For an isothermal process, S = __________?

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.

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)#