Answer
$h_{2}-h_{1}=1869\text{ kJ/kg}$
Work Step by Step
Solving the equation of state for $v$ gives $$
v=\frac{R T}{P}+a
$$ Then, $$
\left(\frac{\partial v}{\partial T}\right)_P=\frac{R}{P}
$$ Using equation 12-35, $$
d h=c_P d T+\left[v-T\left(\frac{\partial v}{\partial T}\right)_P\right] d P
$$ Substituting, $$
\begin{aligned}
d h & =c_p d T+\left(\frac{R T}{P}+a-\frac{R T}{P}\right) d P \\
& =c_p d T+a d P
\end{aligned}
$$ Integrating this result between the two states gives $$
\begin{aligned}
h_2-h_1 & =c_p\left(T_2-T_1\right)+a\left(P_2-P_1\right) \\
& =(5.1926 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(380-20) \mathrm{K}+\\&\left(0.01 \mathrm{~m}^3 / \mathrm{kg}\right)(750-150) \mathrm{kPa} \\
& =1875 \mathrm{~kJ} / \mathbf{k g}
\end{aligned}
$$ For an ideal gas, $$
d h=c_p d T
$$ which when integrated gives $$
h_2-h_1=c_p\left(T_2-T_1\right)=(5.1926 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(380-20) \mathrm{K}=1869\ \mathbf{k J} / \mathbf{k g}
$$
