# For an isothermal process, S = __________?

Jan 4, 2018

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 $\kappa$.

$\Delta {S}_{T} = {\int}_{{V}_{1}}^{{V}_{2}} {\left(\frac{\partial S}{\partial V}\right)}_{T} \mathrm{dV}$ in general.

$= {\int}_{{V}_{1}}^{{V}_{2}} {\left(\frac{\partial P}{\partial T}\right)}_{V} \mathrm{dV}$, for gases.

$= {\int}_{{V}_{1}}^{{V}_{2}} \frac{\alpha}{\kappa} \mathrm{dV}$ 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,

$\mathrm{dA} = - S \mathrm{dT} - P \mathrm{dV}$

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

${\left(\frac{\partial S}{\partial V}\right)}_{T} = {\left(\frac{\partial P}{\partial T}\right)}_{V}$

Proceeding, we note that

$\alpha = \frac{1}{V} {\left(\frac{\partial V}{\partial T}\right)}_{P}$

$\kappa = - \frac{1}{V} {\left(\frac{\partial V}{\partial P}\right)}_{T}$

Using the cyclic rule of partial derivatives:

${\left(\frac{\partial V}{\partial T}\right)}_{P} {\left(\frac{\partial P}{\partial V}\right)}_{T} {\left(\frac{\partial T}{\partial P}\right)}_{V} = - 1$

As a result, since ${\left(\frac{\partial T}{\partial P}\right)}_{V} = \frac{1}{{\left(\frac{\partial P}{\partial T}\right)}_{V}}$:

$- {\left(\frac{\partial P}{\partial T}\right)}_{V} = {\left(\frac{\partial V}{\partial T}\right)}_{P} {\left(\frac{\partial P}{\partial V}\right)}_{T}$

From the definitions above, it follows that

$\textcolor{b l u e}{{\left(\frac{\partial P}{\partial T}\right)}_{V}} = - {\left(\frac{\partial V}{\partial T}\right)}_{P} {\left(\frac{\partial P}{\partial V}\right)}_{T}$

$= - {\left(\frac{\partial V}{\partial T}\right)}_{P} \cdot {\left[{\left(\frac{\partial V}{\partial P}\right)}_{T}\right]}^{- 1}$

$= \frac{1}{V} {\left(\frac{\partial V}{\partial T}\right)}_{P} \cdot {\left[- \frac{1}{V} {\left(\frac{\partial V}{\partial P}\right)}_{T}\right]}^{- 1}$

$= \textcolor{b l u e}{\frac{\alpha}{\kappa}}$