Consider a 20 L sample of moist air at 60C and one atm in which thepartial press

Question

Consider a 20 L sample of moist air at 60C and one atm in which thepartial pressure of water vapor is 0.120 atm.Assume that dry airhas the composition 78 mole percent N2 ,21 mole percentO2 ,and 1 mole percent Ar.   a)what are themole percentages of each of the gases in the sample?  b)the percent relative humidity is defined as %RH=PH2O/P* H2O where P H2Ois the partial pressure of water in the sample and P*H2O=0.197 atm is the equilibrium vapor pressure of waterat 60C .The gas is compressed at 60 C until the relative humidityreaches 100%. When volume does the mixture contain now? c) whatfraction of water vapor will be condensed if the total pressure ofthe mixture is isothermally increased to 200 atm?

Explanation / Answer

PV = nRT n = PV/(RT) = (1 atm)*(20 L)/(0.0821 L*atm/mol/K *333 K) = 0.731547626 moles n_water = P_water*V/RT = (0.120 atm)*(20 L)/(0.0821 L*atm/mol/K*333 K) = 0.0877857151 mole water moles of dry air = n - n_water = 0.731547626 moles - 0.0877857151mole water = 0.643761911 moles dyr air %N2 = 0.78*0.643761911 moles/0.731547626 moles = 68.6% %O2 = 0.21 *0.643761911/0.731547626 = 18.5% ~ %Ar = 0.01*0.643761911/0.731547626 = 0.88% %water = 0.0877857151 /0.731547626 = 12 % b) %RH = 0.120/0.197 = 60.9% c) V = nRT/P = (0.0877857151 moles)*(0.0821 L*atm/mol/K)*(333K)/(0.197atm) = 12.1827411 Liter ~ 12.2 L d) Partial Pressure of water = 0.197 atm Partial pressure of other gases = 200 - 0.197 = 199.803 atm Partial pressure of N2 = 0.78*(199.803 atm) = 155.84634 atm V = n_N2*RT/P_N2 = (0.78*0.643761911 moles dry air)*(0.0821L*atm/mol/K)*(333 K)/155.84634 atm) V = 0.088086504 Liters moles of water n = PV/RT = (0.197 atm) *(0.088086504 L)/(0.0821 L*atm/mol/K * 333K) n = 0.000634728808 moles water moles of water condensed = 0.0877857151 mole water - 0.000634728808moles water                                        = 0.0871509862 = 8.72 x 10-2 moles water

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