How to derive this $Tds$ relation? Given that $$du = Tds - pd\alpha$$ where $u$ is internal energy, $T$ is temperature, $p$ is pressure, $\alpha$ is specific volume, $s$ is entropy. We define specific heat of constant volume as $$c_v\equiv T(\partial s/\partial T)_v = (\partial u/\partial T)_v$$, and define specific heat of constant pressure as $$c_p\equiv T(\partial s/\partial T)_p=(\partial u/\partial T)_p+p(\partial \alpha/\partial T)_p$$
Following above equation to get a $p$ derivative, there is $$T(\partial s/\partial p)_T=(\partial u/\partial p)_T + p(\partial \alpha/\partial p)_T$$
My question is, how to obtain the expression below by "subtracting the $p$ derivative of Eqn. 3 from the $T$ derivative of Eqn. 4 to obtain the expression below? $$(\partial s/\partial p)_T = -(\partial \alpha/\partial T)_p$$
This question is from textbook "Atmosphere-Ocean Dynamics" by Adrian E. Gill" Chapter 3.2
 A: Just follow the literal instructions.
What's the $p$ derivative (at constant $T$) of $$T\left(\frac{\partial s}{\partial T}\right)_p=\left(\frac{\partial u}{\partial T}\right)_p+p\left(\frac{\partial \alpha}{\partial T}\right)_p?$$
It's $$T\frac{\partial}{\partial p}\left[\left(\frac{\partial s}{\partial T}\right)_p\right]_T=\frac{\partial}{\partial p}\left[\left(\frac{\partial u}{\partial T}\right)_p\right]_T+\left(\frac{\partial \alpha}{\partial T}\right)_p+p\frac{\partial}{\partial p}\left[\left(\frac{\partial \alpha}{\partial T}\right)_p\right]_T.$$
What's the $T$ derivative (at constant $p$) of $$T\left(\frac{\partial s}{\partial p}\right)_T=\left(\frac{\partial u}{\partial p}\right)_T+p\left(\frac{\partial \alpha}{\partial p}\right)_T?$$
It's $$\left(\frac{\partial s}{\partial p}\right)_T+T\frac{\partial}{\partial T}\left[\left(\frac{\partial s}{\partial p}\right)_T\right]_p=\frac{\partial}{\partial T}\left[\left(\frac{\partial u}{\partial p}\right)_T\right]_p+p\frac{\partial}{\partial T}\left[\left(\frac{\partial \alpha}{\partial p}\right)_T\right]_p,$$
which equals
$$\left(\frac{\partial s}{\partial p}\right)_T+T\frac{\partial}{\partial p}\left[\left(\frac{\partial s}{\partial T}\right)_p\right]_T=\frac{\partial}{\partial p}\left[\left(\frac{\partial u}{\partial T}\right)_p\right]_T+p\frac{\partial}{\partial p}\left[\left(\frac{\partial \alpha}{\partial T}\right)_p\right]_T$$
because the order of differentiation can be switched. Subtract one from the other to obtain
$$\left(\frac{\partial s}{\partial p}\right)_T=-\left(\frac{\partial \alpha}{\partial T}\right)_p,$$
QED.
But perhaps an easier approach to obtain this Maxwell relation is to recall that since the Gibbs free energy is defined as $G\equiv U-Ts+p\alpha$ (so that $dG=-s\,dT+\alpha\,dp$), then $s\equiv-\left(\frac{\partial G}{\partial T}\right)_p$ and $\alpha\equiv\left(\frac{\partial G}{\partial p}\right)_T$, so $$\left(\frac{\partial s}{\partial p}\right)_T=-\frac{\partial^2 G}{\partial T\,\partial p}=-\frac{\partial^2 G}{\partial p\,\partial T}=-\left(\frac{\partial \alpha}{\partial T}\right)_p,$$
QED.
