1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Quasistatic isothermal expansion
Loading...