Answer
$W = 165.92\ kJ$
Work Step by Step
From steam tables:
At 300 kPa, the specific volume of saturated steam is $v_1 = 0.60582\ m³/kg$.
At 300 kPa and 200°C, the specific volume of the steam is: $v_2 = 0.71643\ m³/kg$.
The work at constant pressure is given by:
$W = mP(v_2-v_1)$
With a mass of steam of $5\ kg$:
$W = 165.92\ kJ$
