Answer
a) $P_2=67.03\ psia$
b) $m_i=85.74\ lbm$
c) $Q_e=80900\ Btu$
Work Step by Step
From tables A-4E to A-6E:
Initial ($x_1=1, T_1=300°F$): $v_1=6.4663\ ft³/lbm,\ u_1=1099.8\ Btu/lbm$
Final ($x_2=?, T_2=300°F$): $v_{2,L}=0.01745\ ft³/lbm,\ v_{2,G}=v_1,\ u_{2,L}=269.51\ Btu/lbm,\ u_{2,G}=u_1$
Entering ($P_i=200\ psia, T_i=400°F$): $h_i=1210.9\ Btu/lbm$
$P_2=P_{sat}=67.03\ psia$
From the material balance:
$m_i=m_2-m_1$
Since $m=V/v,\ V_t=3\ ft³$:
$m_1=0.464\ lbm$
Half of the volume is filled with liquid:
$m_2=V_t/2v_{2,L}+V_t/2v_{2,G}=86.20\ lbm$
Hence $m_i=85.74\ lbm$
Also: $m_2u_2=u_{2,L}V_t/2v_{2,L}+u_{2,G}V_t/2v_{2,G}=23356\ Btu$
From the energy balance:
$Q_i+m_ih_i=m_2u_2-m_1u_1$
Hence $Q_i=-80900\ Btu$
$Q_e=80900\ Btu$
