Answer
The second result.
$T_{2}=1112K$
Work Step by Step
Knowing that:
$P_{1}=P_{2}$
$V_{2}=1.8V_{1}$
$m_{1}=m_{2}$
a) Based on the ideal gas equation:
$\frac{m_{1}RT_{1}}{V_{1}}=\frac{m_{2}RT_{2}}{V_{2}}$
Simplifying and substituting:
$T_{2}=1.8T_{1}=1.8*(300K)=540K$
b) Using the generalized compressibility chart:
$P_{R,1}=\frac{P_{1}}{P_{cr}}=\frac{10MPa}{4.64MPa}=2.16$
$T_{R,1}=\frac{T_{1}}{T_{cr}}=\frac{300K}{191.1K}=1.57$
From Fig A-15
$Z_{1}=0.86$
$\upsilon_{R,1}=0.63$
$P_{R,2}=2.16$
$\upsilon_{R,2}=1.8*0.63=1.134$
From Fig A-15
$Z_{2}=0.42$
$T_{2}=\frac{P_{2}\upsilon_{2}}{Z_{2}R}=\frac{P_{2}\upsilon_{R,2}T_{cr}}{Z_{2}P_{cr}}=\frac{10000kPa*1.134*191.1K}{0.42*4640kPa}=1112K$
Based on the charts, the second temperature is the more accurate.
