Answer
$P_{2}=250psia$
$V_{2}=0.01663ft^3$
Work Step by Step
$\upsilon_{1}=\frac{V_{1}}{m}=\frac{2.4264ft^3}{1lbm}=2.4264\frac{ft^3}{lbm}$
As $upsilon\gt upsilon_{g}$ is a superheated vapor
From table A-6E
$P_{1}=250psia$
Since we have a constat pressure process:
$P_{2}=P_{1}=250psia$
As $T_{2}\lt T_{sat}$ is a compressed liquid
From table A-4E:
$\upsilon_{2}=0.01663\frac{ft^3}{lbm}$
And:
$V_{2}=m\upsilon_{2}=1lbm*0.01663\frac{ft^3}{lbm}=0.01663ft^3$
