2. Binary system phase diagrams
• Temperature – composition diagram
𝑇𝐴 : Boiling point of the more volatile component
𝑇𝐵 : Boiling point of the less volatile component
T
x,y
Liquid
L+V
100% A
0%B
0% A
100%B
Vapour
𝑻𝑩
𝑻𝑨
3. Mole fraction
𝑥𝐴 : mole fraction of component (A) in liquid phase
𝑦𝐴 : mole fraction of component (A) in vapor phase
Dalton’s law:
𝑷𝑨 = 𝒚𝑨 ∗ 𝑷𝑻
Raoult’s law:
𝑷𝑨 = 𝒙𝑨 ∗ 𝑷𝑨
𝒐
From Dalton’s and Raoult’s law:
𝒙𝑨 =
𝑷𝑻 − 𝑷𝑩
𝒐
𝑷𝑨
𝒐
− 𝑷𝑩
𝒐
Total pressure:
𝑷𝑻 = 𝑷𝑨 + 𝑷𝑩
5. Example (1)
The vapor pressures of n-heptane and toluene at 373 K are 106 and 73.7 kPa
respectively, while the total pressure is 101.3 kPa. Determine the mole fractions of n-
heptane in the vapor and in the liquid phase at 373 K.
6. Solution (1)
Find : 𝑥𝐴 & 𝑦𝐴
𝒙𝑨 = 0.85
Substitute in Raoult’s equation
𝑃𝐴 = 90.1 kPa
From Dalton’s law
𝒚𝑨 = 0.889
A :
T :
𝐏𝐀
𝐨
:
n-heptane
373 K
106 kPa
B :
𝐏𝐓 :
𝐏𝐁
𝐨
:
Toluene
101.3 kPa
73.7 kPa
𝒙𝑨 =
𝑷𝑻 − 𝑷𝑩
𝒐
𝑷𝑨
𝒐
− 𝑷𝑩
𝒐
𝑷𝑨 = 𝒙𝑨 ∗𝑷𝑨
𝒐
𝑷𝑨 = 𝒚𝑨 ∗𝑷𝑻
8. Example (2)
Determine the vapor phase composition of a mixture in equilibrium with a liquid
mixture of 0.5 mole fraction benzene and 0.5 mole fraction toluene at 338 K. Do you
think the liquid vaporize at pressure of 101.3 kPa?
The Antoine equation constants are given below:
For benzene: A = 6.906 B = 1211.033 C = 220.79
For toluene: A = 6.953 B = 1343.943 C = 219.377
Solution:
𝑦𝐴 , 𝑦𝐵 ???
𝒚𝑨 = 0.734
𝒚𝑩 = 0.266
The liquid will vaporize if the total pressure is higher than the given pressure
𝑷𝒐
= 𝟏𝟎 𝑨 −
𝑩
𝑪+𝑻
𝑷𝑨 = 𝒚𝑨 ∗𝑷𝑻
9. Bubble point & Dew point
• For a mixture:
The boiling range starts at the bubble point and
ends at the dew point
• To get the values of bubble and dew points:
For Bubble point (Tb) 𝒚𝒊 = 𝟏
For Dew point (Td) 𝒙𝒊 = 𝟏
T
x,y
Liquid
L+V
100% A
0%B
0% A
100%B
Vapour
𝑻𝑩
𝑻𝑨
𝑻𝑩𝒖𝒃𝒃𝒍𝒆
𝑻𝑫𝒆𝒘
10. Example (3)
What is the bubble point of an equimolar mixture of benzene and toluene at 101.3
kPa?
Knowing that:
For benzene: A = 6.906 B = 1211.033 C = 220.79
For toluene: A = 6.953 B = 1343.943 C = 219.377
Liquid phase mixture
12. Example (4)
What is the dew point of an equimolar mixture of benzene and toluene at 101.3 kPa?
Knowing that:
For benzene: A = 6.906 B = 1211.033 C = 220.79
For toluene: A = 6.953 B = 1343.943 C = 219.377
Vapor phase mixture
13. Solution (4)
To get the dew point: 𝒙𝒊 = 𝟏
𝑥𝑖 =
𝑃𝑖
𝑃𝑖
𝑜 =
𝑦𝑖 ∗ 𝑃𝑇
𝑃𝑖
𝑜
𝑇𝐷𝑒𝑤 = 98.78°C
The boiling range is 92.1 - 98.78°C
14. Relative volatility
• For a binary system:
A : More Volatile Component
B : Less Volatile Component
∴ 𝑃𝐴
𝑜
> 𝑃𝐵
𝑜
𝜶 =
𝑷𝑨
𝒐
𝑷𝑩
𝒐
𝛼 : Relative volatility
15. Equilibrium curve
• For binary mixtures there are three ways to get the
equilibrium curve:
1. Using the temperature – composition diagram
2. Using the vapor pressure values and the general
equilibrium relation
𝒙𝑨 =
𝑷𝑻 − 𝑷𝑩
𝒐
𝑷𝑨
𝒐
− 𝑷𝑩
𝒐 & 𝒚𝑨 =
𝑷𝑨
𝒐
𝑷𝑻
. 𝒙𝑨
3. Using the relative volatility (𝛼)
𝒚𝑨 =
𝜶. 𝒙𝑨
𝟏 + (𝜶 − 𝟏) 𝒙𝑨
T
y
x
x
16. • For multi-component systems:
1. The ideal systems
𝒚𝒊 =
𝑷𝒊
𝒐
𝑷𝑻
. 𝒙𝒊 = 𝒌𝒊 . 𝒙𝒊
2. The non-ideal systems
𝒚𝒊 =
𝑷𝒊
𝒐
𝑷𝑻
. 𝜸𝒊 . 𝒙𝒊 = 𝒌𝒊 . 𝜸𝒊. 𝒙𝒊
Equilibrium curve
18. Example (5)
The following vapor pressures were obtained for benzene and toluene:
Construct the following diagrams at 1 atmosphere:
• A temperature-composition diagram
• The equilibrium curve
• Relative volatility against mole fraction of toluene In liquid
Temperature, °C Vapor pressure of benzene, mmHg Vapor pressure of toluene, mmHg
80 760 300
92 1078 432
100 1344 554
110.4 1748 760