Chapter 11, Problem 11.16P

Interpretation Introduction

Interpretation:

To show $\underset{n\to 0}{\lim }\widetilde{\Delta H}=0$

Explanation of Solution

Given Information:

  $\Delta H=x_1+x_2(A_{21}{x}_1+A_{12}{x}_2)$

$\text{ΔH}$ for particular solute (1)/solute (2) is represented by the equation:

  $\Delta H=x_1+x_2(A_{21}{x}_1+A_{12}{x}_2)\mathrm{.....}\text{ (}A)$

Rewrite the equation (A) using the definition of $\tilde{n}$ and $\text{ΔH}$

$\begin{array}{l}{x}_1=\frac{1}{1+\tilde{n}}\\ \Delta H=\Delta H(1+\tilde{n})\\ x_2=1-x_1\\ x_2=\frac{\tilde{n}}{1+\tilde{n}}\end{array}$

Now the equation (A) will become

  $\Delta H=\frac{\tilde{n}}{1+\tilde{n}}\left(\frac{A_{21}}{1+\tilde{n}}+\frac{A_{12}\tilde{n}}{1+\tilde{n}}\right)\mathrm{.........}(1)$

Now show that, $\underset{n\to 0}{\lim }\widetilde{\Delta H}=0$

Thus, Equation 1 will

become,

  $\begin{array}{l}\underset{n\to 0}{\lim }\widetilde{\Delta H}=\underset{n\to 0}{\lim }\left(\frac{\tilde{n}}{1+\tilde{n}}\left(\frac{A_{21}}{1+\tilde{n}}+\frac{A_{12}\tilde{n}}{1+\tilde{n}}\right)\right)\\ \text{ = }\frac{0}{1+0}\left(\frac{A_{21}}{1+0}+\frac{A_{12}0}{1+0}\right)\\ \underset{n\to 0}{\lim }\widetilde{\Delta H}=0\end{array}$

Hence, $\boxed{\underset{n\to 0}{\lim }\widetilde{\Delta H}=0}$

Interpretation Introduction

Interpretation:

To show $\underset{n\to \infty }{\lim }\widetilde{\Delta H}=A_{12}$

Explanation of Solution

Given Information:

  $\Delta H=x_1+x_2(A_{21}{x}_1+A_{12}{x}_2)$

$\text{ΔH}$ for particular solute (1)/solute (2) is represented by the equation:

  $\Delta H=x_1+x_2(A_{21}{x}_1+A_{12}{x}_2)\mathrm{.....}\text{ (}A)$

Rewrite the equation (A) using the definition of $\tilde{n}$ and $\text{ΔH}$

$\begin{array}{l}{x}_1=\frac{1}{1+\tilde{n}}\\ \Delta H=\Delta H(1+\tilde{n})\\ x_2=1-x_1\\ x_2=\frac{\tilde{n}}{1+\tilde{n}}\end{array}$

Now the equation (A) will become

  $\Delta H=\frac{\tilde{n}}{1+\tilde{n}}\left(\frac{A_{21}}{1+\tilde{n}}+\frac{A_{12}\tilde{n}}{1+\tilde{n}}\right)\mathrm{.........}(1)$

Now show that, $\underset{n\to \infty }{\lim }\widetilde{\Delta H}=A_{12}$

As $n\to \infty ,\frac{\tilde{n}}{1+\tilde{n}}\to 1$

Thus, Equation 1 will

become,

  $\begin{array}{l}\underset{\tilde{n}\to \infty }{\lim }\widetilde{\Delta H}=\underset{\tilde{n}\to \infty }{\lim }\left(\frac{\tilde{n}}{1+\tilde{n}}\left(\frac{A_{21}}{1+\tilde{n}}+\frac{A_{12}\tilde{n}}{1+\tilde{n}}\right)\right)\\ \text{ = }1\left(0+A_{12}\right)\\ \underset{\tilde{n}\to \infty }{\lim }\widetilde{\Delta H}=A_{12}\end{array}$

Hence, $\boxed{\underset{\tilde{n}\to \infty }{\lim }\widetilde{\Delta H}=A_{12}}$

Interpretation Introduction

Interpretation:

To show $\underset{\tilde{n}\to 0}{\lim }d\widetilde{\Delta H}/d\tilde{n}=A_{21}$

Explanation of Solution

Given Information:

  $\Delta H=x_1+x_2(A_{21}{x}_1+A_{12}{x}_2)$

Show that, $\underset{\tilde{n}\to 0}{\lim }d\widetilde{\Delta H}/d\tilde{n}=A_{21}$

As we know that

$d\Delta H/d\tilde{n}={\tilde{H}}_{{}_2}^E$

Thus,

  $\underset{\tilde{\tilde{n}}\to 0}{\lim }{\tilde{H}}_2^E=A_{21}$

Here, ${\tilde{H}}_2^E$ is defined as ${\tilde{H}}_2^E=x_1^2\left[A_{21}+2\left(A_{21}+2(A_{12}-A_{21})x_2\right)\right]$

Now,

substituting x₁and x₂expression in terms of $\tilde{n}$ in the above equation.

  $\begin{array}{l}{\tilde{H}}_2^E={\left(\frac{1}{1+\tilde{n}}\right)}^2\left[A_{21}+2\left(A_{21}+2(A_{12}-A_{21})\frac{\tilde{n}}{1+\tilde{n}}\right)\right]\\ \tilde{n}\to \text{0}\\ \underset{\tilde{n}\to \infty }{\lim {\tilde{H}}_2^E}=\underset{\tilde{n}\to 0}{\lim }\left({\left(\frac{1}{1+\tilde{n}}\right)}^2\left[A_{21}+2\left(A_{21}+2(A_{12}-A_{21})\frac{\tilde{n}}{1+\tilde{n}}\right)\right]\right)\\ =1^2\left(\left(A_{21}\right)+2\left(A_{12}-A_{21}\right)0\right)\\ \underset{\tilde{n}\to \infty }{\lim {\tilde{H}}_2^E}=A_{21}\\ \underset{\tilde{n}\to 0}{\lim }d\widetilde{\Delta H}/d\tilde{n}=A_{21}\end{array}$

Hence, $\boxed{\underset{\tilde{n}\to 0}{\lim }d\widetilde{\Delta H}/d\tilde{n}=A_{21}}$