Solution to Problem 10.17(a) Modified Raoult's law yiP = xiγiPisat Known: lnγ1 = P1sat = (10-5) 0.95x22 79.80 kPa lnγ2 = P2sat = 0.95x12 40.50 kPa (a) Make a BUBL P calculation for T = 343.15 K, x1 = 0.05 BUBL P calculation is to find equilibrium pressure and vapor composition for temperature T and liquid compositon x1 From y1P = x1γ1P1sat y2P = x2γ2P2sat Then, sum the two equations (y1+y2)P = x1γ1P1sat + x2γ2P2sat Because of (y1+y2) = 1 and (x1+x2) = 1 P = x1γ1P1sat + x2γ2P2sat = x1γ1P1sat + (1-x1)γ2P2sat = 0.05*exp(0.95*(1-0.05)*(1-0.05))*79.80 + (1-0.05)*exp(0.95*0.05*0.05)*40 = 47.97 kPa sat y1 = x1γ1P1 /P = 0.05*exp(0.95*(1-0.05)*(1-0.05))*79.80/47.97 = 0.1960 05)*40.95*0.50 .5)*exp(0.05*0. 80 kPa lnγ2 = P2sat = 0.95x12 40.17(b) Modified Raoult's law yiP = xiγiPisat Known: lnγ1 = P1sat = (10-5) 0.80) + (1-0.95*0.05)/(exp(0.05/(exp(0.0112)*(1-0. sum the two equations (x1+x2)/P = y1/γ1P1sat + y2/γ2P2sat Because of (y1+y2) = 1 and (x1+x2) = 1 P = 1/(y1/γ1P1sat + (1-y1)/γ2P2sat) = 1/(0.24/(exp(0.25 kPa x1 = y1*P/γ2P1sat = 0.05*42.0105 2nd iteration P = 1/(y1/γ1P1sat + (1-y1)/γ2P2sat) .95*0.95*(1-0.0 = 42.05/(exp(0.05 DEW P calculation is to find equilibrium pressure and liquid composition for temperature T and vapour compositon y1 From y1P = x1γ1P1sat y2P = x2γ2P2sat x1/P = y1/γ1P1sat x2/P = y2/γ2P2sat Then.95*(1-0.05*0. = 42.95*(1-0.50 kPa (a) Make a DEW P calculation for T = 343.0112))*79.80) = 0.95*(1-0.05))*79.15 K.05)/(exp(0.05)*(1-0.0112)*(1-0.05)*(1-0.0112))*79.19 x1 = y1*P/γ2P1sat 0.24/(exp(0.05*42.05))*79.95x22 79.0112 1st iteration 0.80) + (1-0.0112 = = 0.Solution to Problem 10.80) 0. y1 = 0.0112 P = 1/(y1/γ1P1sat + (1-y1)/γ2P2sat) = 1/(0. 0105 3rd interation 0.95*(1-0.0112))*79.0112))*79.0105 = 1/(0.95*(1-0.95*(1-0.0105 = = 0.0.95*0.0 = 42.0112)*(1-0.80) + (1-0.19 x1 = y1*P/γ2P1sat 0.05/(exp(0.80) + (1-0.05)/(exp(0.05*42.0 = 42.0112)*(1-0.95*0.24/(exp(0.05)/(exp(0.0112))*79.0112)*(1-0.80) 0.0105 P = 1/(y1/γ1P1sat + (1-y1)/γ2P2sat) = 1/(0.05/(exp(0.19 . 05)/(exp(0.0112)*40.95*0.1-0.50)) + (1-0.0112*0.50)) .05)/(exp(0.05*0.95*0.05)*40. 05)/(exp(0.95*0.0112*0.95*0.0112)*40.+ (1-0.05)/(exp(0.50)) .50)) + (1-0.0112)*40.0112*0. α12 = 1. an azeotrope does exist. P2satexp and A are given for the temperature of interest.8 = 0.8/(40.15 K Refer to Example 10. (10.95(x22 = 0. (α12)x1=0 = exp(A)P1sat/P2sat (α12)x1=1 = P1sat/P2satexp(A) and Values of P1sat.(10.0 at some intermediate composition.80 kPa lnγ2 = P2sat = 0.9) becomes: γ1az/γ2az = P2sat/P1sat = 40. and γ1 = exp(A). and γ2 = exp(A). y2 = x2.95x22 79.5) yi/xi = γiPisat/P Therefore. by Eq.5*exp(0. because α12 is a continuous function of x1 and must pass through the value of 1.5 = (α12)x1=1 = 79.8/40.95x12 40.95(x2 - 0. γ1 = 1.95x22 = 0.95)) = Since the value at one limit ia greater than 1.17(c) Modified Raoult's law yiP = xiγiPisat Known: lnγ1 = P1sat = (10-5) 0.95x12 x12) x1)(x2 + x1) x1) .95(x2 = 0.95)*79. and α12 = 1. This Calculation is facilitated by using the relative volatility: (y1/x1)/(y2/x2) α12 = (10-8) At an azeotro y1 = x1.50 kPa (c) Find the azeotrope composition and pressure at T = 343. The limiting values of α12 are therefore: (α12)x1=0 = exp(0. In general.5075 The difference between the correlating equations for lnγ1 and lnγ2 provides the general relation: ln(γ1/γ2) = lnγ1-lnγ2 = 0. whereas the value bat the other limit is less than 1. For azeotrope.5/79. γ1P1sat/γ2P2sat α12 = (10-9) The correlating equations for the activity coefficients shows that when x1 = 0. when x1 = 1. and Eq. γ2 = 1. Therefore in these limits.Solution to Problem 10.3(e) on Page 353 of the text First determine whether or not an azeotrope exists at the given temperature. 8570 .5075.95(1-2x1) Therefore: 0.37 kPa Answer: The azeotrope pressure is 81.8 = 81.857) = 0.8570 lnγ1az = 0.67826 and ln(γ1/γ2) = 0.95*(x2az)2 = = 0.67826 x1 = 0. for which: ln(γ1/γ2) = ln(0.01943 Paz = γ1az * P1sat = exp(0.95) + 1/2 0.95(1-2x1) = -0.67826/(2*0.01943)*79. i.95*(1-0.857)*(1-0.e.37 kPa and the azeotrope composition is x1 = y1 = 0.= 0.5075) = -0.95(1-2x1) Thus the azeotropic composition is the value of x1 for which this equation is satisfied when the activity-coefficient ratio has its azeotrope value of 0.. (10.5) ows that when x1 = 0. The 5.given temperature. p(A). This y Eq.7620 value bat the other is a continuous function ediate composition. Therefore in = P1sat/P2satexp(A) f interest. and lnγ2 provides the .0948 0. 5075.ch this equation is trope value of 0. . 24/(exp(0.95*(1-y1)*(1-y1))*79.x1 = y1*P/γ2P1sat = y1*42.80) .