Chapter 10.6, Problem 5E

For a mixture of n-butane (1) + n-pentane (2) at 25°C, what would be the predicted bubble-point pressure using Raoult’s Law if your mixture was 20% n-butane by mole? Would you expect Raoult’s Law to be a good model for this system? Why or why not?

Interpretation Introduction

Interpretation:

To obtain the bubble-point pressure using Raoult’s Law and to predict if the Raoult’s law be expected to be a good model for this system.

Concept Introduction:

The Antoine equation for the 1^st component $n\text{-butane}$ is,

${\log }_{10}\left(P_1{}^{sat}\right)=A-\frac{B}{T+C}$

Here, Antoine constants or coefficients are A, B, and C, vapor pressure is $P^{sat}$, and temperature is T in °C.

The Antoine equation for the 2^nd component $n\text{-pentane}$ is,

${\log }_{10}\left(P_2{}^{sat}\right)=A-\frac{B}{T+C}$

Here, vapor pressure is $P_2{}^{sat}$.

The expression of system pressure using Raoult’s Law is,

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

Here, liquid phase mole fraction of components 1 and 2 is $x_1$ and $x_2$ respectively.

The expression to obtain the vapor phase mole fraction $\left(y_2\right)$ of component 2 is,

$y_2=1-y_1$

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

The expression to obtain the liquid phase mole fraction $\left(x_1\right)$ of components 1 is,

$x_1=\frac{P_{@T=25°C}{y}_1}{P_1^{sat}}$

The expression to obtain the liquid phase mole fraction $\left(x_2\right)$ of component 2 is,

$x_2=1-x_1$

Here, liquid phase mole fraction of component 1 is $x_1$.

Explanation of Solution

Given information:

Temperature is $T=25°\text{C}$.

Write the Antoine equation for the 1^st component $n\text{-butane}$.

${\log }_{10}\left(P_1{}^{sat}\right)=A-\frac{B}{T+C}$        (1)

Refer Appendix E, “Antoine Coefficients”; obtain the constants A, B, and C for $n\text{-butane}$ as 6.808, 935.86, and 238.73 respectively.

Substitute $25°\text{C}$ for T, 6.808 for A, 935.86 for B, and 238.73 for C in Equation (1).

$\begin{array}{l}{\log }_{10}\left(P_1{}^{sat}\right)=6.808-\frac{935.86}{25°\text{C}+238.73}\\ P_1{}^{sat}={10}^{3.259}\\ =1,817.3\text{mm}\text{Hg}\times \left(\frac{1\text{atm}}{760\text{mm}\text{Hg}}\right)\left(\frac{1.013\text{bar}}{1\text{atm}}\right)\end{array}$

$\begin{array}{l}=2.42\text{bar}\left(100\text{kPa}/1\text{bar}\right)\\ =242\text{kPa}\end{array}$

Write the Antoine equation for the 2^nd component $n\text{-pentane}$.

${\log }_{10}\left(P_2{}^{sat}\right)=A-\frac{B}{T+C}$        (2)

Refer Appendix E, “Antoine Coefficients”; obtain the constants A, B, and C for $n\text{-pentane}$ as $\text{6}\text{.852,}1064.84,\text{and}\text{233}\text{.01}$ respectively.

Substitute $25°\text{C}$ for T, $\text{6}\text{.852,}1064.84,\text{and}\text{233}\text{.01}$ for $A,B\text{and}C$ respectively in Equation (2).

$\begin{array}{l}{\log }_{10}\left(P_2{}^{sat}\right)=6.852-\frac{1064.84}{25°\text{C}+233.01}\\ P_2{}^{sat}={10}^{2.724}\\ =530.72\text{mm}\text{Hg}\times \left(\frac{1\text{atm}}{760\text{mm}\text{Hg}}\right)\left(\frac{1.013\text{bar}}{1\text{atm}}\right)\end{array}$

$\begin{array}{l}=0.707\text{bar}\left(100\text{kPa}/1\text{bar}\right)\\ =70.7\text{kPa}\end{array}$

The mixture of n-butane (1) + n-pentane is at the temperature of 25°C. Hence, consider the system pressure is the pressure of corresponding to the temperature of 25°C of steam.

$P=P_{@T=25°C}$

Refer Appendix A-2, “Saturated steam-temperature increments”; obtain the pressure corresponding to temperature of $\text{25°C}$.

$\begin{array}{c}{P}_{@T=25°C}=0.0317\text{bar}\times \frac{100\text{kPa}}{1\text{bar}}\\ =3.17\text{kPa}\end{array}$

Here, pressure at temperature of $\text{25°C}$ is $P_{@T=25°C}$.

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

$y_2=1-y_1$        (3)

It is given that the mixture is composed of 20% of n-butane by mole. It not clear that the given mole fraction is corresponded to liquid phase or vapour phase. Hence, consider that the vapour mole fraction of n-butane is,

$y_1=\frac{20\%}{100}=0.20$

Substitute 0.20 for $y_1$ in Equation (3).

$\begin{array}{c}{y}_2=1-0.20\\ =0.80\end{array}$

Write the expression to obtain the liquid phase mole fraction $\left(x_1\right)$ of components 1.

$x_1=\frac{P_{@T=25°C}{y}_1}{P_1^{sat}}$        (4)

Substitute $0.20$ for $y_1$, $3.17\text{kPa}$ for $P_{@T=25°C}$, and $242\text{kPa}$  for  $P_1^{sat}$ in Equation (4).

$\begin{array}{c}{x}_1=\frac{\left(3.17\text{kPa}\right)\left(0.20\right)}{242\text{kPa}}\\ =0.00262\end{array}$

Write the expression to obtain the liquid phase mole fraction $\left(x_2\right)$ of components 2.

$x_2=1-x_1$        (5)

Substitute $0.00262$ for $x_1$ in Equation (5).

$\begin{array}{c}{x}_2=1-0.00262\\ =0.99738\end{array}$

Write the expression of system pressure using Raoult’s Law.

$P=x_1{P}_1^{sat}+x_2{P}_2^{sat}$        (6)

Substitute $0.00262$ for $x_1$, $0.99738$ for $x_2$, $\text{242}\text{kPa}$ for $P_1^{sat}$, and $\text{70}\text{.7}\text{kPa}$ for $P_2^{sat}$ in Equation (6).

$\begin{array}{c}P=\left[\left(0.00262\right)\left(\text{241}\text{kPa}\right)\right]+\left[\left(0.99738\right)\left(\text{70}\text{.7}\text{kPa}\right)\right]\\ =71.1488\text{kPa}\end{array}$

Similarly using excel spread sheet, calculate the values of $y_2,x_1,x_2,\text{and}P$ by varying the vapour mole fraction $\left(y_1\right)$ of n-butane from 0 to 1 in equations 3, 4, 5, and 6. The obtained values are tabulate below.

$P_1^{sat}$	$P_2^{sat}$	$P=P_{@T=25°C}$
$\text{242}\text{kPa}$	$\text{70}\text{.7}\text{kPa}$	$\text{3}\text{.17}\text{kPa}$

$y_1$	$y_2$	$x_1$	$x_2$	$P$ in kPa (calculated)
0	1	0	1	70.7
0.1	0.9	0.00131	0.99869	70.9244
0.2	0.8	0.00262	0.99738	71.1488
0.3	0.7	0.00393	0.99607	71.3732
0.4	0.6	0.00524	0.99476	71.5976
0.5	0.5	0.00655	0.99345	71.8219
0.6	0.4	0.00786	0.99214	72.0463
0.7	0.3	0.00917	0.99083	72.2707
0.8	0.2	0.01048	0.98952	72.4951
0.9	0.1	0.01179	0.98821	72.7195
1	0	0.0131	0.9869	72.9439

Table 1

The plot of P versus $x_1$ (bubble point curve) and P versus $y_1$ (dew point curve) using Table 1 is shown in Figure 1.

Figure 1

From Figure 1,

The bubble point curve and dew point curve are not meet anywhere. Hence Raoult’s Law is not a good model for this system and the bubble point pressure cannot be predicted.