# How can we calculate DeltaS for an isothermal process?

Jul 27, 2017

Well, there are two ways I can think of. Both show that entropy increases as volume increases at constant temperature.

METHOD 1

You can always start from the first law of thermodynamics:

$\Delta E = q + w$

It is simplest to consider a reversible process, i.e. the 100% efficient process:

$\Delta E = {q}_{r e v} + {w}_{r e v}$,

where:

• ${q}_{r e v}$ is the reversible heat flow.
• ${w}_{r e v} = - \int P \mathrm{dV}$ is the reversible work done from the perspective of the system.
• $\Delta E$ is the change in internal energy.

For an isothermal process, all ideal gases have $\Delta E = 0$ and $\Delta H = 0$, since these are only functions of temperature for ideal gases. Thus,

${q}_{r e v} = - {w}_{r e v} = \int P \mathrm{dV}$

Now, consider the definition of the entropy:

$\mathrm{dS} = \frac{\delta {q}_{r e v}}{T}$,

where $\delta$ indicates a path function and $d$ indicates a state function. Both are differentials and can be integrated.

Thus, by finding ${q}_{r e v}$, one can find $\Delta S$. The change in entropy would have been:

${\int}_{\left(1\right)}^{\left(2\right)} \mathrm{dS} = {\int}_{\left(1\right)}^{\left(2\right)} \frac{1}{T} \delta {q}_{r e v}$

$= \Delta S = {q}_{r e v} / T$

For ideal gases, we know the form of the pressure.

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

$\implies {q}_{r e v} = n R T {\int}_{{V}_{1}}^{{V}_{2}} \frac{1}{V} \mathrm{dV}$

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

So, we have:

$\textcolor{b l u e}{\Delta S \left(\text{const. T}\right) = n R \ln \left({V}_{2} / {V}_{1}\right)}$

METHOD 2

Write the total differential of the entropy as a function of temperature and volume:

$\mathrm{dS} \left(T , V\right) = {\left(\frac{\partial S}{\partial T}\right)}_{V} \mathrm{dT} + {\left(\frac{\partial S}{\partial V}\right)}_{T} \mathrm{dV}$

The constant temperature process, or isothermal process, is the second term, so we seek only:

$\Delta S = {\cancel{{\int}_{{T}_{1}}^{{T}_{2}} {\left(\frac{\partial S}{\partial T}\right)}_{V} \mathrm{dT}}}^{0}$ $+$ ${\int}_{{V}_{1}}^{{V}_{2}} {\left(\frac{\partial S}{\partial V}\right)}_{T} \mathrm{dV}$

$T$ and $V$ are natural variables in the Maxwell Relation for the Helmholtz free energy $A$:

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

Since all thermodynamic functions ($G , A , H , E , S , T , V , P , . . .$) are state functions (whose differentials are "exact"), the cross-derivatives here are equal:

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

So, we have:

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

And for ideal gases, we again know the form of the pressure:

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

So its partial derivative with respect to temperature at constant volume is $n R / V$. This gives:

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

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

just as we got before.