Answer
$T_2=709.2°R=249°F$
$m_2=0.914\ lbm$
Work Step by Step
From the material balance:
$m_i=m_2-m_1$
For ideal gases (air: $V=2\ ft³,\ R=0.3704\ psia.ft³/lbm.°R$):
$PV=mRT$
Initial ($P_1=20\ psia,\ T_1=520°R$): $m_1=0.2077\ lbm$
Final ($P_2=120\ psia,\ T_2=?$): $m_2=647.9/T_2$
From the energy balance:
$m_ih_i=m_2u_2-m_1u_1$
$(m_2-m_1)c_pT_i=c_v(m_2T_2-m_1T_1)$
Given $c_p/c_v=1.4,\ T_i=545°R$:
and solving for $T_2=709.2°R=249°F$
So: $m_2=0.914\ lbm$
