# 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 of the vapor.

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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**