Chapter 5, Problem 5B.3P

(a)

Interpretation Introduction

Interpretation:

The expression for the partial volume of each component at the given temperature has to be derived.

Concept Introduction:

The partial volume of the substance in a mixture is defined as the change in volume per mole of the substance added to the large volume of the mixture.  It is denoted by $V_{\text{J}}$.

Answer to Problem 5B.3P

The partial volume of propionic acid is,

    $V_{\text{J,1}}=V_{\text{m,1}}+a_{\text{o}}{x}_2^2+a_1\left(3x_1-x_2\right)x_2^2$

The partial volume of oxane is,

    $V_{\text{J,2}}=V_{\text{m,2}}+a_{\text{o}}{x}_1^2+a_1\left(3x_1-x_2\right)x_1^2$

Explanation of Solution

The Propionic acid is assumed as component 1 and THP is assumed as component 2.

The excess volume is given as,

    $V^{\text{E}}=x_1{x}_2\left\{a_{\text{o}}+a_1\left(x_1-x_2\right)\right\}\left(1\right)$

Where,

$x_1$ and $x_2$ are the mole fractions of components 1 and component 2, respectively.

$V^{\text{E}}$ is the excess volume.

$a_{\text{o}}$ and $a_1$ are the constants.

The volume of the ideal mixture is,

    ${\text{V}}_{\text{total}}{\text{=x}}_{\text{1}}{\text{V}}_{\text{m,1}}{\text{+x}}_{\text{2}}{\text{V}}_{\text{m,2}}$

Where,

$V_{\text{total}}$ is the total volume of the ideal mixture.

$x_1$ and $x_2$ are the mole fractions of components 1 and component 2, respectively.

$V_{\text{m,1}}$ and $V_{\text{m,2}}$ are the molar volumes of the component 1 and component 2, respectively.

Thus, the partial molar volume of the real mixture is given by the formula,

    ${\text{V}}_{\text{J}}{\text{=V}}_{\text{ideal}}{\text{+V}}_{\text{total}}^{\text{E}}\left(\text{2}\right)$

Where,

$V_{\text{J}}$ is the partial molar volume.

$V_{\text{ideal}}$ is the molar volume of the ideal mixture.

$V_{\text{total}}^{\text{E}}$ is the total excess volume.

The expression for $V_{\text{total}}^{\text{E}}$ is given as,

    $V_{\text{total}}^{\text{E}}=\left(x_1+x_2\right)V^{\text{E}}\left(3\right)$

The mole fraction of component 1 is,

    $x_1=\frac{n_1}{n_1+n_2}\left(4\right)$

Where,

• $n_1$ is the number of moles of component 1.
• $n_2$ is the number of moles of component 2.

The mole fraction of component 2 is,

    $x_2=\frac{n_2}{n_1+n_2}\left(5\right)$

Substitute the equation (1), (4) and (5) in equation (3).

    ${\text{V}}_{\text{total}}^{\text{E}}\text{=}\frac{{\text{n}}_{\text{1}}{\text{n}}_{\text{2}}}{\left({\text{n}}_{\text{1}}{\text{+n}}_{\text{2}}\right)}\left({\text{a}}_{\text{o}}\text{+}\frac{{\text{a}}_{\text{1}}\left({\text{n}}_{\text{1}}{\text{-n}}_{\text{2}}\right)}{{\text{n}}_{\text{1}}{\text{+n}}_{\text{2}}}\right)$

The value of $V_{\text{ideal}}$ is,

  ${\text{V}}_{\text{ideal}}\text{=}\left({\text{x}}_{\text{1}}{\text{V}}_{\text{m,1}}{\text{+x}}_{\text{2}}{\text{V}}_{\text{m,2}}\right)$

Substitute the value of $V_{\text{ideal}}$ and $V_{\text{total}}^{\text{E}}$ in equation (2).

    ${\text{V}}_{\text{J}}{\text{=n}}_{\text{1}}{\text{V}}_{\text{m,1}}{\text{+n}}_{\text{2}}{\text{V}}_{\text{m,2}}\text{+}\frac{{\text{n}}_{\text{1}}{\text{n}}_{\text{2}}}{\left({\text{n}}_{\text{1}}{\text{+n}}_{\text{2}}\right)}\left({\text{a}}_{\text{o}}\text{+}\frac{{\text{a}}_{\text{1}}\left({\text{n}}_{\text{1}}{\text{-n}}_{\text{2}}\right)}{{\text{n}}_{\text{1}}{\text{+n}}_{\text{2}}}\right)$

Thus, the partial volume of component 1 is,

    $\begin{array}{c}{V}_{\text{J,1}}={\left(\frac{\partial {\text{V}}_{\text{total}}}{\partial n_1}\right)}_{{\text{p,T,n}}_{\text{2}}}\\ =V_{\text{m,1}}+\frac{a_{\text{o}}{n}_2^2}{{\left(n_1+n_2\right)}^2}+\frac{a_1\left(3n_1-n_2\right)n_2^2}{{\left(n_1+n_2\right)}^3}\\ =V_{\text{m,1}}+a_{\text{o}}{x}_2^2+a_1\left(3x_1-x_2\right)x_2^2\end{array}$

Thus, the partial volume of component 2 is,

    $\begin{array}{c}{V}_{\text{J,2}}={\left(\frac{\partial {\text{V}}_{\text{total}}}{\partial x_2}\right)}_{{\text{p,T,x}}_1}\\ =V_{\text{m,2}}+a_{\text{o}}{x}_1^2+a_1\left(3x_1-x_2\right)x_1^2\end{array}$

Hence, the partial volume of propionic acid is,

    ${\text{V}}_{\text{J,1}}{\text{=V}}_{\text{m,1}}{\text{+a}}_{\text{o}}{\text{x}}_{\text{2}}^{\text{2}}{\text{+a}}_{\text{1}}\left({\text{3x}}_{\text{1}}{\text{-x}}_{\text{2}}\right){\text{x}}_{\text{2}}^{\text{2}}$

Hence, the partial volume of oxane is,

    ${\text{V}}_{\text{J,2}}{\text{=V}}_{\text{m,2}}{\text{+a}}_{\text{o}}{\text{x}}_{\text{1}}^{\text{2}}{\text{+a}}_{\text{1}}\left({\text{3x}}_{\text{1}}{\text{-x}}_{\text{2}}\right){\text{x}}_{\text{1}}^{\text{2}}$

(b)

Interpretation Introduction

Interpretation:

The partial molar volume for each component in equimolar mixture has to be calculated.

Answer to Problem 5B.3P

The molar volume of the component 1 is $\underset{\_}{75.73c{m}^{3}mo{l}^{-1}}$.  The molar volume of the component 2 is $\underset{\_}{99.07c{m}^{3}mo{l}^{-1}}$.

Explanation of Solution

The Propionic acid is assumed to be component 1 and THP is assumed to be component 2.

The molar volume of component 1 is given by the formula,

    $V_{\text{m,1}}\text{=}\frac{\text{Molar}\text{mass}}{\text{Density}}$

The value of molar mass of component 1 is $74.08\text{g}{\text{mol}}^{-1}$.

The density of the component 1 is $0.97174\text{g}{\text{cm}}^{-3}$.

Substitute the value of molar mass and density in the above expression.

    $\begin{array}{c}{V}_{\text{m,1}}\text{=}\frac{74.08\text{g}{\text{mol}}^{-1}}{0.97174\text{g}{\text{cm}}^{-3}}\\ =76.23{\text{cm}}^{\text{3}}{\text{mol}}^{-1}\end{array}$

The formula for the partial molar volume for component 1 is,

    ${\text{V}}_{\text{J,1}}{\text{=V}}_{\text{m,1}}{\text{+a}}_{\text{o}}{\text{x}}_{\text{2}}^{\text{2}}{\text{+a}}_{\text{1}}\left({\text{3x}}_{\text{1}}{\text{-x}}_{\text{2}}\right){\text{x}}_{\text{2}}^{\text{2}}\left(\text{1}\right)$

Where,

$V_{\text{J,1}}$ is the molar volume of component 1.

$x_2$ is the molar fraction of component 2.

$x_1$ is the molar fraction of component 1.

$V_{\text{m,1}}$ is the molar volume of the component 1.

$a_{\text{o}}$ and $a_1$ are the constants.

The value of $a_{\text{o}}$ is $-2.467{\text{cm}}^{\text{3}}{\text{mol}}^{-1}$.

The value of $a_1$ is $0.0608{\text{cm}}^{\text{3}}{\text{mol}}^{-1}$.

The value of $V_{\text{m,1}}$ is $76.23{\text{cm}}^{\text{3}}{\text{mol}}^{-1}$.

Since, the mixture is equimolar, assume the value of $x_1=x_2$.  Hence, the value of $x_1$ is $0.5$ and the value of $x_2$ is $0.5$.

Substitute the values of $x_1$, $x_2$, $V_{\text{m,1}}$, $a_{\text{o}}$ and $a_1$ in the equation (1).

    $\begin{array}{c}{V}_{\text{J,1}}=76.23{\text{cm}}^{\text{3}}{\text{mol}}^{-1}+\left(-2.467{\text{cm}}^{\text{3}}{\text{mol}}^{-1}\right)\times {\left(0.5\right)}^2+0.5\left(3\times 0.5-0.5\right){\left(0.5\right)}^2\\ =76.23{\text{cm}}^{\text{3}}{\text{mol}}^{-1}-0.616+0.125\\ =\underset{\_}{75.73c{m}^{3}mo{l}^{-1}}\end{array}$

Hence, the molar volume of the component 1 is $\underset{\_}{75.73c{m}^{3}mo{l}^{-1}}$.

The molar volume of component 2 is given by the formula,

    $V_{\text{m,2}}\text{=}\frac{\text{Molar}\text{mass}}{\text{Density}}$

The value of molar mass of component 1 is $86.13\text{g}{\text{mol}}^{-1}$.

The density of the component 1 is $0.86398\text{g}{\text{cm}}^{-3}$.

Substitute the value of molar mass and density in the above expression.

    $\begin{array}{c}{V}_{\text{m,1}}\text{=}\frac{86.13\text{g}{\text{mol}}^{-1}}{0.86398\text{g}{\text{cm}}^{-3}}\\ =99.68{\text{cm}}^{\text{3}}{\text{mol}}^{-1}\end{array}$

The formula molar volume for component 1 is,

    ${\text{V}}_{\text{J,2}}{\text{=V}}_{\text{m,2}}{\text{+a}}_{\text{o}}{\text{x}}_{\text{1}}^{\text{2}}{\text{+a}}_{\text{1}}\left({\text{3x}}_{\text{1}}{\text{-x}}_{\text{2}}\right){\text{x}}_{\text{1}}^{\text{2}}\left(\text{2}\right)$

Where,

$V_{\text{J,2}}$ is the molar volume of component 2.

$x_2$ is the molar fraction of component 2.

$x_1$ is the molar fraction of component 1.

$V_{\text{m,2}}$ is the molar volume of component 2.

$a_{\text{o}}$ and $a_1$ are the constants.

The value of $a_{\text{o}}$ is $-2.467{\text{cm}}^{\text{3}}{\text{mol}}^{-1}$.

The value of $a_1$ is $0.0608{\text{cm}}^{\text{3}}{\text{mol}}^{-1}$.

The value of $V_{\text{m,2}}$ is $99.68{\text{cm}}^{\text{3}}{\text{mol}}^{-1}$.

Since, the mixture is equimolar, assume the value of $x_1=x_2$.  Hence, the value of $x_1$ is $0.5$ and the value of $x_2$ is $0.5$.

Substitute the values of $x_1$, $x_2$, $V_{\text{m,2}}$, $a_{\text{o}}$ and $a_1$ in the equation (2).

    $\begin{array}{c}{V}_{\text{J,2}}=99.68{\text{cm}}^{\text{3}}{\text{mol}}^{-1}+\left(-2.467{\text{cm}}^{\text{3}}{\text{mol}}^{-1}\right){\left(0.5\right)}^2+\left(0.0608{\text{cm}}^{\text{3}}{\text{mol}}^{-1}\right)\\ \left(3\times 0.5-0.5\right){\left(0.5\right)}^2\\ =99.68{\text{cm}}^{\text{3}}{\text{mol}}^{-1}-0.61675+0.0152\\ =\underset{\_}{99.07c{m}^{3}mo{l}^{-1}}\end{array}$

Hence, the molar volume of the component 2 is $\underset{\_}{99.07c{m}^{3}mo{l}^{-1}}$.