Yeap. just as the first answer writes. The constant V or P is just the variable used to describe the ensemble. The state function S depends only on the ensemble.

Entropy is typically reported at given T and P. The specifications constant T and constant P refer to processes (i.e. a transitions from initial state A to final state B), but are irrelevant for tabulated values at a single state. Perhaps you want to rephrase your question.

Thanks everyone for the responses. What motivated this question was this line from a paper I was reading:

"Elements for which data exist to find entropy at con-

stant volume."

So Perhaps this line was misguided?

It is difficult form this single statement to figure what the authors of the paper meant. Can you share the full reference? It is possible that they are calculating a ΔS along a constant volume path.

Sure, I uploaded a pdf and highlighted the relevant part. I've copied it here as well:

"

Let us consider the temperature dependence of thermody- namic properties of the liquid at elevated temperatures. The thermodynamic functions, all being derived from the free energy, contain only one independent function of tempera- ture. We choose to analyze the entropy, and will include only elements in all three of the following categories.

a.) Nearly free electron elements, so we have a reliable theoretical evaluation of SE .

b.)Normal melting elements

c.) Elements for which data exist to find entropy at con-

stant volume, up to temperatures sufficiently high to reveal a meaningful temperature dependence, say up to T/Tm > 2.

"

Just realized I have a bunch of other sections highlighted from that paper. Here's a different version with the relevant section highlighted in blue on page 6 of the pdf

Looking at the reference, it appears that what they are doing is computing the entropy not just in a single state but in a set of states that are defined by varying the temperature while keeping the volume fixed. So, yes, in general this would be a different set of states than the substance would be in if one varied the temperature while keeping the pressure fixed.

To return to your original question: entropy is a function of two independent variables, which we usually take to be temperature and pressure, but we can switch to temperature and volume. If we know the entropy at temperature T1 and volume V, we may calculate it at a different temperature T2 and same volume V along a path of constant V. If have the equation of state we can then calculate the pressure P2 at T2, V and report the entropy in the final state as S(T2,V)= S(T2,P2).

I know it is. Right now I am completing paper regarding completely open non-ideal system (T,P, μ). There are different probability exponents for grand canonical and generalized (completely open) distributions. Difference grows with value of imposed pressure (stress, or other field), so it is general case (Bose or Fermi) open system of non-ideal particles.

“In thermodynamic tables of liquids, S is usually given without mention of constant volume or constant pressure …”

Indeed, and one really has to read the fine print! Thermodynamic states of liquids are usually given at standard pressure (either 1 atm or 0.1 MPa); above the boiling point the vapour pressure is used. If pressure or volume are really controlled variables – this is the case in tabulations appearing in publications of reference equations of state – the control variable appears in the table.

The so-called "entropy " doesn't exist at all.

During the process of deriving the so-called entropy, in fact, ΔQ/T can not be turned into dQ/T. That is, the so-called "entropy " doesn't exist at all.

The so-called entropy was such a concept that was derived by mistake in history.

It is well known that calculus has a definition,

any theory should follow the same principle of calculus; thermodynamics, of course, is no exception, for there's no other calculus at all, this is common sense.

Based on the definition of calculus, we know:

to the definite integral ∫T f(T)dQ, only when Q=F(T), ∫T f(T)dQ=∫T f(T)dF(T) is meaningful.

As long as Q is not a single-valued function of T, namely, Q=F( T, X, …), then,

∫T f(T)dQ=∫T f(T)dF(T, X, …) is meaningless.

1) Now, on the one hand, we all know that Q is not a single-valued function of T, this alone is enough to determine that the definite integral ∫T f(T)dQ=∫T 1/TdQ is meaningless.

2) On the other hand, In fact, Q=f(P, V, T), then

∫T 1/TdQ = ∫T 1/Tdf(T, V, P)= ∫T dF(T, V, P) is certainly meaningless. ( in ∫T , T is subscript ).

We know that dQ/T is used for the definite integral ∫T 1/TdQ, while ∫T 1/TdQ is meaningless, so, ΔQ/T can not be turned into dQ/T at all.

that is, the so-called "entropy " doesn't exist at all.