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Quasistatic isothermal expansion

  1. Feb 9, 2004 #1
    How would I show that during quastatic isothermal expansion of a monatomic ideal gas, the change in entropy is related to the heat input Q by the simple formula: delta S=Q/T
    Show that it is not valid for the free expansion process described.

    Answer: Putting heat into a system always increases its entrophy.
    deltaS=Nk lnVf/Vi (U,N fixed).

    Ireally don't know how to do this. Can someone explain?
  2. jcsd
  3. Apr 7, 2010 #2
    A bit late a reply, but I'm currently busy with this stuff as well so reciting it will help me remembering it, here goes:

    During quasistatic isothermal expansion the inputted heat Q equals the minus work -W done. You derive this by integrating the ideal gas law over the volume change:
    Q = -W = int ( NkT/V *dV ) from Vi to Vf = NkT*ln(Vf/Vi)

    Furthermore, the Sackur-Tetrode equation says that delta S = Nk*ln(Vf/Vi) if you hold U and N fixed, which corresponds to a quasistatic isothermal process. (http://en.wikipedia.org/wiki/Sackur%E2%80%93Tetrode_equation" [Broken])
    So if you derive Q by T you would get delta S.

    A free process is still described by the same function, except the inputted heat is zero. Since spontaneous processes have their entropies increased (second law of thermodynamics) delta S must be growing, and so delta S = Q/T can not be valid.
    Last edited by a moderator: May 4, 2017
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