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Logarithm Review ab = c, (a > 0, a ≠ 1) logac = b log c = log10c

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Presentation on theme: "Logarithm Review ab = c, (a > 0, a ≠ 1) logac = b log c = log10c"— Presentation transcript:

1

2 Logarithm Review ab = c, (a > 0, a ≠ 1) logac = b log c = log10c
Definition ab = c, (a > 0, a ≠ 1) logac = b If a = 10, it is called common logarithm log c = log10c If a = e = ∙ ∙ ∙, it is called natural logarithm ln c = logec Keys on your calculator

3 x > 0, y > 0 ln(xy) = ln(x) + ln(y) ln(x/y) = ln(x) − ln(y)
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

4 Intermolecular Forces
Chapter 11 Liquids, Solids and Intermolecular Forces continued

5 Chemistry, continue on Vapor Pressure

6 Surface Molecules Figure: 11-20

7 Figure: 11-24a,b Temperature: T Temperature: T

8 Figure: 11-21

9 GA, 760 torr = 1 atm Normal Boiling Point H2O 100 °C

10 Tibet, 480 torr < 1 atm Normal Boiling Point H2O 85 °C

11 Figure: 11-21

12 (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)

13 Linear relation: y = kx + C
slope: k = tg θ θ x C: intercept

14 Linear relation: y = kx + C
1/T (K−1) Linear relation: y = kx + C ln P = k(1/T) + C

15 Figure: 11-17

16 Heat of vaporization ∆Hvap: energy needed to convert one mole
of liquid to gas. Unit: J/mol or kJ/mol. ∆Hvap > 0

17 slope k < 0 y x

18

19 ln (P) 1 ln (P1) 2 ln (P2) 1/T1 1/T2 1/T (K−1)

20 Clausius-Clapeyron Equation

21 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. Five: T1, T2, P1, P2, ∆Hvap Four known, calculate the other.

22 PV = nRT Units in ideal gas law P — atm, V — L, n — mol, T — K
Option 1 Chem 1211 R = atm · L · mol−1 · K−1 P — Pa, V — m3, n — mol, T — K Option 2 Clausius-Clapeyron equation R = J · mol−1 · K−1

23 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? ( Please try to work on this question by yourself. Will review next week)

24 Liquid potassium has a vapor pressure of 10.00 torr at 443 °C
and a vapor pressure of torr at 708 °C. Use these data to calculate The heat of vaporization of liquid potassium; The normal boiling point of potassium; The vapor pressure of liquid potassium at 100. °C. ( Please try to work on this question by yourself. Will review next week)

25 Clausius-Clapeyron Equation

26 slope k < 0 y x

27 Linear relation: y = kx + C
slope: k = tg θ θ x C: intercept

28 a > b > 0 > c > d
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

29 What is the order of heat of vaporization
for these three substances?

30 Solids

31

32

33 Glass (SiO2)

34 Crystal Solid Noncrystal

35 Basis Crystal structure

36 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

37

38 Use a point to represent the basis:
=

39 Lattice Lattice point:

40 Unit cell: 2-D, at least a parallelogram
Unit cell is the building block of the crystal

41 How many kinds of 2-D unit cells
can we have?

42

43

44

45                                   

46

47

48

49 Extend the concept of unit cell to 3-D,
the real crystals.

50 : 3-D, at least a parallelepiped
Figure: 11-32 : 3-D, at least a parallelepiped

51 How many kinds of 3-D unit cells
can we have?

52 5. rhombohedral (trigonal) 4. tetragonal
1. triclinic 2. monoclinic c c a ≠ b ≠ c α ≠ β ≠ γ a a ≠ b ≠ c a b b The 14 Bravais lattices 7 crystal systems 3. orthorhombic 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 = γ = 90° a = b ≠ c α = β = 90° ,γ = 120° γ

53 Chem 1212: assume a lattice point is a single atom
Figure: 11-33 (Simple cubic) Chem 1212: assume a lattice point is a single atom

54 Information of a cubic unit cell
• Size of the cell X-ray diffraction

55 The Wave Nature of Light

56 Figure: 11-38C

57 Information of a cubic unit cell
• Size of the cell X-ray diffraction • Size of the atoms Soon • Number of atoms in a cell Now!

58 B C D A

59 B C D E F A Number of atoms in a unit cell = ¼ x 4 = 1

60

61 Number of Atoms in a Cubic Unit Cell
1 2 4

62 The body-centered cubic unit cell of a particular crystalline
form of iron is nm on each side. Calculate the density (in g/cm3) of this form of iron. d = g/cm3 The body-centered cubic unit cell of a particular crystalline form of an element is nm on each side. The density of this element is g/cm3. Identify the element.

63 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

64 Closest Packing

65

66

67 a b c

68 · · · abab · · ·

69 · · · abab · · · Chapter 11, Figure 11.48
Hexagonal Closest-Packing Crystal Structure

70 5. rhombohedral (trigonal) 4. tetragonal
1. triclinic 2. monoclinic a ≠ b ≠ c a ≠ b ≠ c α ≠ β ≠ γ The 14 Bravais lattices 7 crystal systems 3. orthorhombic 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 = γ = 90° a = b ≠ c α = β = 90° ,γ = 120° γ

71 · · · abcabc · · ·

72 abcabc = Cubic Closest Packing
e.g. Ag, 1 atoms (1 lattice point) in a unit cell

73 Information of a unit cell
• Size of the cell X-ray diffraction • Size of the atoms Soon Now! • Number of atoms in a cell Now!

74 L 2r L r Al crystallizes with a face-centered cubic unit cell.
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 2r L

75 What about simple cubic?

76 Simple Cubic r L L = 2r

77 What about body-centered
cubic?

78 Body centered cubic

79 Body diagonal D = 4r L D

80 L F L D L Body diagonal D = 4r Pythagorean theorem 

81 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

82 100-mL container 50 mL 70 mL 50 % 70 %

83 1 2 4 prove 52 % 68 % L = 2r 74 % Packing Efficiency: fraction of volume occupied by atoms

84 Chapter 11, Figure 11.44 The Cubic Crystalline Lattices


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