Answer
a) $P=1062.94kPa$ $(6.29\% error)$
b) $P=1000.37kPa$ $(0.037\% error)$
Work Step by Step
a) Based on the ideal gas equation:
$P=\frac{RT}{\upsilon}=\frac{0.2968\frac{kPam^3}{kgK}*150K }{0.041884\frac{m^3}{kg}}=1062.94kPa$ $(6.29\% error)$
b) Using the Beattie-Bridgeman equation
$v=M\upsilon=28.013\frac{kg}{kmol}*0.041884\frac{m^3}{kg}=1.1733\frac{m^3}{kmol}$
$A=A_{0}(1-\frac{a}{v})=136.2315*(1-\frac{0.02617}{1.1733})=133.193$
$B=B_{0}(1-\frac{b}{v})=0.05046*(1-\frac{-0.00691}{1.1733})=0.05076$
$c=4.2*10^4\frac{m^3K^3}{kmol}$
$P=\frac{R_{u}T}{v^2}(1-\frac{c}{vT^3})(v+B)-\frac{A}{v^2}=\frac{8.314*150}{1.1733^2}(1-\frac{4.2*10^4}{1.1733*150^3})(1.1733+0.05076)-\frac{133.193}{1.1733^2}$
$P=1000.37kPa$ $(0.037\% error)$
