How do you calculate the change in entropy for an isothermal process?

Dec 21, 2016

It depends on which way we change the entropy (pressure or volume?).

Pick one, and you would use one of these:

$\Delta S = n R \ln \left({V}_{2} / {V}_{1}\right)$

$\Delta S = - n R \ln \left({P}_{2} / {P}_{1}\right)$

Actually, if you notice that at constant $T$ and $n$ for a monatomic ideal gas, ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$:

$\ln \left({V}_{2} / {V}_{1}\right) = \ln \left(\frac{{P}_{1} \cancel{{V}_{1}} \text{/} {P}_{2}}{\cancel{{V}_{1}}}\right)$

$= \ln \left[{\left({P}_{2} / {P}_{1}\right)}^{- 1}\right] = - \ln \left({P}_{2} / {P}_{1}\right)$,

which means our two expressions are equivalent for any monatomic ideal gas.

DISCLAIMER: This contains Calculus. I assume that if you read further, you understand partial derivatives, cross-derivatives, and the integral of $\frac{1}{x}$.

Entropy is typically considered a function of temperature and either volume or pressure. When we hold temperature constant (an isothermal process), and change one of the other parameters:

(1) $\Delta S = {\int}_{{P}_{1}}^{{P}_{2}} {\left(\frac{\partial S}{\partial P}\right)}_{T} \mathrm{dP}$

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

If you notice, we don't have an expression for the way entropy changes due to pressure or volume in relation to gases. It would be nice to have a way to get these relationships from known formulas, like the ideal gas law.

It turns out that these can be derived from the Maxwell Relations that contain $T , P$ or $T , V$ as the variables that could potentially change:

(3) $\mathrm{dG} = - S \mathrm{dT} + V \mathrm{dP}$,

where $G$ is the Gibbs' free energy, and

(4) $\mathrm{dA} = - S \mathrm{dT} - P \mathrm{dV}$,

where $A$ is the Helmholtz free energy.

ENTROPY AS A FUNCTION OF TEMPERATURE AND PRESSURE

The first context in which we calculate $\Delta S$ for an isothermal process involves looking at the cross-derivatives in (3); here is how you should examine $S , P , T$:

$\text{ "" "" "uarr "------------------------------------} \downarrow$
$\mathrm{dG} = \stackrel{\text{Numerator")overbrace(-S)" "stackrel("Hold constant")overbrace(dT) + V" "stackrel("Denominator}}{\overbrace{\mathrm{dP}}}$
$\text{ "" "" "" "" "" "" "uarr "---------------------} \downarrow$

Do something similar for $V , T , P$ to get:

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

Then, for a monatomic ideal gas, we can use the ideal gas law ($P V = n R T$) to give an explicit expression for the partial derivative of $V$ with respect to $T$ at a constant $P$:

$V = \frac{n R T}{P}$

$\implies \textcolor{g r e e n}{- {\left(\frac{\partial V}{\partial T}\right)}_{P}} = - \frac{\partial}{\partial T} {\left[\frac{n R T}{P}\right]}_{P}$

$= - \frac{1}{P} \frac{d}{\mathrm{dT}} \left[n R T\right] = \textcolor{g r e e n}{- \frac{n R}{P}}$

That means our first way to get the change in entropy is (plugging back into (1)):

$\textcolor{b l u e}{\Delta S} = {\int}_{{P}_{1}}^{{P}_{2}} {\left(\frac{\partial S}{\partial P}\right)}_{T} \mathrm{dP}$

$= - {\int}_{{P}_{1}}^{{P}_{2}} {\left(\frac{\partial V}{\partial T}\right)}_{P} \mathrm{dP}$

$= - {\int}_{{P}_{1}}^{{P}_{2}} \frac{n R}{P} \mathrm{dP} = - n R {\int}_{{P}_{1}}^{{P}_{2}} \frac{1}{P} \mathrm{dP}$

$= \textcolor{b l u e}{- n R \ln \left({P}_{2} / {P}_{1}\right)}$

ENTROPY AS A FUNCTION OF TEMPERATURE AND VOLUME

The other way to express $\Delta S$ involves equations (2) and (4):

$\text{ "" "" "uarr "------------------------------------} \downarrow$
$\mathrm{dA} = \stackrel{\text{Numerator")overbrace(-S)" "stackrel("Hold constant")overbrace(dT) - P" "stackrel("Denominator}}{\overbrace{\mathrm{dV}}}$
$\text{ "" "" "" "" "" "" "uarr "---------------------} \downarrow$

Do something similar for $P , T , V$ to get:

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

On the ideal gas law, this means:

$P = \frac{n R T}{V}$

$\implies \textcolor{g r e e n}{{\left(\frac{\partial P}{\partial T}\right)}_{V}} = \frac{\partial}{\partial T} {\left[\frac{n R T}{V}\right]}_{V}$

$= \frac{1}{V} \frac{d}{\mathrm{dT}} \left[n R T\right] = \textcolor{g r e e n}{\frac{n R}{V}}$

Substitute this result back into (2) to get:

$\textcolor{b l u e}{\Delta S} = {\int}_{{V}_{1}}^{{V}_{2}} {\left(\frac{\partial S}{\partial V}\right)}_{T} \mathrm{dV}$

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

$= {\int}_{{V}_{1}}^{{V}_{2}} \frac{n R}{V} \mathrm{dV} = n R {\int}_{{V}_{1}}^{{V}_{2}} \frac{1}{V} \mathrm{dV}$

$= \textcolor{b l u e}{n R \ln \left({V}_{2} / {V}_{1}\right)}$

Overall, if you know the initial and final pressure, or the initial and final volume, you can figure out the change in entropy of a monatomic ideal gas at a constant temperature.