Chapter 12, Problem 12.36P

(a)

Interpretation Introduction

Interpretation:

For the given binary mixture, whether one or two liquid phases are present is to be determined. Also, their composition is to be calculated if two phases are present.

Concept Introduction:

The general equation for $G^E/RT$ to predict liquid-liquid equilibrium is

  $\frac{G^E}{RT}=x_1{x}_2\left(A_{21}{x}_1+A_{12}{x}_2\right)$ ..... (1)

Here, $A_{12}\text{ and }{A}_{21}$ are parameters.

The relationship for ${\gamma }_1\text{ and }{\gamma }_2$ deduced from the above equation of $G^E/RT$ are

  $\begin{array}{c}\ln {\gamma }_1=\left[A_{12}+2\left(A_{21}-A_{12}\right)x_1\right]x_2^2\\ \ln {\gamma }_2=\left[A_{21}+2\left(A_{12}-A_{21}\right)x_2\right]x_1^2\end{array}$ ..... (2)

For liquid-liquid equilibrium where two phases, $\alpha \text{ and }\beta$ exists, the relationship between $x_1^{\alpha },x_1^{\beta },{\gamma }_1^{\alpha }\text{ and }{\gamma }_1^{\beta }$ is

  $\ln \left(\frac{{\gamma }_1^{\alpha }}{{\gamma }_1^{\beta }}\right)=\ln \left(\frac{x_1^{\beta }}{x_1^{\alpha }}\right)$ ..... (3)

Also, the relationship between $x_1^{\alpha },x_1^{\beta },{\gamma }_2^{\alpha }\text{ and }{\gamma }_2^{\beta }$ is

  $\ln \left(\frac{{\gamma }_2^{\alpha }}{{\gamma }_2^{\beta }}\right)=\ln \left(\frac{1-x_1^{\beta }}{1-x_1^{\alpha }}\right)$ ..... (4)

Answer to Problem 12.36P

Two phases are present in the given system of binary mixture with phase composition as:

  $\begin{array}{c}{x}_1^{\alpha }=0.095\\ x_1^{\beta }=0.105\end{array}$

Explanation of Solution

Given information:

Excess Gibbs energy for a binary liquid mixture is given by,

  $\frac{G^E}{RT}=2.1x_1{x}_2\left(x_1+2x_2\right)$

Overall composition of the system is given as $z_1=0.2$ .

Rewrite the given equation of $G^E/RT$ as

  $\begin{array}{c}\frac{G^E}{RT}=2.1x_1{x}_2\left(x_1+2x_2\right)\\ \frac{G^E}{RT}=x_1{x}_2\left(2.1x_1+4.2x_2\right)\end{array}$

Compare this equation by equation (1) so that the value of $A_{12}\text{ and }{A}_{21}$ are:

  $\begin{array}{c}{A}_{12}=4.2\\ A_{21}=2.1\end{array}$

Let, the binary mixture contains two phases of liquid and the system is in liquid-liquid equilibrium. Now, use equations set (2) along with equations (3) and (4) to eliminate ${\gamma }_1^{\alpha }\text{, }{\gamma }_1^{\beta },{\gamma }_2^{\alpha }\text{ and }{\gamma }_2^{\beta }$ and substitute the value of $A_{12}\text{ and }{A}_{21}$ as

  $\begin{array}{c}4.2\left({\left(1-x_1^{\alpha }\right)}^3-{\left(1-x_1^{\beta }\right)}^3\right)=\ln \left(\frac{x_1^{\beta }}{x_1^{\alpha }}\right)\mathrm{......}\text{ (6)}\\ 6.3\left({\left(x_1^{\alpha }\right)}^2-{\left(x_1^{\beta }\right)}^2\right)-4.2\left({\left(x_1^{\alpha }\right)}^3-{\left(x_1^{\beta }\right)}^3\right)=\ln \left(\frac{1-x_1^{\beta }}{1-x_1^{\alpha }}\right)\mathrm{......}\text{ (7)}\end{array}$

The value of $x_1^{\alpha }\text{ and }{x}_1^{\beta }$ which satisfy the above equations and lie between $0-0.2$ as the overall composition of the system is $0.2$ are:

  $\begin{array}{c}{x}_1^{\alpha }=0.095\\ x_1^{\beta }=0.105\end{array}$

At this point, there exist equilibrium between two phases for the given system.

Therefore, the assumption that the system is a two-phase system is correct and two phases are present.

(b)

Interpretation Introduction

Interpretation:

For the given binary mixture, whether one or two liquid phases are present is to be determined. Also, their composition is to be calculated if two phases are present.

Concept Introduction:

The general equation for $G^E/RT$ to predict liquid-liquid equilibrium is:

  $\frac{G^E}{RT}=x_1{x}_2\left(A_{21}{x}_1+A_{12}{x}_2\right)$ ..... (1)

Here, $A_{12}\text{ and }{A}_{21}$ are parameters.

The relationship for ${\gamma }_1\text{ and }{\gamma }_2$ deduced from the above equation of $G^E/RT$ are:

  $\begin{array}{c}\ln {\gamma }_1=\left[A_{12}+2\left(A_{21}-A_{12}\right)x_1\right]x_2^2\\ \ln {\gamma }_2=\left[A_{21}+2\left(A_{12}-A_{21}\right)x_2\right]x_1^2\end{array}$ ..... (2)

For liquid-liquid equilibrium where two phases, $\alpha \text{ and }\beta$ exists, the relationship between $x_1^{\alpha },x_1^{\beta },{\gamma }_1^{\alpha }\text{ and }{\gamma }_1^{\beta }$ is:

  $\ln \left(\frac{{\gamma }_1^{\alpha }}{{\gamma }_1^{\beta }}\right)=\ln \left(\frac{x_1^{\beta }}{x_1^{\alpha }}\right)$ ..... (3)

Also, the relationship between $x_1^{\alpha },x_1^{\beta },{\gamma }_2^{\alpha }\text{ and }{\gamma }_2^{\beta }$ is:

  $\ln \left(\frac{{\gamma }_2^{\alpha }}{{\gamma }_2^{\beta }}\right)=\ln \left(\frac{1-x_1^{\beta }}{1-x_1^{\alpha }}\right)$ ..... (4)

Answer to Problem 12.36P

Two phases are present in the given system of binary mixture with phase composition as:

  $\begin{array}{c}{x}_1^{\alpha }=0.095\\ x_1^{\beta }=0.205\end{array}$

Explanation of Solution

Given information:

Excess Gibbs energy for a binary liquid mixture is given by,

  $\frac{G^E}{RT}=2.1x_1{x}_2\left(x_1+2x_2\right)$

Overall composition of the system is given as $z_1=0.3$ .

Rewrite the given equation of $G^E/RT$ as:

  $\begin{array}{c}\frac{G^E}{RT}=2.1x_1{x}_2\left(x_1+2x_2\right)\\ \frac{G^E}{RT}=x_1{x}_2\left(2.1x_1+4.2x_2\right)\end{array}$

Compare this equation by equation (1) so that the value of $A_{12}\text{ and }{A}_{21}$ are:

  $\begin{array}{c}{A}_{12}=4.2\\ A_{21}=2.1\end{array}$

Let, the binary mixture contains two phases of liquid and the system is in liquid-liquid equilibrium. Now, use equations set (2) along with equations (3) and (4) to eliminate ${\gamma }_1^{\alpha }\text{, }{\gamma }_1^{\beta },{\gamma }_2^{\alpha }\text{ and }{\gamma }_2^{\beta }$ and substitute the value of $A_{12}\text{ and }{A}_{21}$ as

  $\begin{array}{c}4.2\left({\left(1-x_1^{\alpha }\right)}^3-{\left(1-x_1^{\beta }\right)}^3\right)=\ln \left(\frac{x_1^{\beta }}{x_1^{\alpha }}\right)\mathrm{......}\text{ (6)}\\ 6.3\left({\left(x_1^{\alpha }\right)}^2-{\left(x_1^{\beta }\right)}^2\right)-4.2\left({\left(x_1^{\alpha }\right)}^3-{\left(x_1^{\beta }\right)}^3\right)=\ln \left(\frac{1-x_1^{\beta }}{1-x_1^{\alpha }}\right)\mathrm{......}\text{ (7)}\end{array}$

The value of $x_1^{\alpha }\text{ and }{x}_1^{\beta }$ which satisfy the above equations and lie between $0-0.3$ as the overall composition of the system is $0.3$ are:

  $\begin{array}{c}{x}_1^{\alpha }=0.095\\ x_1^{\beta }=0.205\end{array}$

At this point, there exist equilibrium between two phases for the given system.

Therefore, the assumption that the system is a two-phase system is correct and two phases are present.

(c)

Interpretation Introduction

Interpretation:

For the given binary mixture, whether one or two liquid phases are present is to be determined. Also, their composition is to be calculated if two phases are present.

Concept Introduction:

The general equation for $G^E/RT$ to predict liquid-liquid equilibrium is

  $\frac{G^E}{RT}=x_1{x}_2\left(A_{21}{x}_1+A_{12}{x}_2\right)$ ..... (1)

Here, $A_{12}\text{ and }{A}_{21}$ are parameters.

The relationship for ${\gamma }_1\text{ and }{\gamma }_2$ deduced from the above equation of $G^E/RT$ are:

  $\begin{array}{c}\ln {\gamma }_1=\left[A_{12}+2\left(A_{21}-A_{12}\right)x_1\right]x_2^2\\ \ln {\gamma }_2=\left[A_{21}+2\left(A_{12}-A_{21}\right)x_2\right]x_1^2\end{array}$ ..... (2)

For liquid-liquid equilibrium where two phases, $\alpha \text{ and }\beta$ exists, the relationship between $x_1^{\alpha },x_1^{\beta },{\gamma }_1^{\alpha }\text{ and }{\gamma }_1^{\beta }$ is

  $\ln \left(\frac{{\gamma }_1^{\alpha }}{{\gamma }_1^{\beta }}\right)=\ln \left(\frac{x_1^{\beta }}{x_1^{\alpha }}\right)$ ..... (3)

Also, the relationship between $x_1^{\alpha },x_1^{\beta },{\gamma }_2^{\alpha }\text{ and }{\gamma }_2^{\beta }$ is:

  $\ln \left(\frac{{\gamma }_2^{\alpha }}{{\gamma }_2^{\beta }}\right)=\ln \left(\frac{1-x_1^{\beta }}{1-x_1^{\alpha }}\right)$ ..... (4)

Answer to Problem 12.36P

Two phases are present in the given system of binary mixture with phase composition as:

  $\begin{array}{c}{x}_1^{\alpha }=0.21\\ x_1^{\beta }=0.29\end{array}$

Explanation of Solution

Given information:

Excess Gibbs energy for a binary liquid mixture is given by,

  $\frac{G^E}{RT}=2.1x_1{x}_2\left(x_1+2x_2\right)$

Overall composition of the system is given as $z_1=0.5$ .

Rewrite the given equation of $G^E/RT$ as:

  $\begin{array}{c}\frac{G^E}{RT}=2.1x_1{x}_2\left(x_1+2x_2\right)\\ \frac{G^E}{RT}=x_1{x}_2\left(2.1x_1+4.2x_2\right)\end{array}$

Compare this equation by equation (1) so that the value of $A_{12}\text{ and }{A}_{21}$ are:

  $\begin{array}{c}{A}_{12}=4.2\\ A_{21}=2.1\end{array}$

Let, the binary mixture contains two phases of liquid and the system is in liquid-liquid equilibrium. Now, use equations set (2) along with equations (3) and (4) to eliminate ${\gamma }_1^{\alpha }\text{, }{\gamma }_1^{\beta },{\gamma }_2^{\alpha }\text{ and }{\gamma }_2^{\beta }$ and substitute the value of $A_{12}\text{ and }{A}_{21}$ as

  $\begin{array}{c}4.2\left({\left(1-x_1^{\alpha }\right)}^3-{\left(1-x_1^{\beta }\right)}^3\right)=\ln \left(\frac{x_1^{\beta }}{x_1^{\alpha }}\right)\mathrm{......}\text{ (6)}\\ 6.3\left({\left(x_1^{\alpha }\right)}^2-{\left(x_1^{\beta }\right)}^2\right)-4.2\left({\left(x_1^{\alpha }\right)}^3-{\left(x_1^{\beta }\right)}^3\right)=\ln \left(\frac{1-x_1^{\beta }}{1-x_1^{\alpha }}\right)\mathrm{......}\text{ (7)}\end{array}$

The value of $x_1^{\alpha }\text{ and }{x}_1^{\beta }$ which satisfy the above equations and lie between $0-0.5$ as the overall composition of the system is $0.5$ are:

  $\begin{array}{c}{x}_1^{\alpha }=0.21\\ x_1^{\beta }=0.29\end{array}$

At this point, there exist equilibrium between two phases for the given system.

Therefore, the assumption that the system is a two-phase system is correct and two phases are present.

(d)

Interpretation Introduction

Interpretation:

For the given binary mixture, whether one or two liquid phases are present is to be determined. Also, their composition is to be calculated if two phases are present.

Concept Introduction:

The general equation for $G^E/RT$ to predict liquid-liquid equilibrium is

  $\frac{G^E}{RT}=x_1{x}_2\left(A_{21}{x}_1+A_{12}{x}_2\right)$ ..... (1)

Here, $A_{12}\text{ and }{A}_{21}$ are parameters.

The relationship for ${\gamma }_1\text{ and }{\gamma }_2$ deduced from the above equation of $G^E/RT$ are:

  $\begin{array}{c}\ln {\gamma }_1=\left[A_{12}+2\left(A_{21}-A_{12}\right)x_1\right]x_2^2\\ \ln {\gamma }_2=\left[A_{21}+2\left(A_{12}-A_{21}\right)x_2\right]x_1^2\end{array}$ ..... (2)

For liquid-liquid equilibrium where two phases, $\alpha \text{ and }\beta$ exists, the relationship between $x_1^{\alpha },x_1^{\beta },{\gamma }_1^{\alpha }\text{ and }{\gamma }_1^{\beta }$ is:

  $\ln \left(\frac{{\gamma }_1^{\alpha }}{{\gamma }_1^{\beta }}\right)=\ln \left(\frac{x_1^{\beta }}{x_1^{\alpha }}\right)$ ..... (3)

Also, the relationship between $x_1^{\alpha },x_1^{\beta },{\gamma }_2^{\alpha }\text{ and }{\gamma }_2^{\beta }$ is:

  $\ln \left(\frac{{\gamma }_2^{\alpha }}{{\gamma }_2^{\beta }}\right)=\ln \left(\frac{1-x_1^{\beta }}{1-x_1^{\alpha }}\right)$ ..... (4)

Answer to Problem 12.36P

Two phases are present in the given system of binary mixture with phase composition as:

  $\begin{array}{c}{x}_1^{\alpha }=0.65\\ x_1^{\beta }=0.05\end{array}$

Explanation of Solution

Given information:

Excess Gibbs energy for a binary liquid mixture is given by

  $\frac{G^E}{RT}=2.1x_1{x}_2\left(x_1+2x_2\right)$

Overall composition of the system is given as $z_1=0.7$ .

Rewrite the given equation of $G^E/RT$ as:

  $\begin{array}{c}\frac{G^E}{RT}=2.1x_1{x}_2\left(x_1+2x_2\right)\\ \frac{G^E}{RT}=x_1{x}_2\left(2.1x_1+4.2x_2\right)\end{array}$

Compare this equation by equation (1) so that the value of $A_{12}\text{ and }{A}_{21}$ are:

  $\begin{array}{c}{A}_{12}=4.2\\ A_{21}=2.1\end{array}$

Let, the binary mixture contains two phases of liquid and the system is in liquid-liquid equilibrium. Now, use equations set (2) along with equations (3) and (4) to eliminate ${\gamma }_1^{\alpha }\text{, }{\gamma }_1^{\beta },{\gamma }_2^{\alpha }\text{ and }{\gamma }_2^{\beta }$ and substitute the value of $A_{12}\text{ and }{A}_{21}$ as:

  $\begin{array}{c}4.2\left({\left(1-x_1^{\alpha }\right)}^3-{\left(1-x_1^{\beta }\right)}^3\right)=\ln \left(\frac{x_1^{\beta }}{x_1^{\alpha }}\right)\mathrm{......}\text{ (6)}\\ 6.3\left({\left(x_1^{\alpha }\right)}^2-{\left(x_1^{\beta }\right)}^2\right)-4.2\left({\left(x_1^{\alpha }\right)}^3-{\left(x_1^{\beta }\right)}^3\right)=\ln \left(\frac{1-x_1^{\beta }}{1-x_1^{\alpha }}\right)\mathrm{......}\text{ (7)}\end{array}$

The value of $x_1^{\alpha }\text{ and }{x}_1^{\beta }$ which satisfy the above equations and lie between $0-0.7$ as the overall composition of the system is $0.7$ are:

  $\begin{array}{c}{x}_1^{\alpha }=0.65\\ x_1^{\beta }=0.05\end{array}$

At this point, there exist equilibrium between two phases for the given system.

Therefore, the assumption that the system is a two-phase system is correct and two phases are present.

(e)

Interpretation Introduction

Interpretation:

For the given binary mixture, whether one or two liquid phases are present is to be determined. Also, their composition is to be calculated if two phases are present.

Concept Introduction:

The general equation for $G^E/RT$ to predict liquid-liquid equilibrium is

  $\frac{G^E}{RT}=x_1{x}_2\left(A_{21}{x}_1+A_{12}{x}_2\right)$ ..... (1)

Here, $A_{12}\text{ and }{A}_{21}$ are parameters.

The relationship for ${\gamma }_1\text{ and }{\gamma }_2$ deduced from the above equation of $G^E/RT$ are:

  $\begin{array}{c}\ln {\gamma }_1=\left[A_{12}+2\left(A_{21}-A_{12}\right)x_1\right]x_2^2\\ \ln {\gamma }_2=\left[A_{21}+2\left(A_{12}-A_{21}\right)x_2\right]x_1^2\end{array}$ ..... (2)

For liquid-liquid equilibrium where two phases, $\alpha \text{ and }\beta$ exists, the relationship between $x_1^{\alpha },x_1^{\beta },{\gamma }_1^{\alpha }\text{ and }{\gamma }_1^{\beta }$ is

  $\ln \left(\frac{{\gamma }_1^{\alpha }}{{\gamma }_1^{\beta }}\right)=\ln \left(\frac{x_1^{\beta }}{x_1^{\alpha }}\right)$ ..... (3)

Also, the relationship between $x_1^{\alpha },x_1^{\beta },{\gamma }_2^{\alpha }\text{ and }{\gamma }_2^{\beta }$ is

  $\ln \left(\frac{{\gamma }_2^{\alpha }}{{\gamma }_2^{\beta }}\right)=\ln \left(\frac{1-x_1^{\beta }}{1-x_1^{\alpha }}\right)$ ..... (4)

Answer to Problem 12.36P

Two phases are present in the given system of binary mixture with phase composition as:

  $\begin{array}{c}{x}_1^{\alpha }=0.65\\ x_1^{\beta }=0.15\end{array}$

Explanation of Solution

Given information:

Excess Gibbs energy for a binary liquid mixture is given by

  $\frac{G^E}{RT}=2.1x_1{x}_2\left(x_1+2x_2\right)$

Overall composition of the system is given as $z_1=0.8$ .

Rewrite the given equation of $G^E/RT$ as:

  $\begin{array}{c}\frac{G^E}{RT}=2.1x_1{x}_2\left(x_1+2x_2\right)\\ \frac{G^E}{RT}=x_1{x}_2\left(2.1x_1+4.2x_2\right)\end{array}$

Compare this equation by equation (1) so that the value of $A_{12}\text{ and }{A}_{21}$ are:

  $\begin{array}{c}{A}_{12}=4.2\\ A_{21}=2.1\end{array}$

Let, the binary mixture contains two phases of liquid and the system is in liquid-liquid equilibrium. Now, use equations set (2) along with equations (3) and (4) to eliminate ${\gamma }_1^{\alpha }\text{, }{\gamma }_1^{\beta },{\gamma }_2^{\alpha }\text{ and }{\gamma }_2^{\beta }$ and substitute the value of $A_{12}\text{ and }{A}_{21}$ as:

  $\begin{array}{c}4.2\left({\left(1-x_1^{\alpha }\right)}^3-{\left(1-x_1^{\beta }\right)}^3\right)=\ln \left(\frac{x_1^{\beta }}{x_1^{\alpha }}\right)\mathrm{......}\text{ (6)}\\ 6.3\left({\left(x_1^{\alpha }\right)}^2-{\left(x_1^{\beta }\right)}^2\right)-4.2\left({\left(x_1^{\alpha }\right)}^3-{\left(x_1^{\beta }\right)}^3\right)=\ln \left(\frac{1-x_1^{\beta }}{1-x_1^{\alpha }}\right)\mathrm{......}\text{ (7)}\end{array}$

The value of $x_1^{\alpha }\text{ and }{x}_1^{\beta }$ which satisfy the above equations and lie between $0-0.8$ as the overall composition of the system is $0.8$ are:

  $\begin{array}{c}{x}_1^{\alpha }=0.65\\ x_1^{\beta }=0.15\end{array}$

At this point, there exist equilibrium between two phases for the given system.

Therefore, the assumption that the system is a two-phase system is correct and two phases are present.