# 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:

$dq_{rev}\;=\left({\frac {\partial U}{\partial T}}\right)_{v}dT+\left[P_{ext}+\left({\frac {\partial U}{\partial V}}\right)_{T}\right]dV$ We assume quasistatic situation, so $P_{ext}=P\;$ .

$dq_{rev}\;=\left({\frac {\partial U}{\partial T}}\right)_{v}dT+\left[P+\left({\frac {\partial U}{\partial V}}\right)_{T}\right]dV$ $=C_{v}dT+\left[P+\left({\frac {\partial U}{\partial V}}\right)_{T}\right]dV$ $=C_{v}dT+\left[P+\left({\frac {a}{{\overline {V}}^{2}}}\right)\right]dV$ $=C_{v}dT+\left({\frac {RT}{{\overline {V}}-b}}\right)dV$ 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).