Answer
a) $P=1588.38kPa$ $(5.54\% error)$
b) $P=1494.71kPa$ $(0.68\% error)$
c) $P=1504kPa$ $(0.07\% error)$
Work Step by Step
a) Based on the ideal gas equation:
$P=\frac{mRT}{V}=\frac{100kg*0.2968\frac{kPam^3}{kgK}*175K }{3.27m^3}=1588.38kPa$ $(5.54\% error)$
b) Using the Van Der Waals equation:
$a=\frac{27R^2T_{cr}^2}{64P_{cr}}=\frac{27*(0.2968\frac{kPam^3}{kgK})^2*(136.2K)^2}{64*3390kPa}=0.175\frac{m^6kPa}{kg^2}$
$b=\frac{RT_{cr}}{8P_{cr}}=\frac{0.2968\frac{kPam^3}{kgK}*126.2K}{8*3390kPa}=0.00138\frac{m^3}{kg}$
$P=\frac{RT}{\upsilon-b}-\frac{a}{\upsilon^2}=\frac{0.2968\frac{kPam^3}{kgK}*175K}{\frac{3.27m^3}{100kg}-0.00138\frac{m^3}{kg}}-\frac{0.175\frac{m^6kPa}{kg^2}}{(\frac{3.27m^3}{100kg})^2}=1494.71kPa$ $(0.68\% error)$
c) Using the Beattie-Bridgeman equation
$v=M\upsilon=28.013\frac{kg}{kmol}*\frac{3.27m^3}{100kg}=0.916\frac{m^3}{kmol}$
$A=A_{0}(1-\frac{a}{v})=136.2315*(1-\frac{0.02617}{0.916})=132.339$
$B=B_{0}(1-\frac{b}{v})=0.05046*(1-\frac{-0.00691}{0.916})=0.05084$
$c=4.2*10^4\frac{m^3K^3}{kmol}$
$P=\frac{R_{u}T}{v^2}(1-\frac{c}{vT^3})(v+B)-\frac{A}{v^2}=\frac{8.314*175}{0.916^2}(1-\frac{4.2*10^4}{0.916*175^3})(0.916+0.05084)-\frac{132.339}{0.916^2}$
$P=1504kPa$ $(0.07\% error)$
