Answer
$\Delta s = -0.01973\ Btu/lbm.°R$
Work Step by Step
$\Delta s = c_p\ln\left(\dfrac{T_2}{T_1}\right)-R\ln\left(\dfrac{P_2}{P_1}\right)$
Given $T_{avg}=150°F,\ c_p=0.241\ Btu/lbm.°R,\ R=0.06855\ Btu/lbm.°R$
$T_1=90°F,\ T_2=210°F,\ P_1=15\ psia,\ P_2=40\ psia$
$\Delta s = -0.01973\ Btu/lbm.°R$
