Chapter 5, Problem 5C.5P

(a)

Interpretation Introduction

Interpretation:

The phases present in the phase diagrams have to be stated.

Concept introduction:

The azeotropic mixture is defined as a mixture of two or more liquids which evaporates without altering the composition of the mixture.  The azeotropic mixture has a constant boiling point because of the same composition in liquid as well as vapor phase.  The components of azeotropic mixture are not separated by distillation processes.

The phase diagram represents the changes in the mixture on changing the parameters like temperature and pressure of the mixture.

Answer to Problem 5C.5P

The phases have been stated in the phase diagrams as,

Explanation of Solution

The given phase diagrams are shown below as,

Figure 1

Figure 2

The region between the liquid phase curve and vapour phase curve has liquid and vapour phase in the equilibrium.  Thus, the number of phases present in that region is $2$.

In the region above the curve, only vapour phase is present.

In the region below the curve, only liquid phase is present.

The regions are stated in the phase diagrams as,

Figure 3

Figure 4

(b)

Interpretation Introduction

Interpretation:

The vapor pressure of the mixture of heptane and hexane at $70°\text{C}$ under the given conditions has to be stated.

Concept introduction:

The azeotropic mixture is defined as a mixture of two or more liquids which evaporates without altering the composition of the mixture.  The azeotropic mixture has a constant boiling point because of the same composition in liquid as well as vapor phase.  The components of azeotropic mixture are not separated by distillation processes.

The phase diagram represents the changes in the mixture on changing the parameters like temperature and pressure of the mixture.

Answer to Problem 5C.5P

The vapour pressure of the mixture of heptane and hexane at $70°\text{C}$ under the given conditions has been stated as $\underset{\_}{620Torr}$.

Explanation of Solution

The number of moles of heptane and hexane present in the mixture is $1.00\text{mol}$.

The mole fraction is calculated by the formula as,

    $x=\frac{n}{n_{\text{total}}}$        (1)

The total number of moles of the mixture is calculated as,

    $\begin{array}{c}{n}_{\text{total}}=n_{\text{Hexane}}+n_{\text{Heptane}}\\ =1+1\\ =2\end{array}$

Substitute the values in the equation (1) to calculate the mole fraction of hexane as,

    $\begin{array}{c}x=\frac{n}{n_{\text{total}}}\\ =\frac{1}{2}\\ =0.50\end{array}$

The pressure corresponding to $x_{\text{Hexane}}=0.5$ is equals to $\text{620}\text{Torr}$ which is taken by drawing the line on the y-axis as shown in phase diagram below.

Figure 5

Thus, the vapour pressure of the mixture of heptane and hexane at $70°\text{C}$ under the given conditions is $\underset{\_}{620Torr}$.

(c)

Interpretation Introduction

Interpretation:

The vapor pressure of the mixture of heptane and hexane at $70°\text{C}$ when just one drop of the liquid remains has to be stated.

Concept introduction:

The azeotropic mixture is defined as a mixture of two or more liquids which evaporates without altering the composition of the mixture.  The azeotropic mixture has a constant boiling point because of the same composition in liquid as well as vapor phase.  The components of azeotropic mixture are not separated by distillation processes.

The phase diagram represents the changes in the mixture on changing the parameters like temperature and pressure of the mixture.

Answer to Problem 5C.5P

The vapour pressure of the mixture of heptane and hexane at $70°\text{C}$ when just one drop of the liquid remains has been stated as more than $\underset{\_}{490Torr}$.

Explanation of Solution

The number of moles of heptane and hexane present in the mixture is $1.00\text{mol}$.

The mole fraction is calculated by the formula as,

    $x=\frac{n}{n_{\text{total}}}$        (1)

The total number of moles of the mixture is calculated as,

    $\begin{array}{c}{n}_{\text{total}}=n_{\text{Hexane}}+n_{\text{Heptane}}\\ =1+1\\ =2\end{array}$

Substitute the values in the equation (1) to calculate the mole fraction of hexane as,

    $\begin{array}{c}x=\frac{n}{n_{\text{total}}}\\ =\frac{1}{2}\\ =0.50\end{array}$

The pressure corresponding to $x_{\text{Hexane}}=0.5$ is equals to $\text{490}\text{Torr}$ which is taken by drawing the line on the y-axis as shown in phase diagram below.

Figure 6

When just one drop of the liquid remains, then the vapour pressure of the mixture is more than $\text{490}\text{Torr}$ as till $\text{490}\text{Torr}$, the amount of liquid phase is high in the mixture.

Thus, the vapour pressure of the mixture of heptane and hexane at $70°\text{C}$ when just one drop of the liquid remains is more than $\underset{\_}{490Torr}$.

(d)

Interpretation Introduction

Interpretation:

The mole fraction of hexane in the liquid and vapour phase for the conditions of part b has to be calculated.

Concept introduction:

The azeotropic mixture is defined as a mixture of two or more liquids which evaporates without altering the composition of the mixture.  The azeotropic mixture has a constant boiling point because of the same composition in liquid as well as vapor phase.  The components of azeotropic mixture are not separated by distillation processes.

The phase diagram represents the changes in the mixture on changing the parameters like temperature and pressure of the mixture.

Answer to Problem 5C.5P

The mole fraction of hexane in the liquid and vapour phase for the conditions of part b has been calculated as $\underset{\_}{0.5}$ each.

Explanation of Solution

The number of moles of hexane present in the mixture is given as $1.00\text{mol}$.

The mole fraction is calculated by the formula as,

    $x=\frac{n}{n_{\text{total}}}$        (1)

The total number of moles of the mixture is calculated as,

    $\begin{array}{c}{n}_{\text{total}}=n_{\text{Hexane}}+n_{\text{Heptane}}\\ =1+1\\ =2\end{array}$

Substitute the values in the equation (1) to calculate the mole fraction of hexane as,

    $\begin{array}{c}x=\frac{n}{n_{\text{total}}}\\ =\frac{1}{2}\\ =\underset{\_}{0.50}\end{array}$

The mole fraction of hexane in the liquid phase is taken by drawing the line on the liquid curve of the phase diagram which is shown by the line AB’ in phase diagram below.

Figure 7

Thus, mole fraction of hexane in the liquid and vapour phase for the conditions of part b is $\underset{\_}{0.5}$ each.

(e)

Interpretation Introduction

Interpretation:

The mole fraction of hexane in the liquid and vapour phase for the conditions of part c has to be calculated.

Concept introduction:

The azeotropic mixture is defined as a mixture of two or more liquids which evaporates without altering the composition of the mixture.  The azeotropic mixture has a constant boiling point because of the same composition in liquid as well as vapor phase.  The components of azeotropic mixture are not separated by distillation processes.

The phase diagram represents the changes in the mixture on changing the parameters like temperature and pressure of the mixture.

Answer to Problem 5C.5P

The mole fraction of hexane in the liquid and vapour phase for the conditions of part c has been calculated as $\underset{\_}{0.8}$ and $\underset{\_}{0.5}$ respectively.

Explanation of Solution

The number of moles of hexane present in the mixture is given as $1.00\text{mol}$.

The mole fraction is calculated by the formula as,

    $x=\frac{n}{n_{\text{total}}}$        (1)

The total number of moles of the mixture is calculated as,

    $\begin{array}{c}{n}_{\text{total}}=n_{\text{Hexane}}+n_{\text{Heptane}}\\ =1+1\\ =2\end{array}$

Substitute the values in the equation (1) to calculate the mole fraction of hexane as,

    $\begin{array}{c}x=\frac{n}{n_{\text{total}}}\\ =\frac{1}{2}\\ =\underset{\_}{0.50}\end{array}$

The mole fraction of hexane in the liquid phase is taken by drawing a parallel line on the liquid curve of the phase diagram which is shown by the line F’G’ in phase diagram below.

Figure 8

The mole fraction of hexane in the liquid phase is found out to be $0.8$ from the phase diagram.

Thus, mole fraction of hexane in the liquid and vapour phase for the conditions of part c is $\underset{\_}{0.8}$ and $\underset{\_}{0.5}$ respectively.

(f)

Interpretation Introduction

Interpretation:

The amounts of substances in the liquid and vapour phase at $85°\text{C}$ and $760\text{Torr}$ having $z_{\text{heptane}}=0.40$ have to be calculated.

Concept introduction:

The azeotropic mixture is defined as a mixture of two or more liquids which evaporates without altering the composition of the mixture.  The azeotropic mixture has a constant boiling point because of the same composition in liquid as well as vapor phase.  The components of azeotropic mixture are not separated by distillation processes.

The phase diagram represents the changes in the mixture on changing the parameters like temperature and pressure of the mixture.

Answer to Problem 5C.5P

The mole fraction of hexane in the liquid and vapour phase at $85°\text{C}$ and $760\text{Torr}$ is $\underset{\_}{0.8}$ and $\underset{\_}{0.3}$ respectively and the mole fraction of heptane in the liquid and vapour phase at $85°\text{C}$ and $760\text{Torr}$ is $\underset{\_}{0.20}$ and $\underset{\_}{0.70}$ respectively.

Explanation of Solution

The overall mole fraction of heptane present in the mixture is given as $0.40$.

Thus, the overall mole fraction of hexane present in the mixture is calculated as,

    $\begin{array}{c}{x}_{\text{Hexane}}=1-x_{\text{Heptane}}\\ =1-0.40\\ =0.60\end{array}$

The mole fraction of hexane in the liquid phase is taken by drawing a line passing through $x_{\text{Hexane}}=0.60$.  This line is then extrapolated onto the liquid curve and vapour curve (RS) as shown in the phase diagram below.

Figure 9

The mole fraction of hexane in the liquid phase is found out to be approximately $\underset{\_}{0.8}$ from the phase diagram.

Thus, the mole fraction of heptane in the liquid phase is calculated as,

    $\begin{array}{c}{x}_{\text{Heptane}}=1-x_{\text{Hexane}}\\ =1-0.80\\ =\underset{\_}{0.20}\end{array}$

The mole fraction of hexane in the vapour phase is found out to be $\underset{\_}{0.3}$ from the phase diagram.

Thus, the mole fraction of heptane in the vapour phase is calculated as,

    $\begin{array}{c}{y}_{\text{Heptane}}=1-y_{\text{Hexane}}\\ =1-0.30\\ =\underset{\_}{0.70}\end{array}$

Thus, mole fraction of hexane in the liquid and vapour phase at $85°\text{C}$ and $760\text{Torr}$ is $\underset{\_}{0.8}$ and $\underset{\_}{0.3}$ respectively and the mole fraction of heptane in the liquid and vapour phase at $85°\text{C}$ and $760\text{Torr}$ is $\underset{\_}{0.20}$ and $\underset{\_}{0.70}$ respectively.