# Thermal expansion and change in enthalpy

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:

$$dh = Tds + vdP$$

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 $$dh$$:

$$dh = c_pdT + \frac{1-\beta T}{\rho} dP$$

Start with $$ds=\left(\frac{\partial s}{\partial T}\right)_PdT+\left(\frac{\partial s}{\partial P}\right)_TdP=C_P\frac{dT}{T}+\left(\frac{\partial s}{\partial P}\right)_TdP$$ Then, using the equation $$dG=-sdT+vdP$$show that $$\left(\frac{\partial s}{\partial P}\right)_T=-\left(\frac{\partial v}{\partial T}\right)_P$$ Finally, make use of the equation $$v=1/\rho$$

• Ah I forgot about the Maxwell relation. That did it.
– Drew
Jan 24 '19 at 3:18