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).

Last modified on 12 November 2010, at 22:46