My classroom contains 250 m^{3} of humid air at 26^{o}C 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: 
V_{tot} 
250 
m^{3} 



T 
26 
^{o}C 

P_{tot} 
105 
kPa 



h_{r} 
44% 











Find: 
m_{H2O} 
??? 
kg H_{2}O 



m_{BDA} 
??? 
kg BDA 










Assumptions: 
 The airwater gas mixture behaves as an ideal
gas. At the
end of the problem we will be able to determine the molar volume of the airwater 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 airwater
gas mixture in the room. 

Ideal
Gas EOS : 




Eqn 1 











Solve Eqn
1 for m_{gas} : 




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 











MW_{BDA} 
29 
g BDA/mole BDA 

MW_{H2O} 
18.016 
g H_{2}O/mole H_{2}O 











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 26^{o}C, we can look up the vapor pressure
of water at this
temperature in the Saturation Temperature Table in the Steam Tables. Unfortunately, because
26^{o}C is not listed in the Saturation Temperature Table,
interpolation is required. 











T_{sat} (^{o}C) 
P_{sat} (kPa) 








25 
3.170 








26 
??? 








30 
4.247 

Interpolation yields : 
P*_{H2O}(26^{o}C) 
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. 







P_{H2O} 
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 : 

y_{H2O} 
0.0142 
mol
H_{2}O / mol gas 











We can calculate y_{BDA} because Sy_{i} = 1 : 

y_{BDA} 
0.986 
mol
BDA / mol gas 











At last, we can use
these mole fractions in Eqn 4
to determine the value of MW_{gas} and then use that in
Eqn 2 to determine
the total mass of gas in the room. 











MW_{gas} 
28.844 
g gas / mol gas 


m_{gas} 
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 


n_{gas} 
10.554 
mol gas 













Eqn 9 


n_{H2O} 
0.150 
mol H_{2}O 







n_{BDA} 
10.405 
mol BDA 












Eqn 10 
















m_{H2O} 
2.70 
kg H_{2}O 



m_{BDA} 
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 











x_{H2O} 
0.00886 
kg H_{2}O /kg gas 

x_{BDA} 
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 











m_{H2O} 
2.70 
kg H_{2}O 



m_{BDA} 
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/molK 







V 
23.69 
L/mol 

The ideal gas
assumption is valid because V > 20 L/mole. 










Answers : 
m_{H2O} 
2.70 
kg H_{2}O 



m_{BDA} 
302 
kg BDA 









