Answer
$P_{2}=91.4psia$
Work Step by Step
$P_{R.1}=\frac{P_{1}}{P_{cr}}=\frac{50psia}{708psia}=0.0706$
$T_{R,1}=\frac{T_{1}}{T_{cr}}=\frac{(100+460)R}{549.8R}=1.019$
$T_{R,2}=\frac{T_{2}}{T_{cr}}=\frac{(540+460)R}{549.8R}=1.8193$
From Fig. A-15:
$Z_{1}=0.977$
$\upsilon_{1}=\frac{Z_{1}RT_{1}}{P_{1}}=\frac{0.977*0.3574\frac{psiaft^3}{lbmR}*560R}{50psia}=3.91\frac{ft^3}{lbm}$
$\upsilon_{1}=\upsilon_{2}$
$\upsilon_{R,2}=\frac{\upsilon_{2}}{\frac{RT_{cr}}{P_{cr}}}=\frac{3.91\frac{ft^3}{lbm}}{\frac{0.3574\frac{psiaft^3}{lbmR}*549.8R}{708psia}}=14.1$
From Fig A-15:
$Z_{2}=1$
Finally:
$P_{2}=\frac{Z_{2}RT_{2}}{\upsilon_{2}}=\frac{1*0.3574\frac{psiaft^3}{lbmR}*1000R}{3.91\frac{ft^3}{lbm}}=91.4psia$
