# What is the entropy change for the isothermal expansion of "0.75 g" of "He" from "5.0 L" to "12.5 L"?

Mar 30, 2017

I got $\text{1.428 J/K}$.

I assume you aren't given an equation and have to derive it.

If we treat entropy as a function of the temperature and volume (that is, $S = S \left(T , V\right)$), and consider a constant-temperature expansion, then the total derivative is:

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

Since we are looking at the change in entropy due to a change in volume, we only consider ${\left(\frac{\partial S}{\partial V}\right)}_{T}$. Take the time to digest this step. This is the key step to understanding where to begin.

Another key step is to relate this to a derivative we are more familiar with. Recall that the Helmholtz free energy $A$ is a function of temperature and volume, and that its Maxwell relation is:

$\mathrm{dA} = - S \mathrm{dT} - P \mathrm{dV} = - \left(S \mathrm{dT} + P \mathrm{dV}\right)$

Since $A$ is a state function, its second derivatives are continuous, so that its cross derivatives are equal.

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

You should get to know this pretty well. The righthand side is a derivative we are familiar with. The ideal gas law can then be used.

$P V = n R T$

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

Now, the derivative is feasible in terms of an equation we know:

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

$= \frac{n R}{V} \frac{\mathrm{dT}}{\mathrm{dT}}$

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

This therefore gives:

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

Next, we can integrate this over the two volumes to get the change in entropy. By moving the $\mathrm{dV}$ in the ${\left(\frac{\partial S}{\partial V}\right)}_{T}$ derivative onto the other side:

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

The integral of $\frac{1}{V}$ is $\ln | V |$, so we have:

$\Delta S = n R {\int}_{{V}_{1}}^{{V}_{2}} \frac{1}{V} \mathrm{dV} = n R | \left[\ln | V |\right] {|}_{{V}_{1}}^{{V}_{2}}$

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

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

So, given $\text{0.75 g}$ of helium, we can get the mols of gas and thus find the change in entropy.

color(blue)(DeltaS) = 0.75 cancel"g He" xx cancel("1 mol")/(4.0026 cancel"g He") xx "8.314472 J/"cancel"mol"cdot"K" xx ln("12.5 L"/"5.0 L")

$=$ $\textcolor{b l u e}{\text{1.428 J/K}}$