Question

For an isothermal process, S = __________?

Answer 1

It is as shown here for ideal gases...
https://socratic.org/questions/can-the-entropy-of-an-ideal-gas-change-during-an-isothermal-process#530118

For anything ever, one would need the particular equation of state or the values of `alpha` or `kappa`.

`DeltaS_T = int_(V_1)^(V_2) ((delS)/(delV))_TdV` in general.

`= int_(V_1)^(V_2) ((delP)/(delT))_VdV`, for gases.

`= int_(V_1)^(V_2) alpha/kappadV` for condensed phases,

where `alpha` is the coefficient of thermal expansion, and `kappa` is the isothermal compressibility.

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PROOF

Starting from the Maxwell relation for the Helmholtz free energy,

`dA = -SdT - PdV`

From this, we find that for any state function, the cross-derivatives are equal, and:

`((delS)/(delV))_T = ((delP)/(delT))_V`

Proceeding, we note that

`alpha = 1/V((delV)/(delT))_P`

`kappa = -1/V((delV)/(delP))_T`

Using the cyclic rule of partial derivatives:

`((delV)/(delT))_P ((delP)/(delV))_T ((delT)/(delP))_V = -1`

As a result, since `((delT)/(delP))_V = 1/(((delP)/(delT))_V)`:

`-((delP)/(delT))_V = ((delV)/(delT))_P ((delP)/(delV))_T`

From the definitions above, it follows that

`color(blue)(((delP)/(delT))_V) = -((delV)/(delT))_P ((delP)/(delV))_T`

`= -((delV)/(delT))_P cdot [((delV)/(delP))_T]^(-1)`

`= 1/V((delV)/(delT))_P cdot [-1/V((delV)/(delP))_T]^(-1)`

`= color(blue)(alpha/kappa)`