MDCAT Chemistry Notes | Nearpeer

Introduction to fundamental concepts of
chemistry. 2
Gases. 23
Liquids. 60
Atomic Structure. 88
Chemical Bonding. 114
Thermochemistry. 154
Chemical equilibrium. 183
Electrochemistry. 220
Reaction kinetics. 258
S and P Block elements. 286
Transition metals. 321
Fundamental principles of
organic chemistry. 340
Chemistry of hydrocarbon. 365
Alkyl halides. 401
Alcohol and phenol. 421
Aldehyde and ketones. 447
TABLE OF CONTENTS
Book Title

Carboxylic acid. 475
Macromolecules. 495
Chemical formula. 520
TABLE OF CONTENTS
Book Title

1
MDACT
Chemistry
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Chemistry
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Chemistry
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2
Introduction of Fundamental Concepts of
Chemistry
 Atomic mass
 Empirical formula
 Molecular formula
 Concept of mole
 Construction of mole ratios as conversion factors in
Stoichiometry calculations
 Avogadro’s number
 Important assumptions of stoichiometric
calculations
 Stoichiometry
 Limiting reactant
 Percentage yield

3
Relative Atomic Mass
The relative atomic mass is the mass of one atom of an element compared with the mass of one atom of
carbon taken as 12.
Why Carbon is taken as a Standard?
(i) It is a stable element.
(ii) Its isotope 12
6C can be found in the purest form.
(iii) It exists abundantly.
Units of Atomic Mass
The unit used for the atomic mass is Atomic mass unit (a.m.u.).
The mass of 1/12th of an atom of carbon 12 is called atomic mass unit."
Molecular and Empirical Formula
Molecular Formula
Actual number of atoms of different elements present in a molecule is called molecular formula.
Empirical Formula
The simplest whole number ratio of atoms of different elements in a compound is called empirical
formula.
Examples
Compouds Molecular Formula Empirical Formula
Glucose C6H12O6
CH
2
O
Benzene C6H6
CH
Methane CH4
CH
4
Water H2O H
2
O
Did you know?
The Empirical Formula
of sand is SiO2

4
Determination of Empirical Formula:
Empirical formulas can be calculated from the given
1. Percentage composition of the compounds.
% of element =
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝑎 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑
x 100
2. Number of gram atoms
No of gram atoms =
𝐌𝐚𝐬𝐬
𝐀𝐭𝐨𝐦𝐢𝐜 𝐌𝐚𝐬𝐬
x NA
3. Atomic ratio
Divide each number of moles by smallest number of moles to get the mole ratio of elements.
4. Whole number ratio
If ratio is simple whole number, then it gives empirical formula, otherwise multiply with a suitable digit to
get the whole number ratio.
Determination of Molecular Formula:
It can be determined by the following expression:
Molecular Formula = n x Empirical Formula
Here
n =
𝐌𝐨𝐥𝐚𝐫 𝐌𝐚𝐬𝐬
𝐄𝐦𝐩𝐢𝐫𝐢𝐜𝐚𝐥 𝐅𝐨𝐫𝐦𝐮𝐥𝐚 𝐦𝐚𝐬𝐬
Combustion Analysis
It is a quantitative analysis that is used to determine empirical formula of those compounds that contain
carbon, hydrogen and oxygen.
Combustion
It is the burning of a compound in excess of oxygen to form carbon dioxide and water Examples:
(1) Methane combusts in the presence of oxygen
CH
4
+ 2O
2
→CO
2
+ 2H
2
O

5
Figure 1.1: Combustion Analysis
Process:
 Take pre weighed sample in combustion tube fitted in furnace.
 Supply oxygen to burn the compound.
 Hydrogen is converted to H
2
O and is absorbed in Magnesium per chlorate, Mg(ClO
4
)
2
.
 Carbon is converted to CO
2
and is absorbed in 50% KOH.
 The difference in the masses of these absorbers gives us the amounts of H
2
O and CO
2
.
Finding the Percentages of Elements
Following formulae are used to get the percentages of carbon hydrogen and oxygen.
% of carbon =
Mass of CO
2
Mass of Compound
x
12
44
x 100
% of Hydrogen =
Mass of H
2
O
Mass of Compound
x
2.016
18
x 100
The percentage of oxygen is determined by the method of difference.
% of Oxygen = 100 – (% of C + % of H)

6
Concept of Mole
Substance No of particles Quantity Special term
H 6.02 x 1023 atoms 1 mole Gram atom
Cl- 6.02 x 1023 ions 1 mole Gram ion
H2O 6.02 x 1023 molecules 1 mole Gram molecule
NaCl 6.02 x 1023 formula units 1 mole Gram formula
Avogadro’s Number
It is the number of atoms, molecules and ions in one gram atom of an element, one gram molecule of a
compound and one gram ion of a substance, respectively.
Examples
1.008g of H = 1 mole of H = 6.02x10
23
atoms of H
23 g of Na = 1 mole of Na = 6.02x10
23
atoms of Na
18 g of H
2
O = 1 mole of H
2
O = 6.02x10
23
molecules of H
2
O
63 g of NO
3
-1
= 1 mole of NO
3
-1
= 6.02x10
23
ions of NO
3
-1
Relationships
No of atoms of an element =
Mass of element
Atomic mass
x NA
No of molecules of a compound =
Mass of 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑
Molecular mass
x NA
No of ions of an ionic specie =
Mass of ion
Ionic mass
x NA

7
Stoichiometry
The branch of chemistry which deals with the quantitative relationship between reactants and products
in a balanced chemical equation is called Stoichiometry.
Conditions to study Stoichiometry
1. All reactants are completely converted into the products.
2. No side reactions occurs.
3. While doing calculations, law of conservation of mass and law of definite proportions are obeyed.
Stoichiometric Amounts:
The amount of reactants and products in a balanced chemical equations are called Stoichiometric
amounts.
e.g. 2H2 + O2 →
2H2O
In above equation 4 g of H2, 32 g of O2 and 36 g of H2O are called Stoichiometric amounts.
Relationships to study stoichiometry
The following type of relationship can be studied with the help of a balanced chemical equation.
(1) Mass - mass relationship:
If we are given the mass of the one substance, we can calculate the mass of the other substance.
(2) Mass - mole relationship or mole - mass relationship:
If we are given the mass of one substance, we can to calculate the moles of others Substance and vice -
versa.
(3) Mass - volume relationship:
If we are given the mass of one substance, we can calculate the volume of the other substance and vice -
versa.
Limiting Reactant
 It is a reactant that is
 Taken in small amount.
 Consumed earlier in a reaction
 Controls (limits) the amount of product.
 Gives minimum amount of product.

8
Example of Sandwiches
1. If we have 30 shami kababs and five breads “having 58 slices”, then we can only prepare 29 sandwiches.
One kabab will be extra. So slices will be the limiting reactant.
2.When 4 g of H2 reacts with 32 g of O2, 36 g of water is produced.
In this reaction there is no limiting reactant because both reactants are in stoichiometric ratio
3.When 6 g of H2 reacts with 32 g of O2, 36 g of water is produced.
In this reaction, formation of H2O is limited by O2 reactant. 2g H2 remains un-reactive therefore in excess
or non-limiting reactant.
Identification of Limiting Reactant
There are three steps to identify the limiting reactant:
 Calculate the number of moles from the given amount of reactants.
 Find out the number of moles of product with the help of balanced chemical equation.
 Identify the reactant which produces the least amount of product as a limiting reactant.
Yield
The amount of product obtained as a result of chemical reaction is called yield.
Theoretical Yield
It is the maximum amount of product that can be produced from given amount of reactants.
A chemical reaction rarely produces the theoretical yield ofproduct.
A chemist determines the actual yield of a reaction through a careful experiment in which the mass of the
product is measured.

9
Actual Yield
It is the amount of product produced when the chemical reaction is carried out in an experiment.
Actual Yield Is Less than Theoretical Yield
Reaction may be reversible.
Side reaction may occur .
Byproduct formation.
Personal error due to inexperienced worker.
Instrumental error due to faulty apparatus.
Mechanical loss of product during separation by filtrations, separation by distillationseparation by
a separating funnel, washing, drying and crystallization is not properlycarried out decreases the
yield.
Efficiency of Reaction
One way of measuring efficiency is by means of percent yield. So efficiency of reaction is expressed in
terms of percentage yield.
Percent yield
Percent yield of product is the ratio of the actual yield to the theoretical yield expressed as a percent.
Percentage yield =
Actual Yield
Theoratical Yield
x 100

10
Assessment 1
1. Mass of 1.5 mole electron is
a. 0.525 mg
b. 0.625 mg
c. 0.725 mg
d. 0.825 mg
2. Glucose and acetic acid have same
a. Molecular mass
b. Molecular formula
c. Empirical formula
d. No of atoms
3. A compound having empirical formula CH2O gives information about
a. Total no of atoms
b. Total number of hydrogen atoms
c. Simple ratio between atoms
d. All of these
4. Which contain Avogadro’s number of particles
a. 23 g of sodium
b. 22 g of carbon dioxide
c. 1 g of hydrogen gas
d. 23 g of calcium
5. A compound having n= 6 and molecular formula mass 78 g/mol, its empirical
formula may be
a. CH2O
b. CHO
c. CH
d. C2H2O
6. No of formula units in 29.25 g of sodium chloride is
a. 6.02x1023
formula units
b. 6.02x1022
formula units
c. 3.01x1023
formula units
d. 3.01x1022
formula units
7. Simple sugars (monosaccharides) generally have the empirical formula
a. CH
b. CH2O
c. C2H6O
d. CHO

11
8. Which has the more number of molecules
a. 1g of hydrogen
b. 1g of carbon dioxide
c. 1g of methane
d. 1g of ammonia
9. The efficiency of the reaction can be checked by percentage yied. Percentage
yield is always found as
a. Theoretical yield/actual yield
b. Actual yield/theoretical yield
c. Actual yield/theoretical yield x 100
d. Calculated from balanced chemical equation
10. Consider the following reaction:
2Mg + O2 →2MgO
Using 0.05 mole of magnesium, how many no of molecules of magnesium oxide
are produced?
a. 3.01x1023
molecules
b. 3.01x1022
molecules
c. 3.01x1021
molecules
d. 3.01x1024
molecules

12
Assessment 2
1. Which of the following is present in 1 mol of CO2:
a. 6.02x1022 molecules of Carbon dioxide.
b. 6.02x1023 atoms of oxygen
c. 1.2x1024 atoms of carbon
d. 3 moles of atoms.
2. Total number of atoms present in one mole of CuSO4.5H2O is:
a. 1 mole atoms
b. 12 mole atoms
c. 21 mole atoms
d. 6 mole atoms
3. The mass of one molecule of water is:
a. 18 g
b. 3x10-26 g
c. 3x10-26 kg
d. Both a and c
4. The no. of covalent bonds present in 9 gm of water are
a. 6.070 x 1022
b. 6.02 x 1023
c. 3.01 x 1024
d. 3.01 x 1024
5. If two compounds have the same empirical formula but different molecular formula,
they may have:
a. Different kind of atoms
b. Different molecular masses
c. Different simple whole number ratio of atoms
d. All of these
6. In which of the following pairs of compounds the ratio of C, Hand O is same:
a. Acetic acid and methyl alcohol
b. Glucose and acetic acid
c. Fructose and benzene
d. All of these
7. Which of the following is the limitation of writing simple chemical equations?
a. Rate of reaction
b. Mass of product produced
c. Prediction of Intermediate
d. Both a and c

13
8. While dealing with the stoichiometry, the following law does not necessarily to be
obeyed?
a. Law of conservation of mass
b. Law of definite proportion
c. Graham’s law of diffusion
d. None of these
9. In a given reaction; 2KOH(aq) + H2SO4(aq) →
K2SO4(aq) + 2H2O(l), which of the
following is not true?
a. 2 mole of KOH produces 2 mole of water
b. In a reaction 2NA of total molecules are produced.
c. In a reaction, 2 moles of water molecules are produced.
d. All of these
10. The percentage composition of carbon in 180 g of glucose is:
a. 40%
b. 6.66%
c. 60%
d. 46.6%

14
Assessment 3
1. 10 moles of H2 reacts with 7 moles of O2 to produce water in the following reaction:
2H2 + O2 →
2H2O
Which of the following is limiting reactant?
a. Only H2
b. Only O2
c. H2O
d. Both H2 and O2
2. Which of the following statement is not true about the yield?
a. Theoretical yield is always greater than actual yield.
b. Theoretical yield is the yield calculated by experiment.
c. Practically inexperience worker may get the lower actual yield.
d. None of these
3. Efficiency of chemical reaction can be checked by:
a. Limiting reactant
b. Theoretical yield
c. Percentage yield
d. All of these
4. For a given reaction, CaCO (s) →
CaO(s) + CO (s); 100g of limestone is decomposed to
give 28g of lime. Percentage yield of the given reaction is:
a. 100%
b. 75%
c. 50%
d. 25%
5. For the given reaction;
Mg (s) + 2HCl (aq)→
MgCl2(aq) + H2(g)
12g of magnesium is made to react with excess of HCl. The volume of hydrogen gas
produced at STP is:
a. 22.414 dm3
b. 2.2414 dm3
c. 11.2 dm3
d. 1.12 dm3
6. A compound (X) having empirical formula C3H3O; molar mass equal to 110 g/mole.
Its molecular formula is:
a. C3H3O
b. C6H6O
c. C6H6O2
d. C6H6
7. No of gram atoms contained in 66 g of C is:
a. 1.82
b. 5.50
c. 11
d. 12

15
8. Hydrogen burns in chlorine to produce hydrogen chloride. The ratio of masses of
reactants in this chemical reaction is:
a. 2:35.5
b. 1:35.5
c. 1:71
d. 2:71
9. Calculate the no of mole of oxygen atoms in 10.6 g of sodium carbonate:
a. 0.1
b. 0.2
c. 0.3
d. 0.4
10. Actual yield is always less than theoretical yield because of the following reason:
a. Production of side product
b. Reversibility of reaction
c. Mechanical loss
d. All of these

16
Assessment 4
1. A polymer of empirical formula CH2 has a molar mass of 14000 g/mol. Its
molecular formula will be:
a. 100 times that of its empirical formula mass
b. 1000 times that of its empirical formula mass
c. 200 times that of its empirical formula mass
d. 2000 times that of its empirical formula mass.
2. In combustion analysis, which of the following can be measured by method of
differences?
a. Percentage of Oxygen
b. Mass of Carbon dioxide
c. Percentage of hydrogen
d. Both a and b
3. In order to find out the molecular formula, n is the ratio of:
a.
Molecular mass
Atomic mass
b.
Molecular mass
Molar mass
c.
Molecular mass
Empirical formula mass
d. All of these
4. 58.5g of the compound (MCl) contains 35.5 g of chlorine. The metal is:
a. Li
b. Na
c. K
d. Rb
5. Which of the following is true about the combustion analysis?
a. The amount of oxygen is determined by the method of difference.
b. 50% KOH is used to absorb CO2
c. Mg(ClO4)2 is used to absorb H2O
d. All of these
6. During combustion analysis, 6g of organic compound is burnt to produce 4g of
carbon dioxide. %age of carbon in it is calculated as:
a. 10%
b. 18%
c. 20%
d. 50%

17
7. 𝑛 =
𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
𝐸𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠
is unity for which of the following
a. Benzene
b. Glucose
c. Ammonia
d. Hydrogen peroxide
8. The molecular mass of glucose is 180 g/mol. Its empirical formula is CH2O. The value of
“n” is:
a. 1
b. 2
c. 6
d. 3
9. One mole of ethanol and one mole of ethane have equal
a. Mass
b. No of atoms
c. No of electrons
d. No of molecules
10. In a given reaction, SiO2 + 3C →
SiC + 2CO; The amount of silicon carbide produced
when 60 g of sand is heated with excess of carbon:
a. 20 g
b. 40 g
c. 10 g
d. 60 g

18
Assessment 2
Key
Assessment 1
1. D
2. C
3. C
4. A
5. C
6. C
7. B
8. A
9. C
10. B
Assessment 2
1. D
2. C
3. C
4. B
5. B
6. B
7. D
8. C
9. B
10. A

19
Assessment 4
1. A
2. B
3. C
4. C
5. C
6. C
7. B
8. B
9. C
10. D
Assessment 3

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Learning Objectives:
Properties of gases
Gas laws
Boyle’s law
Charles’s law
General gas equation
Kinetic molecular theory of gases
Ideal gas equation
Gases

States of Matter
Matter exists in the following four states:
➢ Gas
➢ Liquid
➢ Solid
➢ Plasma
Note:
➢ The simplest form of matter is the gaseous state.
➢ The most of matter around us (on earth) is in the solid state.
➢ Liquid are less common than solids, gases and plasmas because they can exist only
within a relatively narrow range of temperature and pressure.
Properties of Solids, Liquids and Gases
Sr# Property Gases
1 Volume No definite volume
2 Shape No definite shape
3 Density Very low density
4 Diffusion and effusion Rate of diffusion an diffusion is very high
5 Compressibility Much compressible
6
Expansion
Rate of expansion is very high and sudden expansion is
called Joule Thomson Effect
7 Motion of particles Random motion with high energy
8 Intermolecular forces Very weak intermolecular forces of attraction
Pressure
“The force per unit area is called pressure.”
P = F/A
Units of Pressure:
1. Atmosphere
“The pressure of air that can support 760 mm or 76 cm Hg column at sea level at 0o C is
called an atmosphere.”
OR
“The force exerted by 760 mm or 76 cm Hg column on an area of 1 cm2 at sea level at 0oC is
called one atmosphere.”
1 atm = 760 torr = 760 mm of Hg = 101325 Nm–2 =101325 Pa = 101.325 kPa = 14.7 Psi.

Gas law
“The relationships between volume of a given amount of gas and the prevailing
conditions of temperature and pressure are called the gas law.”
Boyle’s law
Statement:
“The volume of a given mass of a gas at constant temperature is inversely proportional
to the pressure applied to the gas.”
Mathematically:
1
V
p
 (at constant temperature and number of moles)
k
V
p
 Here ‘k’ is proportionality constant
PV k
= (When ‘T’ and ‘n’ are constant)
The value of k is different for the different amounts of the same gas.
Another Statement:
“The product of pressure and volume of a fixed amount of a gas at constant
temperature is a constant quantity.”
So, 1 1 2 2
PV k and P V k
= =
Hence 1 1 2 2
PV P V
=
1 1
PV are the initial values of pressure and volume while 2 2
P V are the final values of
pressure and volume.
EXPERIMENTAL VARIFICATION
➢ Take a gas cylinder with a moveable piston.
➢ This cylinder is also attached with a manometer to read the pressure of the gas directly.

➢ Let the initial volume of gas is 1 dm3and its pressure is 2 atmospheres when the piston
has one weight on it.
➢ When the piston is pressed twice with the help of two equal weights, the pressure now
becomes 4 atmospheres.
➢ Similarly, when the piston is pressed with a mass three times greater, then the pressure
becomes 6 atmospheres.
➢ The initial volume of the gas at 2 atmospheres in 1 dm3. It is reduced to 3
1
dm
2
and then
3
1
dm
3
with increase of weights, respectively.
In short,
3 3
1 1
PV 2atm 1dm 2dm atm k
= = =
3 3
2 2
1
P V 4atm dm 2dm atm k
2
=  = =
3 3
3 3
1
P V 6atm dm 2dm atm k
3
=  = =
It is observed that the value of k will remain the same for the same quantity of a gas at the
same temperature. Hence, Boyle’s law is verified.
Graphical explanation of boyle’s law
➢ If we plot a graph between pressure on the x–axis (abscissa) and volume
on the y–axis (ordinate) at a constant temperature, then a curve is
obtained.
➢ This curve is called isotherm (‘Iso’ means same, ‘therm’ mean heat).
➢ On increasing temperature from 0 0C to 25 0C, isotherm moves away
from both axis.
Reason:
➢ At high temperature the volume of the gas has increased.
➢ Similarly, if we increase the temperature further, make it constant and plot another
isotherm, it goes further away from the axes.

Graph between
➢
1
V
(inverse of volume) on x–axis and the pressure P on the y–axis then a straight line is
obtained.
➢ It means the pressure and inverse of volume are directly proportional to each other. By
increasing the temperature of the same gas from T1 to T2and keeping it constant, one
can vary pressure and volume.
Graph between P and PV
➢ If a graph is plotted between pressure on x-axis and the product PV on y-axis, a straight
line parallel to the pressure axis is obtained. This straight line indicates that ‘k’ is a
constant quantity.
➢ This type of straight line will help us to understand the non-ideal behavior of gases.
Boyle’s law is applicable only to ideal gases.

Charles Law
It is a quantitative relationship between temperature and volume of a gas and was given by
French scientist J. Charles in 1787.
Statement:
“The volume of the given mass of gas is directly proportional to the absolute
temperature at constant pressure.”
Mathematically:
V T
 (at constant pressure and number of moles)
Or V kT
=
Or
V
k
T
=
If temperature is changed from T1 to T2,and volume changes from V1 to V2then
1 2
1 2
V V
k and k
T T
= =
1 2
1 2
V V
constant
T T
= =
Another Statement:
“The ratio of volume to temperature remains constant form same amount of gas at
same pressure.”
Experimental Verification
➢ Consider a certain amount of a gas enclosed in a cylinder fitted with a moveable piston.
➢ The volume of the gas is V1 and its temperature T1.
➢ When the gas in the cylinder is heated, both volume and temperature of the gas
increase.
➢ The new values of volume and temperature are V2 and T2 respectively.
Result:
The experiment shows that:
1 2
1 2
V V
k and k
T T
= =

1 2
1 2
V V
constant
T T
= =
Hence, Charles’s law is verified.
Derivation of Absolute Zero
Absolute Zero:
“The hypothetical temperature at which the volume of a gas is supposed to become
zero if the gas remains in gaseous state.”
0K 237.16 C 459 F
= −  = 
Quantitative Definition of Charles’s Law
“At constant pressure, the volume of the given mass of a gas increases or decreases by
1/273 of its original volume at 0o C for every 1o C rises or fall in temperature respectively.”
General Equation for Volume of Gas:
The equation to know the volumes of the gas at various temperatures is:
t o
t
V V 1
273
 
= +
 
 
Where t
V = volume of gas at temperature T.
o
V = volume of gas at 0o C.
t = temperature of centigrade or Celsius scale.
Explanation:
➢ If a gas is warmed by 1o C, it expands by 1/273 of its original volume at 0o C. since,
original volume is 546 cm3.
➢ So for 1o C rise in temperature, 2 cm3increase the volume will take place.
➢ 2 cm3is the 1/273 of 546 cm3.
➢ Similarly for 100o C rise in temperature, a change of 200 cm3will take place.
➢

Volume-Temperature Data for a Given Amount of a Gas at Constant Pressure
Volumes
(cm3)
Celsius Temperature
(oC)
Temperature
(K)
𝐕
𝐓
= K cm3 K–1
1092 273 546 2
846 150 423 2
746 100 373 2
646 50 323 2
566 10 283 2
548 0 274 2
546 0 273 2
544 –1 272 2
526 –10 263 2
400 –73 200 2
346 –100 173 2
146 –200 73 2
0 –273 0 2
Calculation of Volume at 0oC and -273oC:
at
546 1
V 546
273

= +
3
546 2 248 cm
= + =
at 273 C
546 273
V 546
273
− 
−
= +
546 546
= −
3
0 cm
=
Conclusion:
➢ It is observed from above table shows that at 0o C, the volume of the gas taken is 546
cm3.
➢ This is twice of 273 cm3, and is being supposed for the sake of convinced of
understanding.
➢ At 273o C, the volume of the gas has doubled (1092 cm3) and it should become
practically zero at –273o C.
➢ The volume does not increase corresponding to increase in temperature on Celsius
scale. e.g. the increase in temperature from 1 2
T 10 C to T 100 C
=  =  increase the
volume from 3 3
1 2
v 566 cm to v 746cm .
= =

Application of Charles’s Law
By Using Celsius Scale:
Applying Charles’s law:
1 2
1 2
V V
T T
=
➢ It is observed that, two sides of equation
are not equal.
➢ So it is concluded that, Charles’s law is
not being obeyed when temperature is
measured on the Celsius scale.
By using Kelvin Scale:
➢ Charles’s law is obeyed when the
temperature is taken on the Kelvin scale.
➢ For example, at T1 = 283 K (10o C), the
volume v1 = 566 cm3, while at T2 = K
(100oC), the volume is v2 = 746 cm3.
According to Charles’s law:
1 2
1 2
V V
k
T T
= =
566 746
2 k
283 373
= = =
So Kelvin scale was developed to verify Charles’s law.
Graphical Explanation of Charles Law
If we plot a graph between temperature on x-axis and the volume of one mole of an ideal gas
on y-axis, we get a straight line in which cuts the temperature axis at –273.16oC.
Lowest possible temperature:
➢ This can be possible only if we extrapolate the graph upto -273.16oC.
➢ This temperature is the lowest possible temperature, which would have been achieved if
the substance remains in the gaseous state.
➢ Actually, all the gases are converted into liquids above this temperature.

Effect of mass on slope:
➢ Greater the mass of gas taken, greater will be the slop of straight line.
➢ The reason is that greater the number of moles, greater the volume occupied. All these
straight lines when extrapolated meet at a single point of –273.16oC (0K).
Behavior of a Real Gas:
➢ It is apparent that this temperature of –273.16oC will be attained when the volume
becomes zero.
➢ But for a real gas, the zero volume is impossible which shows that this temperature
cannot be attained for a real gas.
➢ This is how we recognize that –273.16oC must represent the coldest temperature.
Scales of Thermometry
The following three scales are used for temperature measurements.
i. Centigrade Scale or Celsius Scale:
➢ It has a zero mark (0oC) for the temperature of ice at one atmospheric pressure.
➢ The mark 100oC indicates temperature of boiling water at one atmospheric pressure.
➢ The space between these temperature marks is divided into 100 equal parts and each
part is 1oC.
ii. Fahrenheit Scale:
➢ The melting point of ice at 1 atmospheric pressure has a mark 32oF and that of
boiling water is 212oF.
➢ The space between these temperature marks is divided into 180 equal parts and
each part is 1oF.
iii. Kelvin Scale or Absolute Scale:
➢ The melting point of ice at 1 at 1 atmospheric pressure 273 K.
➢ Water boils at 373 K or more precisely at 373.16K.

Interconversion of Temperature Scales:
The following relationships help us to understand the interconversion of various scales of
temperature.
➢ oC to K: K = oC + 273.16
➢ oFtooC:oC = 5/9 (oF–32)
➢ oC to oF: oF = 9/5 (oC) + 32
General Equation
According to Boyle’s law:
1
V
P
 (When n and T are constant)
According to Charles’s law:
V T
 (When n and P are constant)
According to Avogadro’s law:
V n
 (When P and T are constant)
By combining all the gas laws:
nT
V
P

nT
V constant
P
=

General Gas Constant:
The constant suggested is R which is called general gas constant.
nT
V R
P
=
PV nRT
=
➢ The general gas equation shows that if we have any quantity of an ideal gas then the
product of its pressure and volume is equal to the product of number of moles, general
gas constant and absolute temperature.
For one mole of a gas:
➢ For one mole of a gas, the general gas equation is:
n 1
=
PV
PV RT or R
T
= =
It means that ratio of PV to T is a constant quantity (molar gas constant).
Hence,
1 1 2 2
1 2
PV P V
R R
T T
= =
Therefore,
1 1 2 2
1 2
PV P V
T T
=
Applications of The General Gas Equation
i. Calculation of the Density of a Gas:
The general gas equation is
PV nRT
=
n = Number of moles of the gas
n =
mass of the gas in g m
molar massofthegas M
=
Hence,
m
PV RT
M
=
m
PM RT
V
=

mass
PM dRT d
Volume
PM
d
RT
 
= =
 
 
=
ii. Calculation of the Mass of a Gas:
PM
d
RT
=
m PM
V RT
=
PMV
m
RT
=
Ideal Gas Constant (R)
Avogadro’s principle is used to calculate values and units of R. According to this principle
“The volume of one mole of an ideal gas at STP (one atmospheric pressure an d273.16K) is
22.414 dm3.”
➢ Its value depends upon the units chose for pressure, volume and temperature.
➢ General gas equation is used to calculate the value of R.
➢ To evaluate ‘R’, the general gas equation can be written as
PV nRT
=
PV
R
nT
=
a. When P in atm and V in dm3:
P = 1 atm n = 1 mole
V = 22.414 dm3 T = 273.16 K
3
PV 1atm 22.414dm
R
nT 1mol 273.16K

= =

3 -1 -1
R = 0.0821 atmdm mol K
Physical meanings of value of R:
➢ If we have one mole of an ideal gas at 273.16 K and one atmospheric pressure and its
temperature is increased by 1K, then it will absorb 0.0821 dm3 atm is the unit of energy
in this situation.

➢ Hence, the value of R is a universal parameter for all the gases.
➢ It tells us that the Avogadro’s number of molecules of all the ideal gases have the same
demand of energy.
b. When P in torr or mmHg and V in dm3:
P = 760 torr n = 1 mole
V = 22.414 dm3 T = 273.16 K
3
PV 1atm 22.414dm
R
nT 1mol 273.16K

= =

s
1 1
K
− −
3
R = 62.4 dm torr or mmHg mol
c. When P in torr or mmHg and V in cm3:
P = 760 torr n = 1 mole
V = 22.414 dm3 T = 273.16 K
3
PV 760torr 22414cm
R
nT 1mol 273.16K

= =

1 1
K
− −
3
R = 62400 cm torr or mmHg mol
d. Value and Units of R in SI:
The SI units of pressure are Nm–2 and of volume are m3. By using Avogadro’s principle.
n = 1mol T = 273.16K
V = 22.414 dm3 = 0.022414 m3
3
PV 1atm 22.414dm
R
nT 1mol 273.16K

= =

-1 -1 -1
R = 8.3143 NmK =8.3143JK mol (1Nm = 1J)
Keep in mind that, wherever the pressure is given in Nm–2 and the volume in m3, then the
value of R used must be 8.3143 JK–1mol–1.
e. Since 1cal. = 4.18 J:
So, 1 1
8.3143
R 1.987 calK mol
4.18
− −
= =
f. When energy is expressed in ergs:
Since Ij 10erg
=
So, 1 1
R 8.3143 J.mol K
− −
=
So, 7 -1 -1
R = 8.3143×10 ergsmol K

Avogadro’s Law
Statement:
“Equal volumes of all the ideal gases at the same temperature and pressure contain
equal number of molecules.”
Explanation:
One mole of an ideal gas at 273.16 K and one atmospheric pressure (STP) has a volume of
22.414 dm3 and one mole of a gas has Avogadro’s number of molecules.
So, 22.414 dm3 of ideal gas at STP will have Avogadro’s number of molecules i.e., 23
6.02 10

molecules.
Examples:
3 23
2
H 2g 1 mole 22.414dm at STP 6.02 10 molecules
= = = = 
3 23
2
O 32g 1 mole 22.414dm at STP 6.02 10 molecules
= = = = 
3 23
2
N 28g 1 mole 22.414dm at STP 6.02 10 molecules
= = = = 
3 23
4
CH 16g 1 mole 22.414dm at STP 6.02 10 molecules
= = = = 
➢ So, one mole of all gases at STP will have same volume of 22.414 dm3 and same number
of molecules i.e., 23
6.02 10
 but their masses are not equal.
➢ Similarly if we have one dm3 of H2, O2, N2 and CH4in separate vessels at STP, then they
have number of molecules i.e., 22
2.68 10
 molecules.
➢ Although, oxygen molecule is 16 times heavier than hydrogen, but this does not disturb
the volume occupied, because molecules of the gases are widely separated from each
other at STP.
➢ One molecule is approximately at a distance of 300 times its own diameter from its
neighbor at room temperature. Analytical chemistry is the science of chemical
characterization.
Kinetic Molecular Theory Of Gases (Kmt)
“A set of postulates that describes the nature and behavior of an ideal gas is called
kinetic molecular theory of gases.”
Fundamental postulates:
➢ Every gas consists of a large number of very small particles called molecules. Gases like
He, Ne, Ar have mono-atomic molecules.
➢ The molecules of a gas move haphazardly (randomly), colliding among themselves and
with the walls of the container and change their directions.

➢ The pressure exerted by a gas is due to collisions of its molecules with the walls of a
container. The collisions among the molecules are perfectly elastic.
➢ The molecules of a gas are widely separated from one another and there are sufficient
empty spaces among them.
➢ The molecules of a gas have no forces of attraction for each other.
➢ The actual volume of molecules of a gas is negligible as compared to the volume of gas.
➢ The motion imparted to the molecules by gravity is negligible as compared to the effect
of the continued collisions between them.
➢ The average kinetic energy of the gas molecules varies directly as the absolute
temperature of the gas.
Clausius’ Kinetic Equation
On the basic of kinetic molecular theory, R.J. Clausius deduced an expression for the pressure of
an ideal gas as:
2
1
PV mNc
3
=
Where P = Pressure, V = Volume, m = Mass of one molecule of the gas
N = Number of molecules of gas in the vessel
2
c = Mean square velocity
Mean Square Velocity:
➢ All the molecules of a gas under the given conditions don’t have the same velocities.
Rather different velocities are distributed among the molecules.
➢ It is explained in Maxwell’s law of distribution of velocities. If there are n1 molecules
with velocity c1, n2 molecules with velocity c2 and so on then:
2 2 2
2 1 2 3
1 2 3
c c c .....
c
n n n .....
+ + +
=
+ + +
In this reference,
1 2 3
n n n ..... N
+ + + =
Root Mean Square Velocity:
➢ 2
c is the average of the squares of all the possible velocities.
➢ When we take the square root of this 2
c , then it is called root mean square velocity
(Crms). So,
2
rms
C c
=
The expression for the root mean square velocity deduced from the kinetic equation is written
as follows:
rms
3RT
C
M
=

Where,
rms
C = Root mean square velocity
M = Molar mass of the gas
T = Temperature
Conclusion:
The above equation is a quantitative relationship between absolute temperature and the
velocities of the gas molecules.
According to this equation, higher the temperature of a gas, greater the velocities.
Explanation of Gas Laws From Kinetic Theory Of Gases
i. Boyle’s Law:
According to one of the postulates of kinetic molecular theory of gases,
“The kinetic energy is directly proportional to the absolute temperature of the gas.”
The kinetic energy of ‘N” molecules is 2
1
mNc
2
So, 2
1
mNc T
2

2
1
mNc kT .....(i)
2
=
Where ‘k’ is the proportional constant.
According to the kinetic equation of gases.
2
1
PV mNC
3
=
Multiplying and dividing by 2 on right hand side
2
2 1
PV mNc
2 3
 
=  
 
2
2 1
PV mNc ........(ii)
3 3
 
=  
 
Putting eq. (i) into eq. (ii)
2
PV kT ........(iii)
3
=
If the temperature (T) is constant then right hand side of eq. (iii)
2
kT
3
is constant.
Let that constant be ‘k’.
PV = k’ (Boyle’s law)
Conclusion:
At the constant temperature and number of moles, the product PV is a constant quantity.

ii. Charles’s Law:
According to one of the postulates of kinetic molecular theory of gases,
“The kinetic energy is directly proportional to the absolute temperature of the gas.”
The kinetic energy of ‘N’ molecules is 2 1
c mN
2
So, 2
1
mNc T
2

2
1
mNc kT .....(i)
2
=
Where ‘k’ is the proportional constant.
According to the kinetic equation of gases
2
1
PV mNc
3
=
Multiply and dividing by 2 on right hand side
2
2 1
PV mNc
2 3
 
=  
 
2
2 1
PV mNc ......(ii)
3 2
 
=  
 
Putting eq. (i) into eq. (ii)
2
PV kT ......(iii)
3
=
2k
V T
3P
 
=  
 
At constant pressure,
2k
k
3P

= (a new constant)
Therefore, V k T

=
Or
V
k (Charles's law)
T

=
Conclusion:
At a constant pressure and number of moles, the ration of V/T is a constant quantity.
iii. Avogadro’s Law
Consider two gases 1 and 2 at same pressure P and having same volume V. Their number of
molecules are N1 and N2, masses of molecules are m1 and m2 and mean square velocities
are 2
1
c and 2
2
c respectively. Their kinetic equation cam be written as follows:
For gas 1: 2
1 1 1
1
PV m N c
3
=

For gas 2: 2
2 2 2
1
PV m N c
3
=
Equalizing:
2 2
1 1 1 2 2 2
1
m N c m N c
3
=
Hence 2 2
1 1 1 2 2 2
m N c m N c ........(i)
=
When the temperature of both gases in the same, their mean kinetic energies per molecule
will also be same, so
2 2
1 1 2 2
1
m c m c
2
=
2 2
1 1 2 2
m c m c ........(ii)
=
Dividing eq. (i) by eq. (ii)
2 2
1 1 1 2 2 2
2 2
1 1 2 2
m N c m N c
m c m c
=
1 2
N = N
Conclusion:
Equal volumes of all the gases at the same temperature and pressure contain equal number
of molecules, which is Avogadro’s law.
iv. Graham’s Law of Diffusion
According to kinetic equation
2
1
PV mNc .....(i)
3
=
If we take one mole of a gas having Avogadro’s number of molecules (N=NA) then the
equation (i) can be written as:
2
A
1
PV mN c
3
=
Or 2
A
1
PV Mc .....(ii) (M mN )
3
= =
Where M is the molecular mass of the gas.
Or 2 3PV
c
M
=
Taking square root.
2 3PV
c
M
=

2 3P
c
M / V
=
2 3P M
c d
d V
 
= =
 
 
‘V’ is the molar volume of the gas at given conditions. Since the root mean square velocity
of the gas is proportional to the rate of diffusion of he gas.
2
c r

So,
3P
r
d
=
At constant pressure,
1
r
d

This Graham’s law of diffusion.
Kinetic Interpretation of Temperature
The kinetic gas equation is givens as:
2
1
PV mNc ......(a)
3
=
Where P = Pressure
V = Pressure
m = Mass of one molecule of the gas
N = Number of molecules of the gas
2
c = Mean square velocity
The average kinetic energy associated with one molecule of a gas due to its translational
motion is given below:
2
K
1
E mc ......(b)
2
=
In the above equation EK represents the average translational kinetic energy of gas molecules.
2
2 1
PV mNc
2 3
= 
2
2 1
PV N mc .....(c)
3 2
 
=  
 
Putting equation (b) into equation (c), we get
K
2
PV NE .....(d)
3
=
If we use one mole of a gas, then N = NA
Now put the value of N in the equation (d)

A K
2
PV N E .....(e)
3
=
According to the general gas equation for one mole of a gas
A K
2
RT N E
3
=
K
A
3R
E
2N T
=
But
A
3R
2N
is a constant quantity. So the above equation can be written as:
K
E Constant T
= 
K
E T

➢ The above equation shows that absolute temperature or Kelvin temperature of a gas is
directly proportional to the average translational kinetic energy its molecules.
➢ This suggests that a change in temperature mean change in the intensity of molecular
motion
Flow of Heat:
➢ When het flows from one body to another body, the molecules in the hotter body give
up some of their kinetic energy through collisions to the molecules in the colder body.
➢ This process of flow of heat continues until the average translational kinetic energies of
all the molecules in both bodies become equal.
Interpretation of temperature of gases, liquids and solids
➢ In gases and liquids, temperature is the measure of average translational kinetic
energies of molecules.
➢ In solids,Where molecules cannot move freely temperature becomes a measure of
vibration kinetic energy.
Interpretation of Absolute Zero:
➢ In the light of kinetic molecular interpretation, Absolute temperature is the temperature
at which molecular motion ceases.
➢ The absolute zero is unattainable.
➢ However current attempts have resulted in a temperature as low as 10–5K.
Non-Ideal Behavior Of Gases

The gases that obey gas laws (Boyle’s law and Charles’s law) and resulting general gas
equation under all conditions of temperature and pressure are called ideal gases.
The gases that donot obey gas laws (Boyle’s law and Charles’s law) and resulting general
gas equation under all conditions of temperature and pressure are called non-ideal gases.
Explanation:
Compressibility Factor:
➢A graph is plotted between pressure on x-axis and PV/RT on y-axis for ideal gas.
The factor PV/RT is called compressibility factor.
➢Its value is unity for 1mole of an ideal gas.
Behavior of Ideal Gas:
For an ideal gas, increase of pressure
decreases the volume such that PV/RT remains
constant at a constant temperature and a straight
line is obtained parallel to x-axis.
Behavior of Real Gases at 0o
C:
All the real gases have been found to
show marked deviations from this behaviour as
discussed below.
He Gas:
Graph for He gas goes along with expected
horizontal line to some extent but goes above this
line at very high pressures. It means that at very
high pressure, the decrease in
volume is not according to general gas equation and the value of PV/RT has increased from the
expected values. With this type of behavior, we would say that the gas is non-ideal.
H2 Gas:
In the case of H2, the deviation starts even at low pressure in comparison to He.
N2 Gas:
N2 shows a decrease in PV/RT value at the beginning and shows marked deviation even
at low pressure than H2.
CO2 Gas:
CO2 gas has a very strange behaviour as it is
evident from the graph.
Limitations For Gases
2.00
1.50
1.00
0.5
PV
RT
0 200 400 600 800 1000
CO
N
2
2
H
He
2
Ideal
gas
T = 0
o
C or 273K

The extent of deviation of these four gases
shows that these gases have their own limitations
for obeying general gas equation. It depends upon
the nature of the gas that at which value of
pressure, it will start disobeying.
Behavior of real gases at 100 o
C
If the behaviour of all these four gases at
elevated temperature (100oC) is studied, then the
graph comes closer to the expected straight line
and the deviations are shifted towards higher
pressure. This means that the increase in
temperature makes the gases ideal.
Conclusion:
This discussion on the basis of experimental observations convinces us that:
1. Gases are ideal at low pressure and non-ideal at high pressure.
2. Gases are ideal at high temperature and non-ideal at low temperature.
Note: The gases are ideal under the conditions where they have negligible forces between their
molecules and vice versa.
Causes for Deviation From Ideality
In 1873, Van der Waal’s attributed the deviation of real gases from ideal behaviour to
two of the eight postulates of kinetic molecular theory of gases.
Faulty Postulates Of Kinetic Molecular Theory:
The faulty postulates of kinetic molecular theory are as follows:
1) There are no forces of attraction among the molecules of a gas.
2) The actual molecular volume of a gas is negligible as compared to volume of the
vessel.
Forces Of Attraction:
When the pressure on a gas is high and the
temperature is low, then the attractive forces
among the molecules become significant, so the
ideal gas equation i.e. PV = nRT does not hold.
Actually, under these conditions, the gas does not
remain ideal.
`
Actual Volume Of Gas Molecules:
Volume of the
molecules of a
gaseous state
(a) (b)
2.00
1.50
1.00
0.5
0 200 400 600 800 1000
P (atm)
CO
N
2
2
H
He
2
Ideal
gas
PV
RT
T = 100 C or 373K
o

The actual volume of the molecules of a gas is usually very small as compared to the
volume of the vessel and hence, it can be neglected. This volume, however, does not remain
negligible when the gas is subjected to high pressure as shown in the figures.
Vander Waal’s Equation For Real Gases
Vander Waal’s pointed out that both pressure and volume factors in ideal gas equation needed
correction to make it applicable to the real gases.
1) Volume Correction:
Volume of the gas molecules cannot be ignored in the vessel. They do occupy certain
space. Because the molecules of a real gas are rigid spherical particles which possess a definite
volume. Hence, the actual volume of molecules cannot be ignored in the highly compressed
gas. This volume is called effective volume of molecules if we have taken initially one mole of
the gas, then the effective volume is represented by ‘b’.
Free Volume:
The volume available to the gas molecules is the volume of the vessel minus the volume
of gas molecules. This available volume is also called free volume. This was considered
to be the ideal volume.
Vfree = Vvessel – b ---------------------- (1)
‘b’ is also called excluded volume. It depends upon the size of the gas molecules.
Vi = V – b for 1 mole.
Vi = V – nb for total no. of moles.
Minimum volume occupied by 1mole of highly compressed gas in gaseous state is called effective
volume. It is also called excluded volume or incompressible volume. It is represented by ‘b’.
The volume occupied by the 1mole gas molecules in their closest approach is called actual volume. It is
represented by Vm.
Effective volume (b) is four times greater than actual volume (Vm).
b = 4Vm
2) Pressure Correction:
A molecule in the interior of the gas is attracted by other molecules on all sides. These
forces of attractions cancel the effect of each other. When a molecule strikes the wall of a
container, it experiences a force of attraction towards other molecules in centre of gas. This
decreases the impact of its force on the wall.
Decrease of pressure due to the attractive forces:
Consider the molecule ‘A’ which is unable to create pressure on the wall due to
presence of attractive forces of ‘B’ type molecules. This is shown in the figure below.
Let the observed pressure on the wall
of container is P. This pressure is less
than the ideal pressure Pi (calculated
A
B
B
B
B
A
B
B
B
Inward
pull on A
Molecular attractions
are balanced
Wall of
vessel
Velocity of A
reduced B
B
B
A

from general gas equation) by an
amount P/. So,
P = Pi – P/
Pi = P + P/-------(i)
Pi is the true kinetic pressure if forces
of attractions would have been
absent.
P is the pressure measured and called actual pressure. As the volume increases, P
decreases.
P/ is the amount of pressure lessened (decreased) due to attractive forces. P/ is expressed
in terms of a constant ‘a’ which accounts for attractive forces. That is why ‘a’ is
called Co-efficient of attraction for one mole of gas. P/ for 1mole of a gas can be
represented as.
P/ = 2
a
V
How to Prove It:
P/ is determined by the forces of attraction between molecules of type A (which are
striking the wall of the container) and molecules of type B (which are pulling them
inward). The net force of attraction is proportional to the concentrations of A type and B
type molecules.
P/  CA.CB
Let n is the number of moles of A and B separately and total volume of both types of
molecules is ‘V’. The n/V is moles dm-3 of A and B, separately.
P/  n n
.
V V
P/ 
2
2
n
V
P/
2
2
an
=
V
P/ = 2
a
V
-----------(ii) (for 1 mole n = 1)
This ‘a’ is a constant of proportionality and is called co-efficient of attraction or
attraction per unit volume. It has a constant value for a particular real gas.
The value of lessened pressure P/ is greater when
(i) Attractive forces among the gas molecules are stronger
(ii) Volume of vessel is smaller

Thus, effective kinetic pressure of a gas is given by Pi, which is the pressure if the gas
would have been ideal.
Put the value of the P/ from eq. (ii) into eq. (i) to get Pi
Pi = P + 2
a
V
Once the corrections for pressure and volume are made, the general gas equation for
one mole of a gas can be constructed by taking pressure as (P + 2
a
V
) and volume as (V -
b).
Now,
(P + 2
a
V
) (V - b) = RT
For ‘n’ moles of a gas
(P +
2
2
a n
V
) (V - nb) = nRT
This is called Van der Waal’s equation. ‘a’ and ‘b’ are called Van der Waal’s constants.
Units of ‘a’:
Since, P/ =
2
2
an
V
So, a =
2
2
P'V
n
i) In common units, pressure is in atm and volume is in dm3
a =
3 2
2
atm×(dm )
(mol)
a = atm.dm6.mol-2
ii) In SI units, pressure is in Nm-2 and volume in m3
a =
-2 3 2
2
Nm ×(m )
(mol)
a = Nm-2. m6.mol-2
a = Nm+4mol-2
Unit of ‘b’:
bis excluded or incompressible volume /mol-1 of gas, b = V / n. Hence, its units should be
i) Common units: b =dm3 mol-1
ii) SI units: b = m3mol-1

Assessment 1
1. At STP, Carbon dioxide bubbles through water and rises up because of:
a. High External pressure
b. Low Density
c. High molar mass
d. Less no of atoms
2. Which of the following has more ability to diffuse?
a. Carbon dioxide gas
b. Red ink
c. Ice vapours
d. Hydrogen gas
3. The pressure of air that can support 760 mmHg column at sea level, is called:
a. Standard pressure
b. Atmospheric pressure
c. Sea level pressure
d. All of these
4. The most widely used unit in Engineering work which is equivalent to one atm
a. 1.47 psi
b. 14.7 psi
c. 17.4 psi
d. 17psi
5. The SI unit of pressure is usually expressed in terms of
a. Nm2
b. torr
c. N/m2
d. atm
6. Intermolecular forces are the cohesive forces of attraction by which molecules cling to each
other. The strength of these forces in gases are
a. Strong
b. Stonger
c. Weak
d. Moderate
7. The human body temperature is 98.6 0F. In centigrade it is
a. 320C
b. 370C
c. 320C
d. 31 0C

8. Robert Boyle investigated the behavior of gases experimentally to explain which of the
following relationships
a. Volume is proportional to pressure at constant temperature
b. Volume is directly proportional to temperature at constant pressure
c. Volume is inversely proportional to pressure at constant temperature
d. Volume is proportional to no of moles at constant P and T
9. Keeping the pressure constant, the temperature of an ideal gas is changed from 10 K to 30 K,
its volume changes from 1 dm3 to
a. 1.2 dm3
b. 2.4 dm3
c. 3.0 dm3
d. 4.8 dm3
10. The number of molecules in 0.0224m3 of oxygen gas at 25°C and 1 atm pressure
a. NA
b. NA/2
c. 2NA
d. 1.5NA

Assessment 2
1. Gases are compressed by applying pressure on them. Why compression is significant and
easy in case of Gases than liquids and solids
a. Stronger attractive forces
b. Low densities
c. Large empty spaces among gas molecules
d. Large available space to gas
2. Keeping the pressure constant, the temperature of gas is changed from 0 0C to 546 K, its
volume changes from 1 dm3 to
a. 2.0 dm3
b. 2.4 dm3
c. 3.6 dm3
d. 4.8 dm3
3. The following curve is obtained when we plot a graph between:
a. Pressure on x-axis and volume on y-axis keeping temperature changed
b. Pressure on x-axis and pv on y-axis keeping temperature unchanged
c. Pressure on x-axis and 1/v on y-axis keeping temperature unchanged
d. Pressure on x-axis and volume on y-axis keeping temperature unchanged
4. General gas equation is used to determine the density of gas. Density of the gas can be
measured as
a. d =PM/R
b. d = nRT
c. d = PM/RT
d. d = P/RT
5. The mass of 22.414 dm3 of ammonia at 00C and 760 torr
a. 0.907g
b. 9.07g
c. 17g
d. 1.7g
6. At constant temperature, density changes with pressure. If pressure is decreased then
density will
a. remains same
b. decreases
c. change but not significantly
d. increases three times

7. Which of the following is true when a gas, enclosed in a cylinder at standard temperature and
pressure having volume 22.414 dm3, is heated to 298K?
a. The kinetic energy of gas molecules increase
b. Collisions between molecules decrease
c. Volume of gas is decreased
d. Mass of gas is increased
8. According to kinetic molecular theory of gases, gas molecules collide with each other and with
the walls of the container. The collisions among them are perfectly elastic. Which of the
following is true when the gas molecules undergo elastic collisions?
a. Energy of system is changed
b. Kinetic energy is changed
c. Potential energy is changed
d. Total energy of system remains same
9. Root mean square velocity is determined by taking square root of the mean square velocity of the
gas molecules. Its expression is:
a. √
3𝑅
𝑀
b. √
3𝑅𝑇
𝑀
c. √
3𝑅𝑇
𝜋𝑀
d. √
3𝑅𝜋
𝑀
10. Which of the following curve can elaborate the ideal behaviour of given mass of gas at
constant temperature?
a.
b.
c.
d.

Assessment 3
1. Kinetic interpretation of temperature was done by Clausius with the help of kinetic gas
equation. According to him, temperature is the measure of _____in case of liquids and gases
is:
a. Average vibrational kinetic energy
b. Average translational kinetic energy
c. Average rotational kinetic energy
d. Both a and b
2. According to kinetic interpretation, the temperature is the measure of average translational
kinetic energy of molecules of gas. The expression of EK is:
a.
2𝑅𝑇
2𝑁𝐴
b.
2𝑅𝑇
3𝑁𝐴
c.
3𝑅𝑇
2𝑁𝐴
d.
2𝑅𝑇
2𝑁𝐴
3. Kinetic interpretation of temperature suggests that a change in temperature has results the
change in:
a. Direction of motion of particles
b. Intensity of motion of molecules
c. Behaviour of molecules
d. Shape of molecules
4. A gas that does not obey gas law and kinetic molecular theory is called non ideal gas. At which
conditions the gas shows such non ideal behavior.
a. Low temperature
b. High pressure
c. High temperature
d. Both a and b
5. Which of the following is an ideal gas?
a. Nitrogen
b. Carbon dioxide
c. Helium
d. None of these
6. The least value of ‘a’ for H2 is due to its:
a. More polar character
b. Small size
c. Non-polar character
d. Both b and c

8. 2.016 g of hydrogen gas will occupy the volume of ______ at the closest approach in the gaseous
state:
a. 133 cm3
b. 266 cm3
c. 22414 cm3
d. 24000 cm3
9. Excluded volume according to Vander Waal’s equation is equal to:
a. Vm
b.
𝑉𝑚
4
c. 4Vm
d.
4
𝑉𝑚
10. The factor
𝑝𝑣
𝑛𝑅𝑇
is known as the compressibility factor which has the value of ____ for the gas
showing ideal behaviour.
a. 1
b. 2
c. 3
d. 4
11. Excluded volume is _____ times the actual volume of molecules:
a. ½
b. Two
c. Three
d. Four

Assessment 4
1. General gas equation needs corrections for gas deviate from ideal behavior
a. Mass correction
b. Volume correction
c. Pressure correction
d. Both b and c
2. The following may be resulted because of non ideal behavior
a. Intermolecular attraction
b. Finite volume
c. Infinite volume
d. Both a and b
3. Which of the following temperature favour more ideal behavior of gases
a. 273K
b. 373K
c. 473K
d. 573K
4. Units of excluded volume b exhibited by non ideal gas at STP
a. volume/mol
b. dm3/mole
c. m3/mole
d. all of the above
5. Among the given non ideal gases which of the following has least value for coefficient of attraction
“a”
a. Hydrogen
b. Oxygen
c. Nitrogen
d. Carbon dioxide
6. The value of excluded volume “b” for hydrogen gas is
a. 0.0266 dm3
b. 0.0318 dm3
c. 0.0391 dm3
d. 0.0428 dm3

12. In general gas equation, “R” is the gas constant whose value depends upon the:
a. composition of gas
b. nature of gas
c. units of measurements of variables
d. pressure of the gas
13. One atmospheric pressure is
a. 76 torr
b. 76 mmHg
c. 101325 pa
d. 101.523 Kpa
14. Which parameter is kept constant while studying the relationship between pressure and volume
of gas
a. Mass of gas
b. No of moles
c. Temperature
d. All of these
15. The weather balloons are filled with helium gas. If a small balloon has one dm3 of gas at rtp,
calculate mass of gas filled in the balloon
a. 1/6 g
b. 1/3 g
c. 1/2g
d. ¼ g

Key
Assessment 1
Assessment 2
1. b
2. d
3. d
4. b
5. c
6. c
7. b
8. c
9. c
10. a
1. c
2. a
3. d
4. c
5. c
6. b
7. a
8. d
9. b
10. c

Key
Assessment 3
Assessment 4
1. b
2. c
3. b
4. d
5. d
6. d
7. b
8. c
9. a
10. d
1. d
2. d
3. d
4. b
5. a
6. a
7. c
8. c
9. d
10. a

Learning Objectives:
Properties of liquids
Intermolecular forces
Hydrogen bonding
Vapor pressure
Boiling point and external pressure
Liquids

Liquids
The existence of matter in our surrounding in different states (solid, liquids and gas) depends upon the
strength of intermolecular forces between the constituent particles. Intermolecular forces are responsible
for a substance to be solid, liquid or gas.
Properties of liquids:
i) Liquids have definite volume.
ii) Liquids donot have definite shape. Instead, they adopt the shape of container because liquids
molecules slip over each other.
iii) Liquid molecules are in constant state of motion. Evaporation and diffusion of liquid
molecules is due to this motion.
iv) Diffusion operates to some extent in liquids.
v) The densities of liquids are closer to solids. But their densities are much greater than gases.
vi) The spaces amongst liquid molecules are negligible.
vii) The intermolecular forces in liquids are intermediate between gases and solids.
viii) Molecules of liquids possess kinetic energy due to their motion.
ix) Molecules of liquids exchange their kinetic energy when they collide with each other. Such
collisions are called inelastic collision.
x) By decreasing their kinetic energy, liquids are converted into solids.
xi) Evaporation in liquids causes cooling.
Intermolecular Forces
The forces of attraction present between the molecules of a substance are called Intermolecular forces.
Example:
The forces of attraction between HCl molecules are:
These forces operate between all kinds of molecules when they are sufficiently close to each other.
Intermolecular forces are called Van der Waal’s forces.
Valence electrons are not involved in these forces. That is why intermolecular forces are much
weaker than intramolecular forces.
The force of attraction that holds the two atoms together within a molecule is called intramolecular
force.
Example:
The covalent bond between Cl and H atoms in HCl molecule is intramolecular force.
Intramolecular force (covalent bond) develops due to mutual sharing of electrons between the two
atoms. This satisfies the outermost shells of both the bonded atoms.
That is why, this is their firm need to stay together. As a result, this force is very strong.
δ+ δ- δ+ δ-
H Cl H Cl
⎯ ⎯
.......

Types of Intermolecular Forces:
There are four main types of intermolecular forces.
(i) Dipole-dipole force
(ii) Ion-dipole force
(iii) Dipole-induced dipole force (Debye force)
(iv) Instantaneous dipole-induced dipole force (London dispersion force)
Dipole-Dipole Force
The intermolecular force of attraction between positive end of a polar molecule (a dipole) and the
negatively charged end of neighbouring polar molecule (another dipole) is called as dipole-dipole force.
Example:
Polar Molecule:
A molecule between two dissimilar atoms (having appreciable difference of electro negativities) is
called polar molecule. There exists dipole-dipole force between such molecules.
Example:
In HCl, Cl being more electronegative develops partial negative charge and hydrogen develops
partial positive charge. As a result, molecule become dipole (polar molecule).
Note: Dipole – dipole forces are 1 % as effective as a covalent bond.
Factors Affecting Dipole-Dipole Forces:
(i) Electronegativity difference
(ii) The inter-molecular distance
(i) Electronegativity difference:
Greater the electronegativity difference ( EN) between the boded atoms, greater will be the
polarity, stronger will be the intermolecular forces. Hence, greater are the values of
thermodynamic properties like:
• melting point
• boiling point
• heat of vaporization
• heat of sublimation …. etc.

(ii) The inter-molecular distance:
In gas phase, the intermolecular forces are weaker due to greater distance between
molecules. But in liquid phase, the forces are reasonably strong due to less distance between
the molecules.
The strength of intermolecular forces affect the physical properties like melting points,
boiling points, heat of vaporization etc. whereas intramolecular forces affect the chemical
properties mainly.
Dipole-Induced Dipole Forces (Debye Forces)
The force of attraction between positive end of the permanent dipole and the negative end of induced
dipole is called dipole-induced dipole force or Debye force.
Example:
Note: Polarity created in a non-polar molecule under the influence of a permanent dipole is called
induced dipole.
Dipole-induced dipole force exists between a polar and a non-polar molecule.
Explanation:
Sometimes, we have a mixture of substances containing polar and non–polar molecules. The positive
end of the polar molecule attracts the mobile electrons of the nearby non-polar molecule. In this way,
polarity is induced in non-polar molecule and both molecules become dipole. These forces are called dipole–
induced dipole forces or Debye forces as shown in figure below:
Instantaneous Dipole-Induced Dipole Force:
The force of attraction between of oppositely charged end of an instantaneous dipole and an induced
dipole is called instantaneous dipole-induced dipole force or London dispersion force.
This is momentary, short lived force that vanishes as quickly as it is formed. But this force extends
and holds when molecules are close to each other and are more polarizable.
Example:
The forces of attraction between chlorine molecules.
Note: • It is the only force that is present between non-polar molecules.
• These force were discovered by a German physicist, Fritz London in 1930.
Instantaneous Dipole:
A temporary dipole created in a non-polar molecule as a result
of repulsion of its electronic cloud with electronic cloud of a
neighbouring non-polar molecule is called instantaneous dipole.
dipole
…

Induced Dipole:
A dipole created in a non-polar molecule under the influence of
a temporary or an instantaneous dipole is called induced dipole.
Explanation:
Let us understand this concept of London force with the help of
example of helium gas.
When electrons of one atom come close to electrons of other atom, they are pushed away from each
other. At this moment, the electron density of atom no more remains symmetrical. It has more negative charge
one side than the other and helium atom becomes a momentary dipole called instantaneous dipole.
The instantaneous dipole disturbs the electronic cloud of nearby atom and induces dipole in it. This
is called induced dipole.
Note: 1. London force is a short-lived attraction because electrons keep on moving. This movement of
electrons causes the dipole to vanish as quickly as they are formed.
2. London forces are present in all the types of molecules whether polar or non-polar but are very
significant between non-polar molecules like H2, Cl2 He, Ne, Ar etc.
Factors Affecting London Forces:
The main factor that affects the strength of London force is polarizability. The quantitative measure of
extent to which electronic cloud can be polarized or distorted or dispersed is called polarizability. The
polarizability in turn depends upon the following factors.
(i) Size of electronic cloud.
(ii) Number of atoms in a molecule. (atomicity)
(1) A large sized atom has large electronic cloud. As a result, its
outermost electrons move away from nuclei and become loosely
bound. Hence, dispersion of electronic cloud becomes more and
more easy. As a result, atoms become more polar and strong forces
are created between such atoms and hence, higher will be the
boiling point.
Greater the size of electronic cloud, greater will be its polarizability
and stronger will be the London force.
Examples:
(i) The boiling points of Noble gases increase down the group
due to increasing.
• Size
• Polarizability
• Strength of London force
All halogens have non-polar diatomic molecules.

(ii) The physical states of halogens change from gas to solid
through liquid and boiling point changes from -188.1oC to
184.4oC. This is due to increasing size, polarizability and
strength of London force down the group.
(2) No. Of Atoms Per Molecule (Atomicity):
Greater the number of atoms, greater is the polarizability and
strength of London force. Actually, when number of atoms in a
molecule increases, molecule becomes longer, the number of polar
sites increases where molecules can attract each other. As a result,
strong forces are created.
Example:
Boiling point of C2H6 and C6H14 are -88.6oC and 68.7oC respectively.
The reason is that longer molecules have more polar sites and
hence strong forces with other molecules.
Note: In hydrocarbons, trend changes from gaseous state to liquid and
finally to solid with increasing molecular mass.
Boiling points and physical states of some hydrocarbons:
Name
B.P
o
C (1atm)
Physical state at
S.T.P
Name
B.P
o
C (1atm)
Physical state at
S.T.P
Methane -16.5 Gas Pentane 36.1 Liquid
Ethane -88.6 Gas Hexane 68.7 Liquid
Propane -42.1 Gas Decane 174.1 Liquid
Butane -0.5 Gas Isodecane 327 Solid
Hydrogen Bonding:
The electrostatic force of attraction between highly partial positive hydrogen and lone pair of highly
electronegative atom is called Hydrogen bonding.
Example: The force of attraction between the lone pair of oxygen and hydrogen in water.
Note: • It is a special type of dipole-dipole force.
• It is much stronger than simple dipole-dipole force.
• Its strength is 20 times less than that of a covalent bond.
Condition For Hydrogen Bonding:
i) Highly partial positive Hydrogen.
ii) Highly electronegative atom.

Explanation:
In water, oxygen being more electronegative withdraws shared
pair strongly from hydrogen. As a result, hydrogen become
highly partial positive and creates strong electric field due to its
small size. This strong electric field attracts the lone pair of
electronegative atom strongly. And resulting electrostatic force
is called hydrogen bonding.
Examples:
1) Hydrogen bonding in NH3:
There is only one hydrogen bond per NH3 molecule.
Hydrogen bonding in NH3 molecules is shown below.
2) Hydrogen bonding in HF:
There is only one hydrogen bond per HF molecule.
HF molecules join each other in Zig-Zag manner due to the
presence of hydrogen bonding.
HF shows exceptionally low acidic strength than HCl, HBr and HI
due to
i. Hydrogen bonding between its molecules.
ii. Hydrogen atom being entrapped between two highly
electronegative atoms (F). As a result, release of hydrogen
atom becomes difficult. This makes it weaker acid.
3) Hydrogen bonding in H2O:
There are two hydrogen bonds per water molecule. This is the reason, why its intermolecular
hydrogen bonding is stronger than that of HF and NH3.
4) Hydrogen bonding between acetone and chloroform molecules.
(i) In Chloroform, three Chlorine atoms and one
hydrogen atom is directly attached to one Carbon.
Chlorine atoms being highly electronegative
withdraw electrons form carbon atom which in turn
withdraws electrons from hydrogen atom. As a
result, hydrogen atom becomes highly partial
positive.
(ii) In acetone, oxygen being more electronegative
withdraws electrons from carbon atom and become
partial negative.
When acetone and Chloroform come closer, highly partial positive hydrogen of chloroform and
lone pair of partial negative oxygen forms hydrogen bond with each other. That is why chloroform
and acetone are miscible into each other.
Cl — C — H O C
Cl
Cl
 +
 +
 −
 −
 −
 −
CH3
CH3
 +

Note: The intermolecular force between acetone and chloroform is hydrogen bonding when they are
mixed. But intermolecular force between Chloroform molecules in pure state or between acetone
molecules in pure state is dipole-dipole force.
Hydrogen bonding is not limited to fluorine, oxygen and nitrogen. But it may be present where
hydrogen atom is bonded to some other element but is highly partial positive
Example: Chloroform.
Properties and applications of compounds Containing Hydrogen Bonding:
1. Thermodynamics properties of covalent Hydrides.
Hydrogen bonding influences the physical properties like melting points and boiling points of
covalent hydrides.
Boiling points of covalent Hydrides of IV – A group elements.
The boiling points of covalent hydrides of group IV-A elements are lower than those of V-A, IV-A
and VII-A group elements due to least electronegativities of IV-A group elements.
CH4 is the hydride of top member of IV-A group. It has lowest boiling point due to its smallest size
and least polarizability. The boiling points of hydrides of this group members increase down the group due
to increasing sizes and polarizabilities of hydrides.
Boiling points of covalent hydrides of V-A group elements.
The boiling point of NH3 is greatest than hydrides of its group members (except AsH3) due to
stronger hydrogen bonding. This in turn is due to greater electronegativity difference between Nitrogen
and Hydrogen.
There is only one hydrogen bond per NH3 molecule due to the presence of only one lone pair on
central nitrogen atom in NH3. There is less elelctronegativity difference (1.1) between N (EN = 3.2) and
H (EN = 2.1).
That is why NH3 is less polar and have weak intermolecular hydrogen bonding between its molecules. As a
result, it exists as a gas with boiling point -33oC.
Boiling points of hydrides of Group VI-A and VII- A.
The boiling points of H2O and HF are greater than those of hydrides of their respective group
members due to strong intermolecular Hydrogen bonding. This in turn is due to greater electronegativities
of oxygen and fluorine respectively.
There is only one hydrogen bond per HF molecule due to the presence of only one hydrogen atom
bonded to fluorine atom. There is greater electronegativity difference (1.9) between F (EN = 4) and
H (EN = 2.1). That is why HF is strongly polar and has strong intermolecular hydrogen bonding between its
molecules. As a result, it exists as a liquid with boiling point 19.9oC.
Comparison of Hydrogen bonding and boiling points of H2O and HF.
There are two hydrogen bonds per H2O molecule due to the presence of two lone pairs and two
hydrogen atoms bonded to oxygen.There is larger electronegativity difference (1.4) between O (EN = 3.5) and
H (EN = 2.1) (Less than HF).
That is why H2O is strongly polar (lesser than HF) and has strong intermolecular hydrogen bonding
(stronger than HF) between its molecules. As a result, it exists as a liquid with boiling point 100oC.

Although water is less polar than HF, still its intermolecular Hydrogen bonding is stronger than HF due
to the presence of two hydrogen bonds per H2O molecule and three dimensional hydrogen bonding.
Comparison of boiling point of HCl, HBr and HI.
HCl is considered border line case between simple
dipole-dipole force and hydrogen bonding.
Boiling point of HBr is greater than that of HCl
Due to greater polarizability of Br atom than that of chlorine
atom. On the similar basis, the boiling point of HI is greater
than HBr.
Note: The order of boiling points of HCl, HBr and HI is as
follows HI > HBr > HI
Comparison of boiling points of Hydrides of 3rd and 4th period.
The boiling points of Hydrides of fourth period (GeH3,
AsH3 , H2Se , HBr) are greater that those of third period (SiH4,
PH3, H2S, HCl) due to larger size and greater polarizabilities.
2. Solubility of Hydrogen bonded molecules:
The compounds that have hydrogen bonding between
their molecules are soluble in water due to Hydrogen bonding
between water molecules and molecules of those compounds.
Example:
i) Ethyl alcohol (C2H5OH) is soluble in water due to
Hydrogen bonding between the molecules of
water and ethyl alcohol.
ii) Small sized carboxylic acids are soluble in water
due to hydrogen bonding between the molecules
of water and carboxylic acids.
iii) Hydrocarbons being non polar are completely
insoluble in water because they will not form
hydrogen bonding with water.
3. Structure of Ice
i) The molecule of water in ice has tetrahedral
structure. Two lone pairs of electrons and two
hydrogen atoms on oxygen atom occupy four
corners of tetrahedron.
The structure of ice is like diamond because each
atom of carbon in diamond is at the center of a
tetrahedron just like oxygen of water in the
structure of ice.
ii) In liquid state, associations of water molecules break and reform because water molecules
are mobile. When temperature of water is decreased below 4oC and ice is formed at 0oC,
then molecules become more regular.

This regularity extends throughout the structure in geometrical patterns (Hexagonal
patterns). As a result, empty spaces are created in the structure of ice.
iii) Due to empty spaces in the structure of ice, it occupies 9% more space than liquid water.
Hence, density of ice is 9% less than liquid water. As a result, ice being lighter than water
floats on surface of water.
iv) When temperature of water reaches below 4oC by fall in temperature in atmosphere, the
water on the surface become less dense. This less denser water stays at the top of slightly
warmer water underneath. A stage reaches, when it freezes at 0oC. This layer of ice insulates
the water which is below it for further heat loss. This is the reason, why water below a layer
of ice stays at 4oC. As a result, fish and plants continue to survive under the blanket of ice.
NOTE
The density of water is maximum at 4oC. When temperature decreases from 100oC to 4oC, its
volume decreases and density increases. Whereas, when temperature decreases from 4oC to
0oC. Volume increases and density decreases by 9% due to Hydrogen bonding and
geometrical arrangement (Hexagonal) of water molecules.
The change in the density of water with a changing temperature is shown below.
Volume Decreases Volume Increases
Density Increases Density Decreases
4. Cleansing action of soaps and detergents.
Soaps and detergents perform their cleansing action due to hydrogen bonding between polar parts
of their molecules and water molecules. The polar parts of the molecules are water soluble and non-polar
parts are water insoluble. Non-polar parts of molecules are either alkyl or aryl groups which remain
outside water.
5. Hydrogen bonding in biological compounds and food materials.
i) In Proteins:
Proteins are the important part of living organisms. Fibers
like those found in the hairs, silk and muscles consist of
long chains of amino acids. These long chains are coiled
around one another into a spiral. This spiral is called a helix
which may be right handed or left handed. In case of right
handed helix, the groups like >NH and >C=O are vertically
adjacent to one another and they are linked together by
hydrogen bonds which link one spiral to the other as
shown in figure.
ii) In Deoxyribonucleic acid (DNA):
DNA has two spiral chains which are coiled about each
other on a common axis. In this way, they give a double
helix. This is 18-20 in diameter. They are linked together

by hydrogen bonding between their sub-units as shown in
the figure.
iii) In Food Materials:
The food materials like carbohydrates include glucose,
fructose and sucrose. They all have –OH group in them
which are responsible for hydrogen bonding in them
6. Hydrogen Bonding in Paints, Dyes and Textile materials
i) One of the most important properties of paints and dyes is
their adhesive action. This property is developed due to
hydrogen bonding.
ii) Hydrogen bonding makes glue and honey as sticky substances.
iii) We use cotton, silk or synthetic fibres for clothing. Hydrogen bonding is of vital importance in
these thread making materials. This hydrogen bonding is responsible for their rigidity and the
tensile strength.
Evaporation
The spontaneous change of high energy liquid molecules into vapours at any temperature is called
evaporation. The escape of high energy molecules from the surface of liquid takes place at all
temperatures is called Evaporation.
Evaporation:
• is surface phenomenon
• is continuous process
• continues at all temperatures.
• is an endothermic process.
Explanation:
The energy of the liquid molecules in not equally distributed. Some of the molecules have low
kinetic energy and move slowly while others have high kinetic energy and move faster. If one of the high
energy molecules reaches the surface, it may break the attractions of neighboring molecules and leaves
the bulk of liquid.
Evaporation causes cooling:
Actually high energy molecules leave the surface leaving behind the low energy molecules. As a
result, temperature of liquid falls and heat moves from surrounding to the liquid. Finally temperature of
surrounding also falls.
Rate of evaporation
The number of vapours formed per unit time is called rate of evaporation
Factors affecting the rate of evaporation:
(i) Surface area
(ii) Intermolecular forces

(iii) Temperature
NOTE:
Evaporation also depends upon external pressure and humidity and is directly proportional to both
these factors
(i) Surface area:
When surface area is increased, then more molecules are able to escape from the surface of
liquid. As a result, liquid evaporates quickly.
Example:
Hot tea placed in a pirch gets cool at a faster rate than in a cup due to greater surface area of
pirch.
(ii) Intermolecular forces:
The molecules of liquids having weaker intermolecular forces have greater chances to
escape from the surface of liquid. As a result, rate of evaporation is greater.
Example:
Gasoline evaporates much faster than water due to weak London forces of attraction
between its molecules.
(iii) Temperature:
The rate of evaporation increases with increasing temperature. At high temperature, greater
number of molecules have high kinetic energy to overcome the attractions of their
neighbouring molecules. As a result, rate of evaporation is greater.
Example:
Wet cloths dry up faster in hot summer in comparison to cold winter due to greater rate of
evaporation at high temperature in summer.
Vapour Pressure:
The pressure exerted by vapours on the surface of the liquid in equilibrium with the liquid at a given
temperature is called vapour pressure.
Explanation:
In a closed container, high energy molecules leave the surface of
liquid and accumulate above the surface. This is called evaporation
(liquid into vapours). The molecules (vapours) collide with the walls of
container as well as on the surface of liquid. The vapours colliding on the
surface are usually recaptured by the surface of liquid. This is called
condensation (vapours into liquid). The process of evaporation and
process of condensation continue until rate of evaporation becomes
equal to the rate of condensation. This is called state of dynamic
equilibrium.
Temperature
(oC)
Vapour
Pressure (Torr)
0
10
20
30
37
40
50
60
70
80
90
100
4.579
9.209
17.54
31.82
47.07
55.32
92.51
149.4
233.7
355.1
527.0
760.0

At the state of dynamic equilibrium, the number of molecules
leaving the surface is just equal to number of molecules coming back
into it at constant temperature.
Liquid Vapours
The pressure exerted by vapours on the surface of liquid at this state is called vapour pressure.
Vapour pressure does not depend upon:
(i) Amount of liquid
(ii) Volume of container
(iii) Surface area of liquid
Factors affecting the vapour pressure:
(i) Intermolecular attractions and sizes of molecules (nature of liquid).
(ii) Temperature
(i) Intermolecular forces
Vapour pressure increases with decreasing
strength of intermolecular forces and vice versa.
Strong intermolecular forces hold the
molecules tightly and chances of liquid
molecules to leave the surface decreases. As a
result, vapour pressure also decreases.
Example:
Vapour pressure of water having stronger
intermolecular forces is lesser (17.5 torr)
than ether (44.2 torr) at 20oC. The
intermolecular forces in ether are weaker.
(ii) Temperature
Vapour pressure increases with the increase in temperature and vice versa. At higher
temperature, weak forces hold molecules loosely and chances of liquid molecules to escape
from surface increases. As a result, vapour pressure also increases.
Example:
i) Vapour pressure of H2O is 92.51 torr at 50oC.
ii) Vapour pressure of H2O is 527 torr at 90oC.
Note: The increase of vapour pressure at high temperature is greater for same difference of temperature.
It is because intermolecular forces are weaker at higher temperature.
Example:
i) When temperature increases from 0oC to 10oC (10oC change of T) then increase of vapour
pressure of is from 4.579 torr to 9.209 torr.
ii) When temperature increases from 90oC to 100oC (10oC change of T), then increase in vapour
pressure is from 527.8 torr to 760 torr.
Name of compound
Vapour pressure at
20oC (torr)
Isopentane 580
Ethyl ether 442.2
Chloroform 170
Carbon Tetrachloride 87
Water 17.54
Mercury 0.012
Glycerol 0.00016

Measurement of Vapour Pressure:
There are many methods but one of the most important method i.e. manometric method is given below.
Manometric method: (Accurate method)
The liquid whose vapour pressure is to be determined is taken in flask placed in the thermostat.
One end of the tube from flask is connected to manometer and other end is connected to vacuum pump.
The air above the liquid is removed in the following steps.
(i) Freezing:
The liquid is frozen with the help of freezing mixture and space above the liquid is evacuated.
In this way, air is removed from the surface of liquid alongwith the vapours of liquid.
(ii) Melting:
The frozen liquid is melted to release entrapped air.
(iii) Re-freezing:
Liquid is again frozen and released air is evacuated. This process is repeated many times till
almost all the air is removed.
Measurement of vapour pressure:
The liquid in the flask is warmed in thermostat at a temperature at which vapour pressure is to be
determined. Difference in the heights of Hg column in two limbs of manometer determines the vapour
pressure of liquid.
The pressure on the surface of liquid is equal to sum of atmospheric pressure and vapour pressure
of liquid. That is the reason why column of manometer facing the liquid is more depressed than facing the
atmosphere.
The vapour pressure of liquid is given by following equation.
P = Pa + h
P = vapour pressure of liquid at one atmosphere.
Pa = atmospheric pressure
h = difference in the heights of Hg column in two limbs of manometer. It gives us the vapour
pressure of liquid.

Boiling Point:
The temperature at which vapour pressure (internal
pressure) of liquid becomes equal to external
pressure is called boiling point.
Examples:
i) Boiling point of H2O = 100oC
ii) Boiling point of C2H5OH = 78.26oC
iii) Boiling point of C2H5OC2H5 = 34.6oC
Explanation:
When a liquid is heated, vapour pressure of liquid
goes on increasing and ultimately becomes equal to
external pressure. On further heating at this stage,
bubbles of vapours which are formed in the interior
of liquid have greater internal pressure than external
pressure on the surface of liquid. This makes the bubble to come out of the liquid and burst on the
surface of liquid. A constant stream of bubbles come out at boiling.
Temperature remains constant at boiling point.
When liquid is heated below the boiling point, kinetic energy of its molecules and temperature
increases. At boiling point, kinetic energy of molecules become maximum and heat given will be
utilized to break the attractions and to convert liquid into the vapours. At this stage, the escaping
vapours carry the absorbed heat alongwith them. As a result, temperature remains constant.
Molar heat of vaporization:
The amount of heat required to vapourize 1mole of a liquid at its boiling point is called its molar heat of
vapourization. It is represented by Hv.
Example:
Molar heat of vapourization of water is 40.6 kJ/mole.
Comparative variation of vapour pressures of different liquids with temperature.
Following graph shows the variation of vapour pressure of
i) water ii) ethyl alcohol iii) ethylene glycol iv) diethyl ether with temperature.
• The vapour pressure-temperature curve for water starts at 4.6 torr and that of diethyl ether
starts around 200 torr. This is due to weak intermolecular forces between the molecules of
diethyl ether than water.
• Vapour pressure-temperature curve for water goes along the temperature axis to a greater
extent at the start than that for diethyl ether. It is due to the reason that it is difficult to
overcome strong intermolecular forces between water molecules at low temperature.
Note:Vapour pressure temperature curve shows that vapour pressure increases rapidly when the
liquids are closer to their boiling points.
Liquids Boiling point (oC)
Acetic Acid 118.50
carbon tetrachloride 76.50
Acetone 56.00
Ethanol 78.26
Aniline 184.4
Naphthalene 218.00
Benzene 80.15
Phenol 181.80
Carbon disulphide 46.30
Water 100.00

Factor affecting the boiling points:
Boiling points of liquids depend upon the following factors.
i) Intermolecular forces
ii) External pressure
(i) Intermolecular forces:
Stronger the intermolecular forces, greater will be the boiling point and vice versa. Greater amount
of heat will be required to
(i) Over come strong intermolecular attractions
(ii) Equalize vapour pressure with external pressure.
Hence, boiling point will be high.
Examples:
The boiling points of some liquid at 1atom.
i) Boiling point of C2H5OC2H5 is 34.6oC (weak dipole-dipole forces)
ii) Boiling point of C2H5OH is 78.26oC (weak hydrogen bonding)
iii) Boiling point of H2O is 100oC (strong hydrogen bonding)
(ii) External pressure:
Greater the external pressure, greater will be the boiling point
and vice versa. Liquid absorbs greater amount of heat to
equalize the vapour pressure with greater external pressure.
Hence, boiling temperature is high.
Example:
i) Boiling point of H2O is 100oC at 760 torr.
ii) Boiling point of H2O is 98oC at 700 torr.
Boiling and External Pressure
Since, a liquid boils at a temperature where vapour pressure becomes equal to external pressure,
hence boiling point varies with external pressure.
Greater the external pressure greater will be the boiling point and vice versa. When external
pressure is high, liquid requires greater amount of heat to equalize its vapour pressure to external
pressure and vice versa.
Examples:
i) Boiling point of H2O is 120oC at 1489 torr.
ii) Boiling point of H2O is 100oC at 760 torr (at sea level).
iii) Boiling point of H2O is 98oC at 700 torr (at Murree hill).
iv) Boiling point of H2O is 69oC at 323 torr (at Mount Everest)
v) Boiling point of H2O is 25oC at 24.7 torr.

Practical applications of effect of external pressure on boiling point:
(i) Pressure cooker (boiling of liquid under increased external pressure)
(ii) Vacuum distillation (boiling of liquid under decreased external pressure)
(i) Pressure cooker (boiling under increased pressure)
Principle: The boiling point increases with external pressure.
Working: Pressure cooker is a closed container. When liquid is heated in it, vapours are formed
which accumulate on the surface of liquid and are not allowed to escape. These vapours exert more
pressure on the surface of liquid. As a result, boiling point increases. As more heat is absorbed by
water, food is cooked quickly under increased pressure.
(ii) Vacuum distillation (boiling under reduced pressure).
Principle: heT boiling point decreases with decreases in external pressure.
Working: The liquids that decompose when they are distilled at their boiling points, their
decomposition can be avoided by distilling at lower temperature and lower external pressure.
Example:
Glycerine boils as well as decomposes at 290oC and 760 torr (1atm). Hence, it cannot be distilled at
290oC. It can be distilled without decomposition at 210oC under reduced pressure of 50 torr. As a
result, it can be purified easily.
Advantages of vacuum distillation:
Vacuum distillation has following advantages.
i) It decreases the time for distillation.
ii) It is economical as it consumes less fuel.
iii) The decomposition of many liquids can be avoided.

Assessment 01
1. Which of the following attractive force has nothing to do with valence electrons?
a. Intramolecular force
b. Ionic bond
c. Covalent bond
d. Intermolecular force
2. Intermolecular forces have no significance when we consider:
a. Melting point
b. Heat of vaporization
c. Oxidation
d. None of these
3. The cohesive force between the molecules of liquids due to electronegativity difference between
atoms is:
a. Dipole-dipole forces
b. London dispersion forces
c. Covalent bond
d. None of these
4. In dipole dipole interaction, the dipoles which can exist are called:
a. Temporary dipoles
b. Permanent dipoles
c. Instantaneous dipoles
d. Induced dipoles
5. The number of poles which are involved when two hydrochloric acid molecules approach each
other:
a. 1
b. 2
c. 4
d. 6
6. In which of the following dipole-dipole interaction exists:
a. HCl
b. H2O
c. CHCl3
d. All of these
7. Which of the following causes the molecules not to have a perfect alignment in dipole-dipole
interaction?
a. Low kinetic energy
b. High potential energy

c. Thermal energy
d. All of these
8. The strength of dipole-dipole forces depends upon the
a. EN difference between atoms
b. Distance between molecules
c. Shape of molecules
d. Both a and b
9. The strength of dipole-dipole force does not influence the following property of polar liquid?
a. Melting point
b. Heat of vaporization
c. Molar mass
d. Boiling point
10. Permanent dipole in case of polar molecules is originated due to:
a. Electronegativity difference
b. Shape of atoms
c. Molar mass of atoms
d. Thermal energy

Assessment 02
1. The intermolecular force that exists between permanent dipole and temporary dipole is:
a. Hydrogen bonding
b. Dipole-Dipole interaction
c. Dipole-Ion dipole force
d. Dipole-induced dipole forces
2. The force of attraction between sodium ion and water molecules is:
a. Dipole-dipole interaction
b. Hydrogen bonding
c. Ion-dipole forces
d. Debye forces
3. Debye force is the special term which is assigned to the following intermolecular force:
b. Ion-dipole forces
c. Dipole-induced dipole force
d. Instantaneous dipole induced dipole forces
4. The intermolecular force that exists within non-polar molecules is:
a. Dipole-Dipole interaction
b. Dipole-induced dipole force
c. Instantaneous dipole-induced dipole forces
d. Hydrogen bonding
5. Which of the following noble gas has more boiling point by keeping external pressure same?
a. Helium
b. Neon
c. Argon
d. Xenon
6. London dispersion force affects the physical state of halogens. The following halogen is solid at
room temperature:
a. F2
b. Cl2
c. Br2
d. I2
7. The following statement corresponds to the strength of London dispersion force:
a. Hexane is hydrocarbon which exists in liquid state at room temperature.
b. Helium has more boiling point among noble gases
c. The molecule with longer chain length experiences stronger attractive force.
d. Both a and c

8. The correct order of strength of intermolecular force is:
a.
b.
c. London dispersion force < Debye force < Dipole-dipole < H-bonding
d.
9. Heat of vaporization of which of the following is greater due to the presence of hydrogen bonding:
a. H2S
b. H2Se
c. H2O
d. All have same
10. The decrease in volume when ice melts to form water is:
a. 1%
b. 9%
c. 6%
d. No change in volume

Assessment 03
1. The accurate order of boiling point for the given hydrides:
H2O, CH4, HF and NH3
a. CH4
b. CH4 <NH3 <HF < H2O
c. NH3
d. CH4
2. The compounds which are soluble in water due to establishing H-bonding with water molecules:
a. Acetic acid
b. Ethyl alcohol
c. Acetone
d. All of these
3. The adhesive property of glue is usually attributed to:
a. Hydrogen bonding
b. High molar mass
c. Large molar mass
d. Low density
4. Two strands of DNA are coiled to each other forming double helix. The attractive force present
between nitrogen bases of two strands is:
a. Ion-dipole forces
b. H-bonding
c. Dipole-Dipole interaction
d. Debye force
5. Water has maximum density at the temperature of:
a. 0 o
C
b. 4 o
C
c. 25 o
C
d. Same at all temperature
6. In extreme cold season fish and aquatic life can survive within frozen water due to:
a. Less density of ice
b. More density of water which sinks
c. Insulation of ice blanket from chilling weather
d. All of these
7. The detergent molecule interacts with grease molecule as well as get soluble in water due to
hydrogen bonding. The following part of soap/detergent interacts with grease that is stick to the
fiber:

a. A only
b. B only
c. Either A or B
d. Both A and B
8. The biological compound which forms helical structure due to hydrogen bonding is:
a. DNA
b. Protein
c. RNA
d. Both a and b
9. Which of the following statement is true about the boiling point of the liquid?
a. Different liquids have same boiling point on the same place.
b. Same liquid has same boiling points on different places
c. Boiling point depends upon the amount of liquid
d. A liquid X has different boiling points on different places.
10. Which of the following intermolecular force is?
a. Dipole-dipole forces
b. London dispersion forces
c. H-bonding
d. Debye forces
B
A
+ -
__
+

Assessment 04
1. At the particular instant, the helium atom when have distorted electron density is called:
a. Permanent dipole
b. Induced dipole
c. Instantaneous dipole
d. Non-polar
2. The intermolecular force that exist between Cl2 molecules:
b. Debye forces
c. Hydrogen bonding
d. London dispersion forces
3. Temperature at which vapour pressure of liquid becomes equal to external pressure is:
a. Boiling point
b. Freezing point
c. Saturation point
d. Melting point
4. In order to boil the water at standard temperature (25 o
C), external pressure should be:
a. 760 mmHg
b. 700 mmHg
c. 23.7 mmHg
d. 1520 mmHg
5. Which of the following statement is true about the boiling point of any liquid?
a. Boiling point of liquid can be changed by changing the external pressure.
b. Boiling point of liquid is high at mountains.
c. Boiling point of liquid is less at sea level and more at land area.
d. Boiling point of water at K-2 is more than at Karachi.
6. The factor which affects the evaporation of water:
a. Temperature
b. Surface area
c. Intermolecular force
d. All of these
7. Which of the following is more volatile in nature?
a. Water
b. Ethanol
c. Diethyl Ether
d. Glycerine

8. The physical phenomenon that can be explained by increasing evaporation:
a. Drying of clothes in summer.
b. More volatile nature of petrol than water
c. Cooling of water in earthen ware vessel
d. All of these
9. Vapour pressure does not depend upon the following factor:
a. Amount of liquid
b. Surface area
c. Intermolecular force
d. Both a and b
10. Water boils at high temperature than ethanol and diethyl ether due to:
a. Stronger intermolecular force
b. More mass
c. Low vapour pressure
d. Both a and c

Key
Assessment 01
1. d
2. c
3. a
4. b
5. c
6. d
7. c
8. d
9. c
10. a
Assessment 02
1. d
2. c
3. c
4. c
5. d
6. d
7. d
8. c
9. c
10. b

Assessment 03
1. b
2. d
3. a
4. b
5. b
6. d
7. a
8. d
9. d
10. d
Assessment 04
1. c
2. d
3. a
4. c
5. a
6. d
7. c
8. d
9. d
10. d

Atomic Structure
Concept of orbital’s
Electronic configuration
Discovery and properties of proton
(positive rays)
Quantum numbers
Shapes of orbital’s

Atomic structure
Discovery Of Proton (Positive Rays):
In 1886, German physicist, Eugene
Goldstein discovered protons or positive
rays or canal rays.
Apparatus:
A discharge tube provided with a
cathode having extremely fine holes in it.
Procedure:
➢ When a large potential difference is applied between electrodes, it is observed that
while cathode rays are traveling away from cathode, there are other rays produced at
the same time moving towards cathode. These are called positive rays.
➢ They are also called as canal rays since they pass through the canals or holes in cathode.
Reason For Production:
When high speed cathode rays strike the residual gas molecules, they knock out
electrons from them and positive ions are produced.
M + 1e- ⎯⎯
→M+ + 2e-
Observation:
These rays after passing through the perforated cathode produce a reddish glow on the
opposite wall.
Properties Of Positive Rays:
1) These rays travel in a straight line towards the cathode.
2) They show deflection by electric field and magnetic field. It shows
that they are positively charged.
3) They cause flash upon ZnS plate.
4) Their e/m ratio varies with the residual gas.
5) Their e/m ratio is maximum in case of hydrogen.
6) Their e/m ratio is smaller than that of cathode rays.
7) These rays consist of tiny particles called protons (when hydrogen gas is used) having
mass 1836 times of electrons.
The e/m value of positive rays is
different for different gases
because the mass of every gas is
different. The positive particle
obtained from hydrogen is lightest
and have maximum e/m value.
Rutherford named this particle as
proton

Planck’s Quantum Theory
Max Planck proposed the quantum theory in 1900 to explain emission and absorption of
radiation.
➢ According to his revolutionary theory, energy travels in discontinuous manner.
➢ Energy is composed of large number of tiny discrete units called quanta.
Postulates of theory:
The main postulate of theory are as follows.
1. Energy is emitted or absorbed in discontinuous manner in the form of energy packets.
Each packet of energy is called quantum which have definite amount of energy. In case
of light radiations quantum is called photon.
2. Energy of quantum is directly proportional to frequency () of radiation.
Number of waves of a radiation passing through a point per second is called frequency.
Its unit is per second (s-1).
E  
E = h………..(i)
E = energy of radiation
 = frequency of radiation
h = Planck’s constant = 6.625  10-34Js.
Planck’s constant is actually ratio of energy and frequency of radiation:
3. A body emits or absorbs energy in the form of quantum of energy (h).
E = h……(i)
Frequency of radiation is inversely proportional to wavelength. Greater the wavelength,
smaller will be the frequency.
……..…… (ii)
Put equation (ii) into (i)
E = hc/
…….. (iii)

This equation (iii) shows that energy inversely proportional to wavelength or radiation.
Greater the wavelength, lesser will be the energy.
1
E


The distance between two adjacent crests or troughs is called wavelength or a wave of
particular radiation.
The units of wavelength are
(i) Angstrom (Ao)
(ii) Nanometer (nm)
(iii) Picometer (pm)
(iv) Meter (m)
1Ao = 10-10m
1nm = 10-9 m
1pm = 10-12 m
4. Number of waves per unit length is called wave number. It is reciprocal of wavelength.
Put equation (iv) into (iii)
E=hc ........(v)
v
−
This equation (v) shows that energy is directly proportional to wave number. Greater
the wave number, greater will be energy of radiation.
Bohr’s Atomic Model
➢ In 1913, Neil Bohr presented the model of an atom of hydrogen.
➢ It is based upon the Planck’s quantum theory.
Postulates:
The postulates of Bohr’s atomic model are as follow.
(i) Electron revolves around the nucleus in some permitted circular paths called
orbits. Each orbit has fixed energy and quantized.
(ii) Energy of an electron is fixed in one orbit. It means it neither emits nor absorbs
energy as long as it is in one orbit.
(iii) The energy of electron changes by E when electron jumps between the orbits.
This energy is equal to energy difference of energies of two orbits (E1 and E2)
between which electron jumps. This energy is given by Planck’s theory.
∆E = E2 - E1
Crest Wave length = 
 = Wave length
trough

Or ∆E = h………. (i)
∆E = energy difference of two orbits.
E2 = energy of higher orbit
E1 = energy of lower orbit
(iv) These electrons can revolve around the nucleus in the orbit having fixed angular
momentum. It is integral multiple of
h
2π
and is given as follows.
nh
mvr =
2π
Where, n = 1, 2, 3 …………….. (orbit number)
The permitted values of angular momentum are , ……
Electron revolves around the nucleus with any one of these values of angular momentum. So,
angular momentum is quantised.
Hydrogen Spectrum
It is an important example of atomic spectrum.
Apparatus:
➢ Hydrogen is filled in a discharge tube at a very low pressure.
➢ A bluish light is emitted from the discharge tube.
➢ This line when viewed through a spectrometer shows several isolated sharp lines.
Spectral Lines:
➢ These sharp lines are called spectral lines.
➢ The wavelengths of these lines lie in the visible, ultraviolet and infrared regions.
Spectral Series:
These sharp lines can be classified into five groups called spectral series. These are
named after their discoverers.
(i) Lyman series (ultraviolet region)
(ii) Balmer series (visible region)
(iii) Paschen series(infrared region)
(iv) Brackett series(Infrared region)
(v) Pfund series (Infrared region)
The lines of Balmer series has been given specific names as H and H……etc.

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