Chapter 14 MASS TRANSFER

Chapter 14 Mass Transfer14-38 A rubber wall separates O 2 and N 2 gases. The molar concentrations of O 2 and N 2 in the wall are to bedetermined.Assumptions The O 2 and N 2 gases are in phase equilibrium with the rubber wall.Properties The molar mass of oxygen and nitrogen are 32.0 and 28.0 kg/kmol, respectively (Table A-1).The solubility of oxygen and nitrogen in rubber at 298 K are 0.00312 and 0.00156kmol/m 3 bar, respectively(Table 14-7).Analysis Noting that 500 kPa = 5 bar, the molar densities of oxygenand nitrogen in the rubber wall are determined from Eq. 14-20 to beCC(0) SPO2, solid sideO2, gasside (0.00312kmol/m3= 0.0156kmol/m(0) SPN2, solid sideN2, gasside (0.00156kmol/m3= 0.0078kmol/m33.bar)(5 bar).bar)(5 bar)That is, there will be 0.0156 kmol of 039 kmol of O 2 and0.0078 kmol of N 2 gas in each m 3 volume of the rubber wall.O 225ºC500 kPaC O2C N2RubberplateN 225ºC500 kPa14-39 A glass of water is left in a room. The mole fraction of the water vapor in the air and the molefraction of air in the water are to be determined when the water and the air are in thermal and phaseequilibrium.Assumptions 1 Both the air and water vapor are ideal gases. 2 Air is saturated since the humidity is 100percent. 3 Air is weakly soluble in water and thus Henry’s law is applicable.Properties The saturation pressure of water at 20C is 2.339 kPa (Table A-9). Henry’s constant for airdissolved in water at 20ºC (293 K) is given in Table 14-6 to be H = 65,600 bar. Molar masses of dry air andwater are 29 and 18 kg/kmol, respectively (Table A-1).Analysis (a) Noting that air is saturated, the partial pressure of water vapor in the air will simply be thesaturation pressure of water at 20C,P vapor P sat @ 20º C 2.339 kPaAssuming both the air and vapor to be ideal gases, the mole fraction ofwater vapor in the air is determined to beyvaporPvapor2.339 kPa 0.0241P 97 kPa(b) Noting that the total pressure is 97 kPa, the partial pressure of dry air isP P P 97 2339 . 94. 7kPa = 0.94 bardry airvaporAir20ºC97 kPaRH=100%EvaporationFrom Henry’s law, the mole fraction of air in the water is determined to beydry air,liquid sidePdry air,gas side 0.947 bar 1.44 10 5H 65,600barWater20ºCDiscussion The amount of air dissolved in water is very small, as expected.14-16