Answer
$w_C=14.8\ Btu/lbm$
$w_P=0.140\ Btu/lbm$
Work Step by Step
From tables A-11E to A-13E:
Inlet ($P_1=15\ psia, x_1=1$): $s_1=0.22717\ Btu/lbm.°R,\ h_1=101.00\ Btu/lbm$
Outlet ($P_2=80\ psia, s_2=s_1$): $h_2=115.81\ Btu/lbm$
From the energy balance for the compressor:
$\dot{W}+\dot{m}h_1=\dot{m}h_2$
$w_C=h_2-h_1=14.8\ Btu/lbm$
If the fluid was a saturated liquid, using a pump would yield:
$w=v_L(P_2-P_1),\quad v_L=0.01164\ ft^3/lbm\mbox{ (Table A-11E)}$
$w_P=0.140\ Btu/lbm$
