Helmholtz Free Energy: Why is F Minimized?

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Discussion Overview

The discussion centers on the minimization of Helmholtz free energy (F) at constant temperature (T) and volume (V) in the context of thermodynamic equilibrium. Participants explore the implications of the mathematical formulation of F and its relevance to closed systems and chemical reactions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions the reasoning behind the minimization of F, seeking clarification on what F is minimized with respect to.
  • Another participant emphasizes that the discussion pertains to closed systems where mass transfer does not occur (dN=0).
  • A different viewpoint suggests that if T, V, and N are constant, then F remains constant, raising doubts about the concept of minimization in such a scenario.
  • One participant asserts that at equilibrium, all parameters are constant, and describes the evolution of a system towards a state where internal energy is minimized and entropy is maximized.
  • Another participant notes that constancy of F is a prerequisite for minimization, but also highlights the need to demonstrate that the second derivative is less than zero.

Areas of Agreement / Disagreement

Participants express differing views on the implications of constant F and the conditions under which F is minimized. There is no consensus on the interpretation of these conditions or the implications for systems undergoing chemical reactions.

Contextual Notes

Participants reference assumptions about closed systems and the constancy of parameters at equilibrium, but these assumptions are not universally accepted or fully resolved in the discussion.

Silviu
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Hello! I read that the Helmholtz free energy is minimized at constant T and V at equilibrium. But I am not sure I understand why. So starting from ##F=U-TS## I got ##dF = \mu dN - pdV -SdT##. So at constant V and T we have ##dF = \mu dN##. Now I am not sure how does this implies that F is minimized and minimized with respect to what? Can someone tell me why is F minimized? Thank you!
 
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I think the statement you are referring to is missing to say that it is for a closed system, that is, there's not mass transferring (so ##dN=0##).
 
Mr rabbit said:
I think the statement you are referring to is missing to say that it is for a closed system, that is, there's not mass transferring (so ##dN=0##).
Thank you for the reply. However, if T, V and N (and hence p) are constant, then nothing changes in the system. Doesn't that means that F is constant, so it doesn't make sense to talk about a minimum? (Also, the context in which I read this is about chemical reactions, so dN changes i assume.)
 
It is assumed (and can be proved) that all parameters are constant at equilibrium. Think, for example, on an expansive system, so you take ##U=U (S, V, N)##; If you "drop" the system from an arbitrary state, always evolves to another state (equilibrium state) until the energy U is minimal and the entropy S is maximal. All parameters evolves too (p, V, T ...) and when the system reaches equilibrium state you get ##p=const ## (mechanical equilibrium), ##T=const## (thermal equilibrium), ##N=const## (chemical equilibrium).

Helmholtz free energy is the internal energy U but depending on temperature instead of entropy.
 
##F## being constant is only one prerequisite to be a minimum; you also need to show that the second derivative is less than zero.
 

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