Answer
a,b) $\Delta S =-14.4\ Btu/°R$
Work Step by Step
$\Delta S = m\left[c_v\ln\left(\dfrac{T_2}{T_1}\right)+R\ln\left(\dfrac{v_2}{v_1}\right)\right]$
Given $m=25\ lbm,\ c_v=0.753\ Btu/lbm.°R,\ 0.4961\ Btu/lbm.°R$
$T_2=700 °R,\ T_1=520°R,\ v_1=50\ ft^3/lbm,\ v_2=10\ ft^3/lbm$
$\Delta S =-14.4\ Btu/°R$
Entropy change is the same in both cases.
