Chemistry. States of matter – SESSION IV Session Objectives.

Chemistry

States of matter – SESSION IV

Session Objectives

Session objectives

Vapour pressure of liquids

Surface tension

Viscosity

Solids and their types

Vapour pressure of liquids

The pressure exerted by vapours in equilibrium with liquid at a giventemperature.

Surface tension

the units of surface tension are Nm–1 or J m–2

A

Bmolecule in the bulk of the liquid

molecule on the surface of the liquid(experiences resultant downward force)

Factors affecting surface tension

(i)Temperature:

• Surface tension decreases with increase in temperature (because increase in temperature decreases the intermolecular forces.)

(ii)Effect of surfactants:The surface active agents decreases the surface tension of water

• For example, soap, detergent, ethanol.

• Addition of surfactants decreases the surface tension of liquid.

Surfactants

Consequences of surface tension

Spherical shape of liquid drops Capillary action

Do you know?

How does water rises in a capillary?

The surface tension pulls the water into the capillary .The surface tension of a fine capillary is very largetherefore, it overcomes the attraction of gravity on water.

Viscosity

The internal resistance which one liquid layer offers to another layer sliding over it, during its flow is known as its viscosity.

av

fx

Viscosity

Where,

f = Force of friction between two layers of liquid

= Coefficient of viscosity

a = Area of one layer of liquid in square centimetre

x = Distance between the two layers of liquid.

v = Difference in velocities of two layers of liquid.

Units of viscosity = dynes cm–2 s

SI units of viscosity = Nm–2 s

What is solid ?

Solids have definite shape, definite volume

and strong force of attraction

among constituting particles.

Crystalline and amorphous solids

Amorphous or pseudo solids Crystalline or true solids

Constituents are arranged in orderly fashion.

Constituents are not arranged in orderly fashion.

AnisotropicIsotropic

Have long range order Have short range order

Undergo clear cleavage Undergo irregular cut

Sharpe melting point Melting over range of temperature

Have definite heat of fusione.g., NaCl

Do not have definite heat of fusione.g., Glass

Ionic crystals

Positive and negative ions arranged in a definite order

Strong electrostaticforces of attraction

Brittle, high melting point,good conductors in the aqueous solutionor fused state, highheats of fusion

Salts like NaCl,KNO3, LiF, BaSO4

Molecular crystals

Small molecules

van der Waal'sforces

Soft, low melting point,volatile, electricalinsulators, poor thermalfusion

Solid CO2 (dry ice),CH4,wax

Covalent crystals

Atoms chemically bounded together in the form of anetwork

Covalentbond forces

Very hard, high meltingpoint, poor conductors of heat and electricity high heats of fusion

Diamond, silicon,quartz.

Metallic crystals

Positive ions andmobile electrons

Electrical attractions(metallic bond)

Very soft to very hard, low to high melting point, good conductors ofelectricity and heat,metallic lustre, malleable and ductile, moderateheats of fusion

All metals and some alloys

Lattice or space lattice or crystal lattice

Regular arrangements of the constituent particles in three dimensional space

Unit cell

Is the smallest portion of the space lattice which when repeated again and again in different directionsgenerates the complete space lattice.

Unit cell

Bravais (1848)

Possible shapes of unit cell

CubicTetragonal

Orthorhombic

Hexagonal

Trigonal or Rhombohedral

Monoclinic

Triclinic

Arrangement of atoms/ ions in unit cell

Primitive or Basic

Body centered

Face centered

End centered

Primitive cubic or simple cubic

Total no. of constituents per unit cell= Total number of corners x contribution by each atom =8 x 1/8

=1C

a

b

Simple or primitive

Rank

Body centered cubic unit cell

Total no. of constituents per unit cell.= Total contribution of constituents at corners + Total contribution of particle at centre

=8 x 1/8 + 1 x 1

=1 + 1

=2

Face centered cubic unit cell

Total no. of constituents per unit cell.= Total contribution by constituents at corners + Total contribution by constituents on the faces =8 x 1/8 + 1/2 x 6

=1 + 3

=4

End centered cubic unit cell

Total no. of constituents per unit cell.= Total contribution by constituents at corners + Total contribution by particles on the faces. =8 x 1/8 + 1/2 x 2

=1 + 1

=2

Co-ordination number

•Number of spheres which are touching a particle sphere

•In ionic crystals Number of oppositely charged ions surrounding a particular ione.g., Co-ordination number of Cl– and Na+ in NaCl molecule is 6each.

Question

Illustrative example

If three elements P,Q and R crystallizein a cubic solid lattice with P atoms at the corners, Q atoms at the cube centre and R atoms at the centre of the faces of the cube, then write the formula of the compound .

Solution:1

Atoms of elements P per unit cell 8 18

Atoms of elements Q per unit cell 1

1Atoms of elements R per unit cell 6 3

2

3The formula of compund PQR

Packing fraction

• Is the fraction of total volume of a cube occupied by constituent particles.

Packing fraction(PF) =

Volume occupied by effective number of particlesVolume of the unit cell

Packing fraction of simple cubic crystalFor simple cubic crystal(scc)

Packing fraction = 3

3

41 r

3

a

3

3

4r

3 a 2r 0.5232r

For bcc, body diagonal, 4r 3a

Packing fraction =

3

3

42 r

3 0.68 i.e., 68%4r

3

aA B

Packing fraction of face cubic crystal

For fcc, face diagonal, 4r = 2a

Packing fraction =

3

3

44 r

3 0.74 i.e., 74%4r

2

A B

C

a

Density

The number of particles present per unit cell.

Suppose the edge length of the unit cell = a

Number of atoms present in one unit cell = Z

Atomic mass of the element = M

Density

Density of unit cell () =Mass contained in one unit cell

Volume of the unit cell

Mass contained in one unit cell = Number of particles in one unit cell × Mass of one particle

Since mass of one particle = Atomic mass

Avogadro ' s number

Density

Therefore, mass contained inone unit cell =

A

MZ

N

Volume of unit cell = (Edge length)3 = a3

A3

MZ

N

a

3

A

Z M

a N

Do you know

density depends on the type of the crystal structure.

A

B

C

DE

F

N

L M8

88

O

X-ray reflection from crystals

Class exercise 1

Which of the following liquids is most difficult to suck into pipette?

(a) Toluene (b) Water

(c) Glycerol (d) Lemon juice

Since it has highest viscosity.

Solution:-

Answer is (c).

Class exercise 2

Which of the following is pseudo solid?

(a) Diamond (b) Common salt

(c) Graphite (d) Plastic

(d) Plastic is pseudo solid or amorphous solid.

Solution:

Class exercise 3

Which one of the following is not the property of crystalline solid?

(a) Isotropic

(b) Sharp melting point

(c) Definite regular geometry

(d) High intermolecular forces

(a) Crystalline solids are an isotropic.

Solution:

Class exercise 4

With increase in temperature, the fluidity of liquids

(a) increases

(b) remains constant

(c) Decreases

(d) None of these

1Fluidity

Coefficient of viscosity

Solution:

Hence, the answer is (a)

Class exercise 5

The rise of liquid in a capillary tube is due to

(a) viscosity (b) effusion

(c) diffusion (d) surface tension

The liquid in capillary rises due to surface tension.

Solution:

Hence, the answer is (d).

Class exercise 6

Which of the following statements is correct?(a) Vapour pressure decreases with increase of temperature

(b) Vapour pressure increases with increase of temperature

(c) Vapour pressure is independent of temperature

(d) None of these

Vapour pressure Temperature

Solution:

Hence, the answer is (b).

Class exercise 7

Particles of quartz are packed by

(a) van der Waals’ forces

(b) covalentely bonded forces

(c) electrical attraction forces

(d) None of these

Quartz is a covalent solid.

Solution:

Hence, the answer is (b).

Class exercise 8

Packing fraction of a scc unit cell is

(a) 52% (b) 74%

(c) 68% (d) 92%

3

3

41 r

3For scc, PF 0.52 since (a 2r)(2r)

Solution:

Hence, the answer is (a).

Class exercise 9

Calculate the packing fraction of a fcc unit cell if two face-centred atom have removed from the unit cell.

For fcc, effective number of atom = 1 18 4 3

8 2

Since two face-centred atoms have lost.

3

3

43 r

3PF 0.555, i.e., 55.5%4r

2

Solution:

Class exercise 10

Metallic gold crystallizes in fcc lattice with edge length 4.07 Å. Find its density. (Au = 197).

3AV

Z MDensity,

N a

3

323 8

4 197g / cm

6.023 10 4.07 10

= 19.4 g/cm3

Solution:

Thank you