Physical Chemistry/State Functions

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:

$d{q}_{rev}={\left(\frac{\partial U}{\partial T}\right)}_{v}dT+[P_{ext}+{\left(\frac{\partial U}{\partial V}\right)}_T]dV$

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

$d{q}_{rev}={\left(\frac{\partial U}{\partial T}\right)}_{v}dT+[P+{\left(\frac{\partial U}{\partial V}\right)}_T]dV$

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

$=C_{v}dT+[P+\left(\frac{a}{\stackrel{2}{\stackrel{‾}{V}}}\right)]dV$

$=C_{v}dT+\left(\frac{RT}{\bar{V}-b}\right)dV$

Now, you take the cross partial derivatives.

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

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

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).

$\frac{d{q}_{rev}}{T}=\frac{C_v}{T}dT+\left(\frac{R}{\bar{V}-b}\right)dV$

Take the cross partial derivatives.

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

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

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