10553869_1520702498236036_4254157203906812411_o.jpg - a)Assume temperature is constant{9000 Calculate the saturation vapor pressure at 9000 Log P1 =

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Unformatted text preview: a)Assume temperature is constant {9000) Calculate the saturation vapor pressure at 9000 Log P1 = 6.90565 61211033931019 = 6.90565 6 3.89662 = 3.0090 P1 = 103.00er =1o2c.9 mmHg Log P2 = 6.59719 01424255730321 2 6.59719 0 4.69256 = 1.89993 P2 = 79.4 mml-Eg The two substances torm ideal solution, so obey)r the Raoultés law. It the vapor pressure P1 of pure component (say B) is known, the partial vapor pressure p1 of B in the mixture of B and EB will be proportional with the mol traction x1 of B in the mixture: p1 = x1.P1 and p2 = x2.P2 The graph p1 versus x1 (Raoult's law vapor pressure of benzene) is a line with two characteristic points: for x1=0 (component B is missing) p1 is 0; and for x1=1 (component EB is missing) p1 = P1. The same for p2 versus x2 {Raoult's law vapor pressure of benzene). But x1 + x2 =1 and you can put both on the same axis. You can superpose the two graphs, like this one, with two vertical axis for p1 and p2 and one horizontal axis for both and and x2 (read x1 from left to right from 0 to 1, and x2 from right to left from 0 to 1). Put a line between P1 (on the left p1 axis) and P2 {on the right p2 axis). It is the sum of previous two lines. This line is the total vapor pressure at the solution (two components), i.e., the sum sum of the two partial pressures. if the actual pressure is above this line, there will be a liquid phase. b) Here you have to put the boiling temperature on a similar diagram (boiling diagram), but the vertical axes are temperature. A pure liquid boils when its vapor pressure equals the ambient pressure (here 760 mmHg). Put 760 as P1 and P2 in the previous relations and calculate Tb1 and Tb2 Put Tb1 and Tb2 on temperature axes and draw two curved lines Log 760 = 6.90565 0 1211.033! (Tb1 + 220.79) {2.8808 - 6.90565).lr 1211.033 = - “(Tb1 + 220.79) - 0.0033235 (Tb1 + 22079) = -1 Tb1 + 220.79 = 300.83 Tb1 = so 60 Log 760 = 6.59719 0 1424.255!(Tb2+ 213.21) (2.8808 6 6.59719)! 1424.255: 1!(-Tb2 0 213.21) (-3.71631'1424255) (-Tb2 - 213.21): 1 Tb2 + 213.21 = 1100026093 Tb2 = 323.2 0 213.2 :12) ac ...
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