Helmholtz Free Energy: Why is F Minimized?

[Silviu]

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!

 

[Mr rabbit]

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 $dN=0$).

 

[Silviu]

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.)

 

[Mr rabbit]

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.

 

[Mapes]

$F$ being constant is only one prerequisite to be a minimum; you also need to show that the second derivative is less than zero.