Answer
$Q=2416\ kJ$
Work Step by Step
Energy balance for the whole process:
$Q-W_b=\Delta U= m(u_3-u_1)$
For constant pressure processes:
$W_{b,1-2}=P_1(V_2-V_1)$
For constant volume processes:
$W_{b,2-3}=0$
For an ideal gas:
$PV=mRT$
Initial state ($P=200kPa,\ m=3\ kg,\ R=0.287\ kJ/kg.K,\ T=300\ K$):
Solving for $V: V_1=1.29m³$
Intermediary state (volume doubles):
$V_2=V_3=2.58\ m³$
Final state (pressure doubles):
Solving for $T: T_3=1200\ K$
Hence,
$W_b=W_{b,1-2} = 258\ kJ$
From Table A-17:
$u_1=214.07\ kJ/kg$
$u_2=933.33\ kJ/kg$
$\Delta U=2158\ kJ$
$Q=2416\ kJ$
