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1. CHAPTER NAME: SOLUTIONS
MODULE NO :03/07(VAPOUR PRESSURE)
SUBJECT : CHEMISTRY
CLASS :XII
SCHOOL : AECS,MYSORE
PREPARED BY : G BALA KRISHNAIAH
PGT(CHEMISTRY)

2. VAPOUR PRESSURE

3. 1. Vapour pressure
2. Factors affect the vapour pressure
3. Vapour pressure of liquid solutions and Raoult’s
law
Case(i). Raoult’s law for volatile solutions(Liquid
in Liquid , L/L)
Case(ii). Raoult’s law for non-volatile
solutions(Solid in Liquid, S/L)
4. Dalton’s law , Relation between Raoult’s and
Dalton’s Laws
5. Similarities and Dissimilarities between Henry’s
and Raoult’s laws
6. Numerical(s)

4. 1. VAPOUR PRESSURE:
Vapour pressure of a liquid is the pressure
exerted by the vapours in equilibrium with the
liquid at a particular temperature.
? How vapour pressure of any liquid/gas is
measured?(Ans.By Manometer.) then
? How Atmospheric pressure is
measured?(Ans.By Barometer)

5. a. Nature of the substance: If intermolecular
forces between molecules of that substance
are weaker then that substance is easily
volatile/evaporate(goes into vapour state)
?Out of water and petrol, which is more
volatile? Why?(Ans. Petrol)
So out of petrol and water , which will have
low boiling point? (Ans.Petrol(40oc)).
Hence more the volatility, lower is boiling
point OR Higher the vapour pressure lower is
BP

6. The boiling point of a substance is
the temperature at which the vapour
pressure of a liquid equals to the
atmospheric pressure

7. Higher is the temperature, greater is the
vapour pressure.(as kinetic energy
increases more number of liquid
molecules goes into vapour state)
?Which is known as rubbing alcohol?
(Ans. Isopropyl alcohol)

8. Case(i):
Raoult’s law for volatile solutions(Liquid in
Liquid L/L)( Vapour Pressure of Liquid-Liquid
Solutions)
If solution contains volatile liquid, as both the
components of the solution(solvent and solute) are
volatile, each component will form vapour above the
solution. When equilibrium is reached, each
component will exert a vapour pressure, called its
partial pressure whose value depends upon the mole
fraction(amount) of the component in the solution
and the vapour pressure of that component in the
pure state.
These results formulated by Raoult’s, known after him
as Raoult’s Law

9.  Let us consider a mixture of two completely miscible
volatile liquids 1(solvent) and 2(solute), having the mole
fractions x1 and x2. Suppose at a certain temperature,
their partial vapour pressures are p1 and p2 and the
vapour pressures in the pure state are p1
o and p2
o.
According to Raoult’s Law:
p1= p1
o x1 and p2= p2
o x2
If P is the total pressure of the system at the same
temperature, then by Dalton’ law of
partial pressures,
Ptotal= sum of the partial pressures
(partial pressure of solvent + partial pressure of
solute)
=(p1 + p2)
Ptotal = p1
o x1 + p2
o x2
Ptotal =(1- x2) p1
o + p2
o x2
Ptotal =( p2
o - p1
o )x2 + p1
o

11.  If there is a non-volatile solute(solid) in
solution it reduces vapour pressure of
the solution because some of the solute
particles occupy the position of the
solvent molecules on the liquid surface,
which retards the solvent molecules to go
into vapour state.

12. From Dalton’ law of partial pressures,
Ptotal= Sum of the partial pressures
(partial pressure of solvent + partial pressure of
solute)
=(p1 + p2)
But here solute is non-volatile, the value p2=0
Hence the Total pressure equal to vapour pressure of
the solvent
i.e. Ptotal=p1=p1
ox1

13. The reduction in the vapour pressure(lowering of vapour
pressure) of solvent (p1) is given as:
p1 = p1
o-p1
p1 = p1
o - p1
o x1
=(1- x1 ) p1
o
p1
o-p1= x2 p1
o (x1+x2=1)
Relative lowering of vapour pressure(p1
o-p1 ) = x2
p1
o
(Here x2 mole fraction of solute)
The relative lowering of vapour pressure of a solution
containing a non-volatile solute is equal to the mole
fraction of the solute in the solution

14. What are applications of these Laws?
 To calculate the composition of gases
present in vapour state, Dalton’s law is
used
 To relate vapour pressure of the
component in the solution to its mole
fraction, Raoult’s law is used

15. Dalton’s law: Mole fractions of component 1 and component
2 in the vapour phase:
Let mole fractions of 1 and 2 are y1 and y2, partial vapour
pressures of components 1 and 2 are p1 and p2. PT is the total
vapour pressure from components 1 and 2
From Dalton’s law
partial vapour pressure of component 1=(mole fraction of
component 1 ). (Total pressure)
i.e. P1=y1 PT, similarly p2=y2 PT
From Raoult’s Law:
P1= p1
o x1 and p2= p2
o x2
Relating both Laws:
y1 PT, = p1
o x1
y1 = p1
o x1 similarly y2 = p2
o x1
PT, PT

16. Similarities Dissimilarities
Both apply to the volatile
component of the solution
In general Henry’s law is
for solute whereas Raoult’s
law is for
solute/solvent/solution
Both state that the vapour
pressure of any component
in the solution is
proportional to the mole
fraction of that component
in the solution.
The two laws differ in the
proportionality constants.
KH in Henry’s law and po in
Raoult’s law

17. Intext Question
2.8 The vapour pressure of pure liquids A and B are 450 and 700 mm
Hg respectively, at 350 K . Find out the composition of the liquid
mixture if total vapour pressure is 600 mm Hg. Also find the
composition of the vapour phase.
Ans. Based on the data given in the numerical
From Raoult’s Law for volatile solute is present
Ptotal =( p2
o - p1
o )x2 + p1
o OR
Ptotal =( pB
o – pA
o )xB + pA
o
Here p1
o =PA
o= 450mm Hg, p2
o = pB
o=700mm Hg
Ptotal =600mm Hg
Here we have to calculate xA,xB,yA and yB
Ptotal =( pB
o – pA
o )xB + pA
o
by substituting values in the above equation
600=(700-450) xB + 450
250xB=150
xB = 0.6, xA=1-0.6=0.4
In vapour phase pA= yA.Ptotal
yA=pA/Ptotal (pA=xA.pA
o)
=180/600
=0.30, yB = 1-0.30=0.70

18. 2.16. Heptane and octane form an ideal solution.
At 373 K, the vapour pressures of the two liquid
components are 105.2 kPa and 46.8 kPa
respectively. What will be the vapour pressure of
a mixture of 26.0 g of heptane and 35 g of octane?
2.17. The vapour pressure of water is 12.3 kPa at 300 K.
Calculate vapour pressure of 1 molal solution of a
non-volatile solute in it.