Example Problem with Complete Solution

2D-9 : Relative Humidity, Partial Pressure and Mole and Mass Fractions 6 pts
My classroom contains 250 m3 of humid air at 26oC and 44% relative humidity. Calculate the mass of dry air (BDA) and the mass of water vapor 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 250 m3 T 26 oC
Ptot 105 kPa hr 44%
Find: mH2O ??? kg H2O mBDA ??? kg BDA
Assumptions: - 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.
Equations / Data / Solve:
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 :
Eqn 1
Solve Eqn 1 for mgas :
Eqn 2
The following equation allows us to calculate the average molecular weight of a gas mixture using the mole fractions and molecular weights of its constituents.
Eqn 3
For our system, Eqn 3 becomes :
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 :
Eqn 5
Since we know the temperature of the system is 26oC, we can look up the vapor pressure of water at this temperature in the Saturation Temperature Table in the Steam Tables.  Unfortunately, because 26oC is not listed in the Saturation Temperature Table, interpolation is required.
Tsat (oC) Psat (kPa)
25 3.170
26 ???
30 4.247 Interpolation yields : P*H2O(26oC) 3.385 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 1.490 kPa
The last key relationship is the one between partial pressure and the mole fraction for ideal gases :
Eqn 6
Solving for the mole fraction yields :
Eqn 7
Plugging numbers into Eqn 7 yields : yH2O 0.0142 mol H2O / mol gas
We can calculate yBDA because Syi = 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 gas in the room.
MWgas 28.844 g gas / mol gas mgas 304.43 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.
Eqn 8 ngas 10.554 mol gas
Eqn 9 nH2O 0.150 mol H2O
nBDA 10.405 mol BDA
Eqn 10
mH2O 2.70 kg H2O mBDA 302 kg BDA
Alternate Ending
Convert mole fractions into mass fractions :
Eqn 11
The following unit analysis shows why Eqn 11 is true.
Eqn 12
xH2O 0.00886 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.
Eqn 13
mH2O 2.70 kg H2O mBDA 302 kg BDA
Verify : The ideal gas assumption needs to be verified.
We need to determine the specific volume and check if :
Eqn 9
R 8.314 J/mol-K
V 23.69 L/mol The ideal gas assumption is valid because V > 20 L/mole.
Answers : mH2O 2.70 kg H2O mBDA 302 kg BDA