Example Problem with Complete Solution

2D-9 : Relative Humidity, Partial Pressure and Mole and Mass Fractions 6 pts
Determine the masses of dry air (BDA) and water vapor contained in a 240 m3 room at 98 kPa, 23°C and 50% relative humidity.  Calculate the partial pressure and mole fraction of water in the gas in the room.
Read : The key to this problem is the definition of relative humidity.  When the relative humidity and temperature are given, we can use data from the Steam Tables to determine the partial pressure and mole fraction of water in the gas phase.  We can convert the mole fraction into a mass fraction.  Then, by assuming the gas phase is an ideal gas, we can determine the total mass of air in the room.  And, finally we can determine the mass of BDA and water in the gas in the room.
Given : Vtot
240
m3 T
23
oC
Ptot
98
kPa hr
50%
Assumption : The air-water gas mixture behaves as an ideal gas.  At the end of the problem we will be able to determine the molar volume of the air-water gas mixture so we can verify this assumption.
Solution : Using the IG EOS and the known P, T and V of the room, we can determine the mass of air-water gas mixture in the room.
Ideal Gas EOS : Equation. Eqn 1
Solve Eqn 1 for mgas : Equation. Eqn 2
From the hints for this problem, we obtain the following relationship between the molecular weight of a gas mixture and the mole fractions and molecular weights of its constituents :
Equation. Eqn 3
For our system, Eqn 3 becomes : Equation. Eqn 4
MWBDA
29
g BDA/mole BDA
MWH2O
18.016
g H2O/mole H2O
So, in order to determine the average molecular weight of the gas, we need to know the composition of the gas.  That is, we need to know the mole fractions of BDA and water in the gas mixture.  Given the relative humidity and the temperature of an ideal gas mixture of air and water, we can determine the composition.
Begin with the definition of relative humidity : Equation. Eqn 5
Since we know the temperature of the system is 23oC, we can look up the vapor pressure of water at this temperature in the Saturation Temperature Table in the Steam Tables.  Unfortunately, because 23oC is not listed in the Saturation Temperature Table, interpolation is required.
Tsat
(oC)
Psat
(kPa)
20.00
2.338
23.00
???
25.00
3.169
Interpolation yields : P*H2O(23oC)
2.837
kPa
We can plug the vapor pressure, along with the given value of the relative humidity into Eqn 5 to determine the partial pressure of water in the gas.
PH2O
1.418
kPa
The last key relationship is the one between partial pressure and the mole fraction for ideal gases : Equation. Eqn 6
Solving for the mole fraction yields : Equation. Eqn 7
Plugging numbers into Eqn 7 yields : yH2O
0.0145
mol H2O / mol gas
We can calculate yBDA because Σyi = 1 : yBDA
0.986
mol BDA / mol gas
At last, we can use these mole fractions in Eqn 4 to determine the value of MWgas and then use that in Eqn 2 to determine the total mass of gass in the room.
MWgas
28.841
g gas / mol gas mgas
275.50
kg gas
Here we can either determine the mass fractions of BDA and water in the gas or we can determine the number of moles of BDA and water in the room.  I will use both methods here.
Equation. Eqn 8 ngas
9.552
mol gas
Equation. Eqn 9 nH2O
0.138
mol H2O
nBDA
9.414
mol BDA
Answers : Equation. Eqn 10 mH2O
2.49
kg H2O
mBDA
273
kg BDA
Alternate Ending
Convert mole fractions into mass fractions : Equation. Eqn 11
The following unit
analysis shows why
Eqn 11 is true.
Equation.
Eqn 12
xH2O
0.00904
kg H2O  /kg gas xBDA
0.991
kg BDA / kg gas
Now, we can determine the mass of each species in the gas by multiplying the total mass by the mass fraction.
Answers : Equation. Eqn 13 mH2O
2.49
kg H2O
mBDA
273
kg BDA