BICOL UNIVERSITY
COLLEGE OF ENGINEERING
Department of Chemical Engineering
Chem 15L: Physical Chemistry for Engineers 2 - Lab
Pressure-Composition and
Temperature-Composition Diagrams
Katherine Trixie V. Astudillo, Prichebelle G. Grafia, Regie A. Zaragoza
I.
Introduction
Phase diagram is a graphical representation of the physical states of a substance under
different conditions of temperature and pressure. A typical phase diagram has pressure on the yaxis and temperature on the x-axis. A binary system has two components; C equals 2, and the
number of degrees of freedom is F=4−P. There must be at least one phase, so the maximum
possible value of F is 3. Since F cannot be negative, the equilibrium system can have no more than
four phases. We can independently vary the temperature, pressure, and composition of the system
as a whole. Instead of using these variables as the coordinates of a three-dimensional phase
diagram, we usually draw a two-dimensional phase diagram that is either a temperature–
composition diagram at a fixed pressure or a pressure–composition diagram at a fixed temperature.
The position of the system point on one of these diagrams then corresponds to a definite
temperature, pressure, and overall composition. The composition variable usually varies along the
horizontal axis and can be the mole fraction, mass fraction, or mass percent of one of the
components.
Raoult’s Law has a significant role in describing ideal mixtures. Raoult’s Law states that
the partial vapor pressure of a component in a mixture is equal to the vapor pressure of the pure
component at that temperature multiplied by its mole fraction in the mixture.
𝑃𝐴 = 𝑥𝐴 𝑃𝐴∗
𝑃𝐵 = 𝑥𝐵 𝑃𝐵∗
The total vapor pressure 𝑃 of the mixture is therefore:
𝑃 = 𝑃𝐴 + 𝑃𝐵
The compositions of the liquid and vapor that are in mutual equilibrium are not necessarily
the same. the vapor should be richer in the more volatile component. Therefore, from Dalton’s
law, the mole fraction in the gas 𝑦𝐴 and 𝑦𝐵 , are:
𝑃
𝑃
𝑦𝐴 = 𝐴, 𝑦𝐵 = 𝐵
𝑃
𝑃
Provided the mixture is ideal, the partial pressures and the total pressure may be expressed
using the equation:
𝑥 𝑃∗
𝐴
𝑦𝑎 = 𝑃∗ +(𝑃𝐴∗ −𝑃
∗ )𝑥 , 𝑦𝐵 = 1 − 𝑦𝐴
𝐵
𝐴
𝐵
𝐴
The total vapor pressure of the mixture varies with the composition of the liquid. Because
we can relate the composition of the liquid to the composition of the vapor through the equation
above, we can now also relate the total vapor pressure to the composition of the vapor:
𝑃=
𝑃𝐴∗ 𝑃𝐵∗
𝑃𝐴∗ + (𝑃𝐵∗ − 𝑃𝐴∗ )𝑦𝐴
Phase diagrams are of considerable commercial and industrial significance, particularly for
semiconductors, ceramics, steels, and alloys. They are also the basis of separation procedures in
the petroleum industry and of the formulation of foods and cosmetic preparations. In this
laboratory activity, students are expected to apply Raoult’s Law in describing ideal mixtures and
prepare a pressure-composition diagram. Students will also plot temperature-composition diagram
and apply it in determining the number of theoretical steps of a fractionating distillation.
II.
Learning Outcomes
After doing the activity, the students are expected to be able to:
1. Apply Raoult’s Law in describing ideal mixtures and prepare a pressure-composition diagram.
2. Plot temperature-composition diagram and apply it in determining the number of theoretical steps
of a fractionating distillation.
III.
Material and Methods
Material
The following resources are needed to complete this activity.
•
•
Laptop/ computers
Calculator
Methods
•
•
•
•
•
Using the calculator, the following data were calculated using the parameters given.
The obtained data from the calculations were presented on the table.
From the table, the data were potted on P-x-y diagram using Microsoft excel for
interpretation.
A point (0.6,60) was trace using lines to determine if the point lies on the equilibrium
curve.
The relative amount of the two phases was determined by creating tie line and tracing
the distance between the interception of the tie line in the saturated liquid and saturated
vapor curve.
•
•
The given temperature-composition data was plotted on the T-x-y diagram using
Microsoft excel.
After plotting the data, theoretical plates for the fractional distillation were determined.
The following calculations were made for the interpretation:
A
B
C
Benzene
13.8594
2773.78
220.07
Toluene
14.0098
3103.01
219.79
The parameter values for the Antoine equation for some liquids are given below:
ln 𝑃 𝑠𝑎𝑡 (𝑘𝑃𝑎) = 𝐴 −
𝐵
𝑇(℃) + 𝐶
Calculations for the vapor pressures of pure benzene and pure toluene at 75°C.
Calculation of the vapor pressure of pure benzene:
ln 𝑃𝑠𝑎𝑡 (𝑘𝑃𝑎) = 13.8594 −
2773.78
75 + 220.07
ln 𝑃 (𝑘𝑃𝑎) = 4.459
𝑒 ln 𝑃 = 𝑒 4.459
𝑃 = 86.40 𝑘𝑃𝑎
Calculation of the vapor pressure of pure toluene:
ln 𝑃𝑠𝑎𝑡 (𝑘𝑃𝑎) = 14.0098 −
3103.01
75 + 219.79
ln 𝑃 (𝑘𝑃𝑎) = 3.484
𝑒 ln 𝑃 = 𝑒 3.484
𝑃 = 32.59 𝑘𝑃𝑎
Calculation of mole fraction of benzene in vapor phase, 𝑦1
*For 𝑥1 = 0.00:
𝑥1 𝑃1∗
𝑦1 = ∗
𝑃2 + (𝑃1∗ − 𝑃2∗ )𝑥1
𝑦1 =
(0)(86.40𝑘𝑃𝑎)
= 0.00
32.59𝑘𝑃𝑎 + (86.40𝑘𝑃𝑎 − 32.59𝑘𝑃𝑎) ∗ 0
*For 𝑥1 = 0.10:
𝑦1 =
𝑦1 =
𝑥1 𝑃1∗
𝑃2∗ + (𝑃1∗ − 𝑃2∗ )𝑥1
(0.10)(86.40𝑘𝑃𝑎)
= 0.23
32.59𝑘𝑃𝑎 + (86.40𝑘𝑃𝑎 − 32.59𝑘𝑃𝑎) ∗ 0.10
*For 𝑥1 = 0.20:
𝑦1 =
𝑦1 =
𝑥1 𝑃1∗
𝑃2∗ + (𝑃1∗ − 𝑃2∗ )𝑥1
(0.20)(86.40𝑘𝑃𝑎)
= 0.40
32.59𝑘𝑃𝑎 + (86.40𝑘𝑃𝑎 − 32.59𝑘𝑃𝑎) ∗ 0.20
*For 𝑥1 = 0.30:
𝑥1 𝑃1∗
𝑦1 = ∗
𝑃2 + (𝑃1∗ − 𝑃2∗ )𝑥1
𝑦1 =
(0.30)(86.40𝑘𝑃𝑎)
= 0.53
32.59𝑘𝑃𝑎 + (86.40𝑘𝑃𝑎 − 32.59𝑘𝑃𝑎) ∗ 0.30
*For 𝑥1 = 0.40:
𝑦1 =
𝑦1 =
𝑥1 𝑃1∗
𝑃2∗ + (𝑃1∗ − 𝑃2∗ )𝑥1
(0.40)(86.40𝑘𝑃𝑎)
= 0.64
32.59𝑘𝑃𝑎 + (86.40𝑘𝑃𝑎 − 32.59𝑘𝑃𝑎) ∗ 0.40
*For 𝑥1 = 0.50:
𝑥1 𝑃1∗
𝑦1 = ∗
𝑃2 + (𝑃1∗ − 𝑃2∗ )𝑥1
𝑦1 =
(0.50)(86.40𝑘𝑃𝑎)
= 0.73
32.59𝑘𝑃𝑎 + (86.40𝑘𝑃𝑎 − 32.59𝑘𝑃𝑎) ∗ 0.50
*For 𝑥1 = 0.60:
𝑦1 =
𝑦1 =
*For 𝑥1 = 0.70:
𝑥1 𝑃1∗
𝑃2∗ + (𝑃1∗ − 𝑃2∗ )𝑥1
(0.60)(86.40𝑘𝑃𝑎)
= 0.80
32.59𝑘𝑃𝑎 + (86.40𝑘𝑃𝑎 − 32.59𝑘𝑃𝑎) ∗ 0.60
𝑦1 =
𝑦1 =
𝑥1 𝑃1∗
𝑃2∗ + (𝑃1∗ − 𝑃2∗ )𝑥1
(0.70)(86.40𝑘𝑃𝑎)
= 0.86
32.59𝑘𝑃𝑎 + (86.40𝑘𝑃𝑎 − 32.59𝑘𝑃𝑎) ∗ 0.70
*For 𝑥1 = 0.80:
𝑥1 𝑃1∗
𝑦1 = ∗
𝑃2 + (𝑃1∗ − 𝑃2∗ )𝑥1
𝑦1 =
(0.80)(86.40𝑘𝑃𝑎)
= 0.91
32.59𝑘𝑃𝑎 + (86.40𝑘𝑃𝑎 − 32.59𝑘𝑃𝑎) ∗ 0.80
*For 𝑥1 = 0.90:
𝑦1 =
𝑦1 =
𝑥1 𝑃1∗
𝑃2∗ + (𝑃1∗ − 𝑃2∗ )𝑥1
(0.90)(86.40𝑘𝑃𝑎)
= 0.96
32.59𝑘𝑃𝑎 + (86.40𝑘𝑃𝑎 − 32.59𝑘𝑃𝑎) ∗ 0.90
*For 𝑥1 = 1.00:
𝑥1 𝑃1∗
𝑦1 = ∗
𝑃2 + (𝑃1∗ − 𝑃2∗ )𝑥1
𝑦1 =
(1.00)(86.40𝑘𝑃𝑎)
= 1.00
32.59𝑘𝑃𝑎 + (86.40𝑘𝑃𝑎 − 32.59𝑘𝑃𝑎) ∗ 1.00
Calculation of total pressure, P
*For 𝑥1 = 0.00 and 𝑦1 = 0.00:
𝑃1∗ 𝑃2∗
𝑃= ∗
𝑃1 + (𝑃2∗ − 𝑃1∗ )𝑦1
𝑃=
(86.40𝑘𝑃𝑎)(32.59𝑘𝑃𝑎)
= 32.59𝑘𝑃𝑎
86.40𝑘𝑃𝑎 + (32.59𝑘𝑃𝑎 − 86.40𝑘𝑃𝑎) ∗ 0
*For 𝑥1 = 0.10 and 𝑦1 = 0.23:
𝑃1∗ 𝑃2∗
𝑃= ∗
𝑃1 + (𝑃2∗ − 𝑃1∗ )𝑦1
𝑃=
(86.40𝑘𝑃𝑎)(32.59𝑘𝑃𝑎)
= 38.04𝑘𝑃𝑎
86.40𝑘𝑃𝑎 + (32.59𝑘𝑃𝑎 − 86.40𝑘𝑃𝑎) ∗ 0.23
*For 𝑥1 = 0.20 and 𝑦1 = 0.40:
𝑃=
𝑃=
𝑃1∗ 𝑃2∗
𝑃1∗ + (𝑃2∗ − 𝑃1∗ )𝑦1
(86.40𝑘𝑃𝑎)(32.59𝑘𝑃𝑎)
= 43.40𝑘𝑃𝑎
86.40𝑘𝑃𝑎 + (32.59𝑘𝑃𝑎 − 86.40𝑘𝑃𝑎) ∗ 0.40
*For 𝑥1 = 0.30 and 𝑦1 = 0.53:
𝑃=
𝑃=
𝑃1∗ 𝑃2∗
𝑃1∗ + (𝑃2∗ − 𝑃1∗ )𝑦1
(86.40𝑘𝑃𝑎)(32.59𝑘𝑃𝑎)
= 48.65𝑘𝑃𝑎
86.40𝑘𝑃𝑎 + (32.59𝑘𝑃𝑎 − 86.40𝑘𝑃𝑎) ∗ 0.53
*For 𝑥1 = 0.40 and 𝑦1 = 0.64:
𝑃1∗ 𝑃2∗
𝑃= ∗
𝑃1 + (𝑃2∗ − 𝑃1∗ )𝑦1
𝑃=
(86.40𝑘𝑃𝑎)(32.59𝑘𝑃𝑎)
= 54.19𝑘𝑃𝑎
86.40𝑘𝑃𝑎 + (32.59𝑘𝑃𝑎 − 86.40𝑘𝑃𝑎) ∗ 0.64
*For 𝑥1 = 0.50 and 𝑦1 = 0.73:
𝑃1∗ 𝑃2∗
𝑃= ∗
𝑃1 + (𝑃2∗ − 𝑃1∗ )𝑦1
𝑃=
(86.40𝑘𝑃𝑎)(32.59𝑘𝑃𝑎)
= 59.76𝑘𝑃𝑎
86.40𝑘𝑃𝑎 + (32.59𝑘𝑃𝑎 − 86.40𝑘𝑃𝑎) ∗ 0.73
*For 𝑥1 = 0.60 and 𝑦1 = 0.80:
𝑃=
𝑃=
𝑃1∗ 𝑃2∗
𝑃1∗ + (𝑃2∗ − 𝑃1∗ )𝑦1
(86.40𝑘𝑃𝑎)(32.59𝑘𝑃𝑎)
= 64.95𝑘𝑃𝑎
86.40𝑘𝑃𝑎 + (32.59𝑘𝑃𝑎 − 86.40𝑘𝑃𝑎) ∗ 0.80
*For 𝑥1 = 0.70 and 𝑦1 = 0.86:
𝑃=
𝑃=
𝑃1∗ 𝑃2∗
𝑃1∗ + (𝑃2∗ − 𝑃1∗ )𝑦1
(86.40𝑘𝑃𝑎)(32.59𝑘𝑃𝑎)
= 70.18𝑘𝑃𝑎
86.40𝑘𝑃𝑎 + (32.59𝑘𝑃𝑎 − 86.40𝑘𝑃𝑎) ∗ 0
*For 𝑥1 = 0.80 and 𝑦1 = 0.91:
𝑃=
𝑃=
𝑃1∗ 𝑃2∗
𝑃1∗ + (𝑃2∗ − 𝑃1∗ )𝑦1
(86.40𝑘𝑃𝑎)(32.59𝑘𝑃𝑎)
= 75.22𝑘𝑃𝑎
86.40𝑘𝑃𝑎 + (32.59𝑘𝑃𝑎 − 86.40𝑘𝑃𝑎) ∗ 0.91
*For 𝑥1 = 0.90 and 𝑦1 = 0.96:
𝑃1∗ 𝑃2∗
𝑃= ∗
𝑃1 + (𝑃2∗ − 𝑃1∗ )𝑦1
𝑃=
(86.40𝑘𝑃𝑎)(32.59𝑘𝑃𝑎)
= 81.05𝑘𝑃𝑎
86.40𝑘𝑃𝑎 + (32.59𝑘𝑃𝑎 − 86.40𝑘𝑃𝑎) ∗ 0.96
*For 𝑥1 = 1.00 and 𝑦1 = 1.00:
𝑃1∗ 𝑃2∗
𝑃= ∗
𝑃1 + (𝑃2∗ − 𝑃1∗ )𝑦1
𝑃=
(86.40𝑘𝑃𝑎)(32.59𝑘𝑃𝑎)
= 86.40𝑘𝑃𝑎
86.40𝑘𝑃𝑎 + (32.59𝑘𝑃𝑎 − 86.40𝑘𝑃𝑎) ∗ 1.00
x1
0.00
0.05
0.15
0.25
0.37
0.44
0.50
0.65
0.82
1.00
y1
0.00
0.12
0.30
0.46
0.60
0.66
0.72
0.83
0.92
1.00
T (°C)
106.5
104
100
96
92
90
88
84
80
76.3
Table 1. Data for Temperature-Composition Diagram
The table above are the data obtained for the liquid-vapor equilibrium compositions of mixtures of
benzene (1) and toluene (2) at 90 kPa.
IV.
Results and Discussion
Mole Fraction of Benzene in Mole Fraction of Benzene in
Total Pressure, P (kPa)
Liquid Phase, x1
Vapor Phase, y1
0.00
0.00
32.59
0.10
0.23
38.04
0.20
0.40
43.40
0.30
0.53
48.65
0.40
0.64
54.19
0.50
0.73
59.76
0.60
0.80
64.95
0.70
0.86
70.18
0.80
0.91
75.22
0.90
0.96
81.05
1.00
1.00
86.40
Table 2. Data for Pressure-Composition Diagram
The above table presents the mole fraction of benzene in liquid phase (x 1) and in vapor
phase (y1) and the total pressure in kPa of the specific mole fraction. As shown on the given data,
at specified pressure, the mole fraction of benzene in vapor phase is greater than of that in liquid
phase. This is due to the fact that in the benzene-toluene mixture, benzene has lower boiling point
than toluene making it more volatile and faster to vaporize. With this, as the mixture vaporizes,
higher amount of benzene becomes vapor compared to toluene giving it a higher mole fraction in
the system.
Pressure-Composition Diagram
90
80
PRESSURE (KPA)
70
60
50
40
30
20
𝑦1 = 0.73
𝑥1 = 0.505
10
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
X1,Y1
Figure 1. Pressure-Composition Diagram
The figure above shows the Pressure-Composition Diagram of benzene-toluene mixture
where the orange curve represents the saturated vapor curve. It suggests that the pressure that lies
on the curve, entails an increased amount of vapor of the mixture with the specified mole fraction
for each component. At this curve, there is a minimal amount of liquid present, but almost
negligible, as it is a turning point of the vapor-liquid equilibrium. Below the orange curve is the
vapor phase of the mixture where no liquid can exist at certain pressure. On the area between the
curves is the vapor-liquid equilibrium where both vapor and liquid phases of the mixture coexist
at certain pressure. A point in the two-phase region indicates not only qualitatively both liquid and
vapor are present but represents quantitatively the relative amounts of each, this can be get using
the lever rule. The blue curve represents the saturated liquid phase where existence of vapor is
very minimal and almost negligible. Going above the blue curve is the liquid phase of the mixture
which implies pressure conditions that only liquid can exist.
With the benzene’s overall composition of 60 mol % at 60 kPa, the system is in equilibrium
where the mixture exists in both vapor and liquid phases. To get the value of mole fractions of
benzene in both vapor and liquid phases, a tie line is drawn by extending the horizontal line up to
the saturated liquid and saturated vapor curves and creating a vertical line downward, the value for
x1 is given as 0.505, therefore x2 is 0.495. The same method can be used to get the value of y1 and
y2 or by using the formula below.
𝑥1 𝑃1∗
𝑦1 = ∗
𝑃2 + (𝑃1∗ − 𝑃2∗ )𝑥1
𝑦1 =
(0.505)(86.40𝑘𝑃𝑎)
= 0.73
32.59𝑘𝑃𝑎 + (86.40𝑘𝑃𝑎 − 32.59𝑘𝑃𝑎) ∗ 0.505
From the calculation, the mole fraction of benzene in vapor phase is y1=0.73. Therefore,
the mole fraction of toluene in vapor phase is y2=0.27.
Pressure-Composition Diagram
90
80
PRESSURE (KPA)
70
𝑛𝛼
60
𝑛𝛽
50
40
𝑙𝛼
30
20
𝑙𝛽
𝑧𝐴 = 0.60
10
𝑥𝐴 = 0.505
0
0
0.1
0.2
0.3
0.4
0.5
𝑦𝐴 = 0.73
0.6
0.7
0.8
0.9
1
X1,Y1
Figure 2. Pressure-Composition Diagram for Calculating Relative Amount Using Lever Rule
At the equilibrium point where the pressure applied to the mixture is 60 kPa and the overall
composition of benzene is 60 mol %, the relative amount of liquid and vapor phase can be
calculated using the lever rule:
𝑛𝛼 𝑙𝛼 = 𝑛𝛽 𝑙𝛽
𝑛𝛼 𝑙𝛽 𝑧𝐴 − 𝑥𝐴 0.60 − 0.505
= =
=
= 0.731
𝑛𝛽 𝑙𝛼 𝑦𝐴 − 𝑧𝐴
0.73 − 0.60
Therefore, the relative amount of liquid in the equilibrium with xA=0.60 at pressure of 60 kPa is
0.731 and the vapor phase is at 0.269. The point at which the mole fraction of benzene is 0.60 is
closer to the liquid curve which means that at this equilibrium, the mixture is composed of greater
amount of liquid than vapor. As we move the mole fraction of benzene to the left, the liquid
composition increases because benzene is more volatile than toluene.
Temperature-Composition Diagram
120
TEMPERATURE (℃)
100
80
60
40
20
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
X1,Y1
Figure 3. Temperature Composition Diagram
On the figure above, it is shown that from the given data plotted on the temperaturecomposition diagram of the mixture with composition of 𝑥1 = 0.15, five (5) theoretical plates are
needed to distill the liquid mixture in fractionation column achieving a product of at least 95 mol
% of benzene. After the distillation, the condensate is composed of 0.95 mol of benzene and 0.05
mol of toluene. Looking into the result of the distillation, benzene has higher concentration in the
condensate because of the fact that benzene at atmospheric pressure has lower bubble temperature
of 353K compared to toluene with 384K bubble temperature. This means that benzene is more
volatile than toluene making it to vaporize first and faster. As the liquid mixture vaporizes, the
vapor has higher mole fraction of benzene than toluene, then condensing the vapor to liquid giving
it the same composition as the vapor. The process is then repeated and as the condensate vaporizes
for the second plate, benzene will vaporize first and faster giving it a higher composition in the
vapor. The repetition of distillation gives purer mixture of benzene.
V.
Conclusion
The activity was carried out using Raoult’s Law. The objective of this activity is to apply the Law
stated in describing Ideal mixtures and prepare a pressure compositions diagram. The following
are the conclusions drawn from the results obtained from the activity:
•
Based on the x-y table, when the mole fraction of Liquid Benzene (x) increases, the mole
fraction of Vapor Toluene (y) also increases. However, the vapor pressure of pure benzene
is higher than the pure toluene at a certain temperature. Because vapor pressure is a
property of liquid and the said liquid has weak intermolecular forces, wherein, a liquid with
weak intermolecular forces evaporates more easily.
•
•
•
•
VI.
The change of phase within the system at a constant temperature is influenced by pressure.
As pressure increases, the molecules of the components compress, resulting in the
formation of a liquid phase. On the contrary, the vapor phase will occur at some time as
the pressure decreases.
Similarly, with constant pressure the change of phase is influenced by the change in
temperature. As temperature increases, the freer the molecules can move and at some time
vapor phase will occur. Conversely, as temperature decreases at some time, the liquid phase
occurs.
The partial vapor pressure of component A at a particular temperature is proportional to its
mole fraction.
Indeed, Raoult’s law plays a significant role in describing ideal system, it assumes optimal
behavior based on the simple microscopic assumption that intermolecular interactions
between dissimilar and comparable molecules are equal: the ideal solution's conditions.
References
(n.d.). Retrieved from speichim: https://speichim.com/en/techniques/main-principes-ofdistillation/
(n.d.). Retrieved from LibreText:
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook
_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Propert
ies_of_Matter/States_of_Matter/Phase_Transitions/Phase_Diagrams
(n.d.). Retrieved from LibreText:
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook
_Maps/DeVoes_Thermodynamics_and_Chemistry/13%3A_The_Phase_Rule_and_Phase
_Diagrams/132_Phase_Diagrams%3A_Binary_Systems