Answer
$Q=-69.57\ kJ$
$W_b=-250.25\ kJ$
Work Step by Step
For an ideal gas:
$PV=mRT$
In the first state ($m=1kg, P_1=100kPa, T_1=25°C=298.15K, R=0.1889\ kJ/kg.K$):
$V_1=0.563m³$
In the second state ($P_2=1000kPa, T_2=300°C=573.15K$):
$V_2=0.108m³$
The boundary work in this case is given by:
$W_b=\frac{1}{2}(P_1+P_2)(V_2-V_1)$
$W_b=-250.25\ kJ$
The energy balance for this system is:
$Q-W_b=\Delta U $
$\Delta U = mc_v\Delta T$
With $c_v=0.657\ kJ/kg.K$ we get to:
$\Delta U = 180.68\ kJ$
Hence $Q=-69.57\ kJ$
