Answer
$T_2=1059\ °R$
$w_s=138\ Btu/lbm$
Work Step by Step
Since this is a reversible and adiabatic processes (isentropic):
$\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^\dfrac{k-1}{k}$
Given $T_1=70\ °F,\ P_2=200\ psia,\ P_1=15\ psia,\ k=1.389$
$T_2=1059\ °R$
From the energy balance:
$\dot{m}h_1+\dot{W}_s=\dot{m}h_2$
$w_s=h_2-h_1=c_p(T_2-T_1)$
Given $c_p=0.245\ Btu/lbm.°R\ T_{avg}=400\ °F$
$w_s=138\ Btu/lbm$
