Question

# A closed container of air and water vapor is at 95 degrees Celsius and 5000 mm...

A closed container of air and water vapor is at 95 degrees Celsius and 5000 mm Hg and contains 12% water by volume. Antoine constants for water are A=7.97, B=1668.2, C=228.0. a) Calculate the dew point temperature and degrees of superheat of the air-water mixture. b) If the air-water mixture is cooled isobarically to a point where the mole fraction of air is .915, what is the temperature and percentage of the water vapor that condenses?

a) Calculate the dew point temperature and degrees of superheat of the air-water mixture.
A=7.97, B=1668.2, C=228.0
Log(pv) = A-B/(T+C), T temperature in degC and pv = vapor pressure of pure substance in mmHg.
mole fraction of water vapor, yH2O = 0.12
Total pressure, P = 5000 mmHg
partial pressure of water vapor, pH2O = yH2O*P = 0.12*5000 = 600 mmHg

At temperature, T=Tdp(dew point) pv = pH2O = 600mmHg

Log(pH2O) =Log(600) = 2.78 = A-B/(Tdp+C) = 7.97-1668.2/(Tdp+228)
2.78 = 7.97-1668.2/(Tdp+228)
7.97-2.78 = 5.19 = 1668.2/(Tdp+228)
(Tdp+228) = 321.4
Dew point temperature, Tdp = 93.4 degC
Temperature, T = 95 degC
Degree of superheat = T - Tdp = 95-93.4 = 1.6 degC

b) If the air-water mixture is cooled isobarically to a point where the mole fraction of air is .915,
what is the temperature and percentage of the water vapor that condenses?

mole fraction of water vapor, yH2O = 1-0.915 = 0.085
partial pressure of water vapor, pH2O = yH2O*P = 0.085*5000 = 425 mmHg

Log(pH2O) =Log(425) = 2.6 = A-B/(Tdp+C) = 7.97-1668.2/(Tdp+228)
2.6 = 7.97-1668.2/(Tdp+228)
7.97-2.6 = 5.37 = 1668.2/(Tdp+228)
(Tdp+228) = 310.65
Dew point temperature, Tdp = 82.65 degC
There is equilibrium between water vapor and condensed liquid.
So, temperature, T = 82.65 degC

Basis: Let total 100 moles of molecules initially present in the container

Balance dry air:

Initially present:
mole fraction of water vapor, yH2O = 0.12
moles of water vapor=100*0.12= 12 mol
Moles of air, n1 = 100 - 12 = 88 mol

Finally present after isobaric compression:
Total moles of molecules = n2
mole fraction of air, yair = 0.915
Moles of air, n1 = n2*0.915
88 = n2*0.915
Total moles of molecules present after compression, n2 = 96.2
Moles of water vapor =n2*(1-0.915)=96.2*0.085=8.2
Moles of molecules condensed = 12-8.2 = 3.8 mol

percentage condensed = (moles H2O condensed)/moles H2O present initially*100 = 3.8/12*100 = 31.67%