Chapter 9

Chapter 9 

Ideal and Real Solutions 

Physical Chemistry 2 nd Edition 

Thomas Engel, Philip Reid 

1

Objectives 

• Introduce ideal and real solution 

• Raoult’s law 

• Henry’s law 

• Introduce the concept of the activity 


© 2010 Pearson Education South Asia Pte Ltd 

2

Outline 

1. Defining the Ideal Solution 

2. The Chemical Potential of a Component in the Gas and 

Solution Phases 

3. Applying the Ideal Solution Model to Binary Solutions 

4. The Temperature– Composition Diagram and Fractional 

Distillation 

5. The Gibbs–Duhem Equation 

6. Colligative Properties 

7. The Freezing Point Depression and Boiling Point 

Elevation 



3

Outline 

8. The Osmotic Pressure 

9. Real Solutions Exhibit Deviations from Raoult’s Law 

10. The Ideal Dilute Solution 

11. Activities Are Defined with Respect to Standard States 

12. Henry’s Law and the Solubility of Gases in a Solvent 

13. Chemical Equilibrium in Solutions 

14. Solutions formed from Partially Miscible Liquids 

15. The Solid-Solution Equilibrium 



4

9.1 Defining the Ideal Solution 

• For a particular ideal solution mixture, the partial 

pressure of each component (i) above the liquid is 

given by 

P 

i 

 

x 

i 

P 

* 

i 

i 

 

1,2 

• It is known as Raoult’s law and is the definition of 

an ideal solution. 

• It is only obeyed if the molecule is in the form of A– 

A, B–B, and A–B interactions and are all equally 

strong. 



5

9.1 Defining the Ideal Solution 

• For a particular ideal solution mixture, the partial 

pressure of each component (i) above the liquid is 

given by 

P 

i 

 

x 

i 

P 

* 

i 

i 

 

1,2 

• Raoult’s law is the definition 

of an ideal solution. 

• It is only obeyed if the molecule is in the form of A– 

A, B–B, and A–B interactions and are all equally 

strong. 



6

Example 9.1 

Assume that the rates of evaporation, R evap , and 

condensation, R cond , of the solvent from the surface of pure 

liquid solvent are given by the expressions 

R Ak 

evap 

evap 

* 

Rcond 

AkevapPsolvent 

where A is the surface area of the liquid and k evap and k cond are 

the rate constants for evaporation and condensation, 

respectively. Derive a relationship between the vapor 

pressure of the solvent above a solution and above the pure 

solvent. 



7

Solution 

For the pure solvent, the equilibrium vapor pressure is found by setting the rates 

of evaporation and condensation equal: 

R 

Ak 

P 

evap 

evap 

* 

solvent 

 

R 

 

 

cond 

Ak 

k 

k 

evap 

cond 

cond 

P 

* 

solvent 



8

Solution 

Next, consider the ideal solution. In this case, the rate of evaporation is reduced 

by the factor x solvent 

and at equilibrium 

R 

evap 

 

Ak 

evap 

x 

solvent 

R 

cond 

 

Ak 

cond 

P 

solvent 

The derived relationship is Raoult’s law. 

R 

Ak 

P 

evap 

evap 

solvent 

 

x 

 

R 

cond 

solvent 

k 

k 

evap 

cond 

 

x 

Ak 

cond 

solvent 

P 

 

solvent 

P 

* 

solvent 

x 

solvent 



9

The molecular basis of Raoult's law 

From Atkins 8e, Fig. 5.13 

The large spheres 

represent solvent 

molecules at the surface 

of a solution (the 

uppermost line of 

spheres), and the small 

spheres are solute 

molecules. The solute 

hinder the escape of 

solvent molecules into 

the vapor, but do not 

hinder their return. 



10

Vapor‐Liquid Equilibrium of Binary Liquid Mixtures 

• Raoult 在 1884 年指出部分分壓與莫耳分率的比例關係 : Pi 

x iP* 

i 

• 理想溶液的溶劑蒸氣壓的值與純溶劑蒸氣壓之比等於溶劑的莫耳分率 

。 p A /p A * = x A 

• 對純物質而言 , P A = P A * , P A * 為液態純物質 A 的平衡蒸氣壓 . 

• 於純液體的理想液體中 , 如果溶液中只有 A, B 兩個組成 , 則 x A +x B = 1, 

P 

A 

 

 

 

1 x P * 或 

B 

A 

• 上式為趨近式 , 當理想混合物成分很相似時 , 最為接近 Raoult’s Law. 當兩 

成份混合時分子間作用力 A-A, A-B, 和 B-B 皆相同或相似 , 則 Raoult’s 

Law 最符合 . 

• 氣相的總壓為 P , y i 為氣態中物種 i 的莫耳分率 : y i P = x i P i * 

• 若氣相為理想氣體時 , 成分的活性等於其莫耳分律 . a i = x i . 

• 此為液相理想溶液的定義 : 

μ 

i 

x 

B 

 

P A*-P 

P * 

A 

A 

l 

μi 

l 

RT 

lnx 

i 



11

From Atkins, 8e. Fig. 5.12 

Two similar liquids, in this case 

benzene and methylbenzene 

(toluene), behave almost 

ideally, and the variation of 

their vapor pressures with 

composition resembles that for 

an ideal solution. 



12


• 兩個液體的理想混合物可按任意比例互溶 , 每個成分的蒸氣壓都服從 Roault 定 

律 , 如此組成的理想的完全互溶雙成份液體系統 , 或稱為理想的液體混合物 

, 如苯和甲苯 , 正己烷與正庚烷等結構相似的化合物可形成這種理想的液體 

混合物。 

• 當系統達到平衡時 , 分佈在不同相中的物種 i 的化學勢能相等 : 

i (g) = i ( l ) 

• 假設蒸氣相為理想氣體 , 化學勢能以分壓表示 : 

• 若是液態混合物中成分 i 的化學勢能以活性 a i 表示 : 

• 合併兩式得 : 

μ 

i 

μ 

i 

μ 

P 

 

P 

 

i 

g 

 

μ g 

RT ln 

 

i 

i 

l 

μi 

l 

RT 

lnai 

P 

P 

 

i 

gRT 

ln 

μi 

l RT 

lnai 



13


Pi 

 

μi 

g 

RT ln 

μi 

l RT lnai 

P 

• 此式亦適用於純液體的理想液體系統 : i *(g) = i *(l ) 

• 於純液體的理想液體中 a i =1 且 P i = P i * 

P i * 為液態純物質的平衡蒸氣壓 . 

• 液態純物質的化學勢能表示為 : 

Pi 

* 

μi 

g 

RT ln 

μi 

 

l 

P 

• 相減合併可得 : P 

i 

RT ln 

RT lnai 

P* 

 

i 

• 假如蒸氣為理想氣體 , 則物種在溶液中的活性值等於物種在溶 

液上方成分的部份分壓與純液體的蒸氣壓比 : Pi 

ai 

 

P * 

i 



14


• 利用 Raoult 定律於理想液體的混合物 , 如果溶液中只有兩個成分 , 總壓 

與莫耳分率的比例關係為 : 

• P = P 1 + P 2 = x 1 P 1 *+ x 2 P 2 * = P 2 * + x 1 (P 1 * - P 2 *) 

• 此方程式亦可稱為 Bubble point line. 雙成份理想混合物上方的蒸氣組 

成亦可由 Raoult 定律推得 : 

解 

或解 

P1 

x1P 

1 

* 

y 

1 

 

 

P1 

P2 

x1P1 

* x 

2P2 

* P 

y1P 

2 

* 

x1 可得 x1 

 

P 1 

* P 

2 

*-P 

1 

* y 

1 

P2 

x 

2P2 

* 

y 

2 

 

 

P P x P * x 

P * P * 

x 



2 

可得 x 

2 

 

2 1 

2 

 

P * 

y 

P* 

P*-P 1 2 

* y 2 

x P * 

1 1 

2 

* 

2 2 

1 2 1 1 2 2 1 

 

15 

P 

1 

* P 

2 

* x 

1 

x 

P * 

P 

2 

* P 1 

* x 

2

From Alberty, 4e. Fig 7(a) 

(a) Plot of total pressure for the benzene(2)‐toluene(1) versus x 1 , as 

given by equation: P =P 2 * + (P 1 *‐P 2 *) x 1 



16


• 可以計算總壓 P 而得到蒸氣相的 dew point line. 

y 

x 

 

 

1 1 1 

y 

 

 

1 

- 

P 

2* 

P* 

1 

P* 

x 

2 

P*P 

1 

* 

P 

2* 

 

y P * P *-P 

P*P * 

1 

1 2 

iP x iP* 

i 解 P P* 1 

y 

1 

P 1 

* y 

1 

 

P 

2 

或 

或 

P 

2 

1 

 

2 

1 

y 

P* 

1 1 

- 

 

P 2* 

P* 

2 

P 

1 

1 

2 

1 2 

* y 2 

P 

2 

*-P 

1 

* 1 



17

From Alberty, 4e. Fig 7(b) 

(b) Plot of the total pressure 

for the benzene(2)‐ 

toluene(1) versus y 1 , as given 

by equation: 

P = (x 1 /y 1 ) P 1 * 

= P 1 *P 2 */ [P 1 *+ (P 2 *‐P 1 *) y 1 ]. 



18

Chap 6 Phase Equilibrium 

• From 

• So 

• Also 

x 

y 

x 

1 

1 

 

x 

 

P 

1 

P 

x P * 

1 1 

2 

* 

1 2 

1 

* 

 

P 

1 

* P 

2 

* 1 

y P * 

P 

2 

*-P 

1 

* y 

1 

P-P* 

P*-P * 

1 

2 

2 

x 

P 

1 

*-P 

2 

1 

1 

P 1 

y 1 

P 2 

* y 1 

* 

P P* 

* x 

P*-P 1 2 

* 1 

2 

x 



19

9.2 The Chemical Potential of a Component in the Gas and 


• The chemical potential of species i in a solution is 

solution * Pi 

i 

i 

RT ln 

* 

P 

• For an ideal solution, the central equation describing 

ideal solutions is obtained: 

i 

 

solution 

i 

* RT ln 

i 

x 

i 

• It is useful in describing the thermodynamics of 

solutions in which all components are volatile and 

miscible in all proportions. 



20

9.2 The Chemical Potential of a Component in the Gas and 


• We can derive relations for the thermodynamics of 

mixing to form ideal solutions as follow: 

G 

S 

mixing 

mixing 

nRT 

 

G 

 

 

T 

i 

x 

i 

mixing 

lnx 

 

 

 

i 

p 

nR 

 

i 

x 

i 

lnx 

i 

V 

mixing 

 

 

 

 

G 

P 

mixing 

 

 

 

T , n 

1 , 

n 

2 

 

0 

and 

H 

mixing 

G 

mixing 

TS 

mixing 

 

nRT 

 

i 

x 

i 

lnx 

i 

T 

 

 

 

nR 

 

i 

x 

i 

lnx 

i 

 

 

 

 

0 



21

Example 9.2 

An ideal solution is made from 5.00 mol of benzene and 3.25 

mol of toluene. Calculate G mixing 

and Smixingat 298 K and 1 bar 

pressure. Is mixing a spontaneous process? 



22

Solution 

The mole fractions of the components in the solutions are 

x benzene =0.606 and x toluene =0.394. 

Gmixing 

nRT 

xi 

ln xi 

8.258.314 

298 

13.710 

S 

mixing 

8.258.314 

298 

46.0JK 

1 

3 

J 

nR 

 

i 

i 

x 

i 

 

ln x 

0.606ln 0.606 0.394ln 0.394 

i 

 

0.606ln 0.606 0.394ln 0.394 

 

 

Mixing is spontaneous because G 

always true that G 

mixing 

0. 

mixing 

0. If 

two liquids are miscible, it is 



23

9.3 Applying the Ideal Solution Model to Binary 

Solutions 

• From Raoult’s law, we have the total pressure as follow: 

1 

* 

* * * * 

P total 

P1 P2 

x1P1 

1 

x1 

P2 

P2 

P1 

P2 

x 

• The mole fraction of each component in the gas phase 

can also be calculated. 



24


Solutions 

• Using the symbols y 1 and y 2 to denote the gas-phase 

mole fractions and the definition of the partial pressure, 

y 

1 

 

P 

P 

1 

total 

 

P 

* 

2 

 

x 

1 

P 

* 

1 

 

* * 

P 

1 

P2 

x 1 

• To obtain the pressure in the vapor 

phase as a function of y 1 , 

we first solve x 1 : 

x 

1 

 

P 

* 

1 

 

y 

1 

P 

* 

2 

 

* * 

P 

2 

P1 

y 1 



25


Solutions 

• And obtain P total from 

P total 

 

P 

* 

1 

 

P 

* 

1 

P 

* 

2 

 

* * 

P 

2 

P1 

y 1 

• It can be rearranged to give an equation for y 1 in terms of 

the vapor pressures of the pure components and the total 

pressure: 

y 

1 

 

P P 

P 

* 

1 total 

 

* 

total 

P1 

P 

 

P 

* * 

1 2 

* 

P 

2 



26

Example 9.3 

An ideal solution is made from 5.00 mol of benzene and 3.25 

mol of toluene. At 298 K, the vapor pressure of the pure 

substances are P* benzene =96.4Tor and P* toluene =28.9Tor. 

a. The pressure above this solution is reduced from 760 Torr. 

At what pressure does the vapor phase first appear? 

b. What is the composition of the vapor under these 

conditions? 



27

Solution 

a. The mole fractions of the components in the solution are x benzene =0.606 and 

x toluene =0.394. The vapor pressure above this solution is 

No vapor will be formed until the pressure has been reduced to this value. 

* 

* 

P x P x P 0.60696.4 

0.39428.9 

69. Torr 

total benzene benzene toulene toulene 

8 



28

Solution 

b. The composition of the vapor at a total pressure of 69.8 

Torr is given by 

y 

 

y 

* 

* 

PbenzenePtotal 

P 

 

* 

Ptotal 

( Pbenzene 

P 

96.4 

69.8 96.4 

28.9 

69.8 

96.4 

28.9 

1 

y 0.163 

benzene 

toulene 

benzene 

benzene toulene 

* 

toulene) 

0.837 

Note that the vapor is enriched relative to the liquid in 

the more volatile component, which has the lower 

boiling temperature. 

 

P 



29


Solutions 

• To calculate the relative amount of material in each of 

the two phases in a coexistence region, we derive the 

lever rule for a binary solution of the components A and 

B. 

n 

tot 

liq 

 

tot 

 

z 

B 

 

x 

B 

n 

vapor 

y 

B 

 

z 

B 

• The lever rule determine 

what fraction of the system 

is in the liquid and vapor phases 



30

Chap 9 Phase Equilibrium 

• Tie lines ( 等壓連接線 ): 

Points on the dew point line and bubble 

point line at the same pressure 

represent the compositions of vapor and 

liquid phases that are in equilibrium. 

These point are connected by a 

horizontal line referred to as a tie line. 

x 

n L x B → n V y B 

in between : (n L +n V ) x TB 

The lever rule: 

(n L +n V ) x TB = n L x B + n V y B 

nL 

yB- 

xTB 

(x- v) 

 

n x 

- x ( l-x) 

V 

TB 

B 



31

Example 9.4 

For the benzene–toluene solution, calculate 

a. the total pressure 

b. the liquid composition 

c. the vapor composition when 1.50 mol of the solution 

has been converted to vapor. 



32

Solution 

The lever rule relates the average composition, 

Z benzene =0.606, and the liquid and vapor compositions: 

n 

vapor 

y 

z 

n z 

x 

benzene 

benzene 

Entering the parameters of the problem, this equation 

simplifies to 

6.75x benzene 

1.50y 

 

The total pressure is given by 

y 

benzene 

 

P 

* 

benzene 

P 

total 

P 

P 

* 

P 

* 

liq 

benzene 

benzene 

5.00 

Ptotal 

 

96.4 

 

Torr 

benzene 

 

2786 

total bemzene toulene 

* 

* 

P 

P 

 

P 

benzene toulene 

total 

67.5 

Torr 



33

Solution 

The solution for x benzene obtained from these two equations 

can be substituted in the first equation to give y benzene . The 

answers are x benzene =0.561, y benzene = 0.810, and P total = 66.8 

Torr. 



34

9.4 The Temperature–Composition Diagram and Fractional 

Distillation 

• The temperature–composition diagram gives the 

temperature of the solution as a function of the average 

system composition for a predetermined total vapor 

pressure, P total . 



35


Distillation 

• The upper red curve shows the boiling temperature as a 

function of y benzene , and the lower red curve shows the 

boiling temperature as a function of x benzene . 

• The area intermediate between the two curves shows the 

vapor–liquid coexistence region. 

• In a boiling point diagram, the liquid and vapor 

composition lines are tangent 

to one another at the 

maximum boiling temperature. 



36


Distillation 

• If the A–B interactions are less attractive than the A–A 

and B–B interactions, a minimum boiling azeotrope can 

be formed. 



37


Distillation 

• Other commonly occurring azeotropic mixtures are 

listed. 



38

9.5 The Gibbs–Duhem Equation 

• The azeotropic composition depends on the total 

pressure, thus pure B can be recovered from the A–B 

mixture by first distilling the mixture under atmospheric 

pressure and, subsequently, under a reduced pressure. 



39

From Alberty, 4e. Fig A.11(a) 

(a) Liquid mixture with 

a negative azeotrope: 

Chloroform(1)- 

Acetone(2) at 35.17 

℃. 



40

From Alberty, 4e. Fig A.11(b) 

(b) Liquid mixture 

with a positive 

azeotrope: Carbon 

disulfide(1)‐ 

Acetone(2) at 

35.17 ℃. 



41

From Alberty, 4e. Fig A.12 

Vapor 

Liquid 

Vapor 

(a)Boiling point curve at 

constant pressure for a 

maximum boiling 

point azeotrope. 

(b)Boiling point curve at 

constant pressure for a 

minimum boiling 

azeotrope. 

Liquid 



42

9.5 The Gibbs–Duhem Equation 

• Gibbs–Duhem equation for a binary solution written in 

either of two forms: 

n1d1 

n2d2 

0 or x 

1d 

1 

x 

2d 

2 

 

0 

• This equation states that the chemical potentials of the 

components in a binary solution are not independent. 



43

Example 9.5 

 

One component in a solution follows Raoult’s law, 

solution * 

1 

1 

RT ln x1 

over the entire range 0 ≤ x 1 ≤ 1 . Using the 

Gibbs–Duhem equation, show that the second component 

must also follow Raoult’s law. 



44

Solution 

We have 

d 

 

 

x d 

 

1 1 1 * 

2 

d 1 

RT ln 

x2 

n2 

n 

 

Because dx 2 = –dx 1 and x 1 + x 2 =1. Integrating this equation, 

one obtains μ 2 = RT ln x 2 + C, where C is a constant of 

integration. This constant can be evaluated by 

examining the limit x 2 →1. This limit corresponds to 

solution * 

the pure substance 2 for which 2 2 

RT ln x. 

2 

We conclude that C= μ 2* and, therefore, μ 2 = μ 2* = RT ln 1 +C 

 

x 

1 

 

RT 

x 

x 

1 

2 

dx 

x 

1 

1 



45

Chapter 9

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