Problem Set 11 Solutions

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43. The normal boiling points of propane and n-butane are –42.1˚C and –0.5˚C, respectively. The followingvapor pressure data have been measured.T (˚C) –31.2 –16.3P*, propane (kPa) 160.0 298.6P*, n-butane (kPa) 26.7 53.3Assume that the substances form ideal solutions.(a) Calculate the liquid phase mole fraction of propane in the solutions in which liquid and vapor are inequilibrium at 1.0 atm pressure at –31.2˚C and –16.3˚C.For each of these temperatures, we can use an approach similar to that taken in <strong>Problem</strong>s 1 and 2 to get themole fraction of propane in the liquid phase. From <strong>Problem</strong> 1 (and the equation of the bubble point line),the liquid phase mole fraction isx 1 =P − P 2*P 1 * − P 2*.Assigning propane as component 1 and n-butane as component 2, at –31.2˚C we have€x propane ==x propane = 0.560.*P − P nbutane**P propane − P nbutane1.0 atm − 0.264 atm1.579 atm − 0.264 atmNote that the pressures in the table above are given in kPa, so the conversion factor 1 atm = 101.32 kPa wasused in the calculation of € the mole fraction.At –16.3˚C, the liquid phase mole fraction of propane isx propane ==x propane = 0.196.*P − P nbutane**P propane − P nbutane1.0 atm − 0.526 atm2.947 atm − 0.526 atm(b) Calculate the vapor phase € mole fraction of propane in each of the solutions.To obtain the vapor phase mole fraction of propane at –31.2˚C, the partial pressure is determined first fromRaoult's Law,*P propane = x propane P propane( )( 1.579 atm)= 0.560P propane = 0.884 atm.€
3 b.) Continued5Then, the mole fraction of propane in the vapor phase is determined from the definition,y propane = P propaneP0.884 atm=1.0 atmy propane = 0.884 .At –16.3˚C, the partial pressure of propane is€*P propane = x propane P propane( )( 2.947 atm)= 0.196P propane = 0.578 atm.The vapor phase mole fraction of propane is determined from the equation,€y propane = P propaneP0.578 atm=1.0 atmy propane = 0.578.€
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Problem Set 11 Solutions

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