Answer
$s_{2}-s_{1}=−0.2386\text{ kJ/kg⋅K}$
Work Step by Step
Solving the equation of state for $v$ gives $$
v=\frac{R T}{P}+a
$$ Then, $$
\left(\frac{\partial v}{\partial T}\right)_P=\frac{R}{P}
$$ The entropy differential is $$
\begin{aligned}
d s & =c_p \frac{d T}{T}-\left(\frac{\partial v}{\partial T}\right)_P d P \\
& =c_p \frac{d T}{T}-R \frac{d P}{P}
\end{aligned}
$$ which is the same as that of an ideal gas. Integrating this result between the two states gives $$
\begin{aligned}
s_2-s_1 & =c_p \ln \frac{T_2}{T_1}-R \ln \frac{P_2}{P_1} \\
& =(5.1926 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}) \ln \frac{573 \mathrm{~K}}{293 \mathrm{~K}}-(2.0769 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}) \ln \frac{600 \mathrm{kPa}}{100 \mathrm{kPa}} \\
& =-0.2386 \mathrm{~kJ} / \mathbf{k g} \cdot \mathbf{K}
\end{aligned}
$$
