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Internal energy or enthalpy are minimized in isoentropic processes?

  1. May 13, 2012 #1
    1. The problem statement, all variables and given/known data

    How to demonstrate that U is minimized at constant V and S, while H at constant P and S?

    2. Relevant equations

    ΔS universe = ΔS system + ΔS environment ≥ 0
    ΔU system = δq reversible + δw reversible = δq irreversible + δw irreversible
    ΔS environment = −∫(δq reversible / T)
    dU = TdS − PdV = δq + δw

    3. The attempt at a solution

    ΔS universe = ΔS system + ΔS environment > 0
    ΔS universe = ΔS system − ∫ ( δq rev /T ) > 0
    ΔS universe = ΔS system − ∫( δq irrev/T + δw irrev − δw rev ) > 0
    TΔS system − T∫( δq irrev/T + δw irrev − δw rev ) > 0
    T∫( δq irrev/T + δw irrev − δw rev ) − TΔS system < 0

    If the process is isoentropic, it follows that it's an adiabatic process with δq irrev = 0

    T∫( δw irrev − δw rev ) < 0

    * If, furthermore, the volume of the system is constant, it follows that irreversible and reversible works are 0, leading to a senseless expression.


    * If not the volume but the pressure is constant, we get
    T∫( δw irrev − δw rev ) < 0
    T∫( δq rev − δq irrev ) < 0 and we recover ∫δq irrev = ΔH

    Now, since reversible −∫PdV is minimum in reversible process, δ rev is maximum.
    It follows that the T∫(positive quantity) < 0
    Which is another senseless expression.

    I guess the flaw comes from the assumption that an isoentropic process is always adiabatic... but how?
     
  2. jcsd
  3. May 14, 2012 #2
    Now I think I have found it: Clausius inequality
    dS ≥ δq reversible/T
    I assumed that isoentropic process implied adiabatic irreversible process; however, it means that δq rev≤0

    So I have that
    ΔS universe = ΔS system − ∫ ( δq rev /T ) > 0

    If the process is isoentropic
    ΔS universe = − ∫ ( δq rev /T ) > 0
    ∫ ( δq rev /T ) < 0
    ∫ ( dU − δw rev ) /T < 0

    If now I assume that the volume is constant, I get δw = 0
    ∫ dU / T < 0

    Which seems like the inequality I have been looking for.

    ∫ ( dU − δw rev ) / T < 0
    ∫ ( dΗ − PdV − VdP − δw rev ) / T < 0

    If I assume that pressure is constant, I get

    ∫ ( dΗ − PdV − δw rev ) / T < 0

    Now If I assume mechanical equilibrium

    ∫ dΗ / T < 0

    Now I guess my reasoning is okay? Someone answer please.
     
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