Answer
a) $\upsilon=0.095352\frac{m^3}{kg}$ $(3.7\% error)$
b) $\upsilon=0.09163\frac{m^3}{kg}$ $(0.36\%error)$
c)$\upsilon=0.09196\frac{m^3}{kg}$
Work Step by Step
a) Based on the ideal gas equation:
$\upsilon=\frac{RT}{P}=\frac{0.4615\frac{kPam^3}{kgK}*(450+273.15)K}{3500kPa}=0.095352\frac{m^3}{kg}$ $(3.7\% error)$
b) Using the generalized compressibility chart:
$P_{R}=\frac{P}{P_{cr}}=\frac{3.5MPa}{22.06MPa}=0.159$
$T_{R}=\frac{T}{T_{cr}}=\frac{723.15K}{647.1K}=1.118$
From Fig A-15
$Z=0.961$
Then $\upsilon=Z\upsilon_{ideal}=0.961*0.995352\frac{m^3}{kg}=0.09163\frac{m^3}{kg}$ $(0.36\%error)$
c)From table A-6:
$\upsilon=0.09196\frac{m^3}{kg}$
