# Example Problem with Complete Solution

2D-6 : Humidity and Partial Pressure in a Humid Ideal Gas 6 pts
The rigid tank shown below contains 5 kg of a non-condensable gas with a molecular weight of 44.1 g/mol. The tank also contains water vapor. The gas in the tank is at 140 kPa and 80oC and the relative humidity is 72%.

Assuming the gas in the tank behaves as an ideal gas, calculate the mass of water vapor in the tank.

Read : The keys here are the definition of relative humidity and the relationship between mole fraction and partial pressure.
We need to assume the humid air behaves as an ideal gas in order to determine the partial pressure of water from the given hr. We can use the mole fraction of water in the gas to determine the mass of water in the gas.
Given: mNCG =  5 kg NCG Ptot =  140 kPa
MWNCG = 44.1 g NCG / mole NCG hr =  72%
T =  80 oC
Find: mH2O = ??? kg
Assumptions: 1- The gas in the tank behaves as an ideal gas. This must be verified.
Equations / Data / Solve:
Let's begin with the definition of relative humidity: Eqn 1
The vapor pressure is equal to the saturation pressure at the system temperature.
We can find this in the saturation temperature section of the steam tables: P*H2O (70o 47.41 kPa
Plug P*H2O and hr into Eqn 1 to get the partial pressure of water from the definition of relative humidity.
PH2O 34.138 kPa
For an ideal gas: PH2O = yH2O  Ptot Eqn 2
or: Eqn 3
Plugging values into Eqn 3 yields: yH2O = 0.244 mol H2O / mol wet gas
For all gases, mole fraction is defined as: Eqn 4
Where : ni = mi / MWi Eqn 5
Now, we solve Eqn 4 for nH2O : Eqn 6
Eqn 7
Now, we can plug the numbers into equations…
Eqn 5 yields :  nNCG = 113.38 moles NCG
Eqn 7 yields :  nH2O = 36.56 moles H2O
Finally, Eqn 5 can be rewritten as : mi = ni  MWi Eqn 8
We can answer the question posed by plugging numbers into Eqn 8 :
Data: MWH20 = 18.016 g H2O / mol H2O mH2O = 658.7 g H2O
Verify: Test if ideal: Ideal Gas EOS : Eqn 3
Solve for the molar volume : Eqn 4
V 20.97 L/mole
Therefore, since V > 20 L/mole, we can treat the wet gas as an ideal gas.
Answers : mH2O = 659 g H2O