Chapter2, Lesson D - Relative Humidity, Partial Pressure and Mole and Mass Fractions

2D-9 :	Relative Humidity, Partial Pressure and Mole and Mass Fractions								6 pts

Determine the masses of dry air (BDA) and water vapor contained in a 240 m3 room at 98 kPa, 23°C and 50% relative humidity. Calculate the partial pressure and mole fraction of water in the gas 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 :	Vtot	240	m3			T	23	oC
	Ptot	98	kPa			hr	50%

Assumption :		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.



Solution :	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.


	Ideal Gas EOS :								Eqn 1

	Solve Eqn 1 for mgas :								Eqn 2

	From the hints for this problem, we obtain the following relationship between the molecular weight of a gas mixture and the mole fractions and molecular weights of its constituents :


									Eqn 3

	For our system, Eqn 3 becomes :								Eqn 4

					MWBDA	29	g BDA/mole BDA
					MWH2O	18.016	g H2O/mole H2O

	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 23oC, we can look up the vapor pressure of water at this temperature in the Saturation Temperature Table in the Steam Tables. Unfortunately, because 23oC is not listed in the Saturation Temperature Table, interpolation is required.



	Tsat (oC)	Psat (kPa)
	20.00	2.338
	23.00	???
	25.00	3.169		Interpolation yields :		P*H2O(23oC)		2.837	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.

							PH2O	1.418	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 :					yH2O	0.0145	mol H2O / mol gas

	We can calculate yBDA because Σyi = 1 :					yBDA	0.986	mol BDA / mol gas

	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 gass in the room.


	MWgas	28.841	g gas / mol gas				mgas	275.50	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			ngas	9.552	mol gas

				Eqn 9			nH2O	0.138	mol H2O
							nBDA	9.414	mol BDA

Answers :				Eqn 10			mH2O	2.49	kg H2O
							mBDA	273	kg BDA

	Alternate Ending
	Convert mole fractions into mass fractions :								Eqn 11

	The following unit analysis shows why Eqn 11 is true.


			Eqn 12



	xH2O	0.00904	kg H2O /kg gas			xBDA	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.


Answers :			Eqn 13				mH2O	2.49	kg H2O
							mBDA	273	kg BDA

Example Problem with Complete Solution