Answer
a) $\upsilon=0.3444\frac{ft^3}{lbm}$
b) $\upsilon=0.3240\frac{ft^3}{lbm}$
c) $\upsilon=0.3240\frac{ft^3}{lbm}$
Work Step by Step
a) Based on the ideal gas equation:
$\upsilon=\frac{RT}{P}=\frac{0.3830\frac{psiaft^3}{lbmR}*(-100+459.67)R }{400psia}=0.3444\frac{ft^3}{lbm}$
b) Using the Benedict-Webb-Rubin equation:
$P=\frac{R_{u}T}{v}+(B_{0}R_{u}T-A{0}-\frac{C_{0}}{T^2})\frac{1}{v^2}+\frac{bR_{u}T-a}{v^3}+\frac{a\alpha}{v^6}+\frac{c}{v^3T^2}(1+\frac{\gamma}{v^2})e^{\frac{-\gamma}{v^2}}$
Substituting all the coefficients for nitrogen of table 3-4 and solving for $v$
$v=0.5666\frac{m^3}{kmol}$
$\upsilon=\frac{v}{M}=\frac{0.5666\frac{m^3}{kmol}}{28.013\frac{kg}{kmol}}*(\frac{16.02\frac{ft^3}{lbm}}{1\frac{m^3}{kg}})=0.3240\frac{ft^3}{lbm}$
c) Using the generalized compressibility chart:
$P_{R}=\frac{P}{P_{cr}}=\frac{400psia}{492psia}=0.813$
$T_{R}=\frac{T}{T_{cr}}=\frac{359.67R}{227.1R}=1.585$
From Fig A-15
$Z=0.94$
Then $\upsilon=Z\upsilon_{ideal}=0.94*0.3447\frac{ft^3}{lbm}=0.3240\frac{ft^3}{lbm}$
