2. PHYSICAL BEHAVIOR OF MATTER UNIT 7

3. SUMMARY OF UNIT 1. PROPERTIES OF SOLIDS, LIQUIDS, AND GASES 2. CALCULATION OF HEAT EXCHANGE DURING PHASE CHANGES 3. KINETIC MOLECULAR THEORY (KMT)

4. PHASES OF MATTER

5. SOLID PHASE Matter in rigid form Definite volume and shape Strong attractive forces hold particles in place –True solids have a crystalline structure

6. CRYSTALLINE STRUCTURE

7. LIQUID PHASE Particles able to move past each other. Mobility of particles prevents definite shape. Definite volume due to attractive forces.

8. GASEOUS PHASE Lack attractive forces to hold them in place No definite shape or volume Spread out indefinitely unless confined in a container.

9. PLASMA STATE IONIZED GAS: –They’re free electrons flying around

10. ENERGY OF MATTER 1. ENERGY A.Capacity to do work 2. 2 TYPES OF ENERGY A.Potential B.Kinetic

11. POTENTIAL ENERGY Stored energy

12. KINETIC ENERGY Energy in motion

13. TYPES OF ENERGY THERMAL – internal energy due to motion of molecules CHEMICAL- due to making/breaking of bonds LIGHT –associated with electromagnetic radiation NUCLEAR – due to change in mass of nuclei HEAT – transfer of energy

14. CONSERVATION OF ENERGY Energy is neither created or destroyed! Energy may be converted from one form of energy to another or transferred from one body to another. Electrical Energy In Light & Heat Energy Out

15. EXOTHERMIC – activity that releases energy, aka, BOND MAKING ENDOTHERMIC- activity that requires energy (absorbs energy), aka, BOND BREAKING

16. EXOTHERMICENDOTHERMIC Precipitation – rain Melting ice cubes Candle Flame Heating water Rusting Iron Baking bread Burning sugar Cooking an egg Formation of snow in clouds Making sugar by photosynthesis

17. TEMPERATURE Measure of average kinetic energy of a substance’s particles. Measured with a THERMOMETER In Chemistry, units are KELVIN (K) or ( O C)

18. TEMPERATURE SCALES

19. HEAT The measure of the amount of energy transferred from one substance to another. UNITS of Heat: Calories (cal) or Joules (J)

20. HEAT vs. TEMPERATURE WHICH REQUIRES MORE ENERGY TO MELT? 1 g 10 g IS THERE A DIFFERENCE IN THE TEMPERATURE?

21. MEASUREMENT OF HEAT ENERGY The amount of heat given off in a reaction is calculated using the equation: q = mC∆T q = heat (in joules) m = mass of substance C = specific heat capacity of substance T = change in temperature

22. CALORIMETER Instrument used to measure the amount of energy given off in a reaction.

23. HEAT ENERGY PROBLEMS EXAMPLE 1: How many joules are absorbed when 50.0g of water is heated from 30.2 o C to 58.6 o C?

24. Measuring Heat Energy KNOWNUNKNOWN m = 50.0g C water = 4.18J/g● o C ∆T = 58.6 o C – 30.2 o C = 28.4 o C = 28.4 o C q = ?J STEP 1: Create a T-Chart of known and unknown information

25. Measuring Heat Energy STEP 2: Plug KNOWN information into heat energy equation: q = mC ∆T q = (50.0g)(4.18J)(28.4 o C) g ● o C g ● o C

26. Measuring Heat Energy STEP 3: Solve the math problem q = (50.0g)(4.18J)(28.4 o C) g ● o C g ● o C q = 5936J = 5.94 x 10 3 J

27. HEAT ENERGY PROBLEMS HEAT OF FUSION PROBLEM HEAT OF FUSION = MELTING HEAT OF FUSION of solid water (ice) at 0 o C and 1 atmosphere (pressure) is 334 J/g 334 J/g HEAT OF FUSION is a phase change. There is NO CHANGE in temperature.

28. HEAT ENERGY PROBLEMS EXAMPLE 2: How many joules are required to melt 255 g of ice at 0 o C? EQUATION: q = mH f (heat of fusion)

29. Measuring Heat Energy STEP 1: Create a T-Chart of known and unknown information KNOWNUNKNOWN m = 255g Heat of fusion = 334 J/g q = ?J

30. Measuring Heat Energy STEP 2: Plug KNOWN information into heat energy equation: q = mH f q = (255g)(334J) g

31. Measuring Heat Energy STEP 3: Solve the math problem. q = (255g)(334J) g q = 85,170J = 85.2kJ

32. HEAT ENERGY PROBLEMS HEAT OF VAPORIZATION PROBLEM HEAT OF VAPORIZATION = Liquid to gas phase HEAT OF VAPORIZATION of water at 100 o C and 1 atmosphere (pressure) is 2260 J/g 2260 J/g HEAT OF VAPORIZATION is a phase change. There is NO CHANGE in temperature.

33. HEAT ENERGY PROBLEMS EXAMPLE 3: How many joules of energy are required to vaporize 423g of water at 100 o C and 1 atm? EQUATION: q = mH v (heat of vaporization)

34. Measuring Heat Energy STEP 1: Create a T-Chart of known and unknown information KNOWNUNKNOWN m = 423 g Heat of vaporization = 2260 J/g q = ?J

35. Measuring Heat Energy STEP 2: Plug KNOWN information into heat energy equation: q = mH v q = (423g)(2260J) g

36. Measuring Heat Energy STEP 3: Solve the math problem. q = (423g)(2260J) g q = 955,980J = 956kJ

37. HOW CAN I REMEMBER ALL THIS INFORMATION????? YOU DON’T HAVE TO!!!!!!!! TAKE OUT YOUR REFERENCE TABLE Equations are on page 12, TABLE T Physical Constants are on page 1, TABLE B

40. KINETIC ENERGY Energy in motion

41. HEATING CURVE

42. AB: Rise in temperature as ice absorbs heat. One phase (solid) Heating Curve of Water 0 20 40 60 80 100 120 140 160 012345678910111213141516 Time (minutes) Temperature ( C) Temperature A BC DE F

43. HEATING CURVE BC: Ice melts – Latent Heat of Fusion Heating Curve of Water 0 20 40 60 80 100 120 140 160 012345678910111213141516 Time (minutes) Temperature ( C) Temperature A BC DE F Constant temp

44. HEATING CURVE CD: Liquid is heating One phase (liquid) Heating Curve of Water 0 20 40 60 80 100 120 140 160 012345678910111213141516 Time (minutes) Temperature ( C) Temperature A BC DE F

45. HEATING CURVE DE: Liquid boils – Latent Heat of Vaporization Heating Curve of Water 0 20 40 60 80 100 120 140 160 012345678910111213141516 Time (minutes) Temperature ( C) Temperature A BC DE F Constant temp

46. HEATING CURVE EF: Gas is heating One phase (gas) Heating Curve of Water 0 20 40 60 80 100 120 140 160 012345678910111213141516 Time (minutes) Temperature ( C) Temperature A BC DE F

47. COOLING CURVE FE: Gas is cooling

48. COOLING CURVE DC: Liquid is cooling

49. COOLING CURVE BA: Solid is cooling

50. COOLING CURVE ED: Condensation; Gas to Liquid

51. COOLING CURVE CB: Freezing; Liquid to Solid

52. OTHER PHASES SUBLIMATION: Solid to gas phase DEPOSITION: Gas to solid

53. BEHAVIOR OF GASES KINETIC MOLECULAR THEORY –Describes the unpredictable behavior of gas particles. STUFF TO CONSIDER TEMPERATURE PRESSURE VOLUME VELOCITY FREQUENCY FORCE OF COLLISION

54. KMT Gases are in constant, random, straight- line motion. Gases collide with each other and the sides of the container. Gases are far away from each other. Gases are not attracted to each other.

55. KMT PRESSURE vs. # OF PARTICLES WHAT HAPPENS IF I ADD PRESSURE TO THE BALLOON? P↑P↑P↑P↑ Collisions ↑

56. KMT PRESSURE vs. VOLUME WHAT HAPPENS IF I ADD VOLUME? P ↓ Collisions ↓

57. KMT PRESSURE vs. TEMPERATURE WHAT HAPPENS IF I DECREASE TEMPERATURE? P ↑

58. COMBINED GAS LAW SHOWS HOW TEMP, VOLUME, AND PRESSURE ARE RELATED. P 1 V 1 T 1 T 1= P 2 V 2 T 2 T 2

59. REVIEW OF UNITS TEMPERATUREPRESSUREVOLUME CELSIUS ( O C) KELVIN (K) ATMOSPHERE (atm) TORR (torr) KILOPASCAL (kPa) LITER (L) (cm 3 ) (cm 3 )

60. STANDARD TEMPERATURE AND PRESSURE (STP) TEMPERATUREPRESSURE 0 o C 273K 1 atm 760 torr 101.3 kPa

61. COMBINED GAS LAW PROBLEM EXAMPLE 1: What volume will a gas occupy if the pressure on 244 cm 3 gas at 4.0 atm is increased to 6.0 atm? Assume the temp remains constant. Assume the temp remains constant.

62. COMBINED GAS LAW PROBLEM NOTE: For problems with two properties involved where the third remains constant, CANCEL OUT the variable representing the constant property P 1 V 1 T 1 T 1 P 2 V 2 T 2 T 2=

63. Measuring Heat Energy STEP 1: Create a T-Chart of known and unknown information KNOWNUNKNOWN V 1 = 244 cm 3 P 1 = 4.0 atm P 2 = 6.0 atm T = constant V 2 = ? cm 3

64. Measuring Heat Energy STEP 2: Rearrange the equation to solve for the unknown = V2V2V2V2 P1V1P1V1P1V1P1V1 P2P2P2P2

65. Measuring Heat Energy STEP 3: Plug in the values and solve the math problem = V2V2V2V2 P1V1P1V1P1V1P1V1 P2P2P2P2 (4.0 atm) (244 cm 3 ) 6.0 atm = 160 cm 3

66. P 2 V 2 T 2 T 2 Measuring Heat Energy STEP 2: Rearrange equation and solve for V 2 P 1 V 1 T 1 T 1= V2V2V2V2 = P 1 V 1 T 2 P 2 T 1 P 2 T 1 P2P2P2P2 P2P2P2P2 T2T2T2T2 T2T2T2T2

67. (348 K) (273 K) Measuring Heat Energy STEP 3: Plug in the values and solve the math problem (760 torr) (75 cm 3 ) = 77 cm 3 (948 torr)

68. COMBINED GAS LAW PROBLEM EXAMPLE 1: If 75 cm 3 of a gas is at STP, what volume will the gas occupy if the temperature is raised to 75 0 C and the pressure is increased to 945 torr?

69. Measuring Heat Energy STEP 1: Create a T-Chart of known and unknown information KNOWNUNKNOWN V 1 = 75 cm 3 P 1 = 760 torr P 2 = 945 torr T 1 = 0 0 C T 2 = 75 0 C V 2 = ? cm 3

70. Measuring Heat Energy STEP 2: Rearrange equation and solve for V 2 = P1V1P1V1P1V1P1V1 T1T1T1T1 P2V2P2V2P2V2P2V2 T2T2T2T2

72. IDEAL VS. REAL GASES KMT (Kinetic Molecular Theory) –Uses the IDEAL GAS LAWS to explain behavior of gases. IDEAL GASES BEHAVE EXACTLY AS PREDICTED!! GASES ARE MOST IDEAL AT: –LOW PRESSURE AND HIGH TEMPERATURES.

73. IDEAL VS. REAL GASES IDEAL GASES FOLLOW BOYLES LAW AND CHARLES LAW BOYLES LAW: P 1 V 1 = P 2 V 2 –As P↑, V↓ http://www.grc.nasa.gov/WWW/K- 12/airplane/boyle.html http://www.grc.nasa.gov/WWW/K- 12/airplane/boyle.html

74. IDEAL VS. REAL GASES CHARLES LAW: V 1 /T 1 = V 2 /T 2 –As V↑, T↑ http://www.grc.nasa.gov/WWW/K- 12/airplane/glussac.html http://www.grc.nasa.gov/WWW/K- 12/airplane/glussac.html

75. SEPARATION OF MIXTURES

76. FILTRATION

77. DISTILLATION

78. CHROMATOGRAPHY

79. HEAT OF FUSION

80. HEAT OF VAPORIZATION

81. BEHAVIOR OF GASES

82. PRESSURE VS. # OF GAS PARTICLES

83. TEMPERATURE VS. GAS PRESSURE

84. TEMPERATURE AND GAS VOLUME