Chapter 11, Problem 11.1P

Interpretation Introduction

Interpretation:

The partial molar volume ${\overline{V}}_1$ and ${\overline{V}}_2$ in a mixture containing 40 mole % of species 1 at the given conditions should be calculated.

Concept Introduction:

$\Delta V$ will be calculated after putting $x_1$ and $x_2$ value in the given expression. Since $x_1$ is given, $x_2$ will be calculated from formula : $x_1+x_2=1$ for binary mixture.

  $V^E=V-{{\displaystyle \sum }}^{\text{}}{x}_i{V}_i$ and $V^E=\Delta V$

Since $\Delta V$, $V_1$, $V_2$, $x_1$ and $x_2$ are known, $V$ can be calculated.

  $\overline{V_1}=V+x_2\frac{\text{d}V}{\text{d}{x}_1}$ and $\overline{V_2}=V-x_1\frac{\text{d}V}{\text{d}{x}_1}$

Absolute value of $V$ is known and it can be expressed as function of variable $x_1$ using the above two equations. Therefore, absolute value of $\frac{\text{d}V}{\text{d}{x}_1}$ can be calculated by differentiation with respect to $x_1$ . So, we can get value of ${\overline{V}}_1$ and ${\overline{V}}_2$ .

Answer to Problem 11.1P

Partial molar volume ${\overline{V}}_1$ =124.76 cm³/mol

Partial molar volume ${\overline{V}}_2$ =93.36 cm³/mol $$

Explanation of Solution

Given Information:

At 25⁰C and atmospheric pressure the volume change of mixing of binary liquid mixtures of species 1 and 2 is given by the following equation:

  $\Delta V=x_1{x}_2\left(45x_1+25x_2\right)$

Where $\Delta V$ is in cm³.mol-1. At this condition $V_1$ = 110 and $V_2$ = 90 cm³.mol-1.

Mole fraction of species 1 ( $x_1$ ) is 0.4.

Calculation:

  $\Delta V$ calculation:

  $x_1=0.4$ (given)

  $x_2=1-0.4=0.6$

  $\Delta V=0.4\times 0.6\left(45\times 0.4+25\times 0.6\right)=7.92c{m}^3/mol$

  $V$ (total molar volume of solution) calculation:

  $\begin{array}{c}V=\Delta V+x_1{V}_1+x_2{V}_2\\ =7.92c{m}^3/mol+0.4\times 110c{m}^3/mol+0.6\times 90c{m}^3/mol\\ =105.92c{m}^3/mol\end{array}$

Calculating $\overline{V_1}$

$\begin{array}{c}\overline{V_1}=V+x_2\frac{\text{d}V}{\text{d}{x}_1}\\ =105.92c{m}^3/mol+x_2\times \frac{\text{d(}{x}_1{V}_1+x_2{V}_2+x_1{x}_2\times (45x_1+25x_2))}{\text{d}{x}_1}c{m}^3/mol\\ =105.92c{m}^3/mol+0.6\times \text{d(110}{x}_1+90(1-x_1)+x(1-x_1)\times (45x_1+25(1-x_1)))/\text{d}{x}_{1}c{m}^3/mol\\ =105.92c{m}^3/mol+0.6\times (45-10x_1-60x_1{}^2)c{m}^3/mol\\ =105.92c{m}^3/mol+0.6\times (45-10\times 0.4-60\times {0.4}^2)c{m}^3/mol\\ =124.76c{m}^3/mol\end{array}$

Calculating ${\overline{V}}_2$

  $\overline{V_2}=V-x_1\frac{\text{d}V}{\text{d}{x}_1}$

  $\begin{array}{l}=105.92c{m}^3/mol-x_1\times \frac{\text{d(}{x}_1{V}_1+x_2{V}_2+x_1{x}_2\times (45x_1+25x_2))}{\text{d}{x}_1}c{m}^3/mol\\ =105.92c{m}^3/mol-0.4\times \text{d(110}{x}_1+90(1-x_1)+x(1-x_1)\times (45x_1+25(1-x_1)))/\text{d}{x}_{1}c{m}^3/mol\\ =105.92c{m}^3/mol-0.4\times (45-10x_1-60x_1{}^2)c{m}^3/mol\\ =105.92c{m}^3/mol-0.4\times (45-10\times 0.4-60\times {0.4}^2)c{m}^3/mol\\ =93.36c{m}^3/mol\end{array}$

Conclusion

Thus, partial molar volume ${\overline{V}}_1$ =124.76 cm³/mol

Partial molar volume ${\overline{V}}_2$ =93.36 cm³/mol $$