Binary Mixture Essay

Residual Curve Mapping and Isobaric Vapor Liquid Equilibrium of tertiary mixture of Hexane –Cyclohexane-Methyl t-butyl ether.

Shraddha H. Motegaonkara, Maya maneb, Shambala N. Shindec, Rahul k. Kulkarnid

a Master student of chemical engineering, Department of chemical Engineering, Bharati Vidyapeeth Deemed University, College of engineering, pune-411046 b Master student of chemical engineering department, Bharati Vidyapeeth college of engineering Pune - 411046 c Associate professor, Department of chemical Engineering, Bharati Vidyaααpeeth Deemed University, College of engineering pune-411046 d Assistant professor, Department of chemical engineering, Bharati Vidyapeeth deemed university, college of engineering pune-411046

Abstract

Isobaric

The VLE data of binary mixtures is often expressed as a plot as shown in above fig. This plot represents the bubble point and dew point of the binary mixture at constant pressure. The particular VLE plot shows a binary mixture that has a uniform vapor liquid equilibrium that is relatively easy to separate the curve line is as equilibrium curve line and gives the compositions of liquid and vapor in equilibrium at some fixed pressure. Azeotrope is also important concept in VLE. If the equilibrium curve crosses the diagonal line then it should be considered that there are azeotropic points where the azeotrope occurs. When thermodynamics is applied to vapor liquid equilibrium, the goal is to find calculation of the Temperatures, pressures, and compositions of phases which are in equilibrium.[3]
VLE can be calculated as two different ways:
At constant temperature:
First consider a system with the two constituents A and B according to Routs law,
PA = XA P0A (1)
PB = XBP0B (2)
The total pressure, according to Dalton’s law, is given by the sum of the partial pressures: P = XA P0A + XBP0B (3)
As the system is binary, it is possible to substitute (1-XA) for XB; after this modification:
P = XA (P0A- P0B)+ P0B (4)
According to Dalton’s law we have for the vapour phase: pA = yAp (5) pB = yBp = (1 – yA) p (6)

( Y_A)/y_B / x_A/( x_B ) = (y_A (1-x_A ))/(x_A (1-y_A ) ) = (P°_A)/(P°_B ) = α