Answer
$W = 135.52\times10^3\ kJ$
Work Step by Step
Assuming the process occurs at a negligible pressure change:
From the steam tables:
The initial point:
Saturated, $T_1=T_2=200°C, P_1=P_2 = 1554.9\ kPa, v_{1,L} = 0.001157 m³/kg, v_{1,G} = 0.12721 m³/kg $
Calculating the specific volume at the final state with steam quality $x=0.8$:
$v_2 = v_{1,L} + x(v_{1,G} -v_{1,L} )$
$v_2 = 0.102 m³/kg$
The mass, from the initial volume of $V_1 = 1\ m³$ is:
$m = V_1/v_{1,L}$
$m = 864.304\ kg$
The work at constant pressure is given by:
$W = mP(v_2-v_{1,G})$
$W = 135.52\times10^3\ kJ$