CHAPTER 13

Liquids and Solids
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1
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Chapter Goals





Kinetic-Molecular Description of Liquids & Solids
Intermolecular Attractions & Phase Changes
Properties of Liquids
– Viscosity, Surface Tension, Capillary Action
– Evaporation, Vapor Pressure,
– Boiling Points & Distillation, Heat transfer
Properties of Solids
– Melting Points, Heat Transfer
– Sublimation & Vapor Pressure
– Phase Diagrams, Amorphous & Crystalline Solids
– Crystal Structures, Bonding in Solids, Band Theory
Synthesis Question
2
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Kinetic-Molecular Description of
Liquids & Solids

Solids & liquids are condensed states
–
–

Liquids & gases are fluids
–

atoms, ions, molecules are close to one another
highly incompressible
easily flow
Intermolecular attractions in liquids &
solids are strong
3
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Kinetic-Molecular Description of
Liquids & Solids
gas
cool
heat
liquid
cool
solid
heat
4
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Kinetic-Molecular Description of
Liquids & Solids
strengths of interactions among particles &
 degree of ordering of particles
Gases< Liquids < Solids
 Miscible liquids diffuse into one another

–

they are soluble in each other
for example:
–
–
water/alcohol
gasoline/motor oil
5
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Kinetic-Molecular Description of
Liquids & Solids

Immiscible liquids do not diffuse into each
other
–

they are insoluble in each other
for example:
–
–
water/oil
water/cyclohexane
6
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Intermolecular Attractions and
Phase Changes

Ion-ion interactions
–
force of attraction between two oppositely
charged ions is determined by Coulomb’s law
q q 

F
+
d
-
2
7
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Intermolecular Attractions and
Phase Changes

energy of attraction between two ions is given
by:
+
-
E =
q q 

Fd =
d
q q 

=
+
d
2
-
d
8
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Intermolecular Attractions and
Phase Changes

Coulomb’s law & the attraction energy
determine:
–
–

melting & boiling points of ionic compounds
the solubility of ionic compounds
Example 13-1: Arrange the following ionic
compounds in the expected order of
increasing melting and boiling points.
NaF, CaO, CaF2
you do it
What important points must you consider?
9
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Intermolecular Attractions and
Phase Changes

Dipole-dipole interactions
–
consider NH3 a very polar molecule
11
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Intermolecular Attractions and
Phase Changes

Hydrogen bonding
–
consider H2O
12
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Intermolecular Attractions and
Phase Changes
13
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Intermolecular Attractions and
Phase Changes

London Forces
very weak
only attractive force in nonpolar molecules
consider Ar
isolated atom
14
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Intermolecular Attractions and
Phase Changes

Group of Ar molecules
temporary dipole induces other dipoles
15
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The Liquid State

Viscosity - resistance to flow
water vs. molasses
you buy oil for your car based on this property
Ostwald viscometer used to measure this
property
16
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The Liquid State
Surface Tension - measure of the unequal
attractions that occur at the surface of a
liquid
 molecules at surface are attracted unevenly

–
–
water bugs
floating razor blades
17
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The Liquid State
Capillary Action - ability of a liquid to rise
(or fall) in a glass tube
 cohesive forces - hold liquids together
 adhesive forces - forces between a liquid
and another surface

–
–
capillary rise implies adhesive > cohesive
capillary fall implies cohesive > adhesive
18
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The Liquid State

Capillary Action
water
mercury
19
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The Liquid State

Evaporation
–
–
process in which molecules escape from the surface
of a liquid
T dependent
20
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The Liquid State

Evaporation
21
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The Liquid State

Vapor Pressure
–

pressure exerted by a liquid’s vapor on its surface
at equilibrium
Vap. Press. (torr) for 3 Liquids
0oC
20oC 30oC
diethyl ether 185 442 647
 ethanol
12 44 74
 water
5 18 32

Norm. B.P.
36oC
78oC
100oC
22
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The Liquid State

Vapor Pressure
23
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The Liquid State

Boiling Points & Distillation
–
–

boiling point is temperature at which the
liquid’s vapor pressure is equal to applied
pressure
normal boiling point is boiling point @ 1 atm
distillation is a method we use to separate
mixtures of liquids based on their
differences in boiling points
24
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The Liquid State

Distillation
–
–
process in which a mixture or solution is
separated into its components on the basis of
the differences in boiling points of the
components
Distillation is another vapor pressure
phenomenon.
25
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The Liquid State

Heat Transfer Involving Liquids
–
from Chapter 1
q = m C T
Example 13-2: How much heat is released by 200
g of H2O as it cools from 85.0oC to 40.0oC?
The specific heat of water is 4.184 J/goC.
you do it
26
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The Liquid State
molar heat capacity - amount of heat required
to raise the T of one mole of a substance one
degree C
 Example 13-3: The molar heat capacity of
ethyl alcohol, C2H5OH, is 113 J/moloC. How
much heat is required to raise the T of 125 g
of ethyl alcohol from 20.0oC to 30.0oC?
1 mol C2H5OH = 46.0 g

you do it
28
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The Liquid State

energy associated with changes of state
heat of vaporization
amount of heat required to change 1 g of a
liquid substance to a gas at constant T
units of J/g
heat of condensation
reverse of heat of vaporization
o
+2260 J
o
1.00 g H2O(l) @ 100 C  1.00 g H2O(g) @ 100 C
-2260 J
30
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The Liquid State
molar heat of vaporization or Hvap
amount of heat required to change 1 mol of a
liquid to a gas at constant T
units of J/mol
molar heat of condensation
reverse of molar heat of vaporization
o
+40.7 kJ
o
1.00 mol H2O(l) @ 100 C  1.00 mol H2O(g) @ 100 C
-40.7 kJ
31
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The Liquid State
32
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The Liquid State

Example 13-4: How many joules of energy
must be absorbed by 500 g of H2O at 50.0oC
to convert it to steam at 120oC? The molar
heat of vaporization of water is 40.7 kJ/mol
and the molar heat capacities of liquid water
and steam are 75.3 J/mol oC and 36.4 J/mol
oC, respectively.
33
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The Liquid State
1 mol H2O
? mol = 500 g H2O 
 278
. mol H2O
18 g H2O
o
1st let's calculate the heat required to warm water from 50 to 100 C


753
. J

o
5
? J = 27.8 mol
1000
.

500
.
C

105
.

10
J

o
 mol C
34
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The Liquid State
2nd let’s calculate the energy required to boil the water
 40.7  103 J 
5
? J = 27.8 mol

1131
.

10
J

mol


35
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The Liquid State
2nd let’s calculate the energy required to boil the water
 40.7  103
? J = 27.8 mol
mol

J
5
.  10 J
  1131

3rd let’s calculate the heat required to heat steam from
100 to 120oC
36.4 J 

o
5
? J =  27.8 mol 
120.0
-100.0
C

0
.
20

10
J

o
 mol C 


36
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The Liquid State
total amount of energy for this process is the sum of
the 3 pieces we have calculated
.  10 J   1131
.  10 J    0.20  10 J  
105
5
5
5
12.56  105 J or 1.26  103 kJ
37
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The Liquid State

Example 13-5: If 45.0 g of steam at 140oC
is slowly bubbled into 450 g of water at
50.0oC in an insulated container, can all the
steam be condensed?
you do it
38
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The Liquid State

Clausius-Clapeyron equation
–
–
–
determine vapor pressure of a liquid at a new T
determine what T we must heat something to get a
specified vapor pressure
way to determine Hvap if we know pressure at 2 T’s
 P2  H vap  1 1 
ln   
  
R  T1 T2 
 P1 
42
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The Liquid State

In Denver the normal atmospheric pressure is 630
torr. At what temperature does water boil in
Denver?
 P2  H vap  1 1 
ln   
  
R  T1 T2 
 P1 
3 J
40
.
7

10
630
torr
1
1


mol 
ln 
 


 760 torr 
8.314 J K mol  373 K T2 

1
ln  0.829  4895 0.002681  
T2 

43
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The Liquid State

 0188
.
1
  0.002681  
4895
T2 

 3.83  10
5
 3.83  10
5

1
  0.002681  
T2 

1
 0.002681  
T2
1
 0.00272  
T2
T2  368 K or 95o C
44
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The Liquid State

Boiling Points of Various Kinds of Liquids
Gas
MW
BP(oC)
He
4
-269
Ne
20
-246
Ar
40
-186
Kr
84
-153
Xe
131
-107
Rn
222
-62
45
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The Liquid State
0
4
20
40
84
131
222
-50
-100
BP-150
-200
-250
-300
MW
46
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The Liquid State
o
Compound MW(amu) B.P.( C)
CH4
16
-161
C2H6
30
-88
C3H8
44
-42
n-C4H10
58
-0.6
n-C5H12
72
+36
47
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The Liquid State
50
0
16
30
44
58
72
-50
BP
-100
-150
-200
MW
48
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The Liquid State
o
Compound
MW(amu)
B.P.( C)
HF
20
19.5
HCl
37
-85.0
HBr
81
-67.0
HI
128
-34.0
49
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The Liquid State
40 HF
20
0
BP
-20 20
37
81
128
HI
-40
-60
HBr
-80
-100
HCl
MW
50
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The Liquid State
Compound
MW(amu)
o
B.P.( C)
H2O
18
100
H2S
34
- 61
H2Se
81
- 42
H2Te
130
-2
51
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The Liquid State
120
100
H2O
80
60
40
20
BP
0
-20 18
34
81
130
H2Te
-40
-60
H2Se
-80
H2S
MW
52
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The Liquid State

Example 13-6: Arrange the following
substances in order of increasing boiling
points.
C2H6, NH3, Ar, NaCl, AsH3
you do it
53
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The Solid State

Normal Melting Point
–
–
T at which the solid melts (liquid and solid in
equilibrium) at 1 atm of pressure
melting point increases as intermolecular
attractions increase
55
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Heat Transfer Involving Solids

heat of fusion
amount of heat required to melt one gram of a
solid at its melting point at constant T
 334 J
1.00 g H2O (s) @ 0 C
1.00 g H2O (l) @ 0 C
o
o
heat of crystallization
reverse of heat of fusion
56
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Heat Transfer Involving Solids
molar heat of fusion
amount of heat required to melt a mole of a
substance at its melting point H fusion
molar heat of crystallization
reverse of molar heat of fusion Hcrystallization
 6012 J
1.00 mol H2O (s) @ 0o C
1.00 mol H2O (l) @ 0o C
57
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Heat Transfer Involving Solids

Summary of heats of transformation of water
o
+2260 J
o
1.00 g H2O(l) @ 100 C  1.00 g H2O(g) @ 100 C
-2260 J
 334 J
1.00 g H2O (s) @ 0 C
1.00 g H2O (l) @ 0 C
o
o
58
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Heat Transfer Involving Solids

Example 11-7: Calculate the amount of heat
required to convert 150.0 g of ice at -10.0oC
to water at 40.0oC.
specific heat of ice is 2.09 J/goC
you do it
59
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Sublimation & Vapor Pressure
of Solids

Sublimation
–
–
solid transforms directly to vapor
solid CO2 or “dry” ice
sublimation
solid

gas
deposition
61
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Phase Diagrams (P vs T)

convenient way to display all of the different
phases of a substance
phase
diagram for
water
–
62
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Phase Diagrams (P vs T)

phase
diagram
for
carbon
dioxide
63
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Amorphous & Crystalline Solids

Amorphous solids do not have a well
ordered structure
paraffin, glasses

Crystalline solids have well defined
structures that consist of extended array of
repeating units
give X-ray difraction patterns
see Bragg equation in book
64
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Structure of Crystals

unit cell - smallest repeating unit of a
crystal
bricks are repeating units for buildings

7 basic crystal systems
65
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Structure of Crystals

Simple
cubic
66
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Structure of Crystals

Simple cubic
–
–
each particle at a corner is shared by 8 unit cells
1 unit cell contains 8(1/8) = 1 particle
67
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Structure of Crystals

Body centered cubic (bcc)
–
–
8 corners + 1 particle in center of cell
1 unit cell contains 8(1/8) + 1 = 2 particles
68
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Structure of Crystals

Face
centered
cubic
(fcc)
69
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Structure of Crystals

Face centered cubic (fcc)
–
–
8 corners + 6 faces
1 unit cell contains 8(1/8) + 6(1/2) = 4 particles
70
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Bonding in Solids

Molecular Solids
–
–
–
molecules occupy unit cells
low melting points,volatile & insulators
examples:
 water,
sugar, carbon dioxide, benzene
71
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Bonding in Solids

Covalent Solids
–
–
atoms that are covalently bonded to one another
examples:
 SiO2
(sand), diamond, graphite, SiC
72
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Bonding in Solids

Ionic Solids
–
–
ions occupy the unit cell
examples:
 CsCl,
NaCl, ZnS
73
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Bonding in Solids

Metallic Solids
–
–
–
positively charged nuclei surrounded by a sea
of electrons
positive ions occupy lattice positions
examples:
 Na,
Li, Au, Ag, ……..
74
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Bonding in Solids
Variations in Melting Points
 Molecular Solids
Compound
Melting Point (oC)
ice
0
ammonia
-77.7
benzene, C6H6
5.5
napthalene, C10H8
80.6
benzoic acid, C6H5CO2H 122.4

75
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Bonding in Solids
Covalent Solids
Substance
sand, SiO2
carborundum, SiC
diamond
graphite

Melting Point (oC)
1713
~2700
>3550
3652-3697
76
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Bonding in Solids
Ionic Solids
Compound
LiF
LiCl
LiBr
LiI
CaF2
CaCl2
CaBr2
CaI2

Melting Point (oC)
842
614
547
450
1360
772
730
740
77
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Bonding in Solids
Metallic Solids
Metal
Na
Pb
Al
Cu
Fe
W

Melting Point (oC)
98
328
660
1083
1535
3410
78
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Unit Cell Problem

A group IVA element with a density of
11.35 g/cm3 crystallizes in a facecentered cubic lattice whose unit cell
edge length is 4.95 A. Calculate the
element’s atomic weight. What is the
atomic radius of this element?
79
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Unit Cell Problem
fcc has 4 atoms per unit cell
 determine the volume of a single unit cell

0
1 A = 10
-8
0
cm  4.95 A  4.95  10
fcc has cubic dimensions so V = l
4.95  10
-8
cm

3
 121
.  10
 22
cm
-8
cm
3
3
80
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Unit Cell Problem



use density to determine the mass of a unit cell
determine the mass of one atom in a unit cell


121
.  1022 cm3 1135
. g / cm3  138
.  1021 g / unit cell
in fcc there are 4 atoms per unit cell
thus the mass of one atom is
138
.  1021 g
 344
.  1022 g / atom
4
81
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Unit Cell Problem


determine the mass of 1 mole of these atoms


344
. 1022 g / atom 6022
. 1023 atoms / mole  2072
. g / mole
Pb has a molar mass of 207.2 g / mol
82
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Unit Cell Problem


determine the radius of a Pb atom requires some
geometry from high school
notice there are 4 radii on the diagonal
83
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Unit Cell Problem

determine the diagonal length then divide by 4 to
get atomic radius
diagonal =


2 4.95  10 cm
-8
 700
.  10 cm
-8
-8
700
.

10
cm
radius =
 175
.  10 cm
4
-8
84
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Band Theory of Metals

Na’s 3s orbitals can interact to produce
overlapping orbitals
85
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Band Theory of Metals

Can also overlap with unfilled 3p orbitals
86
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Band Theory of Metals
Insulators have a large gap - forbidden zone
 Semiconductors have a small gap

87
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Synthesis Question

Maxwell House Coffee Company
decaffeinates its coffee beans using an
extractor that is 7.0 feet in diameter and
70.0 feet long. Supercritical carbon
dioxide at a pressure of 300.0 atm and
temperature of 100.00C is passed through
the stainless steel extractor. The extraction
vessel contains 100,000 pounds of coffee
beans soaked in water until they have a
water content of 50%.
88
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Synthesis Question

This process removes 90% of the caffeine in
a single pass of the beans through the
extractor. Carbon dioxide that has passed
over the coffee is then directed into a water
column that washes the caffeine from the
supercritical CO2. How many moles of
carbon dioxide are present in the extractor?
89
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Synthesis Question
Diameter of vessel  (7.0 ft)(30.48cm/ft)  213.4 cm
Radius of vessel  213.4 cm/2  106.7 cm
Length of vessel  (70.0 ft)(30.48cm/ft)  2134 cm
Volume of vessel   r 2 h  (3.1416)(106.7cm)2 (2134cm)
 (7.633107 cm3 )(1 mL/cm3 )(1 L/1000 mL)
 7.633104 L
90
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Synthesis Question
PV  nRT
n  PV

300 atm 7.63310 L 

RT 0.08206 L atm mol K 373 K 
4
n  748,000 mol of CO 2
91
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Group Question

How many CO2 molecules are there in 1.0
cm3 of the Maxwell House Coffee Company
extractor? How many more CO2 molecules
are there in a cm3 of the supercritical fluid
in the Maxwell House extractor than in a
mole of CO2 at STP?
92
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Chapter Goals
Kinetic-Molecular Description of Liquids &
Solids
 Intermolecular Attractions & Phase Changes
 Properties of Liquids

–
–
–

Viscosity, Surface Tension, Capillary Action
Evaporation, Vapor Pressure,
Boiling Points & Distillation, Heat transfer
Properties of Solids
–
–
–
–
Melting Points, Heat Transfer
Sublimation & Vapor Pressure
Phase Diagrams, Amorphous & Crystalline Solids
Crystal Structures, Bonding in Solids, Band
Theory
93
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End of Chapter 13


Our understanding of
Band Theory was a
major breakthrough
in semiconductor
knowledge.
Why computers
work!
94
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