4. Review
What’s matter?
What are the three common
states of matter?
1) Solids
2) Liquids
3) Gases
What can the states also be called? Phases
So, a phase describes a physical state of matter.

6. Comparison of States
gas liquid solid
Do not have Take the shape Definite shape
SHAPE
definite shape of the container
in which it is
poured
lots of free little free space little free space
DENSITY space between between between
particles particles particles
FLOW particles can particles can rigid - particles
move past one move/slide past cannot
another one another move/slide past
one another

7. THREE GAS LAWS
 BOYLE’SLAW :
RELATE PRESSURE WITH VOLUME
 CHARLES’ LAW
RELATE VOLUME WITH TEMPERATURE
 GAY LUSSAC’S LAW
RELATE PRESSURE WITH TEMPERATURE

8. Robert Boyle (1627-1691)
• Boyle had the good fortune to have Robert Hooke as
an assistant and together they made an air pump.
In 1662, Boyle published what is
now known as Boyle's law:
 At constant temperature the volume of a gas is
inversely proportional to the pressure

9. Boyle’s Law :
The volume of definite quantity of gas is inversely
proportional to it’s pressure at constant temperature.
Mathematically expressed as V ∝ 1 (Constant temp.)
P
∴ V = K/P where K = constant
∴ PV = K
Thus it can also be stated as
“At constant temperature the product of volume and
pressure of definite quantity of gas is constant.
If P1 and V1 is initial pressure and volume of gas at
constant temp and P2, V2 at final state then above equation
can be written as :
P1V1 = P2V2

10. Boyle’s Law
Timberlake, Chemistry 7th Edition, page 253

11. Boyle’s Law
1 atm • As the
pressure on a
2 atm gas increases -
the volume
decreases
4 Liters
2 • Pressure and
volume are
inversely
related

12. Boyle’s Law illustrated

13. Jacques Charles
In the century
following Boyle, a
French physicist,
Jacques Charles
(1746-1823), was
the first person
to fill a balloon
with hydrogen
gas and who
made the first
solo balloon

14. Volume vs. Temperature:
Charles’ Law
• Notice the linear
relationship. This
relationship between
temperature and
volume describes a
“direct relationship”.
This means when
temperature
increases, so does
the volume.

15. Charles' Law :
• ‘The volume of definite quantity of gas is directly
proportional to its absolute temperature at
constant pressure.’
• Mathematically expressed as
V ∝ T (constant pressure)
V = KT (K = constant)
V/T = K
• If V1 and T1 are the volume and temperature of gas
in initial state and V2, T2 at final state then above
equation can be written as :
V1 = V2 or V1 = T1

16. GAY LUSACC LAW
“At constant volume the pressure of the given
quantity 0f the gas is directly proportional to
it’s absolute temperature”.
Mathematically expressed as
P ∝ T (constant pressure)
P = KT (K = constant)
P/T = K
If P1 and T1 are the volume and temperature
of gas in initial state and P2, T2 at final state
then above equation can be written as :
P1 = P2 or P1 = T1
T1 T2 P2 T2

17. Mechanics of
Breathing
Timberlake, Chemistry 7th Edition, page 254

18. Simple gas equation P1 VI T I P2 V 2 T 2
P2 V X T 2
HereV T step –volume isVx T instep – 2 stepsP V T
P1 change in 1 P2 done two
1 1 1 2 2 2
In first step, according to Boyle’s Law
P1V1 = P2Vx (constant temp)
∴Vx = P1V1
P2
In second step, according to Charle’s Law
Vx = V2 ∴ Vx = V2T1
T1 T2 T2
On combining both above steps,
P1V1 = V2T1 ∴P1V1 = P2V2

19. COMBINED GAS EQUATION
P1V1 = P2 V2 .
T1 T2
In simple form combined gas equation can be written as
PV = K (constant)
T
∴ PV = KT
Value of constant K depends on quantity of gas
Here putting K = nR KαV;Vα
where n = quantity of gas in mole n
R = gas constant (does not depend on quantity of
Kαn
gas)
∴ PV = nRT
Above equation is called simple gas equation

20. Derive the value of R
According to Ideal gas equation : PV =
nRT
∴ R = PV = pressure x volume
nT no. of moles x temp.
force x volume
=
area
______________
no. of moles x temp.
force x (length)3
=
(length)2
___________
no. of moles x temp.
= force x length__________

21. • But, force x length = work energy
∴R = work energy________
no. of moles x temp.
• Thus unit of R is work energy/Kelvin mole.
• It is proved by experiment that volume of one mole
of any gas at 0°C and 1 atm pressure is 22.4 litre.

22. It is proved by experiment that volume of one mole of
any gas at 0°C and 1 atm pressure is 22.4 litre
According to simple gas equation
R = PV/nT
where P = 1 atm n = 1 mole
V = 22.4 litre T = 0°C = 273 Kelvin
∴ R = 1 atm x 22.4 litre
1 mole x 273 Kelvin
R = 0.082 litre atm/Kelvin mole

23. Value of gas constant R in
different unit
Value Unit
0.082 litre – atm / Kelvin mole } work
1.987 calorie/Kelvin mole in heat
1.987 x 10-3 Kcal / Kelvin mole energy
8.314 x 107 erg / Kelvin mole (CGS)
8.314 joule / Kelvin mole (MKS)

24. Standard temperature and pressure :
• The temperature of 0°C or 273
Kelvin and pressure of 1 atmosphere
or 760 mm is called standard temp.
and pressure.

25. Dalton’s Law of Partial Pressure :
 The pressure of gaseous mixture is sum of partial
pressure of each component gas’.
 Suppose A and B are the gases filled in two
different vessel of same size and kept at same
temperature.
 Let PA = partial pressure of a gm of gas A
PB = partial pressure of b gm of gas B
 If both this gas are filled in the third container of
same volume and kept at same temperature then
total pressure of gases, according to Dalton law of
partial pressure would be.
 PTotal = PA + PB.
 Here gas A and gas B donot react with each other.

26. Dalton’s Law of Partial Pressure :

27. Application :
• For the gases collected over water, the total
pressure of the gas is equal to partial pressure of
dry gas as well as partial pressure of water
vapour.
• Eg : For oxygen gas collected over water, acc. to
Dalton’s law
• PTotal = Pgas + PH2O Here Pgas = PO2
∀ ∴PO2 = PTotal – PH2O
where PTotal = pressure of gases
PH2O = vapour pressure of water at 25°C
• When volume percentage composition is given,
then
partial pressure of gas P,
•
= Percentage of volume x total pressure
100

28. Graham’s law of gaseous difusion
 Graham in 1928 presented a relation between diffusion
rate of gas and its density in the name of Graham’s law
of diffusion of gases.
 ‘The rate of diffusion of various gases at same conditions
of temperature and pressure is inversely proportional to
the square root of their densities.’
 Suppose the density of any gas is (d) and its rate of
diffusion is (r) then,
r α 1/ d
 The diffusion rate of two gases are compared after
carrying out the experiment at same temperature and
pressure.

29. • Suppose r1 and r2 are the diffusion rate of
gas-1 and gas-2 respectively.
• The densities of these two gases at the
same temperature and pressure are d1
and d2 respectively.
• Acc. to Graham’s law of diffusion of
gases.
r1/r2 = dαM
=
• The ratio of densities of any two gases is
equal to the ratio of the molecular weight
of those two gases.

30. • The diffusion rate of a gas means the
volume of the diffused gas in one second.
• diffusion rate (r) = Volume of gas diffused (V)
Time required for diffusion (t)
r= v/t
• For two gases at the same temperature and
pressure,
r1 = V1/t1 and r2 = V2/t2
• During the experiment, for convenience,
the times required for diffusion of same
volumes of two gases diffusing in the same
time are measured.

31.  
  
Hence the above equation can be written as follows :
 r1/r2 = V1/t1 =
M2 OR V1•t2 = =
d1 M2
V2/t2 M1 V 2•t1 d2 M1
V1 =
V2


 Now if t1 = t2
 M2
M1
then
t2
t1
=



 But if V1 = V2 then
M2
M1

32. Importance of Graham’s law of gaseous
diffusion
 Uranium metal has two isotopes : U235 and U238.
 U235 is very important in production of atomic energy.
 The proportion of U235 in uranium metal is only 0.7%.
 As uranium hexafluoride (UF₆) is a volatile compound
uranium hexafluoride is prepared from uranium metal.
 The difference of molecular weight between 235UF₆ and
238
UF₆ is much less. Hence, the ratio of rate of diffusion
of these gases will be 1.0047.
 Now, if uranium hexafluoride gas is filled in a porous
vessel allowed to have the diffusion, the amount of
235
UF₆ of less density will diffused somewhat more.
Because of the small difference in diffusion rate, a
series of a number of experiments is constructed.

33. This type of work is carried out in a laboratory
extended to kilometer at oak-Ridge in tenessy
state of america.
The experiment of diffusion of this gas through
porous membranes distributed (extended) to
about a kilometer.
After a long time pure 235UF₆ is obtained which is
decomposed to get pure 235U.
In short, the isotopes of uranium 235U and 238U
can be separated.
The importance of this law is in finding the
molecular weights of gases and the densities.
The components gases can also be separated
from the mixture of gases.

34. Avogadro’s Hypothesis
 Avogadro gave a principle in 1811 A.D.
According to it, “Equal volume of the gases
contain equal number of molecules at standard
temperature and pressure .
 Simple gas equation is one of the methods to
presents the Avogadro’s principle.
 One important dimension resulting from
Avogadro’s principle is molar volume.
 Molar volume means the volume occupied by
molecular weight expressed in gram of gas. The
volume of 1 mole at 273 kelvin and 1
atmosphere pressure can be found out by
general gas equation.

36. • PV = nRT where P = 1 atmosphere
V = ? litre
V = nRT n = 1 mole gas
P R = 0.082 lit.a tm./k.mol.
T = 273 K.
V = 1 mole x 0.082 lit.atm./k.mol. 273 ‫ ג‬kelvin
1 atmosphere
V = 22.4 litre
Thus molar volume is also called gram molar

37. Thus molar volume is also called gram molar volume.
The presentation of Avogadro’s principle on the basis
of molar volume can be done as follows :
“In 22.4 litre of any gas at 273 kelvin 1 atmosphere
contains 1 mole molecules.”
This statement can be given alternatively as, “The
weight of 22.4 litre of any gas at 273 kelvin
temperature and 1 atmosphere pressure, is its
molecular weight.”
According to Avogadro’s principle, “ the number of
molecules in 1 molar volume of any gas is 6.022 10 23.”
The weight of one mole of any substance is its
molecular weight.”

38. 1 mole 6.022 x 1023
22.4 litre Gram molecular
weight

39. Kinetic molecular theory
 All the gases are composed of innumerable microscopic particles (atoms or
molecules).
 The volume of molecules is negligible in comparison with the volume
(volume of the vessel) occupied by the gas. All the molecules in each have
same volume and weight.
 The molecules of the gas are in continuous motion.
 The molecules of the gas created (develops) pressure on the wall by striking
with the walls of the vessel.
The molecules in the gas have no attraction or repulsion for each other.
 When the continuously moving molecules strike with one another, they
exchange the kinetic energy
As this process is continuously going on, the molecules in the gas do not
move with uniform velocity. At any time and any temperature, the velocity of
certain molecules will be very less, some will have moderate and will have very
high. Really the average velocity of each molecule

40. What are the different types of solids?
There are four types of crystalline solids --
Ionic solids-- These substances have a definite melting point
and contain ionic bonds. An example would be sodium
chloride (NaCl). View the 3-D structure of a salt crystal.
Covalent solids -- These substance appear as a single giant
molecule made up of an almost endless number of covalent bonds.
An example would be graphite. View the 3-D structure of graphite).
Molecular solids are represented as repeating units made up
of molecules. An example would be ice. View the 3-D
structure of ice.
Metallic solids are repeating units made up of metal atoms. The
valence electrons in metals are able to jump from atom to atom.

41. CRYSTAL LATTICE
• The definate arrangement of constituent
particle (atoms,ions or molecules) shown
by dots in three dimension in crystal is
known as crystal lattice.
UNIT CELL :
• A tiny or smallest part of the crystal lattice
lattice which bears all the characteristics
of the crystal and when repeated in three
dimension forms complete crystal structure

42. UNIT CELL OF NaCl

43. Unit cell of NaCl
It is be observed that each Na+ is surrounded octahedrally by
six chloride icons and similarly each chloride icon by six Na+
ion.
Here the ionic size of Na+ ions is similar and cannot be
arranged in a manner that each ions touches its neighbour
ion.
In this configuration Na+ and Cl- ions are arranged in such a
way that they remain as near as possible with each other.
In this construction Na+ - Na+ ion is maximum. Because of a
such a arrangement, the distance between Cl- ions increased
automatically.
The co-ordination number of Na+ in NaCl crystal is six and the
ratio of Na+ /Cl- radii is 0.53.

44. UNIT CELL OF CsCl

45. Unit cell of CsCl
If we examine the configuration of unit cell of CsCl,
it will be found that each Cs+ is surrounded by eight
Cl- and similarly each Cl- ions is surrounded by eight
Cs+ ions.
If the co-ordination number of metal ion is more in an
ionic crystal, the stability is also more.
Hence the stability of CsCl is more than that of
NaCl .
The co-ordination number of Cs+ in CsCl crystal is
eight and the ratio of Cs+ in CsCl crystal is eight and
the ratio of Cs+ / Cl- radii is 0.92.
It has body centered cubic arrangement.

46. UNIT CELL OF CsCI

47. Unit cell of LiI
In LiI, negative charge possessing I- is much larger in
size as compared to the positive charge possessing
Li+ ion. Hence the negatively charged ions can be
arranged very near to each other.
The positively charged ion can be arranged very
near to each other.
The positively charged ions can be easily placed in
the vacant space formed between these ions. This
type situation arises in the crystal of LiI.
The cross section of layer containing ions is shown
in figure. In this four I- ions are arranged almost
touching each other. I- ions are also above and below
of this central void of this configuration.

48. UNIT CELL OF LiI
• Due to eight I- ions arranged in a manner touching
each other, the shape that evolves is octahedral.
• As I- ion is big, the size of octahedral configuration is
comparatively big.
• Li+ ion being smaller in size can be arranged easily
in the central void (space). In this configuration,
similarly charged I- ions are arranged near each
other in such a manner that the attraction between
them is less and repulsion is more.
• Thus this configuration possesses relatively less
stability. Because of this, the melting point of LiI is
less than that of NaCl.

49. Information about different co-ordination
numbers and ratio of radii
Radii Ratio Co-ordination Arrangement of Examples
(r+/r-) number of positive-negative
positive ions ions
Upto 0.155 2 Linear
0.155 to 0225 3 Planer triangle
0.225 to 0.414 4 Tetrahedral ZnS
0.414 to 0.73 4 Square planer
0.414 to 0.73 6 FCC NaCl
0.73 to 1.0 8 Octahedral-BCC CsCl
Above 1.0 12 HCP

50. CHARACTERISTICS OF THE LIQUID
 Fix volume
 Fluidity
 Non-compressibility
 Diffusion
 Evaporation
 Vapour pressure
 Surface tension
 Viscosity

51. Clearity of the term
 Diffusion :
The property of the liquid to spread
in another liquid.
 Evaporation :
The property of liquids to get
converted of its own into gaseous
state at normal temperature

52.  Vapour pressure :
The vapour exert pressure on the
surface of the liquid at equillibrium.
 Surface tension :
The force exerted by the molecules
on the hypothetical line of unit
length parellel to the surface of the
liquid and perpendicular to the
molecules on the other side of the
molecules.