Answer
$h_{2}-h_{1}=397.2\text{ kJ/kg}$
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 with constant specific heats gives $$
\begin{aligned}
& h_2-h_1=c_p\left(T_2-T_1\right)+a\left(P_2-P\right) \\
& =(1.029 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(420-34) \mathrm{K}+\left(0.01 \mathrm{~m}^3 / \mathrm{kg}\right)(800-100) \mathrm{kPa} \\
& =404.2\ \mathrm{~kJ} / \mathrm{kg} \\
\end{aligned}
$$ For an ideal gas, $$
d h=c_p d T
$$ which when integrated gives
$$
h_2-h_1=c_p\left(T_2-T_1\right)=(1.029 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(420-34) \mathrm{K}=\mathbf{3 9 7 . 2} \mathbf{k J} / \mathbf{k g}
$$
