Chapter 15, Problem 15.2P

(a)

Interpretation Introduction

Interpretation:

To determine sets of miscibility data estimates for A₁₂ and A₂₁ in the Margules Equation

Concept Introduction:

Margules Equations are given as,

${\gamma }_1^{\alpha }=\exp \left\{x_2^{\alpha 2}\times \left[A_{12}+2\times \left(A_{21}-A_{12}\right)\times x_1^{\alpha }\right]\right\}$

${\gamma }_1^{\beta }=\exp \left\{x_2^{\beta 2}\times \left[A_{12}+2\times \left(A_{21}-A_{12}\right)\times x_1^{\beta }\right]\right\}$

  $\begin{array}{l}{x}_2^{\alpha }=1-x_1^{\alpha }\\ x_2^{\beta }=1-x_1^{\beta }\end{array}$

Margules Equations for $x_2^{\alpha },x_2^{\beta }$ are given as,

${\gamma }_2^{\alpha }=\exp \left\{x_1^{\alpha 2}\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times x_2^{\alpha }\right]\right\}$

${\gamma }_2^{\beta }=\exp \left\{x_1^{\beta 2}\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times x_2^{\beta }\right]\right\}$

Answer to Problem 15.2P

A₂₁ = 2.747

A₁₂ = 2.747

Explanation of Solution

  $\begin{array}{l}{x}_1^{\alpha }=0.1\\ x_2^{\alpha }=1-x_1^{\alpha }=0.9\\ x_1^{\beta }=0.9\\ x_2^{\beta }=1-x_1^{\beta }=0.1\end{array}$

Now let’s guess: A₁₂ = 2

and

A₂₁ = 2

${\gamma }_1^{\alpha }=\exp \left\{x_2^{\alpha 2}\times \left[A_{12}+2\times \left(A_{21}-A_{12}\right)\times x_1^{\alpha }\right]\right\}$

${\gamma }_1^{\beta }=\exp \left\{x_2^{\beta 2}\times \left[A_{12}+2\times \left(A_{21}-A_{12}\right)\times x_1^{\beta }\right]\right\}$

${\gamma }_2^{\alpha }=\exp \left\{x_1^{\alpha 2}\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times x_2^{\alpha }\right]\right\}$

${\gamma }_2^{\beta }=\exp \left\{x_1^{\beta 2}\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times x_2^{\beta }\right]\right\}$

As given below:

$x_1^{\alpha }\times {\gamma }_1^{\alpha }=x_1^{\beta }\times {\gamma }_1^{\beta }$

$x_1^{\alpha }\exp \left\{x_2^{\alpha 2}\times \left[A_{12}+2\times \left(A_{21}-A_{12}\right)\times x_1^{\alpha }\right]\right\}=x_1^{\beta }\exp \left\{x_2^{\beta 2}\times \left[A_{12}+2\times \left(A_{21}-A_{12}\right)\times x_1^{\beta }\right]\right\}$

$0.1\exp \left\{{0.9}^2\times \left[A_{12}+2\times \left(A_{21}-A_{12}\right)\times 0.1\right]\right\}=0.9\exp \left\{{0.1}_{}^2\times \left[A_{12}+2\times \left(A_{21}-A_{12}\right)\times 0.9\right]\right\}$

$ln\left[\exp \left\{0.81\times A_{12}+0.162\times \left(A_{21}-A_{12}\right)\right\}\right]=\ln \left[9\exp \left\{0.01A_{12}+0.018\times \left(A_{21}-A_{12}\right)\right\}\right]$

$0.81\times A_{12}+0.162A_{21}-0.162A_{12}=\ln \left[9\right]+0.01A_{12}+0.018A_{21}-0.018A_{12}$

$0.656A_{12}+0.144A_{21}=\ln \left[9\right]$

... Equation (A)

${\gamma }_2^{\alpha }=\exp \left\{x_1^{\alpha 2}\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times x_2^{\alpha }\right]\right\}$

${\gamma }_2^{\beta }=\exp \left\{x_1^{\beta 2}\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times x_2^{\beta }\right]\right\}$

$x_2^{\alpha }\times {\gamma }_2^{\alpha }=x_2^{\beta }\times {\gamma }_2^{\beta }$

$x_2^{\alpha }\exp \left\{x_1^{\alpha 2}\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times x_2^{\alpha }\right]\right\}=x_2^{\beta }\exp \left\{x_1^{\beta 2}\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times x_2^{\beta }\right]\right\}$

$0.9\exp \left\{{0.1}^2\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times 0.9\right]\right\}=0.1\exp \left\{{0.9}_{}^2\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times 0.1\right]\right\}$

$ln\left[9\exp \left\{0.01\times A_{21}+0.018\times \left(A_{12}-A_{21}\right)\right\}\right]=\ln \left[\exp \left\{0.81A_{21}+0.162\times \left(A_{12}-A_{21}\right)\right\}\right]$

$\ln \left[9\right]+0.01\times A_{21}+0.018A_{12}-0.018A_{21}=0.81A_{21}+0.162A_{12}-0.162A_{21}$

$\ln \left[9\right]=0.144A_{12}+0.656A_{21}$

...Equation (B)

Now solving Equation (A) and (B), we get

A₂₁ = 2.747

A₁₂ = 2.747

(b)

Interpretation Introduction

Interpretation:

To determine sets of miscibility data estimates for A₁₂ and A₂₁ in the Margules Equation

Concept Introduction:

Margules Equations are given as,

${\gamma }_1^{\alpha }=\exp \left\{x_2^{\alpha 2}\times \left[A_{12}+2\times \left(A_{21}-A_{12}\right)\times x_1^{\alpha }\right]\right\}$

${\gamma }_1^{\beta }=\exp \left\{x_2^{\beta 2}\times \left[A_{12}+2\times \left(A_{21}-A_{12}\right)\times x_1^{\beta }\right]\right\}$

  $\begin{array}{l}{x}_2^{\alpha }=1-x_1^{\alpha }\\ x_2^{\beta }=1-x_1^{\beta }\end{array}$

Margules Equations for $x_2^{\alpha },x_2^{\beta }$ are given as,

${\gamma }_2^{\alpha }=\exp \left\{x_1^{\alpha 2}\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times x_2^{\alpha }\right]\right\}$

${\gamma }_2^{\beta }=\exp \left\{x_1^{\beta 2}\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times x_2^{\beta }\right]\right\}$

Answer to Problem 15.2P

A₁₂ = 2.148

A₂₁ = 2.781

Explanation of Solution

  $\begin{array}{l}{x}_1^{\alpha }=0.2\\ x_2^{\alpha }=1-x_1^{\alpha }=0.8\\ x_1^{\beta }=0.9\\ x_2^{\beta }=1-x_1^{\beta }=0.1\end{array}$

${\gamma }_1^{\alpha }=\exp \left\{x_2^{\alpha 2}\times \left[A_{12}+2\times \left(A_{21}-A_{12}\right)\times x_1^{\alpha }\right]\right\}$

${\gamma }_1^{\beta }=\exp \left\{x_2^{\beta 2}\times \left[A_{12}+2\times \left(A_{21}-A_{12}\right)\times x_1^{\beta }\right]\right\}$

${\gamma }_2^{\alpha }=\exp \left\{x_1^{\alpha 2}\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times x_2^{\alpha }\right]\right\}$

${\gamma }_2^{\beta }=\exp \left\{x_1^{\beta 2}\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times x_2^{\beta }\right]\right\}$

As given below:

$x_1^{\alpha }\times {\gamma }_1^{\alpha }=x_1^{\beta }\times {\gamma }_1^{\beta }$

$x_1^{\alpha }\exp \left\{x_2^{\alpha 2}\times \left[A_{12}+2\times \left(A_{21}-A_{12}\right)\times x_1^{\alpha }\right]\right\}=x_1^{\beta }\exp \left\{x_2^{\beta 2}\times \left[A_{12}+2\times \left(A_{21}-A_{12}\right)\times x_1^{\beta }\right]\right\}$

$0.2\exp \left\{{0.8}^2\times \left[A_{12}+2\times \left(A_{21}-A_{12}\right)\times 0.2\right]\right\}=0.9\exp \left\{{0.1}_{}^2\times \left[A_{12}+2\times \left(A_{21}-A_{12}\right)\times 0.9\right]\right\}$

$ln\left[\exp \left\{0.64\times A_{12}+0.256\times \left(A_{21}-A_{12}\right)\right\}\right]=\ln \left[4.5\exp \left\{0.01A_{12}+0.018\times \left(A_{21}-A_{12}\right)\right\}\right]$

$0.64\times A_{12}+0.256A_{21}-0.256A_{12}=\ln \left[4.5\right]+0.01A_{12}+0.018A_{21}-0.018A_{12}$

$0.392A_{12}+0.238A_{21}=\ln \left[4.5\right]$

... Equation (C)

${\gamma }_2^{\alpha }=\exp \left\{x_1^{\alpha 2}\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times x_2^{\alpha }\right]\right\}$

${\gamma }_2^{\beta }=\exp \left\{x_1^{\beta 2}\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times x_2^{\beta }\right]\right\}$

$x_2^{\alpha }\times {\gamma }_2^{\alpha }=x_2^{\beta }\times {\gamma }_2^{\beta }$

$x_2^{\alpha }\exp \left\{x_1^{\alpha 2}\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times x_2^{\alpha }\right]\right\}=x_2^{\beta }\exp \left\{x_1^{\beta 2}\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times x_2^{\beta }\right]\right\}$

$0.8\exp \left\{{0.2}^2\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times 0.8\right]\right\}=0.1\exp \left\{{0.9}_{}^2\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times 0.1\right]\right\}$

$ln\left[8\exp \left\{0.04\times A_{21}+0.064\times \left(A_{12}-A_{21}\right)\right\}\right]=\ln \left[\exp \left\{0.81A_{21}+0.162\times \left(A_{12}-A_{21}\right)\right\}\right]$

$\ln \left[8\right]+0.04\times A_{21}+0.064A_{12}-0.064A_{21}=0.81A_{21}+0.162A_{12}-0.162A_{21}$

$\ln \left[8\right]=0.098A_{12}+0.672A_{21}$

...Equation (D)

Now solving Equation (C) and (D), we get

A₁₂ = 2.148

A₂₁ = 2.781

(c)

Interpretation Introduction

Interpretation:

To determine sets of miscibility data estimates for A₁₂ and A₂₁ in the Margules Equation

Concept Introduction:

Margules Equations are given as,

${\gamma }_1^{\alpha }=\exp \left\{x_2^{\alpha 2}\times \left[A_{12}+2\times \left(A_{21}-A_{12}\right)\times x_1^{\alpha }\right]\right\}$

${\gamma }_1^{\beta }=\exp \left\{x_2^{\beta 2}\times \left[A_{12}+2\times \left(A_{21}-A_{12}\right)\times x_1^{\beta }\right]\right\}$

  $\begin{array}{l}{x}_2^{\alpha }=1-x_1^{\alpha }\\ x_2^{\beta }=1-x_1^{\beta }\end{array}$

Margules Equations for $x_2^{\alpha },x_2^{\beta }$ are given as,

${\gamma }_2^{\alpha }=\exp \left\{x_1^{\alpha 2}\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times x_2^{\alpha }\right]\right\}$

${\gamma }_2^{\beta }=\exp \left\{x_1^{\beta 2}\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times x_2^{\beta }\right]\right\}$

Answer to Problem 15.2P

A₁₂ = 2.781

A₂₁ = 2.148

Explanation of Solution

  $\begin{array}{l}{x}_1^{\alpha }=0.1\\ x_2^{\alpha }=1-x_1^{\alpha }=0.9\\ x_1^{\beta }=0.8\\ x_2^{\beta }=1-x_1^{\beta }=0.2\end{array}$

${\gamma }_1^{\alpha }=\exp \left\{x_2^{\alpha 2}\times \left[A_{12}+2\times \left(A_{21}-A_{12}\right)\times x_1^{\alpha }\right]\right\}$

${\gamma }_1^{\beta }=\exp \left\{x_2^{\beta 2}\times \left[A_{12}+2\times \left(A_{21}-A_{12}\right)\times x_1^{\beta }\right]\right\}$

${\gamma }_2^{\alpha }=\exp \left\{x_1^{\alpha 2}\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times x_2^{\alpha }\right]\right\}$

${\gamma }_2^{\beta }=\exp \left\{x_1^{\beta 2}\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times x_2^{\beta }\right]\right\}$

As given below:

$x_1^{\alpha }\times {\gamma }_1^{\alpha }=x_1^{\beta }\times {\gamma }_1^{\beta }$

$x_1^{\alpha }\exp \left\{x_2^{\alpha 2}\times \left[A_{12}+2\times \left(A_{21}-A_{12}\right)\times x_1^{\alpha }\right]\right\}=x_1^{\beta }\exp \left\{x_2^{\beta 2}\times \left[A_{12}+2\times \left(A_{21}-A_{12}\right)\times x_1^{\beta }\right]\right\}$

$0.1\exp \left\{{0.9}^2\times \left[A_{12}+2\times \left(A_{21}-A_{12}\right)\times 0.1\right]\right\}=0.8\exp \left\{{0.2}_{}^2\times \left[A_{12}+2\times \left(A_{21}-A_{12}\right)\times 0.8\right]\right\}$

$ln\left[\exp \left\{0.81\times A_{12}+0.162\times \left(A_{21}-A_{12}\right)\right\}\right]=\ln \left[8\exp \left\{0.04A_{12}+0.064\times \left(A_{21}-A_{12}\right)\right\}\right]$

$0.81\times A_{12}+0.162A_{21}-0.162A_{12}=\ln \left[8\right]+0.04A_{12}+0.064A_{21}-0.064A_{12}$

$0.672A_{12}+0.098A_{21}=\ln \left[8\right]$

... Equation (E)

${\gamma }_2^{\alpha }=\exp \left\{x_1^{\alpha 2}\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times x_2^{\alpha }\right]\right\}$

${\gamma }_2^{\beta }=\exp \left\{x_1^{\beta 2}\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times x_2^{\beta }\right]\right\}$

$x_2^{\alpha }\times {\gamma }_2^{\alpha }=x_2^{\beta }\times {\gamma }_2^{\beta }$

$x_2^{\alpha }\exp \left\{x_1^{\alpha 2}\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times x_2^{\alpha }\right]\right\}=x_2^{\beta }\exp \left\{x_1^{\beta 2}\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times x_2^{\beta }\right]\right\}$

$0.9\exp \left\{{0.1}^2\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times 0.9\right]\right\}=0.2\exp \left\{{0.8}_{}^2\times \left[A_{21}+2\times \left(A_{12}-A_{21}\right)\times 0.2\right]\right\}$

$ln\left[4.5\exp \left\{0.01\times A_{21}+0.018\times \left(A_{12}-A_{21}\right)\right\}\right]=\ln \left[\exp \left\{0.64A_{21}+0.256\times \left(A_{12}-A_{21}\right)\right\}\right]$

$\ln \left[4.5\right]+0.01\times A_{21}+0.018A_{12}-0.018A_{21}=0.64A_{21}+0.256A_{12}-0.256A_{21}$

$\ln \left[4.5\right]=0.392A_{12}+0.238A_{21}$

...Equation (F)

Now solving Equation (E) and (F), we get

A₁₂ = 2.781

A₂₁ = 2.148