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I Helmholtz free energy

  1. Mar 4, 2017 #1
    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!
     
  2. jcsd
  3. Mar 4, 2017 #2
    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##).
     
  4. Mar 4, 2017 #3
    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.)
     
  5. Mar 4, 2017 #4
    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.
     
  6. Mar 13, 2017 #5

    Mapes

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    ##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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