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A Tricky Problem Involving Entropy

  1. Jul 8, 2013 #1
    So we learn in basic thermo that for any system, the derivative of entropy with respect to volume is pressure over temperature:

    ∂S/∂V=P/T.

    Suppose we have a box with two partitions, and a moveable wall in between the partitions. The box and wall are both insulated to prevent heat transfer. The wall will slide until the entropy of the entire system is maximized. This occurs when the change in entropy with volume is equal for the two sides:

    (∂S/∂V)1=(∂S/∂V)2.
    (P/T)1=(P/T)2.

    When both sides are at the same temperature, this result is obvious. The wall simply stops sliding when the pressure is the same on both sides. However, what happens when the temperatures are very different? If you fill one side of the box with a gas at 100K and 1atm, and the other side with a gas at 200K and 2atm, will the wall really not slide, despite there being a pressure difference of 1atm between the two sides?
     
  2. jcsd
  3. Jul 8, 2013 #2

    mfb

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    If your gas is thermally isolated, changing the volume will also change the temperature, which leads to an additional entropy change.
    If the gas is not thermally isolated, the equilibrium has the same temperature on both sides (=obvious).
     
  4. Jul 8, 2013 #3
    Okay, that makes sense. I see now from my thermo references that the relation is only true for [itex](\frac{∂S}{∂V})_{E,N}[/itex]. The wall must be heat-permeable for E to remain constant as volume is shifted.

    However, what if the energy in each compartment is kept constant by other means, such as through heating/cooling elements, while keeping the temperature difference between compartments intact? I realize that now the entire system, including the heating components, may not be tending towards an entropic maximum, but can we still think about the entropy of the subsystem containing only the box? If so, would the input of the heating/cooling elements cause the wall to remain in equilibrium despite a pressure difference between the two sides?
     
  5. Jul 8, 2013 #4

    mfb

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    The entropy of a subsystem does not have to reach a maximum.
    Just consider a refrigerator.

    The equilibrium position for the wall requires pleft=pright.
     
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