Answer
$W = 24.52\ kJ$
Work Step by Step
Since the initial state is saturated water at 90°C, with a quality of 10%, from steam tables we get:
$P_1 = 70.183\ kPa,\ v_{1,L} = 0.001036\ m³/kg,\ v_{1,G} = 2.3593\ m³/kg$
Therefore the initial specific volume:
$v_1 = v_{1,L} + x(v_{1,G} - v_{1,L}) = 0.23686\ m³/kg $
For the final state at $P_2 = 800\ kPa,\ T_2 = 250°C$, from the steam tables: $v_2 = 0.29321\ m³/kg$
Since the piston-cylinder device is spring-loaded, we can expect the pressure and volume increases to be occuring linearly, therefore the work is the area under the trapezoid defined by the points: $(P_1,mv_1), (P_2,mv_2)$:
$W = m(v_2-v_1)(\frac{P_1+P_2}{2})$
$W = 24.52\ kJ$
