# Can the entropy of an ideal gas change during an isothermal process?

Jan 4, 2018

Yes.

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

i.e. at constant temperature, expanding gases increase in entropy.

Yes, $\Delta S$ is not a function of only temperature, so it is not zero.

An isothermal process has $\Delta T = 0$, but one can write a total differential for the entropy as a function of $T$ and $V$:

$\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}$$\text{ "" } \boldsymbol{\left(1\right)}$

In this case, one could say that at constant temperature, $\mathrm{dT} = 0$, so we simplify $\left(1\right)$ down to:

${\mathrm{dS}}_{T} = {\left(\frac{\partial S}{\partial V}\right)}_{T} \mathrm{dV}$$\text{ "" } \boldsymbol{\left(2.1\right)}$

The natural variables associated with this partial derivative are $T$ and $V$, which are found in the Helmholtz Maxwell relation:

$\mathrm{dA} = - S \mathrm{dT} - P \mathrm{dV}$ $\text{ "" } \boldsymbol{\left(3\right)}$

For any state function, the cross-derivatives are equal, so from $\left(3\right)$, we rewrite $\left(2.1\right)$ using the relation:

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

Therefore, in terms of a partial derivative that uses the ideal gas law, we get:

${\mathrm{dS}}_{T} = {\left(\frac{\partial P}{\partial T}\right)}_{V} \mathrm{dV}$ $\text{ "" } \boldsymbol{\left(2.2\right)}$

The right-hand side of $\left(2.2\right)$ from the ideal gas law gives:

${\left(\frac{\partial P}{\partial T}\right)}_{V} = \frac{\partial}{\partial T} {\left[\frac{n R T}{V}\right]}_{V} = \frac{n R}{V}$

Therefore, the change in entropy of an ideal gas at a specific, constant temperature is (by integrating $\left(2.2\right)$):

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

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

So if the gas expands in the isothermal process, then yes, it will have increased entropy.