# Chapter 4:Defining Thermodynamic State:The State Postulate¶

## Example 4.1,Page no:51¶

In :
M=2.5 				#mass of the substance[Kg]
x=0.6 				#mass fraction for vapour phase
P=7 				#pressure [atm]
T=438 				#temperature[K]

Ml=(1-x)*M 			#mass fraction of liquid phase[Kg]
Mg=x*M 				#mass fraction of vapour phase[Kg]
print"M(liquid phase)=",Ml,"Kg\nM(vapour phase)=",Mg,"Kg"

M(liquid phase)= 1.0 Kg
M(vapour phase)= 1.5 Kg


## Example 4.2,Page no:51¶

In :
Vl=0.0177 			#specific volume of saturated liquid[m3/Kg]
Vg=4.43 			#specific volume of saturated vapour[m3/Kg]
P=7 				#pressure[atm]
T=438 				#temperature[K]
x=0.6 				#fraction of vapour phase
M=2.5 				#mass of the substance[Kg]

V=((1-x)*Vl+x*Vg)*M 		#total volume occupied [m3]
print"Total volume occupied =",round(V,2),"m^3"

Total volume occupied = 6.66 m^3


## Example 4.3,Page no:51¶

In :
M=2.5 				#mass of a substance[Kg]
x=0.6 				#fraction of vapour phase
Ug=1105.0 			#specific internal energy of saturated vapour[J/Kg]
Ul=298.0 				#specific internal energy of saturated liquid[J/Kg]
U=M*((1-x)*Ul+x*Ug)
print"The total internal energy of the mixture =",U,"J"
print"\nNOTE:In textbook,it is wrongly calculated as 1950 J"


The total internal energy of the mixture = 1955.5 J

NOTE:In textbook,it is wrongly calculated as 1950 J