Four kilograms of water are placed in an inclosed volume of 1 m^3. Heat is added until the temperature is 150 C.Find the pressure, mass and volume...

We'll calculate the volume of 4 kg of saturated vapor, at
the temperature of 150 C.

0.3928*4 = 1.5712
m^3

The resulted volume is bigger than the given volume of
1 m^3.

If the given volume is less than the resulted
volume, the state is in the quality region, whose pressure is P = 475.8
KPa.

Now, we'll calculate the mass of the vapor. First,
we'll have to determine the quality.

The quality of the
mixture is given by the ratio:

x = mass of saturated
vapor/total mass

v = 1/4

v =
0.25 m^3/Kg

The total volume of the mixture is the sum of
the volume occupied by the liquid and the volume occupied by the
vapor.

v = v1 + x(v2-v1)

0.25
= 0.00109 + x(0.3928 - 0.00109)

We'll remove the brackets
and we'll have:

0.25 = 0.00109 +
0.39171x

0.39171x = 0.24891

x
= 0.24891/0.39171

x
= 0.6354

Mass of saturated vapor = Total
mass*x

m = 4* 0.6354

m = 2.542
Kg

The volume of the vapor
is:

V =
0.3928*2.542

V = 0.998
m^3