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Why Max work is equal to electric work?

  1. Apr 19, 2015 #1
    1. The problem statement, all variables and given/known data
    It was given in my book that:

    Systems are capable of doing pressure-volume (mechanical work) as well as non-mechanical work(like electric work).
    So,(ΔG)P,T=ΔH-TΔS=ΔU+PΔV-TΔS
    And
    ΔU=qrev-wmech-welec
    =qrev-PΔV-welec
    (ΔG)P,T=qrev-PΔV-welec+PΔV-TΔS
    =-welec
    Now in the book its given that:
    (ΔG)P,T=-wmax
    But wmax should comprise of both the mechanical and electric work . Why only electric work is taken as the wmax?
    2. Relevant equations
    None

    3. The attempt at a solution
    -
     
  2. jcsd
  3. Apr 19, 2015 #2

    DrClaude

    User Avatar

    Staff: Mentor

    That's what the Gibbs free energy gives you: the total work available or necessary for the process, considering that the system is at constant T and P. That means that any mechanical work is not taken into account, and is useful when that work is wasted or when you need to include the work done "for free" by the environment.

    If you want to take into account mechanical work, you have to use the Helmholtz free energy.
     
  4. Apr 19, 2015 #3
    Okay.....that means Gibbs free energy doesn't include mechanical work(by definition) whereas helmholtz free energy does. Am I correct here?
     
  5. Apr 19, 2015 #4

    DrClaude

    User Avatar

    Staff: Mentor

    $$
    \begin{align}
    dG &= dU -T dS - P dV \\
    dA &= dU - T dS
    \end{align}
    $$
    So you see that mechanical work is explicitly removed from the chnage internal energy when calculating the Gibbs free energy, but not the Helmholtz free energy.
     
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