Answer
$\Delta \upsilon=0.0172\frac{m^3}{kg}$
Work Step by Step
$P_{R}=\frac{P}{P_{cr}}=\frac{5MPa}{5.12MPa}=0.977$
$T_{R,1}=\frac{T_{1}}{T_{cr}}=\frac{293K}{282.4K}=1.038$
$T_{R,2}=\frac{T_{2}}{T_{cr}}=\frac{473K}{282.4K}=1.675$
From Fig. A-15:
$Z_{1}=0.560$
$Z_{2}=0.961$
The specificic volume change is:
$\Delta \upsilon=\frac{R*(Z_{2}T_{2}-Z_{1}T_{1})}{P}=\frac{0.2964\frac{kPam^3}{kgK}*(0.961*473K-0.56*293K) }{5000kPa}=0.0172\frac{m^3}{kg}$
