Physical Chemistry/State Functions

Current revision (unreviewed)

The following demonstrates what's a state function and what's not a state function.

$q_{rev} \;$ is not exact differential for a gas obeying van der Waals' equation, but $\frac{q_{rev}}{T}$ is as demonstrated below:

$dq_{rev} \;=\left ( \frac{\partial U}{\partial T} \right )_vdT+\left [ P_{ext}+\left ( \frac{\partial U}{\partial V} \right )_T \right ]dV$

We assume quasistatic situation, so $P_{ext} = P \;$ .

$dq_{rev} \;=\left ( \frac{\partial U}{\partial T} \right )_vdT+\left [ P+\left ( \frac{\partial U}{\partial V} \right )_T \right ]dV$

$=C_vdT+\left [ P+\left ( \frac{\partial U}{\partial V} \right )_T \right ]dV$

$=C_vdT+\left [ P+\left ( \frac{a}{\overline V^2} \right ) \right ]dV$

$=C_vdT+\left ( \frac{RT}{\overline V-b} \right )dV$

Now, you take the cross partial derivatives.

$\left ( \frac{\partial C_v}{\partial V} \right )_T=0$

$\left ( \frac{\partial \left ( \frac{RT}{\overline V-b} \right )}{\partial T} \right )_V=\frac{R}{\overline V-b}$

They are not equal; hence, $q_{rev} \;$ is not exact differential (not a state function).

However, if we take $\frac{q_{rev}}{T}$ it will be exact differential (a state function).

$\frac{dq_{rev}}{T}=\frac{C_v}{T}dT+\left ( \frac{R}{\overline V-b} \right )dV$

Take the cross partial derivatives.

$\left ( \frac{\partial \left ( \frac{C_v}{T}\right )}{\partial V} \right )_T=0$

$\left ( \frac{\partial \left ( \frac{R}{\overline V-b} \right )}{\partial T} \right )_V=0$

Both are equal making $\frac{q_{rev}}{T}$ exact differential (a state function).