Derivation of expansion rule

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 \(\partial x/\partial y = \frac{1}{\partial y/\partial x}\).