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What is the new temperature of the system?

  1. Dec 30, 2015 #1
    1. The problem statement, all variables and given/known data
    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

    2. Relevant equations

    Δp = BβΔT
    ΔV = V0βΔT

    3. The attempt at a solution

    2gww58m.jpg
     
  2. jcsd
  3. Dec 30, 2015 #2

    haruspex

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    You do not seem to have taken into account the change of volume of the cylinder as the temperature drops.
     
  4. Dec 30, 2015 #3
    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.
     
  5. Dec 30, 2015 #4

    haruspex

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    Please post your attempt to take into account the cylinder volume change.
     
  6. Dec 30, 2015 #5
    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 V0βc(-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 wich envolves it, and this second compression would cause a new decrease in the temperature wich is given by ΔVL (caused by the second compression) / ΔVL (caused by the first compression). βL ≅ -0.76 °C.
    So the final temperature is 30 °C + [-0.76 + (-9)] °C = 20,24 °C, wich is very close to the answer given by the author: 20,2 °C.
     
  7. Dec 30, 2015 #6

    haruspex

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    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.
     
  8. Dec 30, 2015 #7
    I can't. :( Can you show me how to do this please?
     
  9. Dec 30, 2015 #8

    haruspex

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    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?
     
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