Logarithm Review
Definition
ab = c, (a > 0, a ≠ 1)
logac = b
If a = 10, it is called common logarithm
log c = log10c
If a = e = 2.718281828459045 ∙ ∙ ∙, it is called natural logarithm
ln c = logec
Keys on your calculator
Properties of Logarithm
x > 0, y > 0
ln(xy) = ln(x) + ln(y)
ln(x/y) = ln(x) − ln(y)
ln(xm) = m ln(x)
Also see Appendix I B
Chapter 11
Liquids, Solids and
Intermolecular Forces
continued
Chemistry, continue on
Vapor Pressure
Surface Molecules
Temperature:
T
Temperature:
T
GA, 760 torr = 1 atm
H2O
100 °C
Normal
Boiling
Point
Tibet, 480 torr < 1 atm
H2O
85 °C
Normal
Boiling
Point
(a) The Vapor Pressure of Water, Ethanol, and Diethyl Ether as a
Function of Temperature. (b) Plots of In(Pvap) versus 1/T for
Water, Ethanol, and Diethyl Ether
T is in K!
1/T (K−1)
Linear relation: y = kx + C
y
slope: k = tg θ
θ
x
C: intercept
1/T (K−1)
Linear relation: y = kx + C
ln P = k(1/T) + C
Heat of vaporization ∆Hvap: energy needed to convert one mole
of liquid to gas. Unit: J/mol or kJ/mol.
∆Hvap > 0
ln P  
 H va p 1
R
y
slope k < 0
T
x
C
ln (P)
ln P  
 H va p 1
R
C
T
1
ln (P1)
2
ln (P2)
1/T1
1/T2
1/T (K−1)
Clausius-Clapeyron Equation
ln P  
 H va p 1
R
C
T
 H vap  1
1 
ln
 



P2
R  T1 T 2 
P1
The vapor pressure of water at 25 °C is 23.8 torr, and the heat
of vaporization of water is 43.9 kJ/mol. Calculate the vapor
pressure of water at 50 °C.
 H vap  1
1 
ln
 



P2
R  T1 T 2 
P1
Five: T1, T2, P1, P2, ∆Hvap
Four known, calculate the other.
Units in ideal gas law
PV = nRT
Option 1
Chem 1211
P — atm, V — L, n — mol, T — K
R = 0.082 atm · L · mol−1 · K−1
Option 2
P — Pa, V — m3, n — mol, T — K
R = 8.314 J ·
mol−1 ·
K−1
Clausius-Clapeyron
equation
Carbon tetrachloride, CCl4, has a vapor pressure of 213 torr at
40 °C and 836 torr at 80 °C. What is the normal boiling point of
CCl4?
 H vap  1
1 
ln
 



P2
R  T1 T 2 
P1
( Please try to work on this question by yourself. Will review next week)
Liquid potassium has a vapor pressure of 10.00 torr at 443 °C
and a vapor pressure of 400.0 torr at 708 °C. Use these data
to calculate
(a) The heat of vaporization of liquid potassium;
(b) The normal boiling point of potassium;
(c) The vapor pressure of liquid potassium at 100. °C.
( Please try to work on this question by yourself. Will review next week)
Clausius-Clapeyron Equation
ln P  
 H va p 1
R
C
T
 H vap  1
1 
ln
 



P2
R  T1 T 2 
P1
ln P  
 H va p 1
R
y
slope k < 0
T
x
C
Linear relation: y = kx + C
y
slope: k = tg θ
θ
x
C: intercept
y
a
d
c
b
x
Lines tilt to the right have positive slopes (a and b), left negative
(c and d). Steeper line has greater absolute value of slope. In this
graph, the order of slopes is
a>b>0>c>d
What is the order of heat of vaporization
for these three substances?
Solids
Glass (SiO2)
Crystal
Solid
Noncrystal
Basis
Crystal structure
The basis may be a single atom or molecule, or a
small group of atoms, molecules, or ions.
NaCl:
1 Na+ ion and 1 Cl− ion
Cu:
1 Cu atom
Zn:
2 Zn atoms
Diamond:
2 C atoms
CO2:
4 CO2 molecules
Use a point to represent the basis:
=
Lattice
Lattice point:
Unit cell: 2-D, at least a parallelogram
Unit cell is the building block of the crystal
How many kinds of 2-D unit cells
can we have?
      
      
      
      
      
      
      
      
Extend the concept of unit cell to 3-D,
the real crystals.
: 3-D, at least a parallelepiped
How many kinds of 3-D unit cells
can we have?
1. triclinic
a≠b≠c
α≠β≠γ
2. monoclinic
c
a
a≠b≠c
b
c
a
b
The 14 Bravais lattices
3. orthorhombic
7 crystal systems
α = β = γ = 90°
5. rhombohedral (trigonal)
4. tetragonal
a=b=c
90° ≠ α = β = γ < 120°
a=b≠c
α = β = γ = 90°
7. cubic
6. hexagonal
a=b≠c
α = β = 90° ,γ = 120°
γ
a=b=c
α = β = γ = 90°
(Simple cubic)
Chem 1212: assume a lattice point is a single atom
Information of a cubic unit cell
• Size of the cell
X-ray diffraction
The Wave Nature of Light
Information of a cubic unit cell
• Size of the cell
X-ray diffraction
• Size of the atoms
• Number of atoms in a cell
Soon
Now!
C
D
B
A
C
D
E
B
A
F
Number of atoms in a unit cell = ¼ x 4 = 1
Number of Atoms in a Cubic Unit Cell
1
2
4
The body-centered cubic unit cell of a particular crystalline
form of iron is 0.28664 nm on each side. Calculate the density
(in g/cm3) of this form of iron.
d = 7.8753 g/cm3
The body-centered cubic unit cell of a particular crystalline
form of an element is 0.28664 nm on each side. The density
of this element is 7.8753 g/cm3. Identify the element.
The face-centered cubic unit cell of a particular crystalline
form of platinum is 393 pm on each side. Calculate the
density (in g/cm3) of this form of platinum.
d = 21.4 g/cm3
Closest Packing
a
a
b
c
a
b
c
a
b
c
b
a
b
c
c
b
c
a
b
c
a
a
b
a
c
a
c
b
a
b
a
c
a
a
b
a
c
a
a
a
b
c
a
a
· · · abab · · ·
· · · abab · · ·
1. triclinic
2. monoclinic
a≠b≠c
a≠b≠c
α≠β≠γ
The 14 Bravais lattices
3. orthorhombic
7 crystal systems
α = β = γ = 90°
5. rhombohedral (trigonal)
4. tetragonal
a=b=c
90° ≠ α = β = γ < 120°
a=b≠c
α = β = γ = 90°
7. cubic
6. hexagonal
a=b≠c
α = β = 90° ,γ = 120°
γ
a=b=c
α = β = γ = 90°
· · · abcabc · · ·
abcabc = Cubic Closest Packing
e.g. Ag, 1 atoms (1 lattice point) in a unit cell
Information of a unit cell
• Size of the cell
X-ray diffraction
• Size of the atoms
• Number of atoms in a cell
Soon
Now!
Now!
Example 11.7, page 494
Al crystallizes with a face-centered cubic unit cell.
The radius of a Al atom is 143 pm. Calculate the density
of solid Al in g/cm3.
d = 2.71 g/cm3
L
r
L
8 r
2r
L
r
What about simple cubic?
L
r
Simple Cubic
L = 2r
What about body-centered
cubic?
Body centered cubic
Body diagonal D = 4r
L
D
L
L
D
L
F
L
Body diagonal D = 4r
L 
4
3
r
Pythagorean theorem 
Titanium metal has a body-centered cubic unit cell. The
density of titanium is 4.50 g/cm3. Calculate the edge length
of the unit cell and a value for the atomic radius of titanium
in pm.
L = 328 pm
r = 142 pm
100-mL container
50 mL
70 mL
50 %
70 %
1
52 %
L = 2r
2
4
68 %
74 %
prove
L 
4
3
r
L
8 r
Packing Efficiency: fraction of volume occupied by atoms