Question

By measuring the equilibrium between liquid and vapor phases of a binary solution at 30°C at 1atm, it was found that xA=0.220 when yA=0.314. Calculate the activities and activity coefficients of both components in the solution. The vapor pressures of the pure components at this temperature are 73kPa and 92.1kPa.

Answer 1

Answer:

Explanation:

From the knowledge of liquid and vapor phases

$x_A$ is the mole fraction in the liquid

$y_A$ is the mole fraction in the vapor

Given that :

$x_A = 0.220$

$y_A$ = 0.314

$P^*_A = 73.0 \ kPa$

$P^*_B = 92.1 \ kPa$

using the formula:

$y_A = \frac{p_A}{p_A+p_B}\\ \\ 0.314 = \frac{p_A}{101.3 \ kPa} \\ \\ p_A = (101.3 \ kPa) *(0.314) \\ \\ = 31.8 \ kPa$

$P_B = 101.3 \ kPa - 31.8 \ kPa \\ \\ = 69.5 \ kPa$

To calculate the activities and activity coefficients of both components in the solution; we have:

$a_A = \frac{p_A}{p^*_A} \\ \\a_A = \frac{31.8 \ kPa}{73.0 \ kPa} \\ \\ a_A = 0.436$

$a_B = \frac{p_B}{p^*_B}$

$a_B = \frac{69.5 \ kPa}{92.1 \ kPa}$

$a_B = 0.755$

$y_A = \frac{a_A}{x_A} \\ \\ y_A = \frac{0.436}{0.220} \\ \\ y_A = 1.98$

$y_B = \frac{a_B}{x_B} \\ \\ y_B = \frac{0.755}{0.780} \\ \\ y_B = 0.968$