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Internal entropy change for nonquasistatic process

  1. Feb 13, 2010 #1
    Hi all,

    I read that the internal entropy change of a system ([tex]\Delta[/tex]Si) for a non-quasistatic process where a reservoir cools or heats the system, where N and V are fixed and no work is done is approximately equal to ([tex]\Delta[/tex]T)2, when the difference between TReservoir and Tsystem is small. The book said that expanding the logarithm in:

    ∆Ss= 3/2(Nk(ln(Tr/Ts) - ((Tr – Ts )/ Tr )))

    can achieve this.


    ∆Ss= change in entropy for system
    Tr=Temperature of reservoir
    Ts=Temperature of system

    I have played around with it for a while but dont seem to be able to get this result.

    Thanks everyone :)
  2. jcsd
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