Answer
$P_{1}=51.25kPa$
$T_{2}=66.30^{\circ}C$
Work Step by Step
$\upsilon=\frac{V}{m}=\frac{1.348m^3}{10kg}=0.1348\frac{m^3}{kg}$
From table A-11:
1) $T_{1}=-40^{\circ}C$
$\upsilon_{1}=0.1348\frac{m^3}{kg}$
As $0.0007053\frac{m^3}{kg}\lt \upsilon_{1}\lt 0.36064\frac{m^3}{kg}$ is a saturated mixture
$P_{1}=51.25kPa$
From table A-13:
2) $P_{2}=200kPa$
$\upsilon_{2}=0.1348\frac{m^3}{kg}$
As $\upsilon_{2}\gt \upsilon_{g}$ is a superheated vapor
Interpolating: $T_{2}=66.30^{\circ}C$
