Project ForgeConsider a binary liquid mixture at equilibrium with its vapour at 25°C. Antoine equation: log10(pi^sat)=A−B/(t+C), t in °C, p in Torr. Component 1: A=7.0, B=1210, C=230. Component 2: A=6.5, B=1206, C=223. The vapour phase is ideal; activity coefficients: ln(γ1)=x2²[2−0.6x1], ln(γ2)=x1²[1.7+0.6x2]. If x1=0.11, the mole fraction of component 1 in vapour phase (y1) is _____
Explanation
p1sat: log10=7.0−1210/255=2.255 ⟹ p1sat=179.9 Torr. p2sat: log10=6.5−1206/248=1.637 ⟹ p2sat=43.4 Torr. x2=0.89: ln(γ1)=0.89²(2−0.6×0.11)=0.7921×1.934=1.532 ⟹ γ1=4.63. ln(γ2)=0.11²(1.7+0.6×0.89)=0.0121×2.234=0.0270 ⟹ γ2=1.027. p1=x1γ1p1sat=0.11×4.63×179.9=91.6 Torr. p2=x2γ2p2sat=0.89×1.027×43.4=39.6 Torr. Ptotal=131.2 Torr. y1=p1/Ptotal=91.6/131.2≈0.698.
Source: GATE CH 2014 (2014)
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