For an isothermal process, S = __________?
1 Answer
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
#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
#-((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)#