| 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. |
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| 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. |
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| Given: |
Vtot |
250 |
m3 |
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T |
26 |
oC |
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Ptot |
105 |
kPa |
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hr |
44% |
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| Find: |
mH2O |
??? |
kg H2O |
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mBDA |
??? |
kg BDA |
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| 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. |
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| Equations
/ Data / Solve: |
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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. |
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Ideal
Gas EOS : |
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Eqn 1 |
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Solve Eqn
1 for mgas : |
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Eqn 2 |
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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. |
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Eqn 3 |
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For our system, Eqn 3 becomes : |
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Eqn 4 |
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MWBDA |
29 |
g BDA/mole BDA |
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MWH2O |
18.016 |
g H2O/mole H2O |
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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. |
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Begin with the definition
of relative humidity : |
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Eqn 5 |
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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. |
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Tsat (oC) |
Psat (kPa) |
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25 |
3.170 |
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26 |
??? |
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30 |
4.247 |
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Interpolation yields : |
P*H2O(26oC) |
3.385 |
kPa |
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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. |
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PH2O |
1.490 |
kPa |
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The last key
relationship is the one between partial pressure and the mole fraction for
ideal gases : |
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Eqn 6 |
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Solving for the mole fraction yields : |
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Eqn 7 |
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Plugging numbers into Eqn 7 yields : |
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yH2O |
0.0142 |
mol
H2O / mol gas |
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We can calculate yBDA because Syi = 1 : |
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yBDA |
0.986 |
mol
BDA / mol gas |
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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. |
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MWgas |
28.844 |
g gas / mol gas |
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mgas |
304.43 |
kg gas |
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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. |
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Eqn 8 |
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ngas |
10.554 |
mol gas |
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Eqn 9 |
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nH2O |
0.150 |
mol H2O |
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nBDA |
10.405 |
mol BDA |
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Eqn 10 |
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mH2O |
2.70 |
kg H2O |
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mBDA |
302 |
kg BDA |
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Alternate Ending |
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Convert mole fractions
into mass fractions : |
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Eqn 11 |
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The following unit
analysis shows why Eqn 11 is
true. |
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Eqn 12 |
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xH2O |
0.00886 |
kg H2O /kg gas |
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xBDA |
0.991 |
kg BDA / kg gas |
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Now, we can
determine the mass of each species in the gas by multiplying the total mass
by the mass fraction. |
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Eqn 13 |
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mH2O |
2.70 |
kg H2O |
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mBDA |
302 |
kg BDA |
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| Verify : |
The ideal gas
assumption needs to be verified. |
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We need to determine
the specific volume and check if : |
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Eqn 9 |
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R |
8.314 |
J/mol-K |
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V |
23.69 |
L/mol |
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The ideal gas
assumption is valid because V > 20 L/mole. |
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| Answers : |
mH2O |
2.70 |
kg H2O |
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mBDA |
302 |
kg BDA |
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