### Fundamental Property Relations and the Maxwell Equations

All of thermodynamics is build on the two premises of the first
and second law, which can mathematically be written as

$du = w_{rev} + q_{rev}$

$ds = \frac{q_{rev}}{T}$

Combining these and using the definition of the (reversible)
work, we get

In addition to these two fundamental properties ($u, s$), we
have also defined several "derived properties":

$h \equiv u + Pv$

$a \equiv u - Ts$

$g \equiv h - Ts$

The way that much of the rest of thermodynamics is constructed
is to use these definitions/relations and mathematically manipulate
them. To do so, we need to recall the following two mathematical
truths (note that the subscripts on the ()'s refer to variables
held constant during the differentiation):

##### Note:

For a function $Y$ that depends on two variables $x$ and $z$, we
can write

$dY = \left (\frac{\partial Y}{\partial x} \right )_z dx +
\left (\frac{\partial Y}{\partial z} \right )_x dz$

##### Note:

The **order** of differentiation does not matter,
so :

$\left [\frac{\partial}{\partial z} \left (\frac{\partial
Y}{\partial x} \right )_z \right ]_x= \left [
\frac{\partial}{\partial x} \left (\frac{\partial
Y}{\partial z} \right )_x \right ]_z$

Taking the first of these facts and using it to write an
expression for $u$ as a function of $s$ and $v$, leads to:

$du = \left (\frac{\partial u}{\partial s} \right )_v ds +
\left (\frac{\partial u}{\partial v} \right )_s dv$

Equating the terms in front of the $ds$ and $dv$ to their
counterparts int he earlier expression for $du$ (above) leads
to:

$T = \left (\frac{\partial u}{\partial s} \right )_v$

$-P = \left (\frac{\partial u}{\partial v} \right )_s$

By following the same procedure for $h$ by writing it as a
function of $s$ and $p$, leads to:

$T = \left (\frac{\partial h}{\partial s} \right )_P$

$v = \left (\frac{\partial h}{\partial P} \right )_s$

Finally, doing the same for $g$ and $a$, gives:

$-s = \left (\frac{\partial a}{\partial T} \right
)_v$

$-P = \left (\frac{\partial a}{\partial v} \right )_T$

$-s = \left (\frac{\partial g}{\partial T} \right )_P$

$v = \left (\frac{\partial g}{\partial P} \right )_T$