Answer
$w_s=439\ kJ/kg$
Work Step by Step
Since maximum work is achieved in a reversible processes (isentropic):
$\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^\dfrac{k-1}{k}$
Given $T_1=500\ °C,\ P_2=200\ kPa,\ P_1=3500\ kPa,\ k=1.381$
$T_2=351\ K$
From the energy balance:
$\dot{m}h_1=\dot{m}h_2+\dot{W}_s$
$w_s=h_1-h_2=c_p(T_1-T_2)$
Given $c_p=1.040\ kJ/kg.K\ T_{avg}=550\ K$
$w_s=439\ kJ/kg$
