# Question #7b328

Jun 22, 2017

Well, you haven't specified the conditions (what temperature? What pressure?), but you also haven't said anything about how the temperature changes (or doesn't change??)... I guess I'll have to assume constant temperature... shame on me.

$\Delta {S}_{\text{isothermal}} \left(T , V\right) = {C}_{V} {\cancel{\ln | {T}_{2} / {T}_{1} |}}^{0} + n R \ln | {V}_{2} / {V}_{1} |$

I get $\text{1.44 J/K}$ as a result, but you'll have to supply temperatures if you want a nonisothermal calculation.

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

$\mathrm{dS} = {\left(\frac{\partial S}{\partial T}\right)}_{V} \mathrm{dT} + {\left(\frac{\partial S}{\partial V}\right)}_{T} \mathrm{dV}$ $\text{ } \boldsymbol{\left(1\right)}$

Once we find out what these derivative terms mean, we can integrate $\mathrm{dS}$ from the initial to final state to find $\Delta S$.

The first derivative term is:

${\left(\frac{\partial S}{\partial T}\right)}_{V} = \frac{1}{T} {\left(\frac{\partial U}{\partial T}\right)}_{V} = {C}_{V} / T$, $\text{ } \boldsymbol{\left(2\right)}$

where ${C}_{V}$ is the constant-volume heat capacity and $U$ is the internal energy.

The second term is gotten from the Maxwell Relation for the Helmholtz energy as a function of the natural variables $T$ and $V$:

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

The cross-derivatives are equal because the Helmholtz free energy is a state function:

${\left(\frac{\partial S}{\partial V}\right)}_{T} = {\left(\frac{\partial P}{\partial T}\right)}_{V}$$\text{ } \boldsymbol{\left(3\right)}$

Plugging $\left(2\right)$ and $\left(3\right)$ into $\left(1\right)$, this results in:

$\mathrm{dS} = {C}_{V} / T \mathrm{dT} + {\left(\frac{\partial P}{\partial T}\right)}_{V} \mathrm{dV}$

Now, we can't just leave a derivative term here. Evaluating it first would make this easier. Using the ideal gas law, $P V = n R T$, and assuming helium is ideal:

${\left(\frac{\partial P}{\partial T}\right)}_{V} = \frac{\partial}{\partial T} {\left[\frac{n R T}{V}\right]}_{V} = \frac{n R}{V}$ $\text{ } \boldsymbol{\left(4\right)}$

Plugging $\left(4\right)$ into $\left(1\right)$, this means:

$\mathrm{dS} = {C}_{V} / T \mathrm{dT} + n R \cdot \frac{1}{V} \mathrm{dV}$$\text{ } \boldsymbol{\left(5\right)}$

And finally, integrating $\left(5\right)$ over the initial and final states gives the change in entropy for this process, which $\left(i\right)$ heats/cools the gas at constant volume and then $\left(i i\right)$ expands/contracts the gas at constant temperature.

$\textcolor{g r e e n}{\Delta S \left(T , V\right)} = {\int}_{\left(1\right)}^{\left(2\right)} \mathrm{dS} = {\int}_{{T}_{1}}^{{T}_{2}} {C}_{V} / T \mathrm{dT} + n R {\int}_{{V}_{1}}^{{V}_{2}} \frac{1}{V} \mathrm{dV}$

$= \textcolor{g r e e n}{{C}_{V} \ln | {T}_{2} / {T}_{1} | + n R \ln | {V}_{2} / {V}_{1} |}$

Evidently, we cannot do this without knowing temperature values... So we assume you are somehow implying constant temperature conditions all the way through...

$\implies \textcolor{b l u e}{\Delta {S}_{\text{isothermal}}} = {C}_{V} {\cancel{\ln | {T}_{2} / {T}_{1} |}}^{0} + n R \ln | \frac{2 \cancel{{V}_{1}}}{\cancel{{V}_{1}}} |$

$= \left(1 \cancel{\text{g He" xx cancel"1 mol"/(4.0026 cancel"g"))("8.314472 J/"cancel"mol"cdot"K}}\right) \ln 2$

$=$ $\textcolor{b l u e}{\text{1.44 J/K}}$