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

Raoult's law describes the relationship between vapor pressures of solution components and mole fractions. We can use this law to calculate the vapor pressure and mole fraction of the components in the given solution.

Step-by-step explanation:

Raoult's law describes the relationship between the vapor pressures of solution components and their mole fractions. According to Raoult's law, the partial pressure of a component in an ideal solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution. In this case, the mole fraction of component A (XA) in the liquid phase is given as 0.220 and the mole fraction of component A (yA) in the vapor phase is given as 0.314.

Using the Raoult's law equation, we can calculate the vapor pressure of component A in the solution (PA) by multiplying the vapor pressure of pure A (pA) by the mole fraction of A in the solution (XA).

Since PA = XAPA, we can substitute the given values to find the value of PA. Similarly, we can calculate the mole fraction of component B (XB) in the solution using the equation XB = 1 - XA.

Answer 2

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`.

Step-by-step 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)=(\rho_(A))/(\rho_(A)+\rho_(B))=0.314`

`\rho_(A)=101.3kPa\,*0.314\,=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)=(\rho_(A))/(\rho_(A)^*)`

`=(31.8\,kPa)/(73.0\,kPa)=0.436`

`a_(B)=(\rho_(B))/(\rho_(B)^*)`

`=(69.5.8\,kPa)/(92.1\,kPa)=0.755`

`\gamma _(A)=(a_(A))/(x_(A))`

`=(0.436)/(0.220)=1.98`

`\gamma _(B)=(a_(B))/(x_(B))`

`=(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`.