Answer
$m_e=0.357\ kg$
$Q_i=0\ kJ$
Work Step by Step
For ideal gases:
$PV=mRT$
Given $R=0.287\ kJ/kg.K,\ P=300\ kPa$:
Initial ($V_1=0.2\ m³,\ T_1=293\ K$): $m_1=0.714\ kg$
Final ($V_2=0.1\ m³,\ T_2=T_1$): $m_2=0.357\ kg$
From the material balance:
$-m_e=m_2-m_1$
$m_e=0.357\ kg$
From the energy balance:
$Q_i+W_{b,i}-m_eh_e=m_2u_2-m_1u_1$
Since $\Delta H= \Delta U +\Delta(PV)=\Delta U + W_b$
$Q_i-m_eh_e=m_2h_2-m_1h_1$
At constant temperature $h_e=h_2=h_1=h$
$Q_i=h(m_2-m_1+m_e)=0$
