Find the phase the quality x if applicable and the missing p
Solution
a)
Given, Temperature (T) =1200C = 393 K ;
Volume per unit mass (V/m) = 0.5 m3/Kg;
Given substance is H2O i.e. Molecular Mass (M) = 18 g/mol
From Ideal Gas equation PV=nRT;
where, n= m (mass of given substance )/ M (Molecular mass)
Hence PV=nRT can be rearranged as P*(V/m) = RT/M
therefore P= (8.314* 393) / (0.5*18) = 342.87 K Pa;
where R= 8.314 J/mol/K
b)
P= 100 K Pa;
V/m = 1.8 m3/Kg
M = 18 g/mol
R= 8.314 J/mol/K
Substituting all the values in Ideal Gas Equation i.e. PV=nRT
T= (100 KPa* 18 g/mol *1.8 m3/Kg) / (8.314 J/mol/K) = 389.704 K =116.550C
c)
Same as above, substituting all the values in Ideal gas Equation PV=nRT
P=(8.314*263)/(18*200)= 0.6 KPa
d)
Here M= 17;
Substituting given values in PV=nRT;
T= (800kPa*0.2m3/Kg) / (8.314J/mol/K * 17 g/mol) = 1.13 K = (-272.02 0C)
e) substituting given values in Ideal Gas Equation
P= (8.314 J/mol/K * 293K) / (17 g/mol * 0.1 m3/Kg) = 1432.92 KPa
