A three-component gas phase mixture of benzene (Z), octane (O), and nitrogen (N)
ID: 1056024 • Letter: A
Question
A three-component gas phase mixture of benzene (Z), octane (O), and nitrogen (N) is initially at a total pressure of 760 mm Hg and a temperature of 50 degree C. The initial mole fraction, y_z, of Z in the mixture is 0.1, and the initial mole fraction, y_O, of O in the mixture is also 0.1. The mixture is then compressed, causing a fraction of Z and O to liquefy. The temperature is kept constant at all times at 50 degree C. At the end of the compression, the total mole fraction of Z and O vapor remaining in the gas phase is 0.05; that is, y_O + y_z = 0.05 at the end of the compression. N is not condensable. What is the final pressure P_F to which the mixture was compressed? Assume that Raoult's Law applies to both Z and O, that the liquid and gas at the end of the compression are in equilibrium, and that the gas mixture is ideal at all times. In addition, take the Antoine equation constants for Z to be A = 6.9, B = 1200, and C = 220, and those for O to be A = 6.9, B = 1350, and C = 210 (these constants apply when p* is in mm Hg and temperature is in degree C).Explanation / Answer
Ans.
Initial values:
P1 =760 mm Hg
T = 50 oC = 50+273 = 323 K
P1 = PN+PZ+557*PO
Mole fraction of N = yN1 = 1 -(YO+yZ) = 1 - (0.1+0.1) = 0.8
PN1 = P1*yN1 = 760*0.8 = 608 mm Hg
Final values:
P2 =
T = 50 oC = 50+273 = 323 K
Log(PZ ) = A - B/(T+C) = 6.9 - 1200/(50+220) = 2.46
PZ0 = 288.4 mm Hg
XZ = mole fraction of Z in liquid
PZ = PZ0*XZ = 288.4*XZ
Log(PO ) = A - B/(T+C) = 6.9 - 1350/(50+210) = 1.71
PO0 = 51 mm Hg
PO= PO0*XO = 51*XO = 51*(1-XZ)
Partial pressure of both O and N = PO + PZ = 51*(1-XZ) + 288.4*XZ
PN2 = PN1 = 608 mm Hg
Total pressure P2 = PN2+PO + PZ = 608 + 51*(1-XZ) + 288.4*XZ = 237.4*XZ + 557
Mole fraction of PN in gas, yN = 1-0.05 = 0.95
Mole fraction of PN in gas, yN = PN/P2 = 608/( 237.4*XZ + 557) = 0.95
XZ = 0.35
Total pressure P2 = 237.4*XZ + 557 = 640 mm Hg.
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