Classsification of Matter Phases of Matter Heat Energy Boyle's Law Charles's Law Kinetic Molecular Theory Avogadro's Hypothesis Separation of Mixtures

1. Matter and Energy

2. Matter
 Anything that has mass and occupies
space (volume)
◦ Chemistry is the study of matter.
 Matter is composed of atoms.

3. Classifying Matter

4. Classifying Matter
 Matter
◦ Substances – have constant composition and
constant properties throughout a given
sample. Cannot be separated by physical
means.
 Element – substances that are composed of atoms
that have the same atomic number. Cannot be
broken down by chemical means.
 Examples; Hydrogen (H), Helium (He), Oxygen
(O), Zinc(Zn), etc.
 Compound – a substance composed of two or more
elements that are chemically combined I definite
proportions by mass.
 Examples; H2O, CO2, CH4

5. Classifying Matter
 Matter
◦ Mixtures – a combination of two or more
substances that can be separated by
physical means.
 Homogeneous (a.k.a. Solutions)– a mixture
that is uniform in its properties throughout a
given sample.
 Examples; Coca-Cola, milk, air
 Heterogeneous – consists of physically distinct
parts each with different properties.
 Examples; Soil, salad, cement

6. Properties of Matter
 Physical Change - a change in matter
from one for or another
(rearrangement of particles) without a
change in chemical properties.
◦ Example; Salt-water can easily be
physicall separated back into its
components of solid NaCl crystals and
H2O by a process called distillation.

7. Properties of Matter
 Chemical Change – a change in which
results in the formation of different
kinds of matter with changed
properties.
◦ Example; Rustin of iron – Fe combines
with O2 in the air to from a new material
called rust. The Fe and O2 atoms cannot
be separated by physical means.

8. Phases of Matter
 Solid Phase
 Liquid Phase
 Gaseous Phase
 Plasma Phase

9. Solid Phase
 Rigid form.
 Definite shape.
 Definite volume.
 Strong attractive forces hold atoms
and molecules in fixed locations.
 True solids have a crystalline
structure.

10. Liquid Phase
 Particles not held together as rigidly
and are able to move past one
another.
 Sufficient attractive forces to have
definite volume.
 No definite shape, takes the shape of
the container.

11. Gaseous Phase
 Minimal attractive forces to hold particles
together, no definite volume.
 No definite shape.
 Spread out indefinitely unless confined to
a container where the gas will occupy
the volume of the container.
 Vapor is the gaseous
phase of a substance
that is a liquid or solid
at normal conditions.

12. Particle Diagrams
 Draw the particle diagrams for water
(H2O) in the three phase of matter.
Solid Liquid Gaseous Water
Water Water

13. Diatomic molecules
 HOFBrINCl’s
◦ H = Hydrogen (H2)
◦ O = Oxygen (O2)
◦ F = Flourine (F2)
◦ Br = Bromine (Br2)
◦ I = Iodine (I2)
◦ N = Nitrogen (N2)
◦ Cl = Chlorine (Cl2)

14. Draw the particle diagrams for
the following…
 Solid Sodium (Na)  Liquid Mercury
(Hg)
 Helium Gas (He)
 Nitrogen Gas (N)

15. Draw the particle diagrams for
the following…
 Solid Sodium (Na)  Liquid Mercury
(Hg)
 Helium Gas (He)
 Nitrogen Gas (N)

16. Heating and Cooling Curves
 Describe the changes in energy as
substances are heated and cooled.
◦ Potential Energy – energy of a material as
a result of its position in an
electrical, magnetic, or gravitational field.
◦ Kinetic Energy – energy available
because of the motion of an object.

17. Heating Curve

18. Heating Curve
Boiling
Point Boiling
(Vaporization)
P.E. Increases
K.E. Constant
Meltin
g Melting
Point (Fusion)
P.E. Increases
K.E. Constant

19. Heating Curve Summary
 AB: heating of a solid, one phase
present, kinetic energy increases.
 BC: melting of a solid, two phases
present, potential energy
increases, kinetic energy constant.
 CD: heating of a liquid, one phase
present, kinetic energy increases.
 DE: boiling of a liquid, two phases
present, potential energy
increases, kinetic energy remains
constant.
 EF: heating of a gas, one phase
present, kinetic energy increases.

20. Cooling Curve

21. Cooling Curve
Condensation
P.E. Decreases
Condensin K.E. Constant
g
Point
Freezing
P.E. Decreases
Freezin K.E. Constant
g
Point

22. Cooling Curve Summary
 AB: cooling of a gas (vapor), one phase
present, kinetic energy decreases.
 BC: condensation of a gas (vapor) to
liquid, two phases present, potential
energy decreases, kinetic energy
constant.
 CD: cooling of a liquid, one phase
present, kinetic energy decreases.
 DE: solidification (freezing) of a
liquid, two phases present, potential
energy decreases, kinetic energy
remains constant.
 EF: cooling of a solid, one phase
present, kinetic energy decreases.

23. Heating Curve of H2O

24. Cooling Curve of H2O

25. Temperature
 Temperature – a measure of the
average kinetic energy of the particles
within a substance.
◦ The average kinetic energy depends only
upon the temperature of a substance, not
on the nature or amount of the material.
 Ex; A 10g sample of H2O at 50oC has a greater
average kinetic energy than a 500g sample of
Fe(s) at 20oC.

26. Kelvin Scale
 The SI unit of temperature.
 A measure of absolute temperature.
◦ 0 K = Absolute Zero
K = oC +
273
 A change of one degree Celsius is
equal to a change of one Kelvin.
Celsius to Fahrenheit to
Fahrenheit Celsius

27. Convert the following into Kelvin
 0oC
 25oC
 37oC
 100oC
 200oC

28. Temperature vs. Heat
 Heat – a measure of the amount of
energy transferred from one
substance to another.
◦ Measures in units of calories or joules.
◦ Temperature difference between two
things indicates direction of heat flow.
 Heat flows from an object at higher
temperature to the object at a lower
temperature until both object are at the same
temperature.

29. Temperature vs. Heat
Temperature measures average kinetic energy of particles within object.
Heat measures the amount of energy transferred from one object to
another.

30. Measurement of Heat Energy
Can calculate the amount of heat given off or
absorbed during a reaction using...
q=
mC∆T
q = heat (in joules)
m = mass of the substance
C = specific heat capacity of the
substance
∆T = (Temperaturefinal –
Temperatureinitial)

31. Sample Problem
How many joules are absorbed when 50.0g
of water are heated from 30.2oC to 58.6oC?

32. Sample Problem
How many joules are absorbed when 50.0g
of water are heated from 30.2oC to 58.6oC?
q = mC∆T
m = 50.0g
C = 4.18 J/g•K
∆T = 58.6oC - 30.2oC = 28.4oC
~or~
∆T = 331.6 K – 303.2 = 28.4

33. Sample Problem
How many joules are absorbed when 50.0g
of water are heated from 30.2oC to 58.6oC?
q=
(50.0g)(4.18J/g•oC)(28.4oC)
q = 5936J =
5.94x103J

34. Practice
 When 25.0g of water are cooled from
20oC to 10oC the number of joules of
heat released is?

35. Practice
 How many kilojoules of heat energy
are absorbed when 100.0g of water
are heated from 20oC to 30oC ?

36. Complex Problem
 A 2.70g piece of metal is heated to 98.7oC. It is
then added to a beaker containing 150mL of water
at 23.5oC. The final temperature of the water is
25.2oC. What is the specific heat of this metal?

37. Heat of Fusion ∆Hf
 The amount of heat needed to convert
a mass of a substance from solid to
liquid at its melting point.
◦ The heat of fusion of solid water (ice) at
0oC and 1atmosphere of pressure is 334
J/g
q = mHf

38. Practice Problem
 How many joules are required to melt
255g of ice at 0oC?

39. Practice Problem
 How many joules are required to melt
255g of ice at 0oC?
q = mHf
m = 225 g
Hf = 334 J/g
q = (225g)(334 J/g) = 85,170 J = 85.2
kJ

40. Heat of Vaporization ∆Hv
 Boiling, as substance is converted
from liquid to solid at its boiling point.
◦ The heat of vaporization of water at 100oC
and 1atmosphere of pressure is 2260 J/g
q = mHv

41. Practice Problem
 How many joules of energy are
required to vaporize 423g of water at
100oC and 1atmosphere of pressure.

42. Practice Problem
 How many joules of energy are
required to vaporize 423g of water at
100oC and 1atmosphere of pressure.
q = mHv
m = 423 g
Hv = 2260 J/g
q = (423g)(2260J/g) = 955,980 J = 956

43. Table T – Heat Formulas

44. Behavior of Gases
 Kinetic Molecular Theory (KMT)
◦ Ideal Gases
 Have mass but negligible volume.
 In constant, random, straight-line motion.
 Not subject to attractive or repulsive forces.
 Gas Particles collide with each other and the
walls of their container where a transfer of
energy between particles may occur but there
is no net loss of energy. Collisions are said to
be perfectly elastic.

45. KMT and Pressure
 Gases exert pressure by colliding with
the walls of a container.
◦ The volume occupied by an ideal gas is
essentially the volume of its container.
◦ The greater the number of gas particles
the higher the pressure.
 Example; adding to much air to a balloon
◦ Pressure and number of gas particles are
directly proportional.

46. KMT and Pressure
 Relationship of Pressure and Volume
◦ Boyle’s Law
 Pressure and volume vary inversely.
 As volume decreases pressure increases.
 Temperature and number of particles held constant.
↑ Volume ↓Volume
↓ Pressure ↑ Pressure

47. Boyle’s Law
50 kPa • 6L = 300
kPa•L
100 kPa • 3L = 300 kPa•L
150 kPa • 2L = 300 kPa•L
Pressure • Volume = Constant

48. Boyle’s Law
 Can re-write Boyle’s law to solve
mathematical problems.
P1• V1 = P2• V2
 Problem
◦ According to the graph what will be the
volume of the gas when the pressure is
520 kPa?
P1 = 50 kPa P1• V1 = P2• V2
V1 = 6 L
P2 = 520 kPa 50 kPa • 6 L = 520 kPa • V2
V2 = ? L
V2 = 0.58 L

49. Boyle’s Law Problem
 The volume occupied by a gas at STP is 250 L. At
what pressure (in atmospheres) will the gas occupy
1500 liters, if the temperature and number of
particles remain constant?
P1• V1 = P2• V2

50. KMT and Temperature
 Relationship of Temperature and
Pressure of a gas.
◦ Directly related
 Increase Temp. = Increase Pressure
 Relationship of Temperature and
Volume of a gas.
◦ Charles’s Law
 Temperature and constant directly related at
constant pressure and volume.

51. Charles’s Law

52. Charles’s Law
Pressure and number of particles held constant.
Graph is linear.
Volume and Temperature (K) are directly proportional to each other.

53. KMT and Particle Velocity
 The greater the temperature the
greater the velocity of its particles.
 Increased Kinetic Energy.
 Increased Temperature.

54. Combined Gas Law
 The relationship among
pressure, temperature, and volume can be
mathematically expressed by a single
equation…
◦ The Combined Gas Law

55. Combined Gas Law
Example Problem
 What volume will a gas occupy if the
pressure on 244 cm3 gas at 4.0 atm is
increased to 6.0 atm? Assume temperature
remains constant.

56. Combined Gas Law
Example Problem
 What volume will a gas occupy if the
pressure on 244 cm3 gas at 4.0 atm is
increased to 6.0 atm? Assume temperature
remains constant.
3
P1 = 244 cm P2 = ?
T1 = Constant T2 = Constant
V1 = 4.0 atm V2 = 6.0 atm

57. Combined Gas Law
Example Problem
 What volume will a gas occupy if the
pressure on 244 cm3 gas at 4.0 atm is
increased to 6.0 atm? Assume temperature
remains constant.
3
P1 = 244 cm P2 = ?
T1 = Constant T2 = Constant
V1 = 4.0 atm V2 = 6.0 atm

58. Combined Gas Law
Example Problem
 If 75 cm3 of a gas is at STP, what volume
will the gas occupy if the temperature is
raised to 75oC and the pressure is
increased to 945 torr?
101.3 kPa = 1 atm = 760
torr

59. Ideal vs. Real Gases
 Kinetic Molecular Theory explains the
behavior of gases using “ideal” gases as a
model. Real gases however, do not always
follow this.
◦ Ideal Gases (always behave as predicted; H, He)
 Negligible volume.
 No attractive or repulsive forces.
 Move randomly in straight lines.
 Perfectly elastic collisions.
◦ Real Gases
 Attractive forces cannot always be disregarded.
 Water vapor molecules attract one another to form rain or snow in
the atmosphere.
 Volume of gas particles not negligible under high
pressure.

60. Avogadro’s Hypothesis
 Equal volume of gases.
◦ When volume, temperature, and pressure
of two or more gases are the same, they
contain the SAME number of molecules.
 Ex; 12 L of N2(g) at STP would contain the
same number of molecules as 12 L of O2(g) at
STP

61. Avogadro’s Hypothesis
 22.4 Liters of ANY gas at Standard
Temperature and Pressure (STP) contains
one mole (6.02 x 1023 atoms or molecules)
of that gas.
◦ 22.4 liters of CO2(g) at 0oC and 1atm will
have 6.02 x 1023 CO2 molecules.
◦ 22.4 liters of Ne(g) at 0oC and 1atm will
have 6.02 x 1023 Ne molecules.

62. Separation of Mixtures
 Filtration
 Distillation
 Chromatography

63. Filtration
Allowing some particles to
pass while trapping
others
 Solids suspended in a
liquid.
 Coffee Filter
 Immiscible liquids
 Separatory Funnel
 Mixtures of solids and
gases
 A/C Filters

64. Distilation
Separation by boiling
point.
 Homogeneous
Mixtures
◦ Solid dissolved in
liquid
 Purifying salt water
 Miscible Liquids
◦ Liquids that mix with
each other.
 Gasoline from crude oil.

65. Chromatography
Separation by using the
differing attractive forces of
the components within a
mixture to a transport
medium.
 Paper Chromatography
 Gas Chromatography
 Column
Chomatography

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