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Entropy of isothermal process reversible\irreversible

  1. Sep 26, 2012 #1
    We were shown in class how to get those entropys.

    For reversible isothermal - ΔT=0 thus ΔE=0 thus Q = -W.
    ΔS(sys) = Qrev/T = nR(V1/V2)
    And ΔS(surr) = -nR(V1/V2) because surroundings made opposite work.

    For irreversible isothermal in vacuum - ΔT=0 thus ΔE=0.
    No work is done by surroundings (vacuum) so W=0 then Q=0.
    ΔS(sys) is a function state, so it remains ΔS(sys)=nR(V1/V2).
    ΔS(surr) = Q/T = 0

    So I have 2 questions:
    1 - How come they decide that ΔS(sys) is a state function, but ΔS(surr) is not?? Can't we get ΔS(surr) = -nR(V1/V2) for exactly the same reason??

    2 - I thought the defenition for entropy was integral(Qrev/T), so how come on the irreversible we use Q=0, and not the Q for the reversible process?
     
  2. jcsd
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