Chapter 11.12, Problem 19P

You are interested in finding the pressure at which the first bubble of vapor will form from a liquid mixture of ethanol (1) and benzene (2) (49% by mole ethanol) at 313 K. The Margules parameters for this mixture are A₁₂ = 2.173 and A₂₁ = 1.539, while the Wilson parameters for this mixture are a₁₂/R = 653.13 K and a₂₁/R = 66.16 K. Find the pressure and vapor-phase composition using three ways.

A. The 2-parameter Margules equation

B. The Wilson equation

C. Ideal solution

Interpretation Introduction

Interpretation:

The pressure and vapour-phase composition using 2-parameter Margules equation.

Concept Introduction:

Write the expression to obtain the liquid mole fraction of benzene $\left(x_2\right)$.

$x_2=1-x_1$

Here, liquid mole fraction of ethanol is $x_1$.

Write the expression to obtain the activity coefficient of component 1 $\left({\gamma }_1\right)$.

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

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

Write the expression to obtain the activity coefficient of component 2 $\left({\gamma }_2\right)$.

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

Write the expression of vapor pressure using Antoine equation.

$P^{sat}={10}^{\left[A-\frac{B}{C+T\left(°\text{C}\right)}\right]}$

Here, temperature in degree Celsius is $T\left(°\text{C}\right)$, and Antoine coefficients are $A,B,\text{and}C$.

Write the expression to obtain the pressure $\left(P\right)$ from modified Raoult’s law.

$P={\gamma }_1{x}_1{P}_1^{sat}+{\gamma }_2{x}_2{P}_2^{sat}$

Write the expression to obtain the vapor phase composition.

$y_1=\frac{{\gamma }_1{x}_1{P}_1^{sat}}{P}$

Here, mole fraction of component 1 is $y_1$.

Write the expression to obtain the mole fraction of component 2 $\left(y_2\right)$.

$y_2=1-y_1$

Explanation of Solution

Given information:

Margules parameters are $A_{12}=2.173\text{ and }{A}_{21}=1.539$.

Write the expression to obtain the liquid mole fraction of benzene $\left(x_2\right)$.

$x_2=1-x_1$        (1)

Substitute 49% mole for $x_1$ in Equation (1).

$\begin{array}{c}{x}_2=1-49\%\\ =1-0.49\\ =0.51\end{array}$

Write the expression to obtain the activity coefficient of component 1 $\left({\gamma }_1\right)$.

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

Substitute $0.51$ for $x_2$, $0.49$ for $x_1$, $2.173$ for $A_{12}$, and $1.539$ for $A_{21}$ in Equation (2).

$\begin{array}{l}\ln {\gamma }_1={\left(0.51\right)}^2\left(2.173+2\left[1.539-2.173\right]0.49\right)\\ {\gamma }_1=e^{{\left(0.51\right)}^2\left(2.173+2\left[1.539-2.173\right]0.49\right)}\\ =e^{0.40359}\\ =1.4972\\ \simeq 1.50\end{array}$

Write the expression to obtain the activity coefficient of component 2 $\left({\gamma }_2\right)$.

$\ln {\gamma }_2=x_1^2\left(A_{21}+2\left[A_{12}-A_{21}\right]x_2\right)$        (3)

Substitute $0.51$ for $x_2$, $0.49$ for $x_1$, $2.173$ for $A_{12}$, and $1.539$ for $A_{21}$ in Equation (3).

$\begin{array}{l}\ln {\gamma }_2={\left(0.49\right)}^2\left(1.539+2\left[2.173-1.539\right]0.51\right)\\ {\gamma }_2=e^{{\left(0.49\right)}^2\left(1.539+2\left[2.173-1.539\right]0.51\right)}\\ =e^{0.52478}\\ =1.6900\\ =1.69\end{array}$

Write the expression of vapor pressure using Antoine equation.

$P^{sat}={10}^{\left[A-\frac{B}{C+T\left(°\text{C}\right)}\right]}$        (4)

From Appendix E, “Antoine Equations”, write the following properties for component 1 and 2 as in Table (1).

Parameters	Ethanol	Benzene
A	8.321	6.905
B	1718.10	1211.03
C	237.52	220.79

Convert the unit of temperature.

$\begin{array}{c}T=313\text{K}\\ =\left(313-273.15\right)°\text{C}\\ =39.85°\text{C}\end{array}$

Substitute $39.85°\text{C}$ for T, 8.321 for A, 1718.10 for B, and 237.52 for C in Equation (4).

$\begin{array}{c}{P}_1^{sat}={10}^{\left[8.321-\frac{1718.10}{237.52+39.85}\right]}\\ =133.88\text{mm}\cdot \text{Hg}\end{array}$

Substitute $39.85°\text{C}$ for T, 6.905 for A, 1211.03 for B, and 220.79 for C in Equation (4).

$\begin{array}{c}{P}_2^{sat}={10}^{\left[A-\frac{B}{C+T\left(°\text{C}\right)}\right]}\\ ={10}^{\left[6.905-\frac{1211.03}{220.79+39.85}\right]}\\ =181.39\text{mm}\cdot \text{Hg}\end{array}$

Write the expression to obtain the pressure $\left(P\right)$ from modified Raoult’s law.

$P={\gamma }_1{x}_1{P}_1^{sat}+{\gamma }_2{x}_2{P}_2^{sat}$        (5)

Substitute $1.50$ for ${\gamma }_1$, $0.49$ for $x_1$, $\text{133}\text{.88}\text{mm}\cdot \text{Hg}$ for $P_1^{sat}$, $1.69$ for ${\gamma }_2$, $0.51$ for $x_2$, and $\text{181}\text{.39}\text{mm}\cdot \text{Hg}$ for $P_2^{sat}$ in Equation (5).

$\begin{array}{c}P=\left[\left(1.50\right)\left(0.49\right)\left(\text{133}\text{.88}\text{mm}\cdot \text{Hg}\right)\right]+\left[\left(1.69\right)\left(0.51\right)\left(\text{181}\text{.39}\text{mm}\cdot \text{Hg}\right)\right]\\ =98.4018+156.3458\\ =254.7476\text{mm}\cdot \text{Hg}\\ \simeq 255\text{mm}\cdot \text{Hg}\end{array}$

Thus, the pressure at which the first bubble of vapor will form is $255\text{mm}\cdot \text{Hg}$.

Write the expression to obtain the vapor phase composition.

$y_1=\frac{{\gamma }_1{x}_1{P}_1^{sat}}{P}$        (6)

Substitute $1.50$ for ${\gamma }_1$, $0.49$ for $x_1$, $\text{133}\text{.88}\text{mm}\cdot \text{Hg}$ for $P_1^{sat}$, and $255\text{mm}\cdot \text{Hg}$ for $P$ in Equation (6).

$\begin{array}{c}{y}_1=\frac{\left(1.50\right)\left(0.49\right)\left(\text{133}\text{.88}\text{mm}\cdot \text{Hg}\right)}{255\text{mm}\cdot \text{Hg}}\\ =0.3859\end{array}$

Thus, the mole fraction of component 1 is $0.3859$.

Write the expression to obtain the mole fraction of component 2 $\left(y_2\right)$.

$y_2=1-y_1$        (7)

Substitute $0.3859$ for $y_1$ in Equation (7).

$\begin{array}{c}{y}_2=1-0.3859\\ =0.6141\end{array}$

Thus, the mole fraction of component 2 is $0.6141$.

Interpretation Introduction

Interpretation:

The pressure and vapour-phase composition using the Wilson equation.

Concept Introduction:

Write the expression to obtain the temperature dependent parameters of the Wilson equation $\left({\Lambda }_{12}\right)$.

${\Lambda }_{12}=\frac{{\underset{\_}{V}}_2}{{\underset{\_}{V}}_1}\exp \left(\frac{-a_{12}/R}{T}\right)$

Here, Wilson parameter is $a_{12}/R$, liquid molar volume of component 2 is ${\underset{\_}{V}}_2$, and liquid molar volume of component 1 is ${\underset{\_}{V}}_1$.

Write the expression to obtain the temperature dependent parameters of the Wilson equation $\left({\Lambda }_{21}\right)$.

${\Lambda }_{21}=\frac{{\underset{\_}{V}}_1}{{\underset{\_}{V}}_2}\exp \left(\frac{-a_{21}/R}{T}\right)$

Here, Wilson parameter is $a_{21}/R$.

Write the expression to obtain the activity coefficient of component 1 $\left({\gamma }_1\right)$.

$\ln {\gamma }_1=-\ln \left(x_1+{\Lambda }_{12}{x}_2\right)+x_2\left(\frac{{\Lambda }_{12}}{x_1+{\Lambda }_{12}{x}_2}-\frac{{\Lambda }_{21}}{x_2+{\Lambda }_{21}{x}_1}\right)$

Write the expression to obtain the activity coefficient of component 2 $\left({\gamma }_2\right)$.

$\ln {\gamma }_2=-\ln \left(x_2+{\Lambda }_{21}{x}_1\right)-x_1\left(\frac{{\Lambda }_{12}}{x_1+{\Lambda }_{12}{x}_2}-\frac{{\Lambda }_{21}}{x_2+{\Lambda }_{21}{x}_1}\right)$

Write the expression to obtain the pressure $\left(P\right)$.

$P={\gamma }_1{x}_1{P}_1^{sat}+{\gamma }_2{x}_2{P}_2^{sat}$

Write the expression to obtain the vapor phase composition.

$y_1=\frac{{\gamma }_1{x}_1{P}_1^{sat}}{P}$

Here, mole fraction of component 1 is $y_1$.

Write the expression to obtain the mole fraction of component 2 $\left(y_2\right)$.

$y_2=1-y_1$

Explanation of Solution

Given information:

Wilson parameters are $\frac{a_{12}}{R}=653.13\text{ K and }\frac{a_{21}}{R}=66.16\text{ K}$.

Refer appendix C.1, “Critical point, enthalpy of phase change, and liquid molar volume”, obtain the liquid molar volume $\left(\underset{\_}{V}\right)$ of ethanol and benzene as in Table (2).

Compound	liquid molar volume $\underset{\_}{V}\left({\text{cm}}^{\text{3}}/\text{mol}\right)$
Ethanol (1)	58.68
Benzene (2)	89.41

Write the expression to obtain the temperature dependent parameters of the Wilson equation $\left({\Lambda }_{12}\right)$.

${\Lambda }_{12}=\frac{{\underset{\_}{V}}_2}{{\underset{\_}{V}}_1}\exp \left(\frac{-a_{12}/R}{T}\right)$        (8)

Substitute $89.41{\text{cm}}^{\text{3}}/\text{mol}$ for ${\underset{\_}{V}}_2$, $58.68{\text{cm}}^{\text{3}}/\text{mol}$ for ${\underset{\_}{V}}_1$, $\text{653}\text{.13}\text{K}$ for $a_{12}/R$, and $\text{313}\text{K}$ for $T$ in Equation (8).

$\begin{array}{c}{\Lambda }_{12}=\frac{89.41{\text{cm}}^{\text{3}}/\text{mol}}{58.68{\text{cm}}^{\text{3}}/\text{mol}}\exp \left(\frac{-\text{653}\text{.13}\text{K}}{\text{313}\text{K}}\right)\\ =0.189\end{array}$

Write the expression to obtain the temperature dependent parameters of the Wilson equation $\left({\Lambda }_{21}\right)$.

${\Lambda }_{21}=\frac{{\underset{\_}{V}}_1}{{\underset{\_}{V}}_2}\exp \left(\frac{-a_{21}/R}{T}\right)$        (9)

Substitute $89.41{\text{cm}}^{\text{3}}/\text{mol}$ for ${\underset{\_}{V}}_2$, $58.68{\text{cm}}^{\text{3}}/\text{mol}$ for ${\underset{\_}{V}}_1$, $\text{66}\text{.16}\text{K}$ for $a_{21}/R$, and $\text{313}\text{K}$ for $T$ in Equation (9).

$\begin{array}{c}{\Lambda }_{21}=\frac{58.68{\text{cm}}^{\text{3}}/\text{mol}}{89.41{\text{cm}}^{\text{3}}/\text{mol}}\exp \left(\frac{-\text{66}\text{.16}\text{K}}{\text{313}\text{K}}\right)\\ =0.531\end{array}$

Write the expression to obtain the activity coefficient of component 1 $\left({\gamma }_1\right)$.

$\ln {\gamma }_1=-\ln \left(x_1+{\Lambda }_{12}{x}_2\right)+x_2\left(\frac{{\Lambda }_{12}}{x_1+{\Lambda }_{12}{x}_2}-\frac{{\Lambda }_{21}}{x_2+{\Lambda }_{21}{x}_1}\right)$        (10)

Substitute $0.49$ for $x_1$, $0.51$ for $x_2$, $0.531$ for ${\Lambda }_{21}$, and 0.189 for ${\Lambda }_{12}$ in Equation (10).

$\begin{array}{l}\ln {\gamma }_1=-\ln \left[0.49+\left(0.189\right)\left(0.51\right)\right]+0.51\left[\frac{0.189}{0.49+\left(0.189\right)\left(0.51\right)}-\frac{0.531}{0.51+\left(0.531\right)\left(0.49\right)}\right]\\ \ln {\gamma }_1=0.5338+0.51\left(0.3223-0.6894\right)\\ \ln {\gamma }_1=0.3466\\ {\gamma }_1=e^{0.3466}\end{array}$

${\gamma }_1=1.41$

Write the expression to obtain the activity coefficient of component 2 $\left({\gamma }_2\right)$.

$\ln {\gamma }_2=-\ln \left(x_2+{\Lambda }_{21}{x}_1\right)-x_1\left(\frac{{\Lambda }_{12}}{x_1+{\Lambda }_{12}{x}_2}-\frac{{\Lambda }_{21}}{x_2+{\Lambda }_{21}{x}_1}\right)$        (11)

Substitute $0.49$ for $x_1$, $0.51$ for $x_2$, $0.531$ for ${\Lambda }_{21}$, and 0.189 for ${\Lambda }_{12}$ in Equation (11).

$\begin{array}{l}\ln {\gamma }_2=-\ln \left[0.51+\left(0.531\right)\left(0.49\right)\right]-0.49\left[\frac{0.189}{0.49+\left(0.189\right)\left(0.51\right)}-\frac{0.531}{0.51+\left(0.531\right)\left(0.49\right)}\right]\\ \ln {\gamma }_2=0.2611-0.49\left(0.3223-0.6894\right)\\ \ln {\gamma }_2=0.4409\\ {\gamma }_2=e^{0.4409}\end{array}$

${\gamma }_2=1.55$

Write the expression to obtain the pressure $\left(P\right)$.

$P={\gamma }_1{x}_1{P}_1^{sat}+{\gamma }_2{x}_2{P}_2^{sat}$        (12)

Substitute $1.41$ for ${\gamma }_1$, $0.49$ for $x_1$, $\text{133}\text{.88}\text{mm}\cdot \text{Hg}$ for $P_1^{sat}$, $1.55$ for ${\gamma }_2$, $0.51$ for $x_2$, and $\text{181}\text{.39}\text{mm}\cdot \text{Hg}$ for $P_2^{sat}$ in Equation (12).

$\begin{array}{c}P=\left[\left(1.41\right)\left(0.49\right)\left(\text{133}\text{.88}\text{mm}\cdot \text{Hg}\right)\right]+\left[\left(1.55\right)\left(0.51\right)\left(\text{181}\text{.39}\text{mm}\cdot \text{Hg}\right)\right]\\ =238.3\text{mm}\cdot \text{Hg}\end{array}$

Thus, the pressure at which the first bubble of vapor will form is $238.3\text{mm}\cdot \text{Hg}$.

Write the expression to obtain the vapor phase composition.

$y_1=\frac{{\gamma }_1{x}_1{P}_1^{sat}}{P}$        (13)

Substitute $1.41$ for ${\gamma }_1$, $0.49$ for $x_1$, $\text{133}\text{.88}\text{mm}\cdot \text{Hg}$ for $P_1^{sat}$, and $\text{238}\text{.3}\text{mm}\cdot \text{Hg}$ for $P$ in Equation (13).

$\begin{array}{c}{y}_1=\frac{\left(1.41\right)\left(0.49\right)\left(\text{133}\text{.88}\text{mm}\cdot \text{Hg}\right)}{\text{238}\text{.3}\text{mm}\cdot \text{Hg}}\\ =0.392\end{array}$

Thus, the mole fraction of component 1 is $0.392$.

Write the expression to obtain the mole fraction of component 2 $\left(y_2\right)$.

$y_2=1-y_1$        (14)

Substitute 0.392 for $y_1$ in Equation (14).

$\begin{array}{c}{y}_2=1-0.392\\ =0.608\end{array}$

Thus, the mole fraction of component 2 is $0.608$.

Interpretation Introduction

Interpretation:

The Ideal solution.

Concept Introduction:

Write the expression to obtain the pressure $\left(P\right)$.

$P=x_1{P}_1^{sat}+x_2{P}_2^{sat}$

Write the expression to obtain the vapor phase composition.

$y_1=\frac{x_1{P}_1^{sat}}{P}$

Here, mole fraction of component 1 is $y_1$.

Write the expression to obtain the mole fraction of component 2 $\left(y_2\right)$.

$y_2=1-y_1$

Explanation of Solution

Write the expression to obtain the pressure $\left(P\right)$.

$P=x_1{P}_1^{sat}+x_2{P}_2^{sat}$        (15)

Substitute $0.49$ for $x_1$, $\text{133}\text{.88}\text{mm}\cdot \text{Hg}$ for $P_1^{sat}$, $0.51$ for $x_2$, and $\text{181}\text{.39}\text{mm}\cdot \text{Hg}$ for $P_2^{sat}$ in Equation (15).

$\begin{array}{c}P=\left[\left(0.49\right)\left(\text{133}\text{.88}\text{mm}\cdot \text{Hg}\right)\right]+\left[\left(0.51\right)\left(\text{181}\text{.39}\text{mm}\cdot \text{Hg}\right)\right]\\ =158.11\text{mm}\cdot \text{Hg}\end{array}$

Thus, the pressure at which the first bubble of vapor will form is $158.11\text{mm}\cdot \text{Hg}$.

Write the expression to obtain the vapor phase composition.

$y_1=\frac{x_1{P}_1^{sat}}{P}$        (16)

Substitute $0.49$ for $x_1$, $\text{133}\text{.88}\text{mm}\cdot \text{Hg}$ for $P_1^{sat}$, and $\text{158}\text{.11}\text{mm}\cdot \text{Hg}$ for $P$ in Equation (16).

$\begin{array}{c}{y}_1=\frac{\left(0.49\right)\left(\text{133}\text{.88}\text{mm}\cdot \text{Hg}\right)}{\text{158}\text{.11}\text{mm}\cdot \text{Hg}}\\ =0.415\end{array}$

Thus, the mole fraction of component 1 is $0.415$.

Write the expression to obtain the mole fraction of component 2 $\left(y_2\right)$.

$y_2=1-y_1$        (17)

Substitute 0.415 for $y_1$ in Equation (17).

$\begin{array}{c}{y}_2=1-0.415\\ =0.585\end{array}$

Thus, the mole fraction of component 2 is $0.585$.