Answer
a) $P_{2}=156.41kPa$
b) $P_{2}=156.42kPa$
Work Step by Step
Knowing that:
$V_{1}=V_{2}$
$m_{1}=m_{2}$
a) Based on the ideal gas equation:
$\frac{P_{1}V_{1}}{RT_{1}}=\frac{P_{2}V_{2}}{RT_{2}}$
Simplifying and substituting:
$P_{2}=\frac{P_{1}T_{2}}{T_{1}}=\frac{80kPa*(300+273.15)K}{(20+273.15)K}=156.41kPa$
b) Using the Benedict-Webb-Rubin equation:
$v_{1}=v_{2}=\frac{R_{u}T_{1}}{P_{1}}=\frac{8.314\frac{kPam^3}{kmolK}*(20++273.15)K}{80kPa}=30.466\frac{m^3}{kmol}$
$P_{2}=\frac{R_{u}T_{2}}{v_{2}}+(B_{0}R_{u}T_{2}-A{0}-\frac{C_{0}}{T_{2}^2})\frac{1}{v_{2}^2}+\frac{bR_{u}T_{2}-a}{v_{2}^3}+\frac{a\alpha}{v_{2}^6}+\frac{c}{v_{2}^3T_{2}^2}(1+\frac{\gamma}{v_{2}^2})e^{\frac{-\gamma}{v_{2}^2}}$
Substituting all the coefficients for methane of table 3-4:
$P_{2}=156.42kPa$
