Answer
See the image
Work Step by Step
a) From table A-4
$P=12.352kPa$
As $0.001012\frac{m^3}{kg}\lt v\lt 12.026\frac{m^3}{kg}$ is a saturated mixture.
b) From table A-5
$T=143.61^{\circ}C$
$v=0.46242\frac{m^3}{kg}$
c) From table A-6
$v=0.47443\frac{m^3}{kg}$
As $T\gt T_{sat}$ is a superheated vapor
c) From table A-4
$v=0.001052\frac{m^3}{kg}$
As $T\lt T_{sat}$ is a compressed liquid
