1. Physics 7A – Review Winter 2008 Prof. Robin D. Erbacher 343 Phy/Geo Bldg erbacher@physics.ucdavis.edu Prof. Robin D. Erbacher 343 Phy/Geo Bldg erbacher@physics.ucdavis.edu

2. AnnouncementsAnnouncements 7A-A/B Final Exam Locations: - Last name A-G: Roessler 66 - Last name H-Z: 123 Science Lecture Hall 7A-C/D Final Exam Locations: - Last name A-P: 194 Chemistry - Last name Q-Z: 179 Chemistry Final Exam reviews continue today, see website. No make-up finals! Final Exam Tues. 10:30-12:30. Turn off cell phones and pagers during lecture.

3. Location of 7A-A/B 7A-A/B Final Exam Locations: - Last name A-G: Roessler 66 - Last name H-Z: 123 Science Lecture Hall

4. Location of 7A-A/B 7A-C/D Final Exam Locations: - Last name A-P: 194 Chemistry - Last name Q-Z: 179 Chemistry

5. Review Multiple- Atom Systems: Particle Model of E bond, Particle Model of E thermal

6. Multiple Atom Systems: E bond Typically every pair of atoms interacts Magnitude of E bond for a substance is the amount of energy required to break apart “all” the bonds i.e. we define E bond = 0 when all the atoms are separated We treat bonds as “broken” or “formed”. Bond energy  (per bond ) exists as long as the bond exists. The bond energy of a large substance comes from adding all the potential energies of particles at their equilibrium positions. E bond = ∑ all pairs (PE pair-wise )

7. What is the change in bond energy (∆E bond ) by removing the red atom? 2.2 A 4.4 A 6.6 A -8 x 10 -21 J -0.5 x 10 -21 J ~ 0 J 8.8 A 11 A ~ 0 J Bond energy Separation (10 -10 m) Atom-atom potential for each atom Answer: ~8.5x10 -21 J

8. What is the bond energy E bond for the entire molecule? -8 x 10 -21 J Separation (10 -10 m) Atom-atom potential for each atom -8.5 x 10 -21 J E bond = -42 x 10 -21 Joules

9. What is the bond energy E bond for the entire molecule? Separation (10 -10 m) Atom-atom potential for each atom E bond ≈ -40 x 10 -21 Joules Energy required to break a single pair of atoms apart: +8x10 -21 J =5 bonds. Approximation! (doesn’t include NN neighbors)

10. Recap: Particle Model of E bond E bond for a substance: amount of energy required to break apart “all” the bonds (magnitude only) i.e. we define E bond = 0 when all the atoms are separated The bond energy of a substance comes from adding all the potential energies of particles at their equilibrium positions. E bond = ∑ all pairs (PE pair-wise ) A useful approximation of the above relation is: E bond ~ - (total number of nearest neighbor pairs) x (  )  E bond of the system is negative, determined by: 1) the depth of the pair-wise potential well   (positive) 2) the number of nearest-neighbors.

11. Reminder: E total, E bond, E thermal A pair of atoms are interacting via a atom-atom potential. Only these two atoms are around. In two different situations, the pair have a different amount of total energy. In which situation is E thermal greater? a) Situation A has a greater E thermal b) Situation B has a greater E thermal c) Both have the same E thermal d) Impossible to tell A B

12. A B KE Reminder: E total, E bond, E thermal A pair of atoms are interacting via a atom-atom potential. Only these two atoms are around. In two different situations, the pair have a different amount of total energy. In which situation is E thermal greater? a) Situation A has a greater E thermal b) Situation B has a greater E thermal c) Both have the same E thermal d) Impossible to tell KE We increased E thermal by putting more energy into the system

13. Initial Final Reminder: Individual Atoms Okay. Now let us look at the same problem from the perspective of the KE and PE of individual atoms. Which situation is correct in going from initial to final states?

14. Initial Final Reminder: Individual Atoms Okay. Now let us look at the same problem from the perspective of the KE and PE of individual atoms. Which situation is correct in going from initial to final states?

15. Particle Model of E thermal E thermal is the energy associated with the random microscopic motions and vibrations of the particles. We increased E thermal by putting more energy into the system. KE and PE keep changing into one another as the atoms vibrate, just like in the mass-spring system, so we cannot make meaningful statements about instantaneous KE and PE. We can make statements about average KE and PE. Increasing E thermal increases both KE average and PE average.

16. Particle Model of E thermal The energy associated with the random motion of particles is split between PE oscillation and KE.

17. Mass on a Spring As we increase E tot we increase PE avg and KE avg PE avg = KE avg = E tot /2 (for mass on a spring) Energ y position E tot PE KE

18. Particle Model of E bond and E thermal The energy associated with the random motion of particles is split between PE oscillation, KE. For particles in liquids and solids, let’s not forget the part that corresponds to E bond of the system. E bond of the system is determined by both the depth of the pair-wise potential well and the number of nearest neighbors (# NN).

19. Particle Model of E bond and E thermal E bond of the system is determined by both the depth of the pair-wise potential well and the # NN. Solids/Liquids: KE all atoms = (1/2)E thermal PE all atoms = PE bond + PE oscillation = E bond (PE bond )+ (1/2)E thermal (PE oscillation )  KE all atoms + PE all atoms = E thermal + E bond Gases (monoatomic): The gas phase has no springs: no PE oscillation or PE bond

20. Initial Final Individual Atoms Now: Using our knowledge of the KE and PE of individual atoms. In the final state, what is the average KE for these two atoms? What is Pe avg ? PE avg =KE avg =E thermal /2 E th = E tot -E bond = 5x10 -21 J KE avg = 2.5x10 -21 J = PE avg

21. Summary of E bond and E thermal Potential Energy contributes to both E bond and E thermal. Kinetic Energy contributes to E thermal, but not E bond. Number of bonds is important for E bond, not really for E thermal.

22. Equipartition of Energy

23. Energy associated with the component of motions/vibrations (“modes”) in any particular direction is (1/2)k B T : E thermal per mode = (1/2) k B T a.k.a. Equipartition of Energy Gas Liquids and Solids Equipartition of Energy

24. What are Modes? Modes : Ways each particle has of storing energy. Ex. Mass-spring has one KE mode and one PE mode.

25. Equipartition of Energy: E thermal per Mode = ½ k B T KE mode PE mode Total Solids336 Liquids336 Monatomic gases303 Diatomic gases3+2+117 When energy is added to a system, what does it mean to have more places (modes) to store it in?

26. Answer: Energy for Temperature KE mode PE mode Total Solids336 Liquids336 Monatomic gases 303 Diatomic gases 3+2+117 Does it take more or less energy to raise the temperature of a diatomic gas than a monatomic gas? 1.More C = ∆E thermal / ∆T ∆ E thermal per molecule = number of active modes  (1/2)k B ∆T ∆ E thermal per N atoms = number of active modes  (1/2)k B ∆T  N

27. Molar Specific Heat of Solids All measurements at 25 C unless listed otherwise (-100 C) (0C) k B N A = R = 8.31 J/mole  K : Gas constant or Ideal gas constant k B (Boltzmann constant) = 1.38 x 10 -23 Joule/Kelvin N A (Avogadro’s number) = 6.02 x 10 23 6 modes

28. Molar Specific Heat for Gases All measurements at 25 C unless listed otherwise diatomic (no vibrations) (100 C) (500 C) monatomic Oops! What’s going on??! 3 (5) modes => 3 (5) x ½ k B N A = 12.42 (20.70) J/mol K

29. Discrepancy Explained… Closed box of gas Open box of gas Closed box: all heat goes into the gas’s energy Open box: Some heat goes into pushing air out of the way C V measurement C P measurement

30. Discrepancy Explained… Closed box of gas Open box of gas Closed box: all heat goes into the gas’s energy Open box: Some heat goes into pushing air out of the way C V measurement C P measurement Question So then does it take More energy to raise the Temperature of Closed box of gas Or Open box of gas?

31. Molar Specific Heat for Gases All measurements at 25 C unless listed otherwise diatomic (no vibrations) (100 C) (500 C) monatomic Whew! Now it Makes more sense. Recall, C p = C v + PdV/T, or C p = C v + R!

32. “Constant volume”“Constant pressure” “Process” seems to matter… => Chapter 4 Models of Thermodynamics (definition of heat capacity) In this example of thermal phenomena (i.e.,measuring heat capacity),

33. Summary of Equipartition of Energy Equipartition tells us that the energy per mode Is ½ k b T. For solids and liquids, moving the atom in any direction is like moving a mass on a spring. Three directions means 3 KE modes/atom and 3 PE modes/atom. Nothing to do with # of bonds! For gases things are more complicated: need to count carefully- some KE modes don’t have PE modes.

34. Thermodynamics: Microscopic to Macroscopic

35. State Function Depends only on properties of the system at a particular time Example: E thermal of a gas For an ideal gas, E thermal depends only on: Temperature Number of modes

36. Work and Heat Not a property of a particular object. Instead a property of a particular process, or “way of getting from the initial state to the final state” LHS: depends only on i and f Q,W depend on process between i and f

37. Conservation of Energy (Review) ∆E total must include all changes of energy associated with the system… ∆E total = ∆E thermal + ∆E bond + ∆E atomic + ∆E nuclear + ∆E mechanical Energy associated with the motion of a body as a whole ∆ U First law of Thermodynamics ∆U : Internal energy Energy associated with the atoms/molecules inside the body of material For an ideal gas,  U = E thermal = # modes*1/2 k B T

38. State v. Process-dependent Functions Depend only on what the object is doing at the time. Change in state function depends only on start and end points. Examples: T, P, V, modes, bonds, mass, position, KE, PE... State functions Depend on the process. Not a property of an object Examples: Q, W, learning,...... Process-dependent

39. WorkWork initial final P VV P Along any given segment: W = -P  V (if P constant) OR = - Area under curve (sign positive when V decreases, sign negative when V increases)

40. Reminder: Calculation of Work initial final P V Is the work done in the process to the right positive or negative? A) Negative B) Positive C) Zero D) Impossible to tell.

41. WorkWork initial fina l P V First section: W 1 < 0 (volume expands) Second section: W 2 = 0 (volume constant) Third section: W 3 > 0 (volume contracts)

42. Reminder: Process Comparisons initia l final P V Here are two separate processes acting on two different ideal gases. Which one has a greater magnitude of work? The initial and final points are the same. initia l final P V A)Magnitude of work in top process greater B) Magnitude of work in lower process greater C) Both the same D) Need more info about the gases. Work = 0 (  V=0)

43. Heat and the First Law of Thermo initial final V P We can read work directly off this graph (i.e. don’t need to know anything about modes, U, T, etc.) If we know something about the gas, we can figure out U i, U f and U f - U i For an ideal gas,  U = E thermal = # modes*1/2 k B T

44. Heat and the First Law of Thermo initial final V P

45. ExampleExample 5 moles of an ideal monatomic gas has its pressure increased from 10 5 Pa to 1.5x10 5 Pa. This process occurs at a constant volume of 0.1 m 3. Determine: * work * change in internal energy  U * heat involved in this process. initial final P V HW: There are two ways to solve this using the ideal gas law…

46. CommentsComments Heat depends on the process Work depends on the process  U only depends on initial and final W = 0 for all constant volume processes  U = 0 for all cycles, where the initial and final states are the same.

47. Microstates versus States, and Entropy

48. Microstates and States ConstraintsStatesMicrostates Things we worry about: Constraints: States: Tell us which microstates are allowed. Examples The volume of a box constraints the possible positions of gas atoms. The energy of the box constraints the possible speeds of gas atoms. Groups of microstates that share some average properties, i.e. A collection of states that “look” the same macroscopically. Examples gases: P ~ average density, V~volume filled, T~average KE

49. Microstates v. States Example: Flipping a coin 3 times: Microstates: all possible combinations of coin flips Constraints: some combinations not possible (e.g. HHTHHH) MicrostatesStates States: total number of heads Every microstate is equally likely. is the one with the most microstates Every microstate is equally likely. The state that is most likely is the one with the most microstates Prob.

50. Ten Fair Coin Flips (Example) 1024 microstates, each microstate equal width Define states by “total number of heads” Different states contain different # of microstates Therefore even though each microstate is equally likely, some states are more likely than others.

51. Ten Fair Coin Flips: Divided We can also consider our physical system to be two “sub- systems”: * Sub-system A: the first two coin flips * Sub-system B: the final eight coin flips HH HT TH TT Any of the 256 micro.

52. EntropyEntropy where k B is Boltzman’s constant If our system is composed of two sub-systems A and B: We can add the entropy of the subsystems to get the total entropy.

53. Calculating Entropy We should divide our box up into “atom-sized” chuncks. But how big should our velocity microstates be? ice (0 0 C) Water (0 0 C) How can we get a definite answer for the number of microstates in this system? Relating entropy to microstates is useful for conceptually understanding what entropy is. At this level, it is not useful for calculating the change in entropy

54. Calculating Entropy (Similar to Work Idea) For slow, reversible processes: initial final P VS T initial W > 0 when Delta V 0 Q > 0 when Delta S > 0 Q < 0 when Delta S < 0

55. How to Calculate Entropy? For slow, reversible processes: To get to entropy we can “turn this expression around” If temperature is constant, then we can easily integrate: This last equation is not generally true; as heat enters or leaves a system the temperature often changes. (isothermal only!) Q

56. Equilibrium and Entropy Equilibrium is the most likely state Each microstate is equally likely, so the equilibrium state has the most microstates, or the greatest entropy. Therefore the equilibrium state has the highest entropy. For large (i.e. moles of atoms) systems, the system is (essentially) always evolving toward equilibrium. Therefore the total entropy never decreases: Second law of Thermodynamics

57. Total Entropy This is only for total entropy! ice (0 0 C=273K) air (20 0 C)heat flows this way Heat entering ice: dQ ice = T ice dS ice > 0 =>S ice increasing Heat leaving air: dQ air = T air dS air S air decreasing This is okay, because dS tot = dS ice + dS air > 0

58. What is Entropy? (Summary) Physical systems have an equilibrium state (the one with the most microstates, and therefore the highest entropy) If we wait “long enough” the system will most likely evolve to that equilibrium state. (by the laws of chance) For large (i.e. moles of atoms) systems, the system is (essentially) always evolving toward equilibrium. Therefore the total entropy never decreases: Second law of Thermodynamics

59. Thermal Equilibrium T final Energy leaves hot objects in the form of heat Energy enters cold objects in the form of heat Low tempHigh temp We need to determine the process… and the final state of the system.

60. In Equilibrium T A =150K Ok, now we have found the equilibrium state. T B =150K T A = T B = 150 K From our daily experience, and the activity in DL, we expect this process to be irreversible and spontaneous…. How do we calculate the change in entropy?

61. AB The two blocks A and B are both made of copper, and have equal masses: m a = m b = 100 grams. The blocks exchange heat with each other, but a negligible amount with the environment. a) What is the final state of the system? b) What is the change in entropy of block A? c) What is the change in entropy of block B? d) What is the total change in entropy? (Heat of melting) Temperature!

62. AB Okay, now we have found the equilibrium state. How do we calculate the change in entropy? a)What is the final state of the system? b) What is the change in entropy of block A? c) What is the change in entropy of block B? d) What is the total change in entropy?

63. What is the Entropy Change? In this case T f = 150 K = T f,A = T f,B T i,A = 100 K, T i,B = 200 K ∆S total > 0 Irreversible and spontaneous process ∆S total = 0 Reversible process

64. Calculating Entropy/State Functions Environment system Note: reversible means that (  S system can be +, -, or zero for a reversible process)  S total =0 NOT  S system =0 State functions (S,H,U,P,V,T, …) depend only on the initial and final states of the system Therefore we can join our initial and final states by any process. Most convenient to use reversible processes.

65. Calculating Entropy (Reminder, if TS diagram) For slow, reversible processes: initial final P VS T initial W > 0 when Delta V 0 Q > 0 when Delta S > 0 Q < 0 when Delta S < 0 Rules of thumb, similar to Work case….

66. Back to the Beginning: Three Phase Model

67. Graph of Ice to Steam T bp : Temperature at which a pure substance changes phase from liquid to gas (boiling point). T mp : Temperature at which a pure substance changes phase from solid to liquid (melting point). T bp T mp

68. Phase Changes - a recap You take ice out of the freezer at -30 0 C and place it in a sealed container and slowly heat it on the stove. You would find: the temperature of the ice rises, remains fixed at 0 0 C for an extended time while it is a mixture of ice and water, the temperature rises again after it all melted, remains fixed at 100 0 C for an extended time while it is a mixture of liquid and gas, the temperature rises again after it is all gas (steam).

69. Three Phase Model of Matter Q How do we change the phase of matter? How do we change the temperature of matter? A By adding or removing energy. Often this energy is transferred from, or to, the substance as heat, “Q”. Example H 2 O

70. Heat Capacity The heat capacity, C, of a particular substance is defined as the amount of energy needed to raise the temperature of that sample by 1° C. If energy (heat, Q) produces a change of temperature,  T, then: Heat capacity depends on the amount of a substance we have, since it will take more energy to change the temperature of a larger quantity of something. It is thus called an extensive quantity, or dependent upon the quantity/mass of a substance (kg or mole). Q = C  T

71. Heat capacity C – sort of the slope here of A, C, E Heat of fusion Heat of vaporization Q

72. Specific Heat The specific heat capacity, often simply called specific heat, is a particular number for a given substance and does not depend on quantity. Specific heat is thus an intensive property. The specific heat of water is one calorie per gram per degree Celsius. The specific heat of water is one calorie per gram per degree Celsius. SI units for heat capacity and specific heat: heat capacity J/K specific heatJ/kgK, or J/molK (molar specific heat)

73. Energy Change  E In our notation, we always have  E = E final - E initial.  E negative: Energy is released from the system. (“Neg. energy added.”)  E positive: Energy is put into the system.  Be sure to select the correct sign for all energy transfers! => Note also:  T is always T f - T i.

74. Conservation of Energy, Energy Interaction Model

75. Conservation of Energy Energy is both a thing (quantity) and a process. You & I contain energy, as do the chairs you sit on and the air we breathe. We cannot see it, but we can measure the transformation of energy (or change,  E). Conservation of Energy Energy cannot be created nor destroyed, simply converted from one form to another. Conservation of Energy Energy cannot be created nor destroyed, simply converted from one form to another. If the energy of an object increases, something else must have given that object its energy. If it decreases, it has given its energy to something else. A transfer of energy is when one object gives energy to another. There are 2 types of energy transfers -- Heat and Work.

76. Energy Systems E therma l E bond E movement (KE) E gravit y E electri c E sprin g There are many different types of energies called energy systems:........ For each energy system, there is an indicator that tells us how that energy system can change: E thermal : indicator is temperature E bond : indicator is the initial and final phases

77. 3-Phase Energy System Expressions E thermal = C  T, Temperature is the indicator. Between phase changes, only thermal energy changes. E bond =  |  m  H|,  m is the indicator. At a physical phase change, only the bond-energy system changes.  H is the heat of the particular phase change.  m is the amount that changed phase. In a chemical reaction, there are several bond energy changes corresponding to diff. molecular species (reactants or products). Here  H is the heat of formation for a particular species. E therma l E bond

78. Energy Interaction Diagrams (3-phase) Example: Melting Ice T i = 0°C  T f = room temperature Temperature Energy of substance solid liquid gas l-g coexist s-l coexist Initial T MP T BP Final

79. Energy Interaction Diagrams: 3-Phase Example: Melting Ice Process 1: Ice at T=0ºC  Water at T=0ºC Process 2: Water at T=0ºC  Water at room temperature Temperature Energy of substance solid liquid gas l-g coexist s-l coexist Process 1 Initial T MP T BP Process 1 Final / Process 2 Initial Process 2 Final

80. Energy Interaction Diagrams Example: Melting Ice Process 1: Ice at T=0ºC  Water at T=0ºC Ice ∆T=0 ∆E th =0 Initial phase Solid, Final phase Liquid E therm al E bond Heat

81. Energy Interaction Diagrams Example: Melting Ice Process 1: Ice at T=0ºC  Water at T=0ºC Ice ∆T=0 ∆E th =0 Initial phase Solid, Final phase Liquid ∆E th + ∆E bond = Q+W ∆E bond = Q E therm al E bond Heat

82. Energy Interaction Diagrams Example: Melting Ice Process 2: Water at T=0ºC  Water at room temperature Ice Initial phase Liquid, Final phase Liquid E therm al E bond

83. Energy Interaction Diagrams Example: Melting Ice Process 2: Water at T=0ºC  Water at room temperature Ice Initial phase Liquid, Final phase Liquid ∆E bond = 0 E therm al E bond T Heat ∆E th + ∆E bond = Q+W ∆E th = Q

84. Energy Interaction Diagrams Example: Melting Ice Ice Initial phase Liquid, Final phase Solid E therm al E bond Freezing (Water at T=0°C  Ice at T=0°C) ∆T=0 ∆E th =0 Heat NOTE: Heat is released when bonds are formed! (In general  E is negative)

85. EIM Algebra Review For a closed system: (Is it clear why there’s no Q or W for a closed system?) For an open system: (Q and W can be positive or negative, as can  Es.)

86. Mechanical Energy Systems

87. Three New Energy Systems (Mechanical) Emovement (KE) Egravit y (PEg) E sprin g (PEm- s) Rear shock absorber and spring of BMW R75/5 Motorcycle

88. Kinetic Energy System (KE): Work Kinetic energy is simply E moving. For translational energy, the indicator is speed; the faster an object moves, the more KE it has. There is a quantitative relationship between KE and speed. Also, it is proportional to the mass of the object: The direction of motion of the object is unimportant.  KE trans = ½ m  v 2 Baseba ll WorkKE Speed

89. Potential Energy System: (PE gravity ) Potential energy due to gravity: E height. (There are other types of PE, such as PE in a spring, or chemical PE.) For gravitational PE, the indicator is height; a higher object (with respect to something else) has more PE gravity. Can we show this? The quantitative relationship between PE and height: (g E ~10 m/s 2 is the acceleration due to gravity on Earth.)  PE gravity = mg E  h

90. Conserving Energy: KE and PE gravity PE grav Height  PE gravity =  KE ∆PE gravity depends on two quantities: the change in vertical distance that the object moved, and the mass of the object. Crumpled Paper KE Speed Note: we are neglecting the friction mg E  h = ½ m  v 2 mg E (h f 2 -h i 2 ) = ½ m(v f 2 -v i 2 )

91. KE  PE gravity 1) You throw a ball to the height of the first floor window. 2) Now you want to throw a ball to the height of the 4 th floor. Question: How much faster do you need to throw it? a)  2 times as fast b) Twice as fast c) Thrice as fast d) 4 times as fast e) 16 times as fast

92. Potential Energy: Springs Springs contain energy when you stretch or compress them. We have used them a lot in Physics 7. The indicator is how much the spring is stretched or compressed,  x, from its equilibrium (rest) state. k is a measure of the “stiffness” of the spring, with units [k] = kg/s 2.  x: Much easier to stretch a spring a little bit than a lot!  PE spring = ½ k  x 2 x

93. Sometimes from the conservation of energy: We can also discuss PE mass-spring in terms of conservation of energy: (  KE = ½ mv f 2 – ½ mv i 2 &  KE = ½ mx f 2 – ½ mx i 2 ) Conservation of energy  PE gravity =  KE translational mg  h = ½ m   v 2 ) KE trans = ½ m v 2  PE gravity =  Pe m-s mg  h = ½ m  x 2

94. Potential Energy v. Displacement Displacement from equilibrium y[+][-] direction of force yy PE mass-spring

95. Potential Energy v. Displacement Displacement from equilibrium y[+][-] PE mass-spring Equilibrium Potential Energy curve of a spring:  PE = ½ k (  x) 2 W (work) =  PE = -F ║  x Force ≈ -  PE /  x ≈ - k x Force is always in direction that decreases PE. Force is related to the slope -- NOT the value of PE. The steeper the PE vs r graph, the larger the force. Slope of PE curve Gives Strength of Force

96. Particle Model of Matter

97. “Pairwise” Atom-Atom Potential a.k.a Leonard-Jones Potential separation r Distance between the atoms, r Equilibrium separation r o Potential Energy As the atom-atom separation increases from equilibrium, force from the potential increases. ~ attracting each other when they area little distance apart

98. This is what is meant by a “bond” - the particles cannot escape from one another

99. Potential Energy between two atoms “pair-wise potential” a.k.a. Lennard-Jones Potential Energy r (atomic diameters) r   is the atomic diameter roro   is the well depth r o is the equilibrium separation   ~ 10 -21 J  ~ 10 -10 m = 1Å

100. Molecular Model If the atoms in the molecule do not move too far, the forces between them can be modeled as if there were springs between the atoms. The potential energy acts similar to that of a simple oscillator.

101. Particle Model of E bond and E thermal Example: H 2 O What is E bond in terms of KE and PE of individual atom (atom pair)? What is E thermal in terms of KE and PE of individual atom (atom pair)?

102. Recap: Particle Model of E bond E bond for a substance: amount of energy required to break apart “all” the bonds (magnitude only) i.e. we define E bond = 0 when all the atoms are separated The bond energy of a substance comes from adding all the potential energies of particles at their equilibrium positions. E bond = ∑ all pairs (PE pair-wise ) A useful approximation of the above relation is: E bond ~ - (total number of nearest neighbor pairs) x (  )  E bond of the system is negative, determined by: 1) the depth of the pair-wise potential well   (positive) 2) the number of nearest-neighbors.

103. Good Luck on your Final Exam!

104. Particle Model of E thermal E thermal is the energy associated with the random microscopic motions and vibrations of the particles. We increased E thermal by putting more energy into the system. KE and PE keep changing into one another as the atoms vibrate, just like in the mass-spring system, so we cannot make meaningful statements about instantaneous KE and PE. We can make statements about average KE and PE. Increasing E thermal increases both KE average and PE average.

105. Particle Model of E bond and E thermal The energy associated with the random motion of particles is split between PE oscillation, KE. PE avg = KE avg = ½ E total For particles in liquids and solids, let’s not forget the part that corresponds to E bond of the system. E bond of the system is determined by both the depth of the pair-wise potential well and the number of nearest neighbors (# NN).