# How do I prove that the cyclic rule of partial derivatives applies to P, T, and V?

Jun 16, 2017

The main "tricks" are to write the total derivative of $P \left(T , V\right)$ (pressure as a function of temperature and volume), and recall that the compressibility factor of an ideal gas is $1$.

To derive the cyclic rule of partial derivatives of pressure, volume, and temperature, try writing the total derivative of the pressure as a function of temperature $T$ and molar volume $\overline{V}$.

$\mathrm{dP} \left(T , \overline{V}\right) = {\left(\frac{\partial P}{\partial T}\right)}_{\overline{V}} \mathrm{dT} + {\left(\frac{\partial P}{\partial \overline{V}}\right)}_{T} \mathrm{db} a r V$

Now, we can divide through by $\partial \overline{V}$ at constant $P$ to cancel out the lefthand side (the derivative of pressure at constant pressure is zero).

${\cancel{{\left(\frac{\partial P}{\partial \overline{V}}\right)}_{P}}}^{0} = {\left(\frac{\partial P}{\partial T}\right)}_{\overline{V}} {\left(\frac{\partial T}{\partial \overline{V}}\right)}_{P} + {\left(\frac{\partial P}{\partial \overline{V}}\right)}_{T} {\cancel{{\left(\frac{\partial \overline{V}}{\partial \overline{V}}\right)}_{P}}}^{1}$

$0 = {\left(\frac{\partial P}{\partial T}\right)}_{\overline{V}} {\left(\frac{\partial T}{\partial \overline{V}}\right)}_{P} + {\left(\frac{\partial P}{\partial \overline{V}}\right)}_{T}$

Now all we have to do is subtract the lone derivative to the other side, and cancel it out. Recall that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\frac{\mathrm{dx}}{\mathrm{dy}}}$. Therefore:

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

$- {\cancel{{\left(\frac{\partial \overline{V}}{\partial P}\right)}_{T} {\left(\frac{\partial P}{\partial \overline{V}}\right)}_{T}}}^{1} = {\left(\frac{\partial P}{\partial T}\right)}_{\overline{V}} {\left(\frac{\partial T}{\partial \overline{V}}\right)}_{P} {\left(\frac{\partial \overline{V}}{\partial P}\right)}_{T}$

In the end, we get these partial derivatives in a row:

$\textcolor{b l u e}{{\left(\frac{\partial P}{\partial T}\right)}_{\overline{V}} {\left(\frac{\partial T}{\partial \overline{V}}\right)}_{P} {\left(\frac{\partial \overline{V}}{\partial P}\right)}_{T} = - 1}$

Notice how if you read each column of partial derivatives, it is in a sequence:

$P , T , T , V , V , P$

where the remaining variable not involved in the derivative is held constant. It is in that sense cyclic (we went back to $P$), and that is how you can remember this.

Now, we can use the properties of partial derivatives to evaluate these derivatives (i.e. whatever is held constant can be factored out).

The first one is a straightforward partial derivative:

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

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

$= \frac{1}{\overline{V}} \frac{d}{\mathrm{dT}} \left[R T\right]$

$= \frac{R}{\overline{V}}$ (or $\frac{n R}{V}$.)

The second one involves solving for $T$:

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

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

$= P \frac{d}{\mathrm{db} a r V} \left[\frac{\overline{V}}{R}\right]$

$= \frac{P}{R}$

And lastly, the third partial derivative.

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

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

$= T \frac{d}{\mathrm{dP}} \left[\frac{R}{P}\right]$

$= - \frac{R T}{P} ^ 2$

Lastly, we multiply these three derivatives together to see if they equal $- 1$ (and they should!).

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

$= \frac{\cancel{R}}{\overline{V}} \cdot \frac{\cancel{P}}{\cancel{R}} \cdot - \frac{R T}{{P}^{\cancel{2}}}$

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

And recalling that $Z = \frac{P \overline{V}}{R T}$ is the compressibility factor, if $Z = 1$, the gas is ideal and vice versa. Since we used the ideal gas law, we inherently assumed that the gas was ideal.

Therefore...

$\textcolor{b l u e}{{\left(\frac{\partial P}{\partial T}\right)}_{\overline{V}} {\left(\frac{\partial T}{\partial \overline{V}}\right)}_{P} {\left(\frac{\partial \overline{V}}{\partial P}\right)}_{T}}$

$= - \frac{R T}{P \overline{V}} = - \frac{1}{Z} = - \frac{1}{1} = \textcolor{b l u e}{- 1}$