Chapter 17, Problem 136CP

Interpretation Introduction

Interpretation: The mole fraction of the liquid mixture and vapor above the solution should be determined.

Concept Introduction:

The mole fraction is the number of moles compared with the total number of moles. It is also known as molar fraction that is defined as the amount of the constituents divided by the total amount of the constituents.

Explanation of Solution

The mole fraction is the ratio of the number of moles of a given component to the total number of moles of solution. For a two-component solution, where $n_A$ and $n_B$ represented the number of the moles of the two components.

Mole fraction of component $\text{A}={\text{X}}_{\text{A}}$

This is represented in terms of number of moles as follows:

  $X_A=\frac{n_A}{n_A+n_B}$

Now,

  $P=X_A{P}_T$

Here,

  $P_T=P_A+P_B$

Thus,

  $X_A=\frac{P_A}{P_A+P_B}$

According to question, mole fraction is 30% mole by A thus,

  $0.3=\frac{P_A}{P_A+P_B}$ …… (1)

Given that liquid A has vapor pressure x, and liquid B has vapor pressure y.

The partial pressure and vapour pressure are related to each other as follows:

  $P_A=X_{A}x$

Also,

  $P_B=X_{B}y$

From equation (1):

  $0.3=\frac{P_A}{P_A+P_B}$

This can also be written as:

  $0.3=\frac{X_{A}x}{X_{A}x+X_{B}y}$

Here,

  $X_A+X_B=1$

Thus,

  $0.3=\frac{X_{A}x}{X_{A}x+\left(1-X_A\right)y}$

On rearranging,

  $\begin{array}{l}0.3X_{A}x+0.3y-0.3X_{A}y=X_{A}x\\ 0.3y=X_{A}x-0.3X_{A}x+0.3X_{A}y\\ 0.3y=X_A\left(x-0.3x+0.3y\right)\\ 0.3y=X_A\left(0.7x+0.3y\right)\end{array}$

Thus,

  $X_A=\frac{0.3y}{0.70x+0.30y}$

Also,

  $\begin{array}{l}{X}_B=1-\frac{0.3y}{0.70x+0.30y}\\ =\frac{0.7x+0.3y-0.3y}{0.7x+0.3y}\\ =\frac{0.7x}{0.7x+0.3y}\end{array}$

Now, the mole fraction of the liquid mixture if the vapor given solution is 50% A by moles can be determined as follows:

  $0.5=\frac{X_{A}x}{X_{A}x+\left(1-X_A\right)y}$

On rearranging,

  $\begin{array}{l}0.5X_{A}x+0.5y-0.5X_{A}y=X_{A}x\\ 0.5y=X_{A}x-0.5X_{A}x+0.5X_{A}y\\ 0.5y=X_A\left(x-0.5x+0.5y\right)\\ 0.5y=X_A\left(0.5x+0.5y\right)\end{array}$

Thus,

  $X_A=\frac{0.5y}{0.50x+0.50y}=\frac{y}{x+y}$

Also,

  $\begin{array}{l}{X}_B=1-\frac{y}{x+y}\\ =\frac{x+y-y}{x+y}\\ =\frac{x}{x+y}\end{array}$

The mole fraction of the liquid mixture if the vapor given solution is 80% A by moles:

  $0.8=\frac{X_{A}x}{X_{A}x+\left(1-X_A\right)y}$

On rearranging,

  $\begin{array}{l}0.8X_{A}x+0.8y-0.8X_{A}y=X_{A}x\\ 0.8y=X_{A}x-0.8X_{A}x+0.8X_{A}y\\ 0.8y=X_A\left(x-0.8x+0.8y\right)\\ 0.8y=X_A\left(0.2x+0.8y\right)\end{array}$

Thus,

  $X_A=\frac{0.8y}{0.20x+0.80y}$

Also,

  $\begin{array}{l}{X}_B=1-\frac{0.8y}{0.20x+0.80y}\\ =\frac{0.2x+0.8y-0.8y}{0.20x+0.80y}\\ =\frac{0.2x}{0.20x+0.80y}\end{array}$

Similarly, if liquid A has vapor pressure x, and liquid B has vapor pressure y. The mole fraction of the vapor mixture if the vapor given solution is 30% A by moles will be:

  $\begin{array}{l}{X}_A^V=\frac{0.70y}{0.30x+0.70y}\\ X_B^V=1-X_A^V\\ X_B^V=1-\frac{0.7y}{0.30x+0.70y}=\frac{0.3x}{0.30x+0.70y}\end{array}$

The mole fraction of the liquid mixture if the vapor given solution is 50% A by moles:

  $\begin{array}{l}{X}_A^V=\frac{y}{x+y}\\ X_B^V=1-X_A^V=1-\frac{y}{x+y}\\ =\frac{x}{x+y}\end{array}$

The mole fraction of the liquid mixture if the vapor given solution is 80% A by moles:

  $\begin{array}{l}{X}_A^V=\frac{0.20y}{0.80x+0.20y}\\ X_B^V=1-\frac{0.20y}{0.80x+0.20y}\\ =\frac{0.80x}{0.80x+0.20y}\end{array}$