Answer
a) $T_{2}=597.04K$
b) $T_{2}=460K$
The second result is more accurate.
Work Step by Step
Knowing that:
$m_{1}=m_{2}$
$P_{1}=P_{2}$
$V_{2}=1.6V_{1}$
a) Based on the ideal gas equation:
$\frac{m_{1}RT_{1}}{V_{1}}=\frac{m_{2}RT_{2}}{V_{2}}$
Substituting and simplifying:
$T_{2}=1.6T_{1}=1.6*(100+273.15)K=597.04K$
b) Using the generalized compressibility chart:
$P_{R,1}=\frac{P,1}{P_{cr}}=\frac{10MPa}{4.48MPa}=2.232$
$T_{R,1}=\frac{T,1}{T_{cr}}=\frac{373.15K}{305.5K}=1.221$
From Fig A-15
$Z_{1}=0.61$
$\upsilon_{R,1}=0.35$
$P_{R,2}=P_{R,1}=2.232$
$\upsilon_{R,2}=1.6\upsilon_{R,1}=1.6*0.35=0.56$
From Fig A-15
$Z_{2}=0.83$
$T_{2}=\frac{P_{2}\upsilon_{2}}{Z_{2}R}=\frac{P_{2}\upsilon_{R,2}T_{cr}}{Z_{2}P_{cr}}=\frac{10000kPa*0.56*305.5K}{0.83*4480kPa}=460K$
Based on the charts the second result is more accurate.
