Entropy variables

Question

When we are defining the entropy we have been used two isolated system not a heat bath with a small system, to ensure that entropy is defined as an energy variable dependent. My question why is the entropy defined as (E, V, N) variables not the (T, V, N)? The same as for Helmholtz free energy is defined as (T, V, N) variables.

Answer 1

My question why is the entropy defined as (E, V, N)

In thermodynamics, the entropy $S$ is postulated to exist, and to obey a maximization principle, and to be a function of the macroscopic state, which is determined by the triple: $E$, $V$, and $N$. This is by definition.

In other words, we choose to use the symbol $S$ and the name "entropy" for the thermodynamic potential that is maximized as a function of $E$, $V$, and $N$. That's it!

If we do not want to consider the macrostate in terms of $E$, $V$, and $N$, we can Legendre Transform and consider different thermodynamic potentials, such as the Helmholtz Free Energy.

By convention (or by definition, if you prefer) we agree to use the symbol $F$ and the name "Helmholtz Free Energy" for the Legendre Transformation of the entropy $S$ with respect to $E$: $$ F(T,V,N) \equiv \inf_{E}\left(\frac{1}{T} E - S(E,V,N)\right)\;. $$

Please see Chapter 1 of this book for more details. (N.b., I did not write this book and I am not affiliated with this book. I just like it.)

Answer 2

If we look at the second law of thermodynamics in differential form $ TdS(E, V, N) = dE + pdV - \mu dN$, which tells us what differentials of what variable change the differential of the thermodynamic potential under study.

So to understand what natural units a thermodynamic potential has it is necessary to calculate its differential using its formula. For instance

$dF = dE - d(TS) = TdS - pdV + \mu dN - TdS - SdT = -SdT - pdV + \mu dN; \ F=F(T, V, N)$