A cylinder at T = 300 K and P-2.0 MPa contains two components that are in vapor-

Question

A cylinder at T = 300 K and P-2.0 MPa contains two components that are in vapor-liquid equilibrium. At 298 K, P virial equation of state: = 1.4 MPa. The vapor phase of each component is described by the RT The mixing rule for the virial coefficient of a binary mixture is: In addition, the fugacity coefficient of component 1 in a binary, non-ideal solution described by the virial equation of state is given by: RT At these conditions, the virial coefficients for these components are: Bu =-268 cm'/mol B22 =-7.5 cm?/mol B12 =-75 an3/mol Using this information, estimate the vapor composition at the specified conditions if the solubility of component 2 in the liquid-phase of component 1 is negligible.

Explanation / Answer

By Raoult's Law, For Non ideal solution

Mole fraction of i in vapor phase*Total pressure = Fucacity coefficient of i*Mole fraction of i in liquid phase*Saturation Vapor pressure of i

P1sat = 1.4MPa = ~14 atm

P=2 MPa =~20atm

Mole fraction of 1 in liquid phase = 1 (because Component 2 is not soluble in 1)

y2=1-y (y mole fraction of 1 in vapor phase, y1)

=>y*20=exp(20/.082/300*(2*y*(-268*10^-3)+2*(1-y)*(-75*10^-3)-y^2*(-268*10^-3)-2*y*(1-y)*(-75*10^-3)-(1-y)^2*(-7.5*10^-3)))*1*14

=>Mole fraction of 1st component in Vapor phase = 0.57

=>Mole fraction of 2nd component in Vapor phase = 1-0.57 =.43

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