Chapter 10.7, Problem 15P

A)

Interpretation Introduction

Interpretation:

To plot the $P_{xy}$ and $T_{xy}$ curve for two sets of experimental data.

Concept introduction:

An example of a binary system with constant pressure of vapor liquid equilibrium system is the mixture of n-hexane (1) and ethanol (2).

The experimental vapor liquid equilibrium data is given in Table (1).

$T\left(\text{K}\right)$	$x_1\left(\text{mol}/\text{mol}\right)$	$y_1\left(\text{mol}/\text{mol}\right)$
351.45	0.000	0.000
349.15	0.010	0.095
346.35	0.020	0.193
340.55	0.060	0.365
339.05	0.080	0.420
334.95	0.152	0.532
332.55	0.245	0.605
331.85	0.333	0.630
331.50	0.452	0.640
331.25	0.588	0.650
331.15	0.670	0.660
331.40	0.725	0.670
331.60	0.765	0.675
332.30	0.898	0.710
333.35	0.955	0.745
336.65	0.990	0.840
339.85	0.994	0.935
341.85	1.000	1.000

Table (1)

An example of a binary system with constant temperature of vapor liquid equilibrium system is the mixture of tetrahydrofuran (1) and n-hexane (2).

The experimental vapor liquid equilibrium data is given in Table (2).

$P\left(\text{kPa}\right)$	$x_1\left(\text{mol}/\text{mol}\right)$	$y_1\left(\text{mol}/\text{mol}\right)$
37.69	0	0
39.943	0.098	0.145
42.077	0.185	0.251
43.623	0.275	0.346
44.943	0.385	0.439
45.463	0.448	0.488
45.543	0.488	0.515
45.783	0.520	0.539
45.676	0.535	0.549
45.730	0.573	0.575
45.756	0.620	0.610
45.663	0.665	0.645
45.570	0.713	0.682
45.316	0.738	0.695
44.810	0.803	0.755
44.063	0.843	0.790
43.543	0.883	0.826
42.663	0.925	0.885
41.477	0.975	0.955
40.410	1	1

Table (2)

B)

Interpretation Introduction

Interpretation:

To identify in the $P_{xy}$ and $T_{xy}$ plot the various regions present.

Concept introduction:

An example of a binary system with constant pressure of vapor liquid equilibrium system is the mixture of n-hexane (1) and ethanol (2).

The experimental vapor liquid equilibrium data is given in Table (1).

$T\left(\text{K}\right)$	$x_1\left(\text{mol}/\text{mol}\right)$	$y_1\left(\text{mol}/\text{mol}\right)$
351.45	0.000	0.000
349.15	0.010	0.095
346.35	0.020	0.193
340.55	0.060	0.365
339.05	0.080	0.420
334.95	0.152	0.532
332.55	0.245	0.605
331.85	0.333	0.630
331.50	0.452	0.640
331.25	0.588	0.650
331.15	0.670	0.660
331.40	0.725	0.670
331.60	0.765	0.675
332.30	0.898	0.710
333.35	0.955	0.745
336.65	0.990	0.840
339.85	0.994	0.935
341.85	1.000	1.000

Table (1)

An example of a binary system with constant temperature of vapor liquid equilibrium system is the mixture of tetrahydrofuran (1) and n-hexane (2).

The experimental vapor liquid equilibrium data is given in Table (2).

$P\left(\text{kPa}\right)$	$x_1\left(\text{mol}/\text{mol}\right)$	$y_1\left(\text{mol}/\text{mol}\right)$
37.69	0	0
39.943	0.098	0.145
42.077	0.185	0.251
43.623	0.275	0.346
44.943	0.385	0.439
45.463	0.448	0.488
45.543	0.488	0.515
45.783	0.520	0.539
45.676	0.535	0.549
45.730	0.573	0.575
45.756	0.620	0.610
45.663	0.665	0.645
45.570	0.713	0.682
45.316	0.738	0.695
44.810	0.803	0.755
44.063	0.843	0.790
43.543	0.883	0.826
42.663	0.925	0.885
41.477	0.975	0.955
40.410	1	1

Table (2)

C)

Interpretation Introduction

Interpretation:

To discuss the behavior of the system on the basis of Raoult’s law.

Concept introduction:

Raoult’s law expresses that the incomplete vapor pressure of every segment of a perfect blend of fluids is equivalent to the mixture of vapor pressure of the pure component and the mole portion in the blend

Negative deviation

When the adhesive forces between particles of different type are stronger than the cohesive forces between disparate atoms, the vapor pressure is less than the expected from the Raoult’s law. This is known as negative deviation.

Positive deviation

When the cohesive force between particles of same type is greater than the adhesive forces between atoms of different type, the extremity of dissimilarities leads both segments to escape the arrangement more effortlessly. In such case, the vapor pressure expected is more than that from the Raoult's law. This is called positive deviation.