Answer
a) $P_2 = 450\ kPa,\ T_2=147.90°C$
b)$W = 44.46\ kJ$
Work Step by Step
Assuimg that at the initial state of unsaturated water, the specific volume is approximately equal to the specific volume of saturated water at the same temperature of 25°C: $v_0 = 0.001003 m³/kg$,
with a mass of 50 kg $V_0 = 0.05015\ m³$
Up to the piston reaching the spring, there's an isobaric expansion to 0.2 m³, therefore:
$P_1=P_0 = 250\ kPa,\ W_{0-1}=P_0(V_1 - V_0) = 37.46\ kJ$
The elastic force to displace the spring with 100 kN/m constant by 20 cm is:
$F_s = k\Delta x = 20 kN$
Which increases the pressure by:
$\Delta P = \frac{F_s}{A} = 200 kPa$
Therefore the final pressure is:$P_2 = P_1+\Delta P = 450\ kPa$
The final volume is
$V_2 = V_1 + A\Delta x = 0.22 m³$
$v_2 = \frac{V_2}{m} = 0.0044 m³$
Since the specific volume is intermediary between the saturated liquid and steam at 450 kPa, the final temperature is the saturation temperature: $T_2=147.90°C$
For the spring-loaded expansion, pressure changes linearly with the increase in volume, therefore the work is the area under the trapezoid $(P_1,V_1),(P_2,V_2)$:
$W_{1-2} = (V_2-V_1)\frac{P_1+P_2}{2}$
$W_{1-2} = 7.00\ kJ $
The total work done is:
$W = W_{0-1}+W_{1-2} = 44.46\ kJ$
The P-V diagram would result in a straight horizontal line from $V_0$ to $V_1$ and a straight sloped line between $(P_1,V_1)$ and $(P_2,V_2)$.
