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 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##).
Minimizing Helmholtz free energy is important because it represents the maximum amount of work that can be extracted from a system at a constant temperature and volume. It is a fundamental concept in thermodynamics that helps us understand the stability and equilibrium of a system.
The Helmholtz free energy is related to entropy through the equation F = U - TS, where U is the internal energy, T is the temperature, and S is the entropy. Minimizing F means minimizing the difference between U and TS, which is equivalent to maximizing the entropy of the system.
The fact that F is a state function means that it only depends on the current state of the system and not on the path taken to reach that state. This allows us to calculate the change in F between two different states without needing to know the details of the process that occurred in between.
In a closed system, minimizing F leads to a state of thermodynamic equilibrium, where there is no spontaneous change in the system. This is because at equilibrium, F is at its minimum value and any further change in the system would require an increase in F.
Yes, F can be negative for certain systems at very low temperatures. This indicates that the system has the potential to do work and release energy, and is therefore not at equilibrium. However, as the temperature increases, F becomes positive and approaches zero at equilibrium.