Answer
$T_2=458.4\ K$
$W_{b,e}=120\ kJ$
Work Step by Step
For ideal gases:
$PV=mRT$
Given $R=0.287\ kJ/kg.K,\ P=600\ kPa$:
Initial $V_1=0.25\ m³,\ T_1=300°C$: $m_1=0.9121\ kg$
Final $m_2=0.25m_1=0.2280\ kg$
Hence the final temperature ($V_2=0.05\ m³$): $T_2=458.4\ K$
$W_{b,e}=-P\Delta V$
$W_{b,e}=120\ kJ$
