What is the new temperature of the system?

[kent davidge]

Homework Statement

A liquid is enclosed in a metal cylinder that is provided with a piston of the same metal. The system is originally at a pressure of 1.00 atm (1.013 x 10⁵ Pa) and at a temperature of 30 °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 cylinder is sufficiently strong so that its volume is not altered by changes in pressure, but only by changes in temperature.

Compressibility of liquid: k = 8.5 x 10^-10 Pa^-1
Coeff of volume expansion of liquid: β = 4.8 x 10^-4 K^-1
Coeff of volume expansion of metal: β = 3.9 x 10^-5 K^-1

Homework Equations

Δp = BβΔT
ΔV = V₀βΔT

The Attempt at a Solution

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[haruspex]

You do not seem to have taken into account the change of volume of the cylinder as the temperature drops.

 

[kent davidge]

Yes. I attempted a solution considering the change in volume of cylinder but it didnt work. Please give me the correct answer and tell me if my argument in the image above is correct.

 

[haruspex]

Yes. I attempted a solution considering the change in volume of cilyinder but it didnt work. Please give me the correct answer and tell me if my argument in the image above is correct.

Please post your attempt to take into account the cylinder volume change.

 

[kent davidge]

As I mentioned in the image above, the decrease in the temp is -9 °C. But that decrease would change the volume of the cylinder by V₀β_c(-9)°C.
The effect of this change would be to compress the liquid, decreasing its volume. Then the volume of liquid suffered two compressions, one caused by the pressure and the other caused by the cylinder which envolves it, and this second compression would cause a new decrease in the temperature which is given by ΔV_{L (caused by the second compression)} / ΔV_{L (caused by the first compression)}. β_L ≅ -0.76 °C.
So the final temperature is 30 °C + [-0.76 + (-9)] °C = 20,24 °C, which is very close to the answer given by the author: 20,2 °C.

 

[haruspex]

As I mentioned in the image above, the decrease in the temp is -9 °C. But that decrease would change the volume of the cylinder by V₀.β_c(-9)°C.
The effect of this change would be to compress the liquid, decreasing its volume. Then the volume of liquid suffered two compressions, one caused by the pressure and the other caused by the cylinder which envolves it, and this second compression would cause a new decrease in the temperature which is given by ΔV_{L (caused by the second compression)} / ΔV_{L (caused by the first compression)}. β_L ≅ -0.76 °C.
So the final temperature is 30 °C + [-0.76 + (-9)] °C = 20,24 °C, which is very close to the answer given by the author: 20,2 °C.

That's an iterative approach, which in principle goes on forever. But there's a more direct way.
Consider the differential shrinkage between the metal and the liquid.

 

[kent davidge]

Consider the differential shrinkage between the metal and the liquid.

I can't. :( Can you show me how to do this please?

 

[haruspex]

I can't. :( Can you show me how to do this please?

You start with, in effect, an excess volume of liquid, i.e. the volume it has been compressed by.
If the temperature goes down $\Delta T$, how much does the liquid shrink by? How much does the cylindrical cavity shrink by? What is the net effect on the excess volume?