Answer
$P_{1}=51.25kPa$
$T_{2}=50^{\circ}C$
Work Step by Step
$\upsilon_{1}=\frac{V}{m}=\frac{0.09m^3}{1kg}=0.09\frac{m^3}{kg}$
As $0.0007053\frac{m^3}{kg}\lt \upsilon\lt 0.36064\frac{m^3}{kg}$ is a saturated mixture.
From table A-11:
$P_{1}=51.25kPa$
$\upsilon_{2}=\upsilon_{1}$
As $\upsilon_{2}\gt \upsilon_{g}$ is a superheated vapor
From table A-13
$T_{2}=50^{\circ}C$
