For the ethyl ethanoate- (n)-heptane system at (343.15 mathrm{~K}), assuming the system to follow modified Raoult's law,

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For the ethyl ethanoate- \(n\)-heptane system at \(343.15 \mathrm{~K}\), assuming the system to follow modified Raoult's law, predict whether an azeotrope gets formed or not. If it is so, calculate the azeotrope composition and pressure at \(T=343.15 \mathrm{~K}\). We are given that

\[ \ln \gamma_{1}=0.95 x_{2}^{2} \]

\[ \begin{aligned} \ln \gamma_{2} & =0.95 x_{1}^{2} \\ P_{1}^{\text {Sat }} & =79.80 \mathrm{kPa} \\ P_{2}^{\text {Sat }} & =40.50 \mathrm{kPa} \end{aligned} \]

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