I came across the following equation for differential change in enthalpy:
$$ dh = c_p dT + \frac{1-\beta T}{\rho}dP$$
where $\beta = -\frac{1}{\rho}(\partial \rho / \partial T)_P$ is the thermal expansion coefficient.
I would like to prove this equation for myself, but I'm stuck.
I start from
$$ dh = Tds + vdP$$
and use $c_p = T\frac{dS}{dT}$ which gives
$$ dh = c_pdT + vdP$$
but this implies that
$$ v = \frac{1-\beta T}{\rho}$$
if I want to approach the first equation. I don't see why this is true, if it's even true at all. How can I arrive at the first equation?
EDIT:
I'm gonna start with
$$ dh = Tds + vdP$$
and use a total differential for $ds$:
$$ds=C_P\frac{dT}{T}+\left(\frac{\partial s}{\partial P}\right)_TdP$$
Then use a Maxwell relation $\left(\frac{\partial s}{\partial P}\right)_T=-\left(\frac{\partial v}{\partial T}\right)_P$:
$$ds=C_P\frac{dT}{T}-\left(\frac{\partial v}{\partial T}\right)_PdP$$
where $\left(\frac{\partial v}{\partial T}\right)_P = \beta/\rho$. Now substitute this $ds$ into $dh$:
$$ dh = c_pdT + \frac{1-\beta T}{\rho} dP$$