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What does it mean when people say that Gibb's free energy is the 'maximum amount of non P-V work'

Physics Asked by Buraian on October 8, 2020

The maximum amount of work makes complete sense for me as from the premise of the derivation being that we are considering a reversible process. However, what does it mean to say maximum amount of non-expansion work? (*)


My personal research attempts:

  • In a chemistry discussion forum (**), they say that non expansion work is all the work which is not related to expansion. The problem I have is this is what other kind of work can be there? I think the expression would evaluate to zero if we constrain the systems pressure and temperature since the volume is implicitly controlled in single substance system

The kind of answer I want:

One which illustrates the concept using an actual physical application rather than an abstract description.


References:

*:Derivation for gibb’s free energy being maximum non expansion work

**: The chemistry discussion forum post

One Answer

In "The Principles of Chemical Equilibrium," by K.G. Denbigh, in section 2.4 discusses, in a very precise way, how this applies to (a) a closed system at constant temperature and pressure and (b) an open system operating at steady state in contact solely with a constant temperature reservoir. In the former case he discusses a specific example of non-PV work being done by an ionic reaction involving electrochemical potential.

He explains this all much better than I could ever do at paraphrasing it, particularly for the closed system case.

However, for the open system case, he shows that the maximum shaft work possible is equal to the decrease in Gibbs free energy per unit mass passing through the system, and this is much easier to understand. Moreover, the open system case does not require constant pressure, which, in my judgment, is particularly appealing.

For an open system operating at steady state (with negligible changes in kinetic and potential energy of the flow), the first law gives us $$Q-W_s=Delta H$$where $W_s$ is the shaft work, and all quantities are per unit mass of flowing stream. In addition, if there is a constant temperature reservoir at temperature T' as the only means of heat transfer, the 2nd law gives us $$Delta S=frac{Q}{T'}+sigma $$where $sigma$ is the entropy generated in the control volume per unit mass of flowing stream passing through the control volume. If we combine these two equations, we obtain: $$W_s=-Delta H+T'Delta S-T'sigma$$ If the steady flow system is operated reversibly and such that the inlet and outlet temperatures from the system T are the same as the reservoir temperature (i.e., T' = T), this equation gives as the maximum shaft work for isothermal operation, $$W_s=-Delta G$$

Answered by Chet Miller on October 8, 2020

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