Answer
a) $P=1.861MPa$
b) $P=1.583MPa$
c) $P=1.6MPa$
Work Step by Step
a) Based on the ideal gas equation:
$P=\frac{mRT}{V}=\frac{1kg*0.08149\frac{kPam^3}{kgK}*(110+273.15)K}{0.016773m^3}=1.861MPa$
b) Using the generalized compressibility chart:
$T_{R}=\frac{T}{T_{cr}}=\frac{383.15K}{374.2K}=1.024$
$\upsilon_{R}=\frac{\upsilon}{\frac{RT_{cr}}{P_{cr}}}=\frac{\frac{0.016773m^3}{1kg}}{\frac{0.08149\frac{kPam^3}{kgK}*374.2K}{4059kPa}}=2.233$
From Fig A-15
$P_{R}=0.39$
$P=P_{R}P_{cr}=0.39*4059kPa=1.583MPa$
c) From table A-13
$P=1.6MPa$
