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?


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 $dh$:

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


1 Answer 1


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 $$v=1/\rho$$

  • $\begingroup$ Ah I forgot about the Maxwell relation. That did it. $\endgroup$ Commented Jan 24, 2019 at 3:18

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