A vapor stream of composition y_1F = 0.75 and y_2F = 0.25 is cooled to a tempera

Question

A vapor stream of composition y_1F = 0.75 and y_2F = 0.25 is cooled to a temperature T in the two-phase region and flows into a separation chamber at a pressure of 1 bar. If the composition of the liquid product is to be x_1 = 0.50, what is the required value of T, and what is the values of y_1? For the liquid mixtures of species 1 and 2, molar excess Gibbs free energy is given as G^ex/RT = 1.2 x_1 x_2 The vapor pressures of the pure species are given by; In P^vap _1 (bar) = 10.00 - 2950/T (K) - 36.0 In P^vap _2 (bar) = 11.70 - 3840/T (degree C) - 44.8

Explanation / Answer

Given Gex /RT = 1.2x1x2

So,

ln gamma 1 = 1.2x22  = 1.2 x 0.5 x0.5 = 0.3

gamma 1 = 1.35

ln gamma 2 = 1.2x12  = 1.2 x 0.5 x0.5 = 0.3

gamma 2 = 1.35

Total pressure = 1 bar.

we know,

P = x1 gamma-1 P1sat + x2 gamma-2 P2sat

1 bar = 0.5 x 1.35 x exp ( 10- 2950/T-36 ) + 0.5 x 1.35 x exp ( 11.7 - 3840/T-44.8 )

On solving for T,

The temperature is 70.5 oC

Note while calculating temperature observe that T in P sat is in Kelvin for componenet-1 and is in C for component-2

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