Answer
$u_{2}-u_{1}=205\text{ kJ/kg}$
Work Step by Step
Solving the equation of state for $P$ gives $$
P=\frac{R T}{v-a}
$$ Then, $$
\left(\frac{\partial P}{\partial T}\right)_v=\frac{R}{v-a}
$$ Using equation $12-29$, $$
d u=c_v d T+\left[T\left(\frac{\partial P}{\partial T}\right)_v-P\right] d v
$$ Substituting, $$
\begin{aligned}
d u & =c_v d T+\left(\frac{R T}{v-a}-\frac{R T}{v-a}\right) d v \\
& =c_v d T
\end{aligned}
$$ Integrating this result between the two states with constant specific heats gives
$$
u_2-u_1=c_v\left(T_2-T_1\right)=(0.731 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(300-20) \mathrm{K}=\mathbf{2 0 5} \mathbf{k J} / \mathbf{k g}
$$ The ideal gas model for the air gives $$
d u=c_\nu d T
$$ which gives the same answer.
