# How do you calculate DeltaS_(sys) for the isothermal expansion of 1.5 mol of an ideal gas from 20.0 L to 22.5L?

Dec 12, 2016

I got $\text{1.47 J/K}$. This should make sense to be positive since a larger volume allows more particle movement, which increases the extent of energy dispersal in the system.

We begin by realizing that an isothermal expansion is one where $\Delta T = 0$. The change in entropy due to a change in volume at a constant temperature is denoted as ${\left(\frac{\partial S}{\partial V}\right)}_{T}$.

To get from ${V}_{1}$ to ${V}_{2}$, we integrate the following:

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

We denote molar entropy as $\overline{S} = \frac{S}{n}$ and molar volume as $\overline{V} = \frac{V}{n}$, so that we then have our starting equation:

$\textcolor{g r e e n}{\Delta {\overline{S}}_{\text{sys}} = {\int}_{{\overline{V}}_{1}}^{{\overline{V}}_{2}} {\left(\frac{\partial \overline{S}}{\partial \overline{V}}\right)}_{T} \mathrm{db} a r V}$

From the Maxwell Relation for $\overline{A} \left(T , \overline{V}\right)$ (the molar Helmholtz Free Energy as a function of temperature and molar volume)

$\mathrm{db} a r A = - \overline{S} \mathrm{dT} - P \mathrm{db} a r V$,

we use the cross-derivative relationship to get

${\left(\frac{\partial \overline{S}}{\partial \overline{V}}\right)}_{T} = \textcolor{g r e e n}{{\left(\frac{\partial P}{\partial T}\right)}_{\overline{V}}}$,

meaning that the change in entropy due to the change in volume at constant temperature is equal to the change in pressure due to the change in temperature at a constant volume.

Next, we would have to figure out what this new derivative gives for the ideal gas law. Recall that the ideal gas law is:

$P V = n R T$

or

$P \overline{V} = R T$

$\implies P = \frac{R T}{\overline{V}}$

The partial derivative of pressure with respect to temperature at constant molar volume is then:

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

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

$= \frac{R}{\overline{V}} \cancel{\frac{\mathrm{dT}}{\mathrm{dT}}}$

$= \frac{R}{\overline{V}}$

So, what's left is to integrate the result by substituting $\frac{R}{\overline{V}}$ back in for ${\left(\frac{\partial \overline{S}}{\partial \overline{V}}\right)}_{T}$:

$\textcolor{b l u e}{\Delta {\overline{S}}_{\text{sys}}} = {\int}_{{\overline{V}}_{1}}^{{\overline{V}}_{2}} \frac{R}{\overline{V}} \mathrm{db} a r V$

$= R {\int}_{{\overline{V}}_{1}}^{{\overline{V}}_{2}} \frac{1}{\overline{V}} \mathrm{db} a r V = R | \left[\ln \left(\overline{V}\right)\right] {|}_{{\overline{V}}_{1}}^{{\overline{V}}_{2}}$

$= R \left[\ln \left({\overline{V}}_{2}\right) - \ln \left({\overline{V}}_{1}\right)\right] = R \ln \left({\overline{V}}_{2} / \left({\overline{V}}_{1}\right)\right) = R \ln \left(\left({V}_{2} \text{/"cancel(n))/(V_1"/} \cancel{n}\right)\right)$

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

Notice how the molar volumes have canceled out to give the regular volume.

Therefore, the change in entropy for the system at constant temperature is:

color(blue)(DeltaS_"sys") = nDeltabarS_"sys" = nRln(V_2/V_1)

= ("1.5 mols")("8.314472 J/mol"cdot"K")ln(("22.5 L")/("20.0 L"))

$=$ $\textcolor{b l u e}{\text{1.47 J/K}}$