Derivation of expansion rule
Let’s assume we have 5 variables, \(x,y,u,v,z\), such that all 5 are state
variables and we can write an exact differentiable for any of the
quantities. Then we could write:
\[dz = \left ( \frac{\partial z}{\partial x}\right )_{y} dx + \left ( \frac{\partial z}{\partial y}\right )_{x} dy \]
or
\[dz = \left ( \frac{\partial z}{\partial u}\right )_{v} du + \left ( \frac{\partial z}{\partial v}\right )_{u} dy \]
Thermodynamically, if \(z\) is equivalent to some thermodynamic potential,
like \(H\), then \(x,y,u,v\) are either intensive or extensive parameters.
We can equivalently write expressions for \(x\) and \(y\) in terms of \(u\)
and \(v\):
\[\begin{split}\begin{aligned}
dx &= \left ( \frac{\partial x}{\partial u}\right )_{v} du + \left ( \frac{\partial x}{\partial v}\right )_{u} dv \\
dy &= \left ( \frac{\partial y}{\partial u}\right )_{v} du + \left ( \frac{\partial y}{\partial v}\right )_{u} dv
\end{aligned}\end{split}\]
Substituting these two expressions into the expressions for \(dz\) as a
function of \(dx\) and \(dy\) and again collecting like terms gives:
\[\begin{split}\begin{aligned}
dz &= \left ( \frac{\partial z}{\partial x}\right )_{y} \left ( \frac{\partial x}{\partial u}\right )_{v} du + \left ( \frac{\partial z}{\partial x}\right )_{y} \left ( \frac{\partial x}{\partial v}\right )_{u} dv + \left ( \frac{\partial z}{\partial y}\right )_{x} \left ( \frac{\partial y}{\partial u}\right )_{v} du + \left ( \frac{\partial z}{\partial y}\right )_{x} \left ( \frac{\partial y}{\partial v}\right )_{u} dv \\
&= \left [ \left ( \frac{\partial z}{\partial x}\right )_{y} \left ( \frac{\partial x}{\partial u}\right )_{v} + \left ( \frac{\partial z}{\partial y}\right )_{x} \left ( \frac{\partial y}{\partial u}\right )_{v} \right ] du + \\\nonumber
& \left [ \left ( \frac{\partial z}{\partial x}\right )_{y} \left ( \frac{\partial x}{\partial v}\right )_{u} + \left ( \frac{\partial z}{\partial y}\right )_{x} \left ( \frac{\partial y}{\partial v}\right )_{u} \right ] dv
\end{aligned}\end{split}\]
Comparing this expression to the second expression for \(dz\) in terms of
\(du\) and \(dv\), we see that:
\[\left ( \frac{\partial z}{\partial u}\right )_{v} = \left ( \frac{\partial z}{\partial x}\right )_{y} \left ( \frac{\partial x}{\partial u}\right )_{v} + \left ( \frac{\partial z}{\partial y}\right )_{x} \left ( \frac{\partial y}{\partial u}\right )_{v}\]
We can now find a couple of different relationships. First, assume that
there are only 4 variables under consideration, and \(u = y\). Then this
expression reduces to:
\[\begin{split}\begin{aligned}
\left ( \frac{\partial z}{\partial y}\right )_{v} &= \left ( \frac{\partial z}{\partial x}\right )_{y} \left ( \frac{\partial x}{\partial y}\right )_{v} + \left ( \frac{\partial z}{\partial y}\right )_{x} \left ( \frac{\partial y}{\partial y}\right )_{v} \\
&= \left ( \frac{\partial z}{\partial x}\right )_{y} \left ( \frac{\partial x}{\partial y}\right )_{v} + \left ( \frac{\partial z}{\partial y}\right )_{x}
\end{aligned}\end{split}\]
Note that we can also prove another relation using this same approach,
this time letting \(u = y\) and \(v = z\):
\[\begin{split}\begin{aligned}
\left ( \frac{\partial z}{\partial y}\right )_{z} &= \left ( \frac{\partial z}{\partial x}\right )_{y} \left ( \frac{\partial x}{\partial y}\right )_{z} + \left ( \frac{\partial z}{\partial y}\right )_{x} \left ( \frac{\partial y}{\partial y}\right )_{z} \\
0 &= \left ( \frac{\partial z}{\partial x}\right )_{y} \left ( \frac{\partial x}{\partial y}\right )_{z} + \left ( \frac{\partial z}{\partial y}\right )_{x} \\
\left ( \frac{\partial z}{\partial y}\right )_{x} &= -\left ( \frac{\partial z}{\partial x}\right )_{y} \left ( \frac{\partial x}{\partial y}\right )_{z} \\
\left ( \frac{\partial z}{\partial y}\right )_{x} \left ( \frac{\partial y}{\partial x}\right )_{z} \left ( \frac{\partial x}{\partial z}\right )_{y} &= -1
\end{aligned}\end{split}\]
In this last line, we used the reciprocal relation that
\(\partial x/\partial y = \frac{1}{\partial y/\partial x}\).