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Question on Thermodynamics (adiabatic and isothermal expansions)

  1. Jan 29, 2008 #1
    1. The problem statement, all variables and given/known data

    A cubic metre of air at 0degreesC and 1 atm is compressed reversibly to 10 atm.

    (a) What is the final temperature if it is compressed adiabatically?

    (b) How much heat must be removed if it is compressed isothermally?

    I understand what the two different terms mean, but I just don't seem to be able to answer this for some reason. If someone could explain how to go about answering it that would be really helpful!


    2. Relevant equations

    3. The attempt at a solution

    For (a) I tried to use the relationship that T^gamma x p ^ (1-gamma) is constant, where gamma = Cp/Cv, but that didn't work at all.

    For (b) I thought I'd take dU = -pdV and integrate it with respect to dV with p = 1/V but (obviously I guess) this didn't really work.

    Any help in the way of what method to use, etc, is appreciated!

  2. jcsd
  3. Jan 29, 2008 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    It should work. Try using:

    [tex]PV^{\gamma} = K[/tex]

    to find the new volume and then apply PV=nRT to work out the new temperature.

    What are you using for [itex]\gamma[/itex] ?

    I get T2 = (10/5.18)*273 = 527 K.

    If it is isothermal, does U change? So how is the heat released related to the work done? Use the first law: dQ = dU + PdV

    The question is not clear whether you are to compress it to the same volume or pressure as in the adiabatic case. If you assume it is to be compressed to the 10 atm, it is just a matter of applying PV=nRT to find V where T is constant. If you compress to the same volume as in the adiabatic case, use that volume.

    Possible shortcut to consider: If the volume is to be compressed to 5.18 L, (same as the adiabatic case) how much heat is removed to bring the air down to 273K after you adiabatically compress it? Is that equivalent to an isothermal path (assuming the heat capacity does not depend on temperature)?

    Last edited: Jan 29, 2008
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