# Show that for a van der Waals gas, ((delC_V)/(delV))_T = 0, where C_V = ((delU)/(delT))_V?

## Jan 22, 2017

By definition, the constant-volume heat capacity was:

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

where $U$ is the internal energy, and $T$ and $V$ are temperature and volume, respectively, as defined in the ideal gas law and other gas laws.

To show that for a van der Waals gas, the constant-volume heat capacity does not change due to a change in volume at a constant temperature, i.e. ${\left(\frac{\partial {C}_{V}}{\partial V}\right)}_{T} = 0$, we first write out the van der Waals (vdW) equation of state:

$\boldsymbol{P = \frac{R T}{\overline{V} - b} - \frac{a}{{\overline{V}}^{2}} = \frac{n R T}{V - n b} - \frac{a {n}^{2}}{{V}^{2}}}$, $\text{ "" } \boldsymbol{\left(2\right)}$

where $\overline{V} = \frac{V}{n}$ is the molar volume, $R$ is the universal gas constant, $a$ and $b$ are the vdW constants for intermolecular forces of attraction and excluded volume (respectively), and $P$ is the pressure.

This means we'll need to have an expression for $\mathrm{dU}$ that includes $P$ somehow, or even a partial derivative.

We can use the Maxwell relation for the internal energy in a closed system undergoing a reversible process:

$\mathrm{dU} = T \mathrm{dS} - P \mathrm{dV}$,

If we divide by ${\left(\partial V\right)}_{T}$ for the entire equation, we get:

${\left(\frac{\partial U}{\partial V}\right)}_{T} = T {\left(\frac{\partial S}{\partial V}\right)}_{T} - P {\cancel{{\left(\frac{\partial V}{\partial V}\right)}_{T}}}^{1}$ $\text{ "" } \boldsymbol{\left(3\right)}$

This will be the main expression we'll work with. Note that $T$ and $V$ are the natural variables of the Helmholtz free energy, $A$, whose Maxwell relation is:

$\mathrm{dc} o l \mathmr{and} \left(red\right) \left(A\right) = - S \textcolor{red}{\mathrm{dT}} - P \textcolor{red}{\mathrm{dV}}$

Using the fact that the Helmholtz free energy is a state function (just like $G$, $H$, $S$, etc), there is a cyclic relationship we can use:

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

Plugging into $\boldsymbol{\left(3\right)}$ gives:

${\left(\frac{\partial U}{\partial V}\right)}_{T} = T {\left(\frac{\partial P}{\partial T}\right)}_{V} - P$ $\text{ "" } \boldsymbol{\left(4\right)}$

which is something we can relate back to the vdW equation of state, using ${\left(\frac{\partial P}{\partial T}\right)}_{V}$. We're almost there.

Given that $U$ is a state function (and this is very important!), the order of partial differentiation does not matter:

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

or

${\left(\frac{{\partial}^{2} U}{\partial V \partial T}\right)}_{V , T} = {\left(\frac{{\partial}^{2} U}{\partial T \partial V}\right)}_{T , V}$

This will become relevant if we take the partial derivative of ${C}_{V}$ with respect to $V$ at constant $T$ and plug in the definition of ${C}_{V}$ from $\boldsymbol{\left(1\right)}$:

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

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

$= \frac{\partial}{\partial T} {\left[{\left(\frac{\partial U}{\partial V}\right)}_{T}\right]}_{V}$$\text{ "" } \boldsymbol{\left(5\right)}$

Now consider the right side of $\boldsymbol{\left(5\right)}$ and plug in $\boldsymbol{\left(4\right)}$:

${\left(\frac{\partial {C}_{V}}{\partial V}\right)}_{T}$

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

$= \frac{\partial}{\partial T} {\left[T {\left(\frac{\partial P}{\partial T}\right)}_{V}\right]}_{V} - {\left(\frac{\partial P}{\partial T}\right)}_{V}$ $\text{ "" } \boldsymbol{\left(6\right)}$

Now, we should figure out what ${\left(\frac{\partial P}{\partial T}\right)}_{V}$ is so we can use it! Plugging in the vdW equation of state from $\boldsymbol{\left(2\right)}$:

${\left(\frac{\partial P}{\partial T}\right)}_{V} = \frac{\partial}{\partial T} {\left[\frac{n R T}{V - n b} - \frac{a {n}^{2}}{{V}^{2}}\right]}_{V}$

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

Plugging this back into $\boldsymbol{\left(6\right)}$, we get:

$\textcolor{b l u e}{{\left(\frac{\partial {C}_{V}}{\partial V}\right)}_{T}} = \frac{\partial}{\partial T} {\left[\frac{n R T}{V - n b}\right]}_{V} - \frac{n R}{V - n b}$

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

$= \textcolor{b l u e}{0}$