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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
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