Answer
$T_2=351\ K$
$m_i=1.11\ kg$
Work Step by Step
$PV=mRT,\ R=0.287\ kJ/kg.K$
Initial ($P_1=150\ kPa,\ V_1=0.11\ m^3,\ T_1=295\ K$): $m_1=0.1949\ kg$
Final ($P_2=600\ kPa,\ V_2=0.22\ m^3,\ T_2=?$): $m_2=459.9/T_2$
$m_i=m_2-m_1$
$m_ih_i-W_b=m_2u_2-m_1u_1$
$(m_2-m_1)c_pT_i-W_b=m_2c_vT_2-m_1c_vT_1$
Linear spring:
$W_b=\frac{P_1+P_2}2(V_2-V_1)=41.25\ kJ$
$T_i=22°C,\ c_p=1.005\ kJ/kg.°C,\ c_v=0.718\ kJ/kg.°C,\ $
solving for the final temperature: $T_2=351\ K$
$m_2=1.309\ kg$
$m_i=1.11\ kg$
