Question

By measuring the equilibrium between liquid and vapour phases of a solution at 30°C at 1.00 atm, it was found that xA = 0.220 when yA = 0.314. Calculate the activities and activity coefficients of both components in this solution on the Raoult’s law basis. The vapour pressures of the pure components at this temperature are: pA = 73.0 kPa and pB = 92.1 kPa. (xA is the mole fraction in the liquid and yA the mole fraction in the vapour.)

Answer 1

Answer:

The activity coefficients of both components in this solution are $a_{A}=0.436,a_{B}=0.755,\gamma _{A}=1.98 \,and\,\gamma _{B}=0.968$.

Explanation:

from the given,

Pressure = 1 atm = 101.3 kPa

$x_{A}=0.220$

$y_{A}=0.314$

The vapour pressures of pure components

$\rho _{A}=73.0\,kPa$

$\rho _{B}=92.1\,kPa$

$y_{A}=\frac{\rho_{A}}{\rho_{A}+\rho_{B}}=0.314$

$\rho_{A}=101.3kPa\,\times0.314\,=31.8\,kPa$

$\rho_{B}=Total\,pressure- \rho_{A}$

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

Let's calculate the each activity coefficient:

$a_{A}=\frac{\rho_{A}}{\rho_{A}^*}$

$=\frac{31.8\,kPa}{73.0\,kPa}=0.436$

$a_{B}=\frac{\rho_{B}}{\rho_{B}^*}$

$=\frac{69.5.8\,kPa}{92.1\,kPa}=0.755$

$\gamma _{A}=\frac{a_{A}}{x_{A}}$

$=\frac{0.436}{0.220}=1.98$

$\gamma _{B}=\frac{a_{B}}{x_{B}}$

$=\frac{0.755}{0.780}=0.968$

Therefore, The activity coefficients of both components in this solution are $a_{A}=0.436,a_{B}=0.755,\gamma _{A}=1.98 \,and\,\gamma _{B}=0.968$.