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The document discusses the effect of adding two miscible liquids on vapor pressure, primarily through Raoult's law, which states that the vapor pressure of a component in a solution is proportional to its mole fraction and its pure vapor pressure. It also covers the behavior of non-volatile solutes in volatile liquids, explaining how the vapor pressure decreases due to fewer solvent particles at the surface. Additionally, the document includes various examples and calculations related to vapor pressures and ideal versus non-ideal solutions.

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0% found this document useful (0 votes)

22 views61 pages

Ls 1

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ishanaplayz

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What happens to

vapour pressure
when we add two
miscible liquids
According to Raoult's law
According to Raoult's law

Vapour pressure contributed by a volatile component of a liquid solution at a

given temperature is the product of its vapour pressure in pure state and its
mole fraction in solution at equilibrium

Where,
P𝐀 = 𝐏𝐀∘ 𝝌𝐀
χA, χB → Mole fraction of A and B in liquid
P𝐁 =𝐏𝐁° 𝝌𝐁 solution at equilibrium.
𝐏𝐀° , 𝐏𝐁° → Vapour pressure of pure A and B
PTotal = PA + PB PA, PB → Vapour pressure contribution of A
and B respectively.

PTotal = PA + PB = 𝐏𝐀° χA + 𝐏𝐁° χB

In a given phase, sum of the mole fractions of all components = 1

χA + χB =1

So, PTotal = χA PA° + (1–χA) PB°

PS = PB° – χAPB° + χAPA°

PS = PB° + (PA° – PB° ) χA

Y = c + mX
Graph of Raoult’s Law

PA =PA° XA PB =P𝐵° XB

PA° P𝐵°

PA PB

0 1 0 1
XA XB
PB°

Vapour Pressure
PA°

χA = 1 Mole Fraction χA = 0
χB = 0 χB = 1
PB°

Vapour Pressure
d

PA° c

χA = 1 a χA = 1
Mole Fraction
χB = 0 χB = 1
Q.1 Solubility of a specific non-volatile salt is 4 gm in 100 gm H2O at 25°. If 2 gm,
4 gm and 6 gm added to 100 gm water at 25°C in system x, y, z. The order of
V.P. is

Ans : y = z > x
Q.2 INCORRECT statement is:
(A) (KH)O2 > (KH)H2 at T K
(B) Henry constant not applicable on HClg
(C) KH , T 
(D) Henry law apply on very dilute solution.

Ans : (A)
Q.3 Henry constant for solubility of N2 gas in H2O at 298 K is 105 atm. The mole
fraction of N2 in Air is 0.6 the number of mole of N2 from air dissolved in
10 mole of H2O at 298 K and 5 atm is

Ans : 3 × 10–4
Q.4 T K, P = 1 atom
(mass)O3 = M gm (Vol)O3 = V lit dissolved into 1 lit H2O if at same
temperature (Henry Applicable) mass of Vol of O3 dissolved into 2 ⊥ water
and 5 atom will be.

Ans : V2 = 2V
Q.5 Two liquid A and B have vapor pressure in two Ratio PA° ; PB° = 1 : 3 at a
certain temperature A and B from ideal solution the ratio of mole fraction
A to B in vapor phase is 4 : 3 then the mole fraction of B in solution at same
temperature is (1/5).

Ans : 1/5
Q.6 An ideal solution has two component A and B. A is more volatile than B. &
PA° > PB° and also PA° > PT if XA and YB are mole fraction in liquid phase and
vapor phase then correct option is:
(A) XA = YA (B) XA > YA
(C) XA < YA (D) None of these
Q.7 An ideal solution of two liquid A and B has mole fraction 0.3 and 0.5 for the
vapor phase and liquid phase respectively. What would be the mole
fraction of B in vapor phase when mole fraction at A in liquid is 0.25 ?
A B C
Q.8
Vapor

x x x
y y y

distilate

T = 100°C Px° = 300

Py° = 100
Find out mole fraction of x at last if x = 1 mole, y = 1 mole present at
initially at 100°C.
Q.9 One more of component A and 2 mole of component B are mixed at 27° to
form an ideal solution calculate Hmix , Gmix & Smix . (R = 8.314)
Example

Two liquid A and B are mixed to form an ideal solution. If one mole of A
and 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 𝐏𝐀° &
𝐏𝐁° .
(A) 348 mm, 420 mm (B) 600 mm, 400 mm
(C) 588 mm, 348 mm (D) 348 mm, 588 mm

Solution

Ans. (B)
Example

Consider the graph of total vapour pressure (PS) for ideal solution of
A and B at Temperature (T) versus mole fraction of A (XA)

200

PS (mm Hg)

100
Solution 0 XA 1

Select the correct statement.

(1) Boiling point of pure A is more than B
(2) Vapour pressure of pure A at temperature T is 100 mm Hg
(3) Vapour pressure of pure B at temperature T is 200 mm Hg
(4) Boiling point of pure A is less than B
Ans. (D)
Example

The following graph is plotted between the vapour pressures of two

volatile liquids against their respective mole fractions
𝐏𝐁𝐨

𝐏𝐀𝐨

χA= 1 χB= 1
χB= 0 χA= 0
X

Which of the following combinations are correct?

(1) When χA = 1, P = 𝐏𝐀𝐨 (2) When χB = 1, P > 𝐏𝐀𝐨
(3) When χA = 1, P < 𝐏𝐀𝐨 (4) When χB = 1, P > 𝐏𝐁𝐨

Solution Ans. (1,2)

Example

STATEMENT-1 : Henry’s Law and Raoult’s Law are not independent,

i.e., one can be derived from the other.
STATEMENT-2 : The partial pressure is directly proportional to the
mole fraction of concerned species for ideal solutions.

Solution

Ans. (B)
Example

Consider the following vapour pressure-composition graph SP is equal

to :
a. PQ + RS b. PQ + QR
c. SR + SQ d. PQ + QR + RS
Total vapour pressure
Solution
P PB°

PA° PA PB
Q

PB R
PA

0 χ S 1

Ans. (C)
Calculation of YA & YB from Dalton’s law

Vapour phase
YA+YB=1
Liquid phase
χA+χB=1

Dalton’s law of partial pressure

Partial pressure of gas = Mole fraction in vapour phase (Y) × Total pressure of
gas (PT )
PA = YA PT PB = YB PT
Calculation of YA & YB from Dalton’s law

Combining Raoult’s law and Dalton’s law

χA PA° = YA PT

χB PB° = YB PT

χ𝐀 P𝐀∘
Y𝐀 =
P𝐓

χ𝐁 P𝐁∘
Y𝐁 =
P𝐓
In a liquid phase, sum of the mole fractions of all components = 1

χA + χB = 1

YA PT YB PT
+ =1
P°A P°B

YA YB 1
+ =
PA° PB° PT
𝐘𝐀 𝐘𝐁 𝟏
+ =
𝐏𝐀° 𝐏𝐁° 𝐏𝐓

𝐏𝐁°

𝐏𝐀°

0 1
Mole fraction of B
in vapour phase
100% liquid
𝐏𝐁°

𝐏𝐀°

P 100% vapours

0 1
Mole fraction of B
1. Vapour Pressure V/S Vapour Composition

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

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

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

changing the pressure.

At any pressure in between PA° and PB° , the physical state of system may
be changed by changing the composition.
“ Example “
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 “
“ Solution “

Ans : (1) 20mm ,17mm of Hg

(2) 37 mm of Hg
(3) Y(CH3OH)= 20/37
“ Example “
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 𝐏𝐀° & 𝐏𝐁° . If x is
the mole fraction of B in solution.

“ Solution “

Ans : (1) 𝐏𝐀° = 235 mm Hg

(2) 𝐏𝐁° = 100 mm of Hg
“ Example “
Two liquid A and B are mixed to form an ideal solution. If one mole of A
and 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 𝐏𝐀° &
𝐏𝐁° .
(A) 348 mm, 420 mm (B) 600 mm, 400 mm
(C) 588 mm, 348 mm (D) 348 mm, 588 mm

“ Solution “
“ Solution “

Ans : (1) 𝐏𝐀° = 348 mm

(2) 𝐏𝐁° = 588 mm
“ Example “
A solution has two liquids A and B and vapour pressures of A and B in
pure state are 𝐏𝐀° = 80 atm and 𝐏𝐁° = 120 atm. Then find out mole fraction
of A in vapour phase if initially equal moles of A and B are taken.

“ Solution “

Ans : 0.4
“ Example “
Ratio of vapour pressures of A and 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 “

Ans : 0.2
“ Example “
° °
Calculate PT in following cases (𝐠𝐢𝐯𝐞𝐧 𝐏𝐂𝐇𝟑 𝐎𝐇 = 𝟔𝟎, 𝐏𝐂 𝟐 𝐇𝟓 𝐎𝐇
= 𝟖𝟎)
(1) 1 mol CH3OH & 1 mol C2H5OH
(2) 1 mol CH3OH & 3 mol C2H5OH
(3) 3 mol CH3OH & 1 mol C2H5OH

“ Solution “

Ans : (1) 70 mm
(2) 75 mm
(3) 65 mm
What happens when
a non-volatile solute
is added to a volatile
liquid ?
Non-volatile solute :
Glucose, Sugar, Urea, NaCl
etc.
When a non-volatile solute is added to a volatile liquid

It's vapour pressure decreases because less vapours are formed.

Less number of solvent particles are present at the surface of the

solution.
Calculation of vapour pressure

For a non-volatile solute added to a solution, it’s vapour pressure is given

by Raoult’s law :

PS = PA° χA + PB° χB

PB° = 0

𝐏𝐒 = PA° 𝛘𝐀 = 𝐏 ° 𝛘𝐒𝐨𝐥𝐯𝐞𝐧𝐭
At constant temperature vapour pressure of solution containing non-volatile
solute is proportional to mole fraction of solvent.

P𝐀°

0 𝜒A 1
Solutions

Ideal Non-Ideal

Positive Deviation Negative Deviation

Ideal Solutions

Those solutions which obey Raoult's law over the entire range of
concentration are known as ideal solutions.

For Ideal Solutions :

A ••••• A B ••••• B A ••••• B

+ =
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.

For Positive Deviation :

A ••••• A B ••••• B A •••••••••• B

+ =
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.