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Change in Entropy for Irreversible Processes

  1. Apr 5, 2014 #1
    I was following along in my Thermodynamic textbook and began playing with some definitions. In the following formulation, I somehow managed to prove (obviously incorrectly) that [itex]dq = TdS[/itex] for even irreversible processes. I was hoping someone could point out where in the proof I'm going wrong.

    First, consider a both reversible and adiabatic process.

    Since [itex]dw = -p_{ext}dV[/itex] for all processes and [itex]dq = 0[/itex] for adiabatic processes: [itex]dU = -p_{ext}dV[/itex]

    We also know for a reversible and adiabatic process, U is a function of V only and not S, so: [itex]dU = \frac{∂U}{∂V}dV[/itex]

    Setting the coefficients equal: [itex]\frac{∂U}{∂V}=-p_{ext}[/itex]

    This equation involves only state variables and is therefore valid for all process, reversible or irreversible.


    Next consider a general process (either reversible or irreversible).

    As before [itex]dw = -p_{ext}dV[/itex] but now U is a function of S and V, so: [itex]dq = dU - dw = (\frac{∂U}{∂S}dS + \frac{∂U}{∂V}dV) - (-p_{ext}dV)[/itex]

    Simplifying: [itex]dq = \frac{∂U}{∂S}dS + \frac{∂U}{∂V}dV + p_{ext}dV[/itex]

    Finally, since we know [itex]\frac{∂U}{∂V}=-p_{ext}[/itex], the last two terms cancel, leaving: [itex]dq = \frac{∂U}{∂S}dS = TdS[/itex] for any processes, reversible or irreversible.

    But clearly, this is not true since [itex]dq >TdS[/itex] for irreversible processes.

    At which point, then, should I have needed to invoke irreversibility?

    Thank you in advance.
     
  2. jcsd
  3. Apr 5, 2014 #2

    UltrafastPED

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