Example Problem with Complete Solution

2D-6 : Humidity and Partial Pressure in a Humid Ideal Gas 6 pts
An unknown amount of water and 5 kg of a non-condensable gas (Molecular Weight = 40) are produced in an industrial plant at a temperature of 70oC.  These products are sent to a rigid storage tank.  The pressure and relative humidity in the tank are 2 atm and 65% respectively.  Determine the mass of water present in the tank.  Assume that the gas behaves as an ideal gas.
      figure 1        
               
Given: mNCG =  5 kg NCG
MWNCG = 40 g NCG / mole NCG
T =  70 oC
Ptot =  2 atm       = 202.7 kPa
hr =  65%
Find: mH2O = ??? kg
Assumption: The gas in the tank behaves as an ideal gas.
Solution : Let's begin with the definition of relative humidity:
equation 1 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 (70oC) = 31.19 kPa
Plug P*H2O and hr into Eqn 1 to get the partial pressure of water from the definition of relative humidity.
PH2O = 20.274 kPa
For an ideal gas: PH2O = yH2O  Ptot Eqn 2
or: equation 2 Eqn 3
yH2O = 0.100 mol H2O / mol gas
For all gases, mole fraction is defined as:
equation 3 Eqn 4
Where : ni = mi / MWi Eqn 5
Now, we solve Eqn 4 for nH2O : equation 4 Eqn 6
equation 5 Eqn 7
Now, we can plug the numbers into equations…
Eqn 5 yields :  nNCG = 125 moles NCG
Eqn 7 yields :  nH2O = 13.90 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 = 250 g H2O