Relating Helmholtz & Gibbs Free Energy: F=G-PdV+VdP

AI Thread Summary
The discussion centers on the relationship between Helmholtz free energy (F) and Gibbs free energy (G), specifically the equation F=G-PdV. The initial assumption that PdV represents the total work due to pressure is challenged, revealing that the correct expression includes both PdV and VdP, which simplifies to d(PV). Participants clarify that V is not constant when dV is non-zero, indicating that pressure can indeed perform work in a dynamic system. The confusion arises from the interpretation of V as constant while varying pressure, leading to a deeper understanding of how work is defined in thermodynamic contexts. Ultimately, the conversation highlights the intricacies of thermodynamic equations and the conditions under which they apply.
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i'm trying to relate the helmholtz and the gibbs free energy together. I thought that since the gibbs was just the W(other) and the helmholtz was the total work, that the following would be true.

F=G-PdV

because.. i thought that PdV was the work due to pressure, but it turns out that it's actually PdV+VdP which is just d(PV)

but i thought PdV was the only term that was the work due to pressure, how is VdP work due to pressure? V is constant!
 
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iScience said:
it turns out that it's actually PdV+VdP which is just d(PV)

but i thought PdV was the only term that was the work due to pressure, how is VdP work due to pressure? V is constant!

V is not constant if dV is non-zero, right? Indeed, if V were constant the pressure wouldn't be doing any work (just as a weight sitting stationary on a rigid table does no work on the table).
 
i know dV isn't zero, this is what gives rise to a displacement in the system which ultimately gives rise to work.
i was referring to V, not dV. for the VdP term, i thought we were keeping V constant and varying P.
 
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