pressure is dropped from 50 atm back down to 1 atm? This new volume V₂ of the liquid must match the new volume V₂ of the cylinder/piston, so set the expressions for these two quantities equal to one another. This will give you a linear equation for calculating the temperature T.

 

[haruspex]

It seems to me, from the problem statement, that, during the initial compression phase, the liquid and cylinder/piston were both held at constant 30 C temperature. If the volume of the liquid initially were V₀, what would be the liquid volume V₁ after raising the pressure to 50 atm? Now, in phase 2, the temperature of the liquid and the cylinder/piston are both raised to temperature T.

I read it as the temperature being lowered to something below 30C in order to get back to 1 atm. The liquid will shrink more than the vessel.

 

[Chestermiller]

I read it as the temperature being lowered to something below 30C in order to get back to 1 atm. The liquid will shrink more than the vessel.

Yes. You are right. I didn't notice that the coeff of thermal expansion for the liquid was higher than that of the metal. Of course, if you set the problem up the way I recommended, you will still get the right answer.

 

[Chestermiller]

I read it as the temperature being lowered to something below 30C in order to get back to 1 atm. The liquid will shrink more than the vessel.

Yes. You are right. I didn't notice that the coeff of thermal expansion for the liquid was higher than that of the metal. Of course, if you set the problem up the way I recommended, you will still get the right answer.