Answer
a) $T_{2}=985.17R$
b) $T_{2}=957.86R$
Work Step by Step
a) Based on the ideal gas equation:
$T_{2}=T_{1}(\frac{P_{2}}{P_{1}})^{\frac{n-1}{n}}=(300+459.67)R*(\frac{2000psia}{1000psia})^{\frac{1.6-1}{1.6}}=985.17R$
b) Using the Beattie-Bridgeman equation:
$A=A_{0}(1-\frac{a}{v})=130.7812*(1-\frac{0.02328}{v})$
$B=B_{0}(1-\frac{b}{v})=0.03931*(1-\frac{0}{v})$
$c=5.99*10^4\frac{m^3K^3}{kmol}$
$P_{1}=6894.76kPa$
$T_{1}=422.04K$
$R_{u}=8.314\frac{kPam^3}{kmolK}$
$P=\frac{R_{u}T}{v^2}(1-\frac{c}{vT^3})(v+B)-\frac{A}{v^2}$
$6894.76=\frac{8.314*422.04}{v^2}(1-\frac{5.99*10^4}{v*422.04^3})(v+0.03931)-\frac{130.7812*(1-\frac{0.02328}{v})}{v^2}$
Solving for $v$:
$v_{1}=0.511774\frac{m^3}{kmol}$
$v_{2}=v_{1}(\frac{P_{1}}{P_{2}})^{\frac{1}{n}}=0.511774\frac{m^3}{kmol}*(\frac{1}{2})^{\frac{1}{1.6}}=0.331844\frac{m^3}{kmol}$
$P_{2}=2000psia=13789.51kPa$
$P=\frac{R_{u}T}{v^2}(1-\frac{c}{vT^3})(v+B)-\frac{A}{v^2}$
$13789.51=\frac{8.314*T}{0.331844^2}(1-\frac{5.99*10^4}{0.331844*T^3})(0.331844+0.03931)-\frac{130.7812*(1-\frac{0.02328}{0.331844})}{0.331844^2}$
Solving for $T$:
$T_{2}=532.1453K=957.86R$
