According to Raoult's law

At constant temperature partial vapour pressure of component is directly

proportional to mole fraction of component in solution.

P∝

For a solution of volatile liquids the partial vapour pressure of any component
at constant temperature is equal to vapour pressure of pure component
multiplied by mole fraction of that component in the solution.

P ∝χ so P = P∘ χ

P ∝χ so P =P ° χ
According to Dalton's law

Ptotal or Psolution = PA + PB + .....

° °
Psolution = PA + PB = χA + χB

since, (χA + χB =1)

So, PS = (1–χB) P ° + χB P °

PS = P° – χBP ° + χBP °

PS = P° + (P ° – P ° ) χB

Y = c + mX
 PA°

Vapour Pressure
PB°

χA=0 Mole Fraction χA=1

χB=1 χB=0
 Calculation of YA & YB from Dalton’s law

Partial pressure of gas = Mole fraction in vapour phase (Y) × Total pressure of
gas (PT )

PA = YA PT PB = YB PT
χ P ° =Y P χ P ° =Y P

χ P∘ χ P∘
Y = and Y =
P P
Vapour phase
YA+YB=1

YA and YB =Mole fraction in vapour phase Liquid phase

χA and χB =Mole fraction in liquid phase. χA+χB=1
∵ χA + χB = 1

° + ° =1

Y Y 1
+ =
P° P° P
 PA°

Vapour Pressure
PB° c

χA=0 a χA=1
χB=1 Mole fraction of solute in liquid χB=0
Graphs fort ideal binary solution of liquid A and liquid B

01 Vapour pressure v/s liquid composition

PA°
d

Vapour Pressure
So, PA = χAP ° = (1–χB) P ° PB° c

PA = P ° – χBP ° b

PB = χBP ° a
χA=0 χA=1
Mole fraction of solute in liquid
PT = χAP ° – χBP ° = (1–χB) P° + χB P ° χB=1 χB=0

 PT = P ° + χB(P ° – P ° )
As PT = PA + PB , ad = ab + ac
02 Vapour pressure v/s composition

(i) The V.P. of ideal solution always lie in between the V.P. of pure components.

(ii) Below vaporous curve, the system will by 100% vapour & above liquidous
and curve, 100% liquid. Both the physical states exists only in between the
curves.

(iii) At any composition, the physical state of system may be changed by

changing the pressure.

(iv) At any pressure in between PA° and PB° , the physical state of system may
be changed by changing the composition.
 100% liquid
PB°

P 100% vapors

0 1
Mole fraction of B
Vapour pressure v/s vapour composition

= ∘ + ∘
PB°

= ∘ + ∘ − ∘

PA°
= +

P
PB PA

YA = 1 Mole fraction in vapours YA = 0

YB = 0 YB = 1
32 g of methanol (Mw = 32) is added in 23g of ethanol (Mw = 46) to form
an ideal solution. If vapour pressure of CH3OH & C2H5OH in pure state are
30 & 51 mm of Hg. Then Calculate
(1) Partial vapour pressure of CH3OH & C2H5OH.
(2) Total vapour pressure of solution.
(3) Mole fraction of methanol and ethanol in vapour phase

Solution
Two liquids A & B are mixed to form an ideal solution. The total vapour
pressure of solution is 235 – 135 x. Then find the value of ° & ° . If x is
the mole fraction of B in solution.

Solution
Two liquid A and B are mixed to form an ideal solution. If one mole of A
& 4 moles of B are mixed then the total vapour pressure of solution
becomes 540 mm of Hg. If 1 moles of B is further added then total vapour
pressure of solution increases by 8 mm of Hg, Then find the value of ° &
°
.
(A) 348 mm, 420 mm (B) 600 mm, 400 mm
(C) 588 mm, 348 mm (D) 348 mm, 588 mm

Solution
A solution has two liquids A & B and vapour pressures of A and B in pure
state are ° = 80 atm, ° = 120 atm. Then find out mole fraction of A in
vapour phase if initially equal moles of A & B are taken.

Solution
Ratio of vapour pressures of A & B in pure state is 1:2 and ratio of moles of
A and B is also 1:2, then find out mole fraction of A in vapour phase.

Solution
 ° °
Calculate PT in following cases ( = , = )
(1) 1 mol CH3OH & 1 mol C2H5OH
(2) 1 mol CH3OH & 3 mol C2H5OH
(3) 3 mol CH3OH & 1 mol C2H5OH

Solution
Ideal & Non-ideal Solutions

Solutions

Ideal Non-Ideal

Positive Deviation Negative Deviation

An ideal solution possesses the following characteristics :

1. Vapor pressure

P = P∘ χ
P = P∘ χ
P = ( P∘ χ + P∘ χ )

2. Intermolecular interactions

A-----A or B-----B = A-----B

3. Change in volume (Vmix) = Zero

Ex. A + B = A-----B
10 ml 10ml 20ml
Intermolecular distances are same.

4. Change in enthalpy (Hmix) = Zero

No heat is absorbed or evolved as a result of mixing.

5. Change in Gibb’s energy (Gmix) = Negative(–)

Mixing of solutions is spontaneous process.

6. Change in entropy (Smix) = Positive (+)

Entropy increases as a result of mixing.

7. Boiling point

(B.P.)th = (B.P.)exp
Pressure of A and B are equal to expected pressures.
Graph of ideal solution

PA°

Vapour Pressure
PB°

χA=0 Mole Fraction χA=1

χB=1 χB=0
Example of ideal solution

(i) Benzene and toluene

(ii) CCl4 and SiCl4

(iii) n-hexane and n-heptane

(iv) C2H5Br and C2H5Cl

(v) PhCl and PhBr

(vi) n-butylchloride and n-butylbromide

 Non-Ideal Solutions

Those solutions which do not obey Raoult's law over the entire range of
concentration, then they are called non-ideal solutions.
PA ¹ P∘
P ¹ P∘
PS  P ∘ χA + P∘ χB
Non ideal solutions show either positive or negative deviations from Raoult's law.

(a) A positive deviation solution possesses the following characteristics :

1. Vapour pressure

P > P∘ χ
P > P∘ χ
P > ( P∘ χ + P∘ χ )

2. Intermolecular interactions

A-----A or B-----B > A-----B

3. Change in volume (Vmix) = Positive (+)

Ex. A + B = A-----B
10 ml 10ml 20.2 ml
Intermolecular distance increases.

4. Change in enthalpy (Hmix) = Positive (+)

Attraction force decrease

5. Change in Gibb's energy (Gmix) = Negative (–)

Mixing of solutions is spontaneous process.

6. Change in entropy (Smix) = Positive (+)

Entropy increases as a result of mixing.

7. Boiling point

(B.P.)th > (B.P.)exp

'A' and 'B' escape easily showing higher vapour pressure than the
expected value.
Graph of non-ideal solution having positive deviation.

PA°

Vapour Pressure
PB°

χA=0 Mole Fraction χA=1

χB=1 χB=0
Example of positive deviation
(i) Ethanol and cyclohexane
(ii) Ethanol and Water
(iii) Ethanol and Acetone
(iv) Methanol and H2O
(v) CCl4 and Benzene
(vi) CCl4 and Toluene
(vii) CCl4 and CHCl3
(viii) CCl4 and Methanol
(ix) Benzene and Acetone
(x) CS2 and Acetone
(xi) CS2 and Acetaldehyde
(b) A negative deviation solution possesses the following characteristics :

1. Vapour pressure

P < P∘ χ

P < P∘ χ

P < ( P∘ χ + P∘ χ )

2. Intermolecular interactions

A-----A or B-----B < A-----B

3. Change in volume (Vmix) = Negative (–)

Ex. A + B = A-----B
10 ml 10ml 19.8 ml
Intermolecular distance decreases.

4. Change in enthalpy (Hmix) = Negative (Exothermic)

Heat is evolved as result of mixing.

(attraction force increases)
5. Change in Gibb's energy (Gmix) = Negative (–)

Mixing of solutions is spontaneous process.

6. Change in entropy (Smix) = Positive (+)

Entropy increases as a result of mixing.

7. Boiling point

(B.P.)th < (B.P.)exp

'A' and 'B' escape not easily showing less vapour pressure than the
expected value.
Graph of non-ideal solution having negative deviation.

PA°

Vapour Pressure
PB°

χA=0 χA=1
Mole Fraction
χB=1 χB=0
Example of negative deviation

(i) CHCl3 and CH3COCH3

(ii) CHCl3 and C6H6

(iii) CHCl3 and C2H5OC2H5

(iv) CHCl3 and HNO3

(v) CHCl3 and CH3COOH

(vi) H2O and HCl

(vii) H2O and HNO3

(viii) CH3COOH and CH3OH

(ix) CH3COOH and C5H5N

(x) CH3COCH3 and Aniline

Difference between ideal and non ideal solutions

Positive Negative
Properties Ideal Solutions
Deviation deviation

Intermolecular
A-----A or B-----B = A-----B A–A or B–B > A–B A–A or B–B < A–B
interactions

Vapour
Ps = P ∘ χA + P ∘ χB Ps > P ∘ χA + P ∘ χB Ps < P ∘ χA + P ∘ χB
Pressure

Positive (+) Negative (–)

Zero
A + B = A-----B A + B = A-----B
A + B = A-----B
10ml 10ml 20.2ml 10ml 10ml 19.8ml
10ml 10ml 20ml
Vmix
Intermolecular Intermolecular
Intermolecular distances
distances distances
are same.
increase. decreases.
Properties Ideal Solutions Positive Deviation Negative deviation

Zero Positive Negative(Exothermic)

(Endothermic)
No heat is Heat is evolved as result
Hmix
absorbed or Heat is absorbed as of mixing.
evolved as a result result of mixing.
of mixing.

Gmix Negative(–) Negative(–) Negative(–)

Smix Positive (+) Positive (+) Positive(+)