Answer
$d h=c_p d T+v d P$
Work Step by Step
The change in enthalpy is expressed as $$
d h=c_P d T+\left(v-T\left(\frac{\partial v}{\partial T}\right)_P\right) d P
$$ For an ideal gas $v=R T / P$. Then, $$
v-T\left(\frac{\partial v}{\partial T}\right)_P=v-T\left(\frac{R}{P}\right)=v-v=0
$$ Thus, $$
d h=c_p d T
$$ To complete the proof we need to show that $c_p$ is not a function of $P$ either. This is done with the help of the relation $$
\left(\frac{\partial c_p}{\partial P}\right)_T=-T\left(\frac{\partial^2 v}{\partial T^2}\right)_P
$$ For an ideal gas, $$
\left(\frac{\partial v}{\partial T}\right)_P=\frac{R}{P} \text { and }\left(\frac{\partial^2 v}{\partial T^2}\right)_P=\left(\frac{\partial(R / P)}{\partial T}\right)_P=0
$$ Thus, $$
\left(\frac{\partial c_P}{\partial P}\right)_T=0
$$ Therefore we conclude that the enthalpy of an ideal gas is a function of temperature only.
For an incompressible substance $v=$ constant and thus $\partial v \partial \mathrm{T}=0$. Then, $$
d h=c_p d T+v d P
$$ Therefore we conclude that the enthalpy of an incompressible substance is a function of temperature and pressure.
