Question

Components \( A \) and \( B \) form an azeotrope. The saturation vapor pressures of \( A \) and \( B \) at the boiling temperature of the azeotrope are 87 kPa and 72.7 kPa, respectively. The azeotrope composition is _________ mol% \( A \) (rounded off to the nearest integer).
% Given GIVEN: \[ \ln \left( \frac{\gamma_A}{\gamma_B} \right) = 0.9 \left( x_B^2 - x_A^2 \right) \] where \( x_i \) and \( \gamma_i \) are the liquid phase mole fraction and activity coefficient of component \( i \), respectively.

Hint:

For an azeotrope, use Raoult’s Law and the activity coefficient equation to solve for the azeotrope composition.

Answer 1

At the azeotrope, the total vapor pressure is equal to the sum of the partial vapor pressures of the components in the liquid phase: \[ P_{{total}} = P_A x_A + P_B x_B \] For an azeotrope, the mole fraction of component \( A \) in the vapor phase is equal to the mole fraction in the liquid phase: \[ y_A = x_A \] Using Raoult’s Law, the mole fraction of \( A \) in the vapor phase is given by: \[ y_A = \frac{P_A x_A}{P_{{total}}} \] Since at the azeotrope \( y_A = x_A \), we can write: \[ x_A = \frac{P_A x_A}{P_{{total}}} \] This equation can be rearranged to find \( x_A \) at the azeotrope. The given equation for the activity coefficients is: \[ \ln\left(\frac{\gamma_A}{\gamma_B}\right) = 0.9(x_B^2 - x_A^2) \] Substitute \( x_B = 1 - x_A \) into the equation and solve for \( x_A \). The mole fraction of \( A \) in the azeotrope is approximately 0.608, or 61 mol% \( A \).