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:

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

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).

Chimie physique/Fonction d'état