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Physics 7A – ReviewWinter 2008
Physics 7A – ReviewWinter 2008
Prof. Robin D. Erbacher343 Phy/Geo Bldg
Prof. Robin D. Erbacher343 Phy/Geo Bldg
• 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.
•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
ReviewMultiple- Atom
Systems:Particle Model of Ebond ,Particle Model of Ethermal
ReviewMultiple- Atom
Systems:Particle Model of Ebond ,Particle Model of Ethermal
• Typically every pair of atoms interacts
• Magnitude of Ebond for a substance is the amountof energy required to break apart “all” the bonds i.e. we define Ebond = 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.
Ebond = ∑all pairs(PEpair-wise)
What is the change in bond energy (∆Ebond) by removing the red atom?
2.2 A
4.4 A
6.6 A
-8 x 10-21J
-0.5 x 10-21J
~ 0 J8.8 A
11 A~ 0 J
~ 0 J
Bond energy
Separation (10-10m)
Answer:~8.5x10-21 J
What is the bond energy Ebond for the entire molecule?
-8 x 10-21J
Separation (10-10m)
-8.5 x 10-21J
-8.5 x 10-21J -8.5 x 10-21J
-8.5 x 10-21J
Ebond = -42 x 10-21 Joules
What is the bond energy Ebond for the entire molecule?Separation (10-10m)
Ebond ≈ -40 x 10-21 Joules
Energy required to break a single pair of atoms apart: +8x10-21 J
=5 bonds.
• Ebond for a substance: amount of energy requiredto break apart “all” the bonds (magnitude only) i.e. we define Ebond = 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. Ebond = ∑all pairs(PEpair-wise)
A useful approximation of the above relation is:Ebond~ - (total number of nearest neighbor pairs) x ()
Ebond of the system is negative, determined by:1) the depth of the pair-wise potential well (positive) 2) the number of nearest-neighbors.
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 Ethermal greater?a) Situation A has a greater Ethermal
b) Situation B has a greater Ethermal
c) Both have the same Ethermal
d) Impossible to tell
A
B
A
BKE
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 Ethermal greater?a) Situation A has a greater Ethermal
b) Situation B has a greater Ethermal
c) Both have the same Ethermal
d) Impossible to tell
KE
We increased Ethermal by putting more energy into the system
Initial
Final
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?
Initial
Final
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?
• Ethermal is the energy associated with the random microscopic motions and vibrations of the particles.
• We increased Ethermal 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 Ethermal increases both KEaverage and PEaverage .
The energy associated with the random motion of particles is split
between PEoscillation and KE .
As we increase Etot we increase PEavg and KEavg
PEavg = KEavg = Etot/2 (for mass on a spring)
En
erg
y
position
Etot
PE
KE
• The energy associated with the randommotion of particles is split between PEoscillation , KE.
• For particles in liquids and solids, let’s not forgetthe part that corresponds to Ebond of the system.
• Ebond of the system is determined by both the depth of the pair-wise potential well and the number of nearest neighbors (# NN) .
Initial
Final
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 Peavg?
PEavg=KEavg=Ethermal/2
Eth= Etot-Ebond = 5x10-21 JKEavg= 2.5x10-21 J = PEavg
• Potential Energy contributes to both Ebond andEthermal.
• Kinetic Energy contributes to Ethermal, but not Ebond.
• Number of bonds is important for Ebond,not really for Ethermal.
Equipartition of Energy
Equipartition of Energy
Energy associated with the component of Energy associated with the component of motions/vibrations (“modes”) in any motions/vibrations (“modes”) in any particular direction is (1/2)kparticular direction is (1/2)kBBT :T :
EEthermal per modethermal per mode = (1/2) k = (1/2) kBBTT
a.k.a. Equipartition of Energya.k.a. Equipartition of Energy
GasLiquids and Solids
Modes : Ways each particle has of storing energy.
Ex. Mass-spring has one KE mode
and one PE mode.
KE modeKE mode PE modePE mode TotalTotalSolids 3 3 6
Liquids 3 3 6Monatomic gases 3 0 3Diatomic gases 3+2+1 1 7
When energy is added to a system, what does it mean to have more places (modes) to store it in?
KE KE modemode
PE PE modemode
TotalTotal
Solids 3 3 6Liquids 3 3 6
Monatomic gases
3 0 3
Diatomic gases
3+2+1 1 7
Does it take more or less energy to raise the temperature of a diatomic gas than amonatomic gas?
1. More
C = ∆Ethermal / ∆T ∆ Ethermal per molecule = number of active modes (1/2)kB∆T
∆ Ethermal per N atoms = number of active modes (1/2)kB∆T N
All measuremen
ts at 25 Cunless listed
otherwise
(-1
00
C)
(0C
)
kBNA = R = 8.31 J/moleK : Gas constant or Ideal gas constant
kB(Boltzmann constant) = 1.38 x 10-23 Joule/Kelvin
NA(Avogadro’s number) = 6.02 x 1023
6 modes
All measuremen
ts at 25 Cunless listed
otherwise
diatomic(no vibrations)
(10
0 C
)
(50
0 C
)monatomic
Oops! What’s going on??!
3 (5) modes => 3 (5) x ½ kBNA = 12.42 (20.70) J/mol K
Closed box of gasOpen 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
CV measurement
CP measurement
Closed box of gasOpen 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
CV measurement
CP measurement
Question
So then does it takeMore energy to raise the
Temperature of
Closed box of gasOr
Open box of gas?
All measuremen
ts at 25 Cunless listed
otherwise
diatomic(no vibrations)
(10
0 C
)
(50
0 C
)
monatomic
Whew! Now itMakes more sense.
Recall, Cp = Cv + PdV/T, orCp = Cv + R!
“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) ,In this example of thermal phenomena (i.e.,measuring heat capacity) ,
• Equipartition tells us that the energy per modeIs ½ kbT.
• For solids and liquids, moving the atom in anydirection is like moving a mass on a spring. Threedirections means 3 KE modes/atom and 3 PEmodes/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.
Thermodynamics: Microscopic to Macroscopic
Thermodynamics: Microscopic to Macroscopic
Depends only on properties of the system at a particular time
Example: Ethermal of a gas For an ideal gas, Ethermal depends only on:
• Temperature • Number of modes
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
∆Etotal must include all changes of energy associated
with the system…
∆Etotal = ∆Ethermal+ ∆Ebond+ ∆Eatomic+ ∆Enuclear+ ∆Emechanical
Energy associated with the motion of a body
as a whole
∆ U
First law of Thermodynamics
∆U : Internal energyEnergy associated with the atoms/molecules inside the body of material
For an ideal gas,U = Ethermal = # modes*1/2 kBT
• 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
initial
initial
finalfina
l
P
VV
P
Along any given segment:
W = -PV(if P constant)
OR
= - Area under curve (sign positive when V decreases, sign negative when V increases)
initial
final
P
V
Is the work done in the process to the right positive or negative?
A) NegativeB) Positive C) Zero D) Impossible to tell.
initial
final
P
VFirst section: W1 < 0 (volume expands)Second section: W2 = 0 (volume constant)Third section: W3 > 0 (volume contracts)
initial
finalP
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.
initial
finalP
V
A) Magnitude of work in top
process greaterB) Magnitude of work in
lower process greaterC) Both the sameD) Need more info about the
gases.
Work = 0(V=0)
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 Ui, Uf and Uf - UiFor an ideal gas,
U = Ethermal = # modes*1/2 kBT
initial
final
V
P
5 moles of an ideal monatomic gas has its pressure increased from 105 Pa to 1.5x105 Pa. This process occurs at a constant volume of 0.1 m3. Determine: * work * change in internal energy U * heat involved in this process.
initial
final
P
V
HW: There are twoways to solve this using the ideal gas law…
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.
Microstates versusStates, and EntropyMicrostates versusStates, and Entropy
ConstraintsConstraints StatesStatesMicrostatesMicrostates
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.
Examplesgases: P ~ average density, V~volume filled, T~average KE
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 Every microstatemicrostate is equally likely. is equally likely. The state that is most likely is the one with the most is the one with the most
microstatesmicrostates
Prob.
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.
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
HHHHHTHTTHTHTTTT
Any of the 256 micro.Any of the 256 micro.
Any of the 256 micro.Any of the 256 micro.
Any of the 256 micro.Any of the 256 micro.
Any of the 256 micro.Any of the 256 micro.
where kB 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.
We should divide our box up into “atom-sized” chuncks. But how
big should our velocity microstates be?
ice (0ice (000 C)C)
Water (0Water (000 C) 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
For slow, reversible processes:
initial
final
final P
VS
T
initial
W > 0 when Delta V < 0W < 0 when Delta V > 0
Q > 0 when Delta S > 0Q < 0 when Delta S < 0
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
• 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
This is only for total entropy!
iceice(0(000C=273KC=273K
))
air (200C)heat flows this way
Heat entering ice: dQice = Tice dSice > 0 =>Sice increasing Heat leaving air: dQair = Tair dSair < 0 =>Sair decreasing
This is okay, because dStot = dSice + dSair > 0
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
Tfinal
Energy leaves hot objects in the form of heat Energy enters cold objects in the form of heat
Low temp High temp
We need to determine the process… and the final state of the system.
TA=150K
Ok, now we have found the equilibrium state.
TB=150K
TA = TB = 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?
AA BB
The two blocks A and B are both made of copper, and have equal masses: ma = mb = 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!
AA BB
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?
In this case Tf = 150 K = Tf,A = Tf,B
Ti,A = 100 K, Ti,B = 200 K
∆Stotal > 0 Irreversible and spontaneous process
∆Stotal = 0 Reversible process
EnvironmentEnvironment
system
Note: reversible means that
(Ssystem can be +, -, or zero for a reversible process)
Stotal=0 NOT Ssystem=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 anyprocess. Most convenientto use reversible processes.
For slow, reversible processes:
initial
final
final P
VS
T
initial
W > 0 when Delta V < 0W < 0 when Delta V > 0
Q > 0 when Delta S > 0Q < 0 when Delta S < 0
Rules of thumb, similar to Work case….
Back to the Beginning:
Three Phase Model
Back to the Beginning:
Three Phase Model
• Tbp: Temperature at which a pure substance changes phase from liquid to gas (boiling point).• Tmp: Temperature at which a pure substance changes phase from solid to liquid (melting point).
Tbp
Tmp
You take ice out of the freezer at -300C 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 00C for an extended time while it is a mixture of ice and water,
• the temperature rises again after it all melted,
• remains fixed at 1000C for an extended time while it is a mixture of liquid and gas,
• the temperature rises again after it is all gas (steam).
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 H2O
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 TQ = C T
Heat capacity C – sort of the slope here of A, C, E
Heat of fusion Heat of vaporization
Q
€
C = Q
ΔT
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 heat J/kg•K, or J/mol•K (molar specific heat)
In our notation, we always have E = Efinal - Einitial .
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 Tf - Ti .
Conservation of Energy,
Energy Interaction Model
Conservation of Energy,
Energy Interaction Model
• 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 EnergyEnergy cannot be created nor destroyed, simply
converted from one form to another.
Conservation of EnergyEnergy 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.
Etherma
l
EbondEmovement
(KE)
Egravit
y
Eelectri
c
Esprin
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:
Ethermal: indicator is temperatureEbond: indicator is the initial and final phases
•Ethermal = C T, Temperature is the indicator.
• Between phase changes, only thermalenergy changes.
• Ebond = |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.
Etherma
l
Ebond
Example: Melting IceTi= 0°C Tf = room temperature
Tem
pera
ture
Energy of substance
solid
liquid
gas
l-g coexist
s-l coexist
Initial
TMP
TBP
Final
Example: Melting IceProcess 1: Ice at T=0ºC Water at T=0ºC
Process 2: Water at T=0ºC Water at room temperatureTem
pera
ture
Energy of substance
solid
liquid
gas
l-g coexist
s-l coexist
Process 1Initial
TMP
TBP
Process 1Final /
Process 2Initial
Process 2Final
Example: Melting IceProcess 1: Ice at T=0ºC Water at T=0ºC
Ice
∆T=0
∆Eth=0
Initial phase Solid, Final phase Liquid
Etherm
al
Ebond
Heat
Example: Melting IceProcess 1: Ice at T=0ºC Water at T=0ºC
Ice
∆T=0
∆Eth=0
Initial phase Solid, Final phase Liquid
∆Eth + ∆Ebond= Q+W
∆Ebond= Q
Etherm
al
EbondHeat
Example: Melting IceProcess 2: Water at T=0ºC Water at room temperature
Ice
Initial phase Liquid, Final phase Liquid
Etherm
al
Ebond
Example: Melting IceProcess 2: Water at T=0ºC Water at room temperature
Ice
Initial phase Liquid, Final phase Liquid
∆Ebond= 0
Etherm
al
Ebond
T
Heat
∆Eth + ∆Ebond= Q+W
∆Eth= Q
Example: Melting Ice
Ice
Initial phase Liquid, Final phase Solid
Etherm
al
Ebond
Freezing (Water at T=0°C Ice at T=0°C)
∆T=0
∆Eth=0
Heat
NOTE: Heat is released when bonds are formed! (In general E is negative)
• 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.)
€
E total = ΔE1 + ΔE 2 + ΔE 3 + ... = 0
€
E total = ΔE1 + ΔE 2 + ΔE 3 + ... = Q + W
Mechanical Energy Systems
Mechanical Energy Systems
EEmovementmovement
(KE)(KE)
EEgravitgravit
yy
(PEg)(PEg)
EEsprinsprin
gg
(PEm-s)(PEm-s)
Rear shock absorber and spring of
BMW R75/5Motorcycle
• Kinetic energy is simply Emoving.
• 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.
KEtrans = ½ m v2KEtrans = ½ m v2
Baseball
WorkKEKE
SpeedSpeed
• Potential energy due to gravity: Eheight. (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 PEgravity. Can we show this?
• The quantitative relationship between PE and height:
(gE~10 m/s2 is the acceleration due to gravity on Earth.)
PEgravity = mgEhPEgravity = mgEh
PEgrav
Height
PEgravity = KE PEgravity = KE
• ∆PEgravity depends on two quantities: the change in vertical distance that the object moved, and the mass of the object.
Crumpled PaperKE
SpeedNote: we are neglecting the friction
mgEh = ½ mv2 mgE(hf
2-hi2) = ½ m(vf
2-vi2)
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 4th floor.
Question: How much faster do you need to throw it?
a) 2 times as fastb) Twice as fast• Thrice as fast• 4 times as fast• 16 times as fast
• 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/s2.
• x: Much easier to stretch a spring a little bit than a lot!
PEspring = ½ kx2PEspring = ½ kx2
x
• Sometimes from the conservation of energy:
• We can also discuss PEmass-spring in terms of conservation of energy:
• (KE = ½ mvf2 – ½ mvi
2 & KE = ½ mxf2 – ½ mxi
2)
PEgravity = KEtranslational
mgh = ½ m v2)
KEtrans = ½ m v2 KEtrans = ½ m v2
PEgravity = Pem-s
mgh = ½ mx2
Displacement from equilibrium y[+][-]
direction of force
y
PEmass-spring
Displacement from equilibrium y[+][-]
PEmass-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 ofPE curveGives Strength of Force
Particle Model of Matter
Particle Model of Matter
separation
r
Distance between the atoms, r
Equilibrium separation
ro
Po
ten
tial
En
erg
y
As the atom-atom separation increases from equilibrium, force from the potential increases. ~ attracting each other when
they area little distance apart
This is what is This is what is meant by a “bond” - meant by a “bond” - the particles cannot the particles cannot escape from one escape from one anotheranother
Potential Energy between two atomsPotential Energy between two atoms“pair-wise potential” a.k.a. Lennard-“pair-wise potential” a.k.a. Lennard-
Jones PotentialJones Potential
Potential Energy between two atomsPotential Energy between two atoms“pair-wise potential” a.k.a. Lennard-“pair-wise potential” a.k.a. Lennard-
Jones PotentialJones PotentialEnergy
r (atomic diameters)
r
is the atomic diameter
ro
is the well depth
ro is the equilibrium separation
~ 10-21 J
~ 10-10m = 1Å
• 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.
Example: H2O
What is Ebond in terms of KE and PE of individual atom (atom pair)?
What is Ethermal in terms of KE and PE of individual atom (atom pair)?
• Ebond for a substance: amount of energy requiredto break apart “all” the bonds (magnitude only) i.e. we define Ebond = 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. Ebond = ∑all pairs(PEpair-wise)
A useful approximation of the above relation is:Ebond~ - (total number of nearest neighbor pairs) x ()
Ebond of the system is negative, determined by:1) the depth of the pair-wise potential well (positive) 2) the number of nearest-neighbors.
Good Luck on your Final Exam!
Good Luck on your Final Exam!
• Ethermal is the energy associated with the random microscopic motions and vibrations of the particles.
• We increased Ethermal 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 Ethermal increases both KEaverage and PEaverage .
• The energy associated with the randommotion of particles is split between PEoscillation , KE.PEavg = KEavg = ½ Etotal
• For particles in liquids and solids, let’s not forgetthe part that corresponds to Ebond of the system.
• Ebond of the system is determined by both the depth of the pair-wise potential well and the number of nearest neighbors (# NN) .
