Answer
a) $T_1=113.3°C$, $T_3=675°C$
b) $m_{2,L} = 0\ kg$
c) $W_b=218\ kJ$
Work Step by Step
a) Since the initial state is saturated at $P_1=160\ kPa$:
$T_1=113.3°C,\ v_{1,L}= 0.001054\ m³/kg,\ v_{1,G} = 1.0915\ m³/kg$
Given the initial masses of 1 kg of liquid and 2 kg of steam:
$V_1=m_{1,L}v_{1,L}+m_{1,G}v_{1,G} = 2.184\ m³$
Thus the final volume: $V_3=1.2V_1=2.621\ m³$
and $v_3=\frac{V_3}{m}=0.8736\ m³/kg$
With $P_3=500\ kPa, V_3$, from table A-6: $T_3=675°C$
b) Since $V_2=V_1$
$v_2=\frac{V_1}{m}=0.7280\ m³/kg > 0.37483\ m³/kg=v_{2,g,sat}$
Therefore there's no liquid in the second state: $m_{2,L} = 0\ kg$
c) Since from 1-2 the volume is constant there's no boundary work, hence
$W_b=W_{b,2-3}=P_2(V_3-V_2)$
$W_b=218\ kJ$
