Answer
$m_e=4.655\ lbm$
$W_{e,i}=185\ Btu$
Work Step by Step
From Table A-17E:
$T_i=T_2=T_1=580°R$:$h_i=138.66\ Btu/lbm,\ u_2=u_1=98.90\ Btu/lbm$:
For ideal gases:
$Pv=mRT$, $R_{air}=0.3704\ psia.ft³/lbm.°R,\ V_{tank}=40\ ft³$
Initial ($P_1=50\ psia,\ T_1$): $m_1=9.310\ lbm$
Final ($P_2=25\ psia,\ T_2$): $m_2=4.655\ lbm$
From the material balance:
$-m_e=m_2-m_1$
$m_e=4.655\ lbm$
From the energy balance:
$W_{e,i}-m_eh_e=m_2u_2-m_1u_1$
$W_{e,i}=185\ Btu$
