Example Problem with Complete Solution

2D-5 : Relative and Absolute Humidity of Air 4 pts
A pressure gauge on a rigid steel tank reads 50 kPa. The tank holds 2.1 kg of air and 0.250 kg of water vapor at 70oC. Calculate the relative humidity of the air in the tank.
Read : The keys here are the definition of relative humidity and the relationships between mass, moles, molecular weight, mole fraction and partial pressure.
We need to assume the humid air behaves as an ideal gas in order to determine the partial pressures from the given mwater and mBDA.
Given: T 70 oC mtot 2.100 kg wet air
Ptot 50 kPa mH2O 0.250 kg H2O
Find: hr = ??? %
Assumptions: 1- Air is a non-condensable gas.
2- Humid air behaves as an ideal gas.
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
section of the steam tables:
P* (80oC) = 47.39 kPa
For an ideal gas: PH2O = yH2O  Ptot Eqn 2
Test if ideal: Ideal Gas EOS :
Eqn 3
Solve for the molar volume :
Eqn 4
V 57.06 L/mole
Therefore, since V > 20 L/mole, we can treat the wet gas as an ideal gas.
For all gases, mole fraction is defined as :
Eqn 5
Where : ni = mi / MWi Eqn 6
Data: MWH20 = 18.016 g H2O / mol H2O
MWBDA = 28.96 g bone-dry air / mol bone-dry air
Since we know the mass of water and bone-dry air in the tank, as well as their molecular weights, we can calculate the number of moles of water and BDA in the tank using Eqn 6.
Then, we can calculate the mole fraction of water in the gas in the tank, using Eqn 5.  Next, we can use the given total pressure to calculate the partial pressure of water in the gas using Eqn 2.
Here are the numerical results: nH2O 13.88 mol H2O
mBDA 1.850 kg BDA
nBDA 63.88 mol BDA
yH2O 0.178 mol H2O / mol wet gas
PH2O 8.92 kPa
Finally, we can calculate the relative humidity using Eqn 1:
hr = 18.8% %
Verify: 1 - The assumption is sound since the normal boiling points of oxygen and nitrogen are 68 K and 70 K and our system is at 343 K.
2 - The ideal gas assumption was verified above because V = 57 L/mol  which is greater than 20 L/mole. The limit for non-diatomic gases applies because of the water in the air.
Answers : hr = 19% %