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

Step-by-step 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 = (p_A)/(p_A+p_B)\\ \\ 0.314 = (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 = (p_A)/(p^*_A) \\ \\a_A = (31.8 \ kPa)/(73.0 \ kPa) \\ \\ a_A = 0.436`

`a_B = (p_B)/(p^*_B)`

`a_B = (69.5 \ kPa)/(92.1 \ kPa)`

`a_B = 0.755`

`y_A = (a_A)/(x_A) \\ \\ y_A = (0.436)/(0.220) \\ \\ y_A = 1.98`

`y_B = (a_B)/(x_B) \\ \\ y_B = (0.755)/(0.780) \\ \\ y_B = 0.968`