1. Physics 7A – Lecture 7 Winter 2009
Prof. Robin D. Erbacher
343 Phy/Geo Bldg

2. Announcements No Quiz today, Quiz 5 is next week.
Join this Class Session with your PRS clicker!
Check Physics 7 website frequently for updates.
Turn off cell phones during lecture.

3. Review Multiple- Atom Systems: Particle Model of Ebond , Particle Model of Ethermal

4. Particle Model of Ebond and Ethermal
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)?

5. Multiple Atom Systems: Ebond
Typically every pair of atoms interacts
Magnitude of Ebond for a substance is the amount
of 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 e (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)

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

7. Continued: Particle Model of Ebond
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 nearest neighbor pairs per atom)/2 xN] x ()
Ebond of the system is negative, determined by:
1) the depth of the pair-wise potential well e (positive)
2) the number of nearest-neighbors.

8. Clicker: Etotal, Ebond, Ethermal
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 Etot greater?
a) Situation A has a greater Etot
b) Situation B has a greater Etot
c) Both have the same Etot
d) Impossible to tell

9. Clicker: Etotal, Ebond, Ethermal
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 Ebond greater?
a) Situation A has a greater Ebond
b) Situation B has a greater Ebond
c) Both have the same Ebond
d) Impossible to tell

10. Clicker: Etotal, Ebond, Ethermal
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 Ebond greater?
a) Situation A has a greater Ebond
b) Situation B has a greater Ebond
c) Both have the same Ebond
d) Impossible to tell
We did not break a bond - Ebond did not change!

11. Clicker: Etotal, Ebond, Ethermal
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

12. Clicker: Etotal, Ebond, Ethermal
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

13. Clicker: Individual Atoms
Initial
Final
Clicker: 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. Clicker: Individual Atoms
Initial
Final
Clicker: 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 Ethermal
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 .

16. Particle Model of Ethermal
The energy associated with the random
motion of particles is split
between PEoscillation and KE .

17. As we increase Etot we increase PEavg and KEavg
Mass on a Spring
Energy
position
Etot PE KE
As we increase Etot we increase PEavg and KEavg
PEavg = KEavg = Etot/2

18. Particle Model of Ebond and Ethermal
The energy associated with the random
motion of particles is split between PEoscillation , KE.
For particles in liquids and solids, let’s not forget
the 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) .

19. Particle Model of Ebond and Ethermal
Solids/Liquids:
Ethermal = KE + PEoscillation:
KEall atoms = (1/2)Ethermal
PEall atoms = PEbond + PEoscillation
= Ebond (PEbond )+ (1/2)Ethermal (PEoscillation)
Etotal = KEall atoms + PEall atoms = Ethermal + Ebond
Gases (monoatomic):
The gas phase has no springs:
no PEoscillation or PEbond

20. Summary of Ebond and Ethermal
Potential Energy contributes to both Ebond and
Ethermal.
Kinetic Energy contributes to Ethermal, but
not Ebond.
Number of bonds is important for Ebond,
not really for Ethermal.

21. Equipartition of Energy

22. Intro to Equipartition of Energy
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.
Each particle in a solid or liquid oscillates in 3 dimensions about its equilibrium positions as determined by its single-particle potential.

23. Intro to Equipartition of Energy
Another way of stating: Each particle has six “ways” to store the energy associated with its random thermal motion.
We call this “way” for a system to have thermal energy a “mode”.

24. Temperature IS Thermal Energy!? ?
Question
What is Temperature
in terms of Ethermal?
Answer:
Temperature IS Thermal Energy!

25. kB = 1.38  10-23 Joule per degree Kelvin
But Wait a Minute…
[Energy] = [Joule] [Temperature] = [Kelvin]
Answer revised:
Temperature is proportional to
Thermal Energy Ethermal.
The constant of proportionality is kB: Boltzmann’s Constant
kB = 1.38  Joule per degree Kelvin

26. Equipartition of Energy
Gas
Liquids and Solids
To be precise, energy associated with the component of motions/vibrations (“modes”) in any particular direction is (1/2)kBT :
Ethermal per mode = (1/2) kBT
a.k.a. Equipartition of Energy

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

28. Equipartition of Energy: Restated
Gas
Liquids and Solids
Equipartition of Energy Restated
In thermal equilibrium, Ethermal is shared equally among all the “active” modes available to the particle. In other words,each “active” mode has the same amount of energy given by : Ethermal per mode = (1/2) kBT

29. Modes of an atom in solid/liquid
Every atom can move in three directions
3 KEtranslational modes

30. Modes of an atom in solid/liquid
Every atom can move in three directions
3 KEtranslational modes
Plus 3 potential energy modes along three directions
3 PE modes
Total number of modes is 3PE + 3KE = 6
Ethermal = 6(1/2)kBT

31. Modes of an atom in a Monoatomic Gas
Every atom can move in three directions
3 KEtranslational modes
0 PE modes
(no bonds)
Gas:
No bonds,
i.e. no springs
Total number of modes is 3KE = 3:
KEavg = 3(1/2)kBT= Ethermal
PEavg = 0

32. Modes of a Molecule in a Diatomic Gas
3 KEtranslational modes
2 KErotational modes
2 vibrational modes (1 KE, 1PE)
(associated with atom-atom interaction
within the molecule)

33. Modes of a Molecule in a Diatomic Gas
3 KEtranslational modes
2 KErotational modes
2 vibrational modes (1 KE, 1PE)
(associated with atom-atom interaction
within the molecule)
Total number of modes is 6KE + 1PE = 7
Ethermal = 7(1/2)kBT
Sometimes (at lower temperatures), however, not all the modes are “active”. (Freezing out of modes)

34. Equipartition of Energy: Ethermal per Mode = ½ kBT
KE mode
PE mode
Total
Solids 3 6
Liquids
Monatomic gases
Diatomic gases
3+2+1 1 7

35. Equipartition of Energy: Ethermal per Mode = ½ kBT
KE mode
PE mode
Total
Solids 3 6
Liquids
Monatomic gases
Diatomic 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?

36. Clicker Question: Energy for Temperature
Does it take more or
less energy to raise
the temperature of a
diatomic gas than a
monatomic gas?
KE mode
PE mode
Total
Solids 3 6
Liquids
Monatomic gases
Diatomic gases
3+2+1 1 7
More
Less
Equal
I don’t even know how to start guessing

37. Real Heat Capacity Measurements…
diatomic (no vibrations)
monatomic
All measurements at 25 C unless listed otherwise
(100 C)
(500 C)

38. Answer: Energy for Temperature
Does it take more or
less energy to raise
the temperature of a
diatomic gas than a
monatomic gas?
KE mode
PE mode
Total
Solids 3 6
Liquids
Monatomic gases
Diatomic gases
3+2+1 1 7
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

39. Measuring Heat Capacity
Thermometer
Stirring rod
Heating element
Substance of interest
Thermometer
Stirring rod
Substance of interest
Heating element

40. Specific Heat of Solids
All measurements at 25 C unless listed otherwise
(-100 C)

41. Molar Specific Heat of Solids
All measurements at 25 C unless listed otherwise
(0C)
(-100 C)

42. Molar Specific Heat of Solids
6 modes
All measurements at 25 C unless listed otherwise
kBNA = R = 8.31 J/moleK : Gas constant or Ideal gas constant
kB(Boltzmann constant) = 1.38 x Joule/Kelvin
NA(Avogadro’s number) = 6.02 x 1023
(0C)
(-100 C)

43. Specific Heat of Liquids
All measurements at 25 C unless listed otherwise

44. Molar Specific Heat of Liquids
All measurements at 25 C unless listed otherwise
6 modes

45. Molar Specific Heat of Liquids
For polyatomic substances ,
the values of molar specific heat
of liquids are greater than
the values for solids.
Limitation of our model…
For monatomic substances,
the value of molar specific
heat of liquids is similar
to the values for solids.
Our model works well here!
All measurements at 25 C unless listed otherwise
6 modes

46. Specific Heat for Gases
All measurements at 25 C unless listed otherwise

47. Molar Specific Heat for Gases
Oops!
What’s going on??!
All measurements at 25 C unless listed otherwise
diatomic (no vibrations)
monatomic
The dotted lines are the values you’d get if you calculated using 1/2kT or ½ nRT. Doesn’t match the measured values as it does with solids. Why?
(500 C)
(100 C)

48. Discrepancy Explained…
Closed box of gas
Open box of gas
CV measurement
CP measurement
Closed box: all heat goes into the gas’s energy
Open box: Some heat goes into pushing air out of the way

49. Discrepancy Explained…
Closed box of gas
Open box of gas
Question
So then does it take
More energy to raise the
Temperature of
Closed box of gas Or
Open box of gas?
CV measurement
CP measurement
Takes more energy to heat an open box of gas, which explains why the Cp values are higher than you would calculate using ½ kT
Closed box: all heat goes into the gas’s energy
Open box: Some heat goes into pushing air out of the way

50. Molar Specific Heat for Gases
Whew! Now it
Makes more sense.
diatomic (no vibrations)
All measurements at 25 C unless listed otherwise
monatomic
(500 C)
(100 C)

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

52. Summary of Equipartition of Energy
Equipartition tells us that the energy per mode
Is ½ kbT.
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.

53. Summary of the Particle Model
Atom-atom interactions determine how much energy we need to
 pull our system apart. Sum over all pairs to find Ebond
Ebond depends on the strength of the interactions (well-depth) and
 the number of interactions (determined by packing)
Ethermal depends on the number of modes. 

Equipartition tells us that each (active) mode has the same 
 amount of energy, and that the energy per mode depends 
 only on temperature.

Our mode counting works well for solids and gases. For gases 
 we need to distinguish between work and heat carefully (next)

Our mode counting does not work for liquids.

54. Thermodynamics: Microscopic to Macroscopic

55. What is Thermodynamics?
In a nutshell, we study
the transfer of energy
between systems and how
the energy instills
movement, i.e., how the
system responds.
Ex. If we heat something,
it expands.
Sadi Carnot ( )
Father of Thermodynamics
Lord Kelvin
( )
James Joule
( )

56. Thermodynamics: Goals
Carefully distinguish between heat and work.
Learn what a state function is.
Think about things that depend on the
process, and ask about processes.
(e.g. ice melts/water freezes, eggs fry)
 Why doesn’t the fried egg turn into raw egg again?
 Why can carbon exist as a diamond as well
as graphite but not ice and vapor?

57. State Function 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

58. 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

59. Conservation of Energy (Review)
∆Etotal must include all changes of energy associated
with the system…
∆Etotal = ∆Ethermal+ ∆Ebond+ ∆Eatomic+ ∆Enuclear+ ∆Emechanical
∆U : Internal energy
Energy associated with the
atoms/molecules inside
the body of material
Energy associated with
the motion of a body
as a whole
So then, if there’s no change in ∆Emechanical
∆ U
…If no change in
DEmechanical
First law of Thermodynamics

60. State v. Process-dependent Functions
State functions
Process-dependent
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...
Depend on the process.
Not a property of an object
Examples: Q, W, learning ,

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

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

63. Work P V First section: W1 < 0 (volume expands)
final
initial V
First section: W1 < 0 (volume expands)
Second section: W2 = 0 (volume constant)
Third section: W3 > 0 (volume contracts)

64. Clicker: Process Comparisons
final
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. P
initial V
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. P
final
initial V

65. Heat and the First Law of ThermoP
initial
final V
We can read work directly off this graph (i.e. don’t need to know anything about modes, U, T, etc.)
PV = nRT
U = 5/2 nRT
We have PiVi from the graph…
If we know something about the gas, we can figure out Ui, Uf and Uf - Ui

66. Heat and the First Law of ThermoP
initial
final V

67. Example
5 moles of a 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 DU * heat involved in this process.
initial
final P V
W=0
dU=3/2nRdT = 3/2dPV = 3/2 0.5x10^5 (0.1) = 7500 J = dQ
= 3/2 (5)RdT = 3/2(5)R(120) =3/2 (5)(8.314)(120) = J = dQ
HW:
There are two
ways to solve
this using the ideal gas law…

68. Comments Work depends on the process Heat depends on the process
DU only depends on initial and final
W = 0 for all constant volume processes

69. Enthalpy Why is this useful?!? H = U + PV
Is it a state function? - U depends only on state of system - P depends only on state of system - V depends only on state of system => H depends only on state of system
(Hess’s law)
Why is this useful?!?

70. Enthalpy Constant volume Constant pressure P V P V W = 0
initial
final P V
initial
final P V
Enthalpy useful for constant pressure processes:
dH = dU + PdV
dU = dQ – dW = dQ – PdV
dH = dQ – PdV + PdV OR dH = dQ (constant pressure process and only PV work)
W = 0
*Derivation in P.84
Note: nothing about gases used –
works for solids and liquids too!

71. Unit 1 Revisited In unit #1 we taught you:
But the gas is expanding, so work is being done!
Now we know better….
But in a large room, pressure is roughly constant…

72. Unit 1 Revisited

73. Next Time: Microstates versus States, and Entropy