1

These lecture notes are based uponMatter and Interactions 2nd

Edition, Volume 1

Ruth Chabay

and Bruce SherwoodChapters …

6. Energy in Macroscopic Systems 7. Energy Quantization

11. Entropy: Limits on the Possible12. Gases and Engines

PHY2009S 2009Intermediate PhysicsThermodynamics and statistical models in physics

… and are derived in some places from slides produced by David Aschman

and Indresan Govender.

Prof Andy BufflerRoom 503 RW Jamesandy.buffler@uct.ac.za

2

PHY2009S Thermodynamics and statistical models in physicsPart 2

Entropy:Limits on the

possibleMatter & Interactions

Chapter 11

3

Thermal Physics: Thermodynamics

Thermodynamics is a framework for relating macroscopic properties of a system to one another, for example:

How does pressure of a gas depend on the temperature and volume of its container?

How does a refrigerator work? What is its maximum efficiency?

How much energy do we need to add to a kettle of water to change it to steam?

Applications: for example thermal engines (internal combustion engine, steam engine,

…)

4

Thermal Physics: Statistical Mechanics:

Why are the properties of water different from those of steam, even though water and steam consist of the same type of molecules?

How are the molecules arranged in a liquid?

The goal of statistical mechanics is to begin with the microscopic

laws of physics that govern the behaviour of the

constituents of the system and deduce the properties of the system as a whole, for example:

Statistical mechanics is the bridge between the microscopic and macroscopic worlds.

5

Einstein and thermodynamics

“A theory is the more impressive the greater the simplicity of its premises, the more different kinds of things it relates, and the more extended its area of applicability. Therefore the deep impression that classical thermodynamics made upon me. It is the only physical theory of universal content which I am convinced will never be overthrown, within the framework of applicability of its basic concepts.”

–

Albert Einstein

6

Consider a ball bouncing on the floor.If we consider the closed system to be the ball and floor, what is the cause of the ball eventually coming to rest?

1. Friction2. Energy “loss”3. Momentum “loss”4. None of the above

1 2 3 4 5

7

Consider a ball bouncing on the floor.If we consider the closed system to be the ball and floor, which of the following principles are violated?

1. Conservation of momentum2. Conservation of energy3. Neither

1 2 3 4 5

8

The bouncing ball

Initially (before 1st

bounce) all energy is associated with centre

of mass of ball, i.e. one degree of freedom.After several bounces energy associated with both the ball and the individual molecules of floor, i.e. many degrees of freedom

(both ball and floor feel warm).

How do we quantify these “degrees of freedom”?

9

Statistical issues

Very notion of temperature: average kinetic energy of molecules

Large numbers of molecules: NA

= 6.023 ×

1023

mole-1

Thermal energy flow: hotter to colder (on average)

Reversibility

M&I11.1

10

We observe that energy spontaneously flows from hot to cold

Hightemperature Low

temperature

Energy transfer Q

11

Hightemperature Low

temperature

Energy transfer Q ??

But we do not observe the spontaneous flow of energy from cold to hot

12

Reversible processes

… cannot tell forward movie from backward

… collisions possible in both directions

… laws of physics (strong, electromagnetic, gravitational interactions) completely reversible

Many processes in physics are reversible, e.g. atomic collisions, or frictionless 2-d elastic collisions …

13

Apparent paradox in thermodynamics

All fundamental processes exhibit time reversal invariance, i.e. both forward and backward movie of fundamental interactions

look possible.However macroscopic change has a preferred direction, for example …

Bouncing balls come to rest …

and never reverse their motion Energy moves from hot to cold …

never the reverse direction

If the sub-microscopic world exhibits time reversal invariance, surely the macroscopic world (which is made up of the sub-microscopic) should also be time reversal invariant

!?

Better to think of macroscopic change statistically

…… a bouncing ball increasing in height with each bounce is not

impossible, just wildly (ridiculously) improbable.

14

Modelling strategy

Use a very simple model of a solid to understand the distribution of energy between the atoms.

Determine probability of particular energy (speed) distributions.

Use this probability to understand why certain distributions are very unlikely and others a sure bet.

Test microscopic predictions with macroscopic measurements.

15

Practical example on probability

What is the probability of picking out a:a)

Yellow ball?

b)

Blue ball?c)

Green ball?

d)

Red ball?

Consider the following collection of balls in a box:

5000 red500 green50 blue5 yellow

If you reached into the bag (without looking), which colour ball are you most likely to pick up?

16

Model of a solid: ball and springs

We model a solid as a large number of tiny masses (the atoms) connected to their neighbours by springs (the inter-atomic bonds).

M&I11.2

17

Simplified model: Einstein model of a solid

Each atom is a 3D quantum mechanical oscillator …

… based on Planck’s quantisation assumption.

Each atom moves independently (connected to imaginary rigid walls …

not other atoms)

… no mechanism for energy exchange between atoms.

All atoms vibrate with the same frequency.

Ignore collective motion of groups of atoms.

18

We consider each atom in a solid to be connected by springs to immovable walls. Each isolated atom is a 3D spring-mass system where is the 3D vector displacement away from equilibrium.

2 2 2 2x y zp p p p= + + 2 2 2 2

x y zs s s s= + +

s

22 22 2 2

vib1 1 1

2 2 2 2 2 2yx z

s s x s y s z

pp pK U k s k s k sm m m

⎛ ⎞⎛ ⎞ ⎛ ⎞+ = + + + + +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠

… mathematically equivalent to replacing each 3D oscillator (an atom) with three ordinary 1D oscillators …We can think of a block as containing N

1D

oscillators, corresponding to N/3 atoms.

Simplified model: Einstein model of a solid

19

Quantised energy levels of a 1D oscillator

0E

1 0 01E Eω= +

3 0 03E Eω= +

4 0 04E Eω= +

2 0 02E Eω= +

Energy can only be added in multiples of where 0ω 0 ,s i ak mω =

,s ik : “inter-atomic spring stiffness”

20

x y z

The quantized 3D oscillator

wells_oscillator.py

21

Simple case: One atom equivalent to three

oscillators.

Imagine that the atom has four

quanta of energy.

If we want to distribute these 4

quanta over the 3

oscillators …

How many

ways to do it?

Here are three ways …

how many more are there?

4 0 0 2 1 1 0 3 1

Quantised energy levels of a 3D oscillator

22

23

Microstates and macrostates

There are 15 microstates

…… each microstate is a way to distribute the energy.

All microstates belong to the same macrostate

…… which is characterized by the total energy being equal to 4 quanta of energy, no matter how distributed.

In the previous example …

24

Fundamental assumption of statistical mechanics

Over time, an isolated system in a given macrostate

is equally likely to be found in any one

of the microstates

states accessible to it.

This assumption allows all possible configurations of the system (macrostate) if we observe it for long enough.

However, you might have to wait a very long time to observe certain configurations.

25

Two blocks in thermal contact

In fact, take two atoms! …

the smallest possible “block”. Count the ways to distribute 4 quanta of energy between the two atoms (six independent oscillators).

Take two small blocks in thermal contact.

26

All 4

quanta to one atom, none

to the other15

microstates.

3

quanta to one atom, and 1

to

the other

Atom 1 Atom 2 # ways

4 0 (15)(1)0 4 (1)(15)

Atom 1 Atom 2 # ways

2 2 (6)(6)

Atom 1 Atom 2 # ways

3 1 (10)(3)1 3 (3)(10)

30

microstates.

36

microstates.

2

quanta to one atom, and 2

to

the other

15

microstates.

30

microstates.

126

microstates.Total:

27

Histogram of the 126 ways to distribute four quanta of vibrational

energy between two atoms (each consisting

of three independent oscillators).

15 15

30

36

30

number of ways

0 1 2 3 4q1

(# quanta in first atom)

28

You put an ice cube into a styrofoam

cup containing hot coffee. You would probably be surprised if the ice cube got colder and the coffee got hotter.Would this be a violation of the energy principle?

1. Yes.2. No.3. The energy principle does not apply in this situation.

1 2 3 4 5

29

Consider 3 quantized oscillators (a model for 1 atom). They share 4 "quanta" of energy. One way to arrange the energy among these oscillators can be written as "400" (4 in the first oscillator, none in the others). Another is "103" (1 in the first oscillator, none in the second, 3 in the third). List all the ways you can arrange these 4 quanta of energy among the 3 oscillators; how many arrangements are there?

1. 3 ways2. 9 ways3. 12 ways4. 15 ways5. 18 ways6. 21 ways

1 2 3 4 5

30

Consider 4 quantized oscillators (a model for 1 and 1/3 atoms!). They share 2 “quanta”

of energy.

List all the ways you can arrange these 2 quanta of energy among the 4 oscillators (such as “2000”); \

how many arrangements are there?

1. 42. 73. 104. 155. 20

1 2 3 4 5

31

One system many times, or many systems once?

Make many observations of our system.In 29%

(36/126) of these we will have a 2|2

split.

Make many copies of our system (macrostate).Observe each one.In 29%

of these we will have a 2|2

split.

or

In either case, counting states for real systems involving many atoms is both tedious and impractical …

need a better way

32

Sequences of removing five numbered balls from a bag

How many sequences or permutations are there to remove the balls from the bag ?

5 ×

4 ×

3 ×

2 ×

1 = 5! = 120

Why ??

33

Permutations of three numbered balls

The number of permutations is 3 ×

2 ×

1 = 6 = 3!

For n objects, there are n! permutations.

Note 0! = 1

1 2 3

1 2

1

1

1

1

1

2

2

2

2

2

3

3

3

3

3

3

34

Making fewer distinctions

Of the five balls, imagine three are red, and two green.Say we do not care about the ball number, just the colour

sequence,

eg.There are 5! = 120 numbered sequences.There are 3! = 6 permutations of red balls, andthere are 2! = 2 permutations of green balls.Thus the number of different colour

sequences is:

5! 120 103!2! (6)(2)

= =

Check:

35

Arranging quanta among oscillators

6 objects: 4 quanta and 3 −

1 = 2 boundaries.

Number of arrangements =

=

=

2 1 1

1 2 1

6! 720 154!2! (24)(2)

= =

36

A general formula for the number of microstates

Number of microstates with q quanta divided amongst N oscillators:

For q

= 100 and N

= 300 , Ω

= …

= 1.7 ×

1096

Probability that all 100 quanta are on oscillator number 3 isP(0, 0, 100, 0, . . . , 0) = 1/ Ω ≈ 0.6 ×

10−96

… not impossible, but very

unlikely!

( )( )

1 !! 1 !

q Nq N+ −

Ω =−

… gets very big very fast!

37

How do we use the formula?

Program syntax Program

combin(a, b) Vpython

combin(a, b) Excel

exp(gammaln(a+1) + gammaln(b+1) + gammaln(a-b+1))

Matlab/Octave

a = q + N -

1b = q

a-b = N -

1

38

Noticing that is much smaller than other terms for large n

…

( )12! 2 n nn n n eπ −=

( ) ( )12ln ! ln 2 lnn n n n nπ= + −

( )12 ln 2 nπ

One form of Stirling’s theorem is that for large n

If n

is very large we can simplify this by taking logs

…we can simply further as

Stirling’s Theorem

( )ln ! lnn n n n= −

… is needed to handle large numbers computationally.

39

Thermal equilibrium of two blocks in contact

Blocks insulated from environment, but can exchange energy between them, whereq1

+ q2

= 100.

300oscillators

(100 atoms)

200oscillators

(~67 atoms)

How will the energy be distributed?What principle will govern this distribution?Does it matter what the initial distribution is?

Some important questions …

M&I11.3

Take two blocks of the same material …

40

Distribution of quanta for two block system

q1 q2

= 100 –

q1

Total

0 100 1 2.772 E+81 2.772 E+81

1 99 300 9.271 E+80 2.781 E+83

2 98 4.515 E+04 3.080 E+80 1.391 E+85

3 97 4.545 E+06 1.016 E+80 4.619 E+86

4 96 3.443 E+08 3.331 E+79 1.147 E+88

… … … … …

( )( )

11

1

300 1 !! 300 1 !

qq+ −

Ω =−

( )( )

22

2

200 1 !! 200 1 !

qq

+ −Ω =

− 1 2Ω Ω

41

Histogram of Ω

for two block system

Number of ways of distributing 100 quanta of vibrational

energy

between two blocks having 300 and 200 oscillators respectively. q1

is the number of quanta in 1st (larger) block

1 2Ω Ω

q1

42

Ω

for two block system

Most probable (peak of histogram) is to have 60 quanta in first block.Note 60/100 = 3/5 = 300/(300 + 200). Nice and plausible.Number of ways to have 60 quanta in first block = 7 ×

10114

Number of ways to have 0 quanta in first block = 2.772 ×

1081

Close to

zero chance to have 0 quanta in first block.

Note … … we haven’t said anything about what the starting distribution was.

43

… distribute 1000 quanta amongst two blocks containing 3000 and 2000 oscillators …

Fractional width of distribution depends on larger of or .1 N

1 q

Width of the distribution

q1

1 2Ω Ω

44

Peak narrows as q

increases

45

For large systems most probable is the only one

For macroscopic objects with 1023

atoms or more, the peak is

very narrow …

a spike. The most probable distribution is very, very much more probable than any other.

The most probable distribution is the only real possibility.

In macroscopic systems, large fluctuations from the most probable are very

rare.

46

Two objects share a total energy E

= E1

+E2

. There are 10 ways to arrange an amount of energy E1

in the first object and 15 ways to arrange an amount of energy E2

in the second object. How many different ways are there to arrange the total energy E

= E1

+E2

so that there is E1

in the first object and E2

in the other?

1. 102. 153. 254. 1505. 1E15

1 2 3 4 5

47

A system of 300 oscillators contains 100 quanta of energy. What is the physical meaning of this model?

1) one atom oscillating in 300 dimensions2) 300 atoms, each in the 100th energy level3) 300 atoms with 100 joules of energy distributed among them4) 100 atoms with 300 joules of energy distributed among them5) 100 atoms with joules of energy among them6) something else

1 2 3 4 5

100 sk m

48

Which arrangement is most probable?

1. A2. B3. C4. D5. They’re equally probable

1 2 3 4 5

49

Entropy and equilibrium

At thermal equilibrium energy is distributed in the most probable manner, and there are no big fluctuations.

Why is equilibrium attained?

… has to do with the number of microstates.

Entropy will be a measure of this.

What is the link between temperature and entropy and energy transfer?

50

Approach to thermal equilibrium

If the initial energy distribution between two systems in thermal contact is not

the most probable

energy distribution, then energy will be exchanged until the most probable distribution is reached.

51

Two smallish systems in thermal equilibrium

Number of ways of distributing 100 quanta of vibrational

energy

between two blocks having 300 and 200 oscillators respectively. q1

is the number of quanta in 1st (larger) block

q1

1 2Ω Ω

We can’t see what is going on here (it’s not zero here).

52

Use of (natural) logarithms

Natural logarithms …

If y

= ex

then x = loge

y

= ln

y

… then …

y

= ex

… then …

y

= ln

x

Useful property of logarithms: ln

(ab) = ln a + ln

b

dy ydx

=

1dydx x

=

53

ln

Ω

for 100 quanta over (300, 200) system

Take

ln

to get this

54

Entropy

Entropy is proportional to the logarithm of the number of microstates

where Boltzmann’s constant k

= 1.4 ×

10−23

J K−1.

Entropy is a thermodynamic quantity (see later).

In equilibrium, the most probable energy distribution

is that which maximizes the total entropy

lnS k≡ Ω

tot 1 2S S S= +

tot 1 2 1 2 1 2ln( ) ln lnS k k k S S= Ω Ω = Ω + Ω = +

dQdST

=

55

First Law of Thermodynamics (conservation of energy) E W QΔ = +

Second law of thermodynamics

If a closed system is not in equilibrium, the most probable consequence is that the entropy will increase.A closed system tends towards maximum entropy.

Vague statement: A closed system tends towards increasing disorder.

M&I11.4

56

Irreversibility

Any process in which the entropy of the Universe increases is called “irreversible”.

Any process in which the entropy of the Universe does not change is in principle “reversible”

…

… although 100% reversible processes do not exist in nature.A nearly reversible process …… a steel ball bouncing on a steel plate.

Reversible process:

Irreversible process:

system surroundings 0S SΔ + Δ =

system surroundings 0S SΔ + Δ >

57

Irreversibility …2

Does this violate the principle of conservation of energy?

58

What is temperature?

Associated with the average energy of a molecule.

Entropy allows a deeper connection between macroscopic measurements of temperature and fundamental (atomic) statistical view of matter and energy.

We will develop statistically based definition of temperature.

M&I11.5

59

Entropy of the two block systemtot 1 2S S S S= = +

At maximum S, the slope is zero…

1 2

1 1 1

0dS dSdSdq dq dq

= + =

2 1 2 1100q q dq dq= − ⇒ = −

or 1 2dq dq= −

1 2

1 2

0dS dSdq dq

∴ − =

1 2

1 2

dS dSdq dq

∴ =

(at equilibrium and maximum total entropy)1 2T T=

back to the …

Write

60

Progress towards equilibrium

Move 1 quantum from block 1 to block 2 …q1

→89 and q2

→11

S2

increases more than S1

decreases

Therefore total entropy S

increases

Temperature of block 1 must be higher than block 2 since it gives up energy on average.

Smaller the slope, the larger the temperature, i.e.

Initially 1 2θ θ<

1

1 1

1 dST dq∝

61

Definition of temperature

Magnitude of one quantum varies for different systems.

Therefore use Eint

instead of q where

Thus defineint

1 dST dE≡

int sE q k m=

int

1 ST E

∂≡∂

Actually

… we hold the volume of the system constant … (do no work on the system)

Write int

1

V

ST E

⎛ ⎞∂= ⎜ ⎟∂⎝ ⎠

62

Work (W), thermal transfer of energy (Q), and entropy (S)

For a reversible change, doing work on system does not change entropy, e.g. compressing a blockEnergy transfer due to temperature differences is “disorganised”

energy transfer, e.g. 2 blocks (different temperatures) in contact.

Work input W

alters energy levels without altering which level the system is in, and does not alter the entropy.

Raise U0

Heat (thermal transfer of energy) Q alters which state the system is in, and alters entropy S

.

63

Entropy change with small heat transfer

For small thermal transfer of energy ∆Eint

= Q,

there is an entropy change S given by .

For small Q, if T is nearly constant, then .

int

1 S ST E Q

Δ Δ= =Δ

QST

Δ =

If two objects (like the two blocks) are in thermal contact, then the entropy of one can spontaneously decrease, and entropy of the other increase, such that overall there is an increase. The initial 90 –

10 energy split spontaneously moved to thermal

equilibrium with a 60 –

40 split. This was irreversible. Chance that the 60 –

40 energy split goes back to 90 –

10 is so

small we say it never

happens.

64

Inside an insulated container two aluminum blocks, 1 kg and 2 kg, have been in contact for a long time. What physical property is the same for the two blocks?

1. the mass2. the temperature3. the volume4. the thermal energy5. the weight

1 2 3 4 5

65

The slope of a graph of entropy vs. energy (dS/dE) in a metal block is related to the temperature of the block. From the graph of entropy for two blocks in contact, we see that the block with the larger slope tends to gain energy from the block with the smaller slope. Therefore, which of these statements is true?

1. Big dS/dE

means high temperature2. Small dS/dE

means high temperature

1 2 3 4 5

66

There is thermal transfer of energy of 5000 J into a system. The entropy of the system increases by 50 J K-1.What is the approximate temperature of the system?

1. 5000 K2. 100 K3. 50 K4. 0.01 K5. 0.0002 K

1 2 3 4 5

67

Initially the entropy of object A is 100 J K-1, and the entropy of object B is also 100 J K-1. Then both objects are immersed in large vats of hot water.When the thermal energy of A has increased by 1000 joules, its entropy is 200 J K-1.

When the thermal energy of B has

increased by 2000 joules, its entropy is 300 J K-1.Which object is at a higher temperature?

1. A is at a higher temperature than B.2. B is at a higher temperature than A.3. Their temperatures are the same.

1 2 3 4 5

68

341.05 10 J s−= ×231 mol = 6 10 atoms×

Consider a 3 kg block of aluminum. One mole of aluminum has a mass of 27 grams (0.027 kg). From Young's modulus we determined that the stiffness of the interatomic

bond is 16 N m-1,

but in the Einstein model the x, y, and z

oscillations each involve 2 half-length springs, so the effective stiffness is 64 N m-1.

What is the energy in joules of one quantum of energy?

1. 1.62E-34 J2. 4.85E-34 J3. 5.11E-33 J4. 1.25E-22 J5. 3.96E-21 J

1 2 3 4 5

69

Here is a table of the number of ways to arrange energy in a certain microscopic object, as a function of the energy in the object:

E, J 4E-21 6E-21 8E-21 10E-21 12E-21 14E-21 16E-21

#ways 6 20 37 60 90 122 148

When the energy is 12E-21 J, what is the temperature? (k

= 1.38E-23 J K-1.)

1. 193.2 K2. 357.4 K3. 408.4 K4. 476.4 K5. 2114.0 K

1 2 3 4 5

70

Specific heat capacity of a solid

Hard to measure the total energy in a block. However, we can measure the change in temperature of the block as we add a known amount of energy to the block (by heating).

For an N atom system, the specific heat capacity

per atom

mheater

<< msolid

system atomsatom E NECT T

ΔΔ= =

Δ Δ

M&I11.6

71

Specific heat capacity of a solid

Dulong

and Petit (1819):Observed heat capacity C

at room temperature was 24.9 J mol-1

K-1.

Explained by classical theory:Principle of equipartition

of energy

Modelled atom as linear oscillator with 6 degrees of freedom, each with energy ½

kT

Measurement also showed that C:

Varies with temperatureDecreases to zero as T 0 KWas much smaller than 24.9 J mol-1

K-1

for beryllium,

boron, carbon, and silicon at room temperature.

72

Quantum theory allows only . Einstein (1907, quantum theory of solids) obtains agreement with

heat capacity data at all temperatures.

This agrees with data at high temperature, where energy of oscillator is much more than a quantum

… but disagreed with data at low temperature.

Classical theory: oscillator can have any energy;predicted energy for an oscillator.

Heat capacity per atom (Law of Dulong

and Petit).

Energy quantization and heat capacity

0E ω>>

0nE n ω=

3 3kTC kT

= =

kT

73

Heat capacity per atom

1. What is the heat capacity per molecule for water?

2. A 100 gram block of metal at a temperature of 20°C is placed into an insulated container with 400 grams of water at a temperature of 0 °C. What is the specific heat capacity of this metal, per gram?

3. Worked example in M&I

, p 392. “A lead nanoparticle.”

74

We have two blocks, one aluminum (Al) and one lead (Pb), each containing 6e23 atoms (one mole). The aluminum block has a mass of 27 grams, and the lead block has a mass of 207 grams. Which of the following pictures shows the blocks in the correct relative sizes?

1.

2.

3.

1 2 3 4 5

75

Initially the two blocks are at a temperature very near absolute zero (0 K). We will add 1 J of energy to the aluminum block, and 1 J of energy to the lead block, and see which block has the larger increase in temperature. We will step through a chain of reasoning using statistical mechanics to answer this question, which will let us determine whether aluminum or lead has the higher heat capacity at low temperatures.

1 2 3 4 5

76

From Young’s modulus we found that the effective stiffness of the interatomic

bond for Al is about 16 N m-1

and for Pb

is

about 5 N m-1. A mole of Al is 27 grams, and a mole of Pb

is 207 grams. Here are energy level diagrams for the quantized harmonic oscillators used in the Einstein solid. Which diagram represents Al and which represents Pb?

A B

1. A

is Al and B

is Pb2. A

is Pb

and B

is Al

3. B

is both Al and Pb

(they are the same)

1 2 3 4 5(a)

77

We add 1 J of energy to each block. Given the fact that Al has the greater energy-level spacing, which block now has the larger number of quanta of energy, q?

1. The number of quanta q

is greater in the Al2. The number of quanta q

is greater in the Pb

3. The number of quanta

q

is the same for Al and Pb

1 2 3 4 5(b)

78

What about the number N

of quantized oscillators in the two blocks?

1. N

is greater in the Al2. N

is greater in the Pb

3. N

is the same for Pb

and Al

1 2 3 4 5(c)

79

The Pb

block has more quanta corresponding to the 1 J of thermal energy. Therefore, in which block is there a larger number of ways Ω

of arranging the thermal energy?

1. The number of ways Ω

is greater in the Al.2. The number of ways Ω

is greater in the Pb.

3. The number of ways Ω

is the same in the Pb

and the Al.

1 2 3 4 5(d)

80

The Pb

block has the larger number of ways Ω

to arrange the energy. So which block now has the larger entropy S?

1. The entropy S

is now greater in the Al.2. The entropy S

is now greater in the Pb.

3. The entropy S

is the same in the Al and the Pb.

1 2 3 4 5(e)

81

Originally the temperature of the blocks was near absolute zero, with almost no thermal energy in the blocks. How many ways are

there to arrange zero energy in a block? Just 1. So what was the original entropy in a block?

1. 0 J K-1

2. 1 J K-1

3. infinite

1 2 3 4 5(f)

82

We found that after adding 1 J to each block, the entropy S

is now greater in the Pb

block. Both blocks started with zero

entropy. Therefore which block experienced a larger change in entropy ΔS?

1. The entropy change ΔS

was greater in the Al.2. The entropy change ΔS

was greater in the Pb.

3. The entropy change ΔS

was the same in the Pb

and the Al.

1 2 3 4 5(g)

83

We added the same amount of energy ΔE

= 1 J to each block, and the entropy change ΔS

was greater in the Pb

block. Which

block now has the higher temperature?

1. The temperature of the Al is now higher.2. The temperature of the Pb

is now higher..

3. The temperature of the Al and Pb

are the same.

1 2 3 4 5(h)

84

The original temperature was 0 K, and the final temperature of the Al block is higher than that of the Pb

block, so the Al

block has the larger change in temperature, ΔT. At low temperatures, which block has the greater heat capacity per atom, C

= (ΔE/ ΔT)/6e23?

1. The low-temperature heat capacity per atom of Al is greater.2. The low-temperature heat capacity per atom of Pb

is greater.

3. The low-temperature heat capacity per atom is the same for Pb

and Al.

1 2 3 4 5(i)

85

Strategy for constructing C versus T graph

Lets take aluminium as an example with N

= 200 and use 100 quanta, i.e. q

= 100.

Useful properties of Al: ks

= 16 N m-1; 1 mol = 27 grams = 6 ×

1023

atoms

Starting at T

= 0 (zero quanta added ⇒ T0

= 0, S0

= 1)

Add 1 quantum at a time:For each addition, compute entropy S1

, S2

, …

Sn

(n

= 1, 2, 3, …100)For each addition, compute Tn

= ∆E /∆S

= Eq

/ (Sn

-

Sn-1

)

The heat capacity per oscillator for each addition is: Cn

= ∆E

/∆T

= Eq

/ (Tn

–

Tn-1

)

or Cn

= ∆E

/∆T

= Eq

/ (Tn+1 –

Tn

)

23

40.027 / 6 10

sq

kE =×

86

Implementation of strategy for aluminium and lead (N = 500; q = 800)

87

The Boltzmann distribution

Many systems in nature have a small system (like a molecule) in contact with a large system (like the earth or its atmosphere).

Large system is called a “thermal reservoir”, because if energy ∆E

flows from it to a small system, and the energy of the large

system doesn’t change much.

Understanding the probability of these changes occurring provides insight into many phenomena in nature …… for example …

Why do chemical reaction rates depend on temperature?Why is there less oxygen on top of Mount Everest?

M&I11.7

88

The Boltzmann distribution …2

We want to derive a very general, useful result …… the probability, at temperature T, of finding a microscopicsystem in a state of energy ∆E above the ground state isproportional to

In general, if there are two energy levels E1

and E2

, so∆E = E2

−E1

, the populations in them are N1

and N2

where

EkTeΔ

−

( )2

2 1

1

1

2

EE kTE E kT kT

E kT

N e e eN e

Δ− −− −−= = =

89

The Boltzmann distribution …3

Consider a small system in thermal contact with a large system (a “reservoir”), with both systems isolated from their surroundings.

Systems share a fixed energy Etot

= Eres

+ E.

reservoir

smallsystem

ΩresΩ

Say that reservoir has microstates when energy is Eres

. ( )res resEΩ

… and small system has microstates when energy is E. ( )EΩ

resE E … since reservoir is so big.

Total number of microstates for combined system is .( )tot totEΩ

90

The Boltzmann distribution …4

( ) ( )( )

res res

tot tot

( )E E

P EE

Ω Ω=

Ω

Probability P(E) of finding the energy split between the reservoir and the small system so that there is energy E

in the

small system is

Take logs and multiply by k:

( )( ) ( )( ) ( )( )res res tot totln ( ) ln ln lnk P E k E k E k E= Ω + Ω − Ω

Most probable value of the energy E

to be found in the small system is zero. Why?

How fast does P(E) decrease as more energy moves into the small system?

91

The Boltzmann distribution …5

tot resE E E= −

Entropy Sres

vs

Eres

for the reservoir …

Sres

falls linearly for small butincreasing

Slope of expanded bit (straight line) is tot resE E E= −

tot resE E E= −

resE

resE

Expanded scale

res reslnS k= Ω

res reslnS k= Ω

res

res

dSdE

Line goes through where the energy in the reservoir is Etot

. res tot( )S E

Therefore write

resres res res tot

res

( ) ( ) dSS E S E EdE

= −

92

The Boltzmann distribution …6

resres res res tot

res

( ) ( ) dSS E S E EdE

= −

The entropy of the reservoir decreases as more energy E

is shifted into the small system.

But res

res

1dSdE T

= where T

is the temperature of the reservoir

Therefore res res res res res tot( ) ln ( ) ( ) ES E k E S ET

= Ω = −

Then res tottot tot

( )ln ( ) ln ( ) ln ( )S E EP E E Ek kT

= − + Ω − Ω

All terms that don’t involve E

are all constant.

93

The Boltzmann distribution …7

res tottot tot

( )ln ( ) ln ( ) ln ( )S E EP E E Ek kT

= − + Ω − Ω

Raising to e

… ln ( ) constant ln ( )E

P E E kTe e e e−Ω=

or ( ) ( )EkTP E A E e

−= Ω where A

is some constant.

This is the Boltzmann distribution

… the probability of finding E

in a small system in contact with a large reservoir.

EkTe

−is the Boltzmann factor, and

( )EΩ is the number of microstates corresponding to energy E.

94

Boltzmann factor

∆E

= 0 (ground state) gives maximum probability

∆E

> 0 probability drops exponentially

EkTe

−

95

E versus kT

If E1

<< kT

then an excited state can be populated.

If E1

>>

kT

then

an excited state is very unlikely to be populated.

At room temperature T

= 300 K1 eV40

kT

96

Boltzmann factor versus temperature

97

The surface temperature of the Sun is about 6000 K. Approximately what is the probability of finding a hydrogen atom in its first excited state (10.2 eV

above the ground state)?

k

= 1.38 ×

10-23

J K-1

1. 1e-102. 3e-103. 1e-94. 3e-95. 1e-8

1 2 3 4 5

98

At approximately what temperature would the probability of finding a hydrogen atom in its first excited state (10.2 eV

above

the ground state) be about 1%? k

= 1.38 ×

10-23

J K-1

1. 10000 K2. 15000 K3. 20000 K4. 25000 K5. 30000 K

1 2 3 4 5

99

The energy spacing for quantized oscillations in aluminum is

4 skm

≈

At approximately what temperature is there a 10% probability of finding one quantum of energy in one of these oscillators?k

= 1.38 ×

10-23

J K-1

1. 60 K2. 120 K3. 240 K4. 360 K5. 420 K

4 ×10-21

J or 0.025 eV.

1 2 3 4 5

100

Boltzmann distribution of energies in a gas

Reasonable at room temperatureEsurroundings

<< Enuclear/electronic

Boltzmann distribution also applies to gases.We will focus on ideal gases …

molecules don’t interact.

For real gases we assume very dilute, i.e. low density, few interactions

Energy of molecule in an ideal gas in the gravitational field near the Earth’s surface:

molecule trans vib rot cmE K E E Mgy= + + +

We ignore:rest energynuclear energyelectronic energy

M&I11.8

101

Probability of having energy E

T

constant (thermal equilibrium)

If temperature T

is constant everywhere in the (ideal) gas, then the probability of a molecule having a certain amt. of energy E

is proportional to:trans vib rot cm

( )K E E Mgy

kTE e+ + +

−Ω

Note that we have two systems interacting:… the molecule of interest, and… the rest of the molecules

From here on: cmy y=

cmv v=

102

Separating the various factors

Distributionof

kineticenergy

Distributionof

vibrationalenergy

Distributionof

rotationalenergy

Distributionof

positionalenergy

trans KkTe

− MgykTe

− vib EkTe

− rot EkTe

−

103

Height distribution in a gas

How does the density of the air vary with height above sea level?

What is air composed of?

78% N2

(1 mol = 28 grams)21% O2

(1 mol = 32 grams)1% argon & 0.03% CO2

1 mol of dry air ≈

0.78*28 + 0.21*32 = 28.56 ≈

29 grams

1 mol (6 ×1023

molecules) of a gas at STP

occupies 22.4 dm3

and

104

Height distribution in a gas

Energy (E = Mgy) being considered is significantly larger than a quantum of energy. E

is nearly continuous.

P

is the probability of finding a molecule between

x

& x + dx, y

& y + dy

and z

& z + dz

where dx, dy

and dz

are large compared to a molecule but small compared to the size of the system.

Easier to think of probability density functions:

( , ; , ; , )

MgykTP x x dx y y dy z z dz e dxdydz

−+ + + ∝

105

Height distribution in a gas

For an ideal gas at constant temperature T, the probability density P(y) is related to the number density n(y) …

... the number of molecules N

per unit volume V :

( )

MgykTP y e dxdydz

−∝

In fact:

( )( 0)

MgykTn y e

n y−

==

y

106

Distribution of velocities in a gas

P

is the probability, at temperature T

, of finding a molecule with velocity in the range (vx

+dvx

,

vy

+dvy

,

vz

+dvz

) .

21cm2K Mv=Write translational kinetic energy as

where 2 2 2 2cm x y zv v v v= + +

trans ( , , )

KkT

x y z x y zP v v v e dv dv dv−

∝Then

… can be written as22 211 1

22 2 ( , , )

yx zMvMv MvkT kT kT

x y z x y zP v v v e dv e dv e dv− − −⎡ ⎤⎡ ⎤ ⎡ ⎤

∝ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

107

Distribution of vx for helium at room temperature

Graph follows a Gaussian shape

A similar curve is produced for vy

and vz

.

Most probable velocity is vx

= 0 … molecule is equally likely to

move in any direction!

212 xMv

kTe−

108

Velocity probability distribution in polar coordinates

Volume element (dvx

)(dvy

)(dvz

) in rectangular coordinatesis

equivalent to

(4πv2

dv) in polar coordinates …

v

dv2 21 1

2 2 2 4Mv MvkT kT

x y ze dv dv dv e v dvπ− −

=

… a spherical shell of radius v

and thickness dv

Then write

Normalisation factor

212

32 2( ) 4

2

MvkTMP v v e

kTπ

π−⎛ ⎞= ⎜ ⎟

⎝ ⎠

Maxwell-Boltzmannspeed distribution

(in a low density gas)

109

Maxwell-Boltzmann speed distribution for helium atoms at two temperatures

212

32 2( ) 4

2

MvkTMP v v e

kTπ

π−⎛ ⎞= ⎜ ⎟

⎝ ⎠

110

Recall the use of probability density functions to describe to knowledge gained from a measurement.

Probability density functions in measurement

0

m (g)

p(m)g -1

2

2u = = 0.4 g

3.2 3.4 3.6 3.8 4.0 4.2 4.4a

6a

Probability of finding the mass between 3.3 g and 4.3 g is 1.0.Probability of finding the mass between 3.4 g and 3.5 g is ... (shaded area).Most probable value of the mass is 3.8 g .

Standard uncertainty u

= 12 (4.3 3.3) 0.4 g

6−

=

111

Distribution of speeds in helium gas at 293 K

Fraction of helium atoms, at 293 K, with speeds between 500 m s-1

and 600 m s-1.

112

Maxwell-Boltzmann speed distributions for helium at different temperatures

113

Measurement of molecular speeds

Collimating holes

Number of molecules striking various locations along drum is directly related to speed distribution inside gas.

N

is total number of atoms (molecules)∆N

is number in a particular

speed range v + dv

114

Average translational kinetic energy in a gas

Energy depends on a quadratic quantity ω2, e.g.21

trans cm2K Mv= 212s sU k x= 21

rot 2E Iω=

Average value of ω2

is denoted 2ω

2

2

22 0

0

1......2

kT

kT

e dkT

e d

ω

ω

ω ωω

ω

−∞

−∞= = =∫

∫If , average of a quadratic energy term is0kT ω 1

2kT

The number of quadratic terms in the expression for the energy is often called the “degrees of freedom.”For an ideal monatomic gas:

( )2 2 21trans 2

1 332 2x y zK M v v v kT kT⎛ ⎞= + + = =⎜ ⎟

⎝ ⎠

(integrate by parts)

115

Average speed and rms

speed

Root mean square speed: 2rmsv v≡

rms3kTvm

=

1 1 2 2 ...N v N vvN

+ +=

Sum becomes integral for continuous distribution

( )21

2

32 2

rms0

8 34 0.922 3

mvkT

m kTv v v dv vkT meπ

π π

∞−⎛ ⎞= = =⎜ ⎟

⎝ ⎠∫

2 2rms

1 1 32 2 2

mv mv kT= =

Average speed of molecules in a gas:

write or

… average speed is smaller than the rms

speed.

Calculate the average and rms

average of the following numbers:10, 20, 13, 44, 57, 62. [Average = 34.33 rms

average = 40.20]

116

Retaining a gas in the atmosphere

Why does the Moon not have any atmosphere at all?

Calculate the escape speed from the moon and compare with typical rms

speeds of helium, nitrogen and oxygen.

Mass of moon is 7 ×

1022

kg, and its radius is 1.75 ×

106

m.

rms3kTvm

=

117

Vibrational

energy in a diatomic gas molecule

Two vibrational

degrees of freedom (kinetic and potential).

0E

1 0 01E Eω= +

3 0 03E Eω= +

4 0 04E Eω= +

2 0 02E Eω= +

2 221 2

vib1 2

12 2 2 sp pE k sm m

= + +

1 2p p= for momenta

relative to the centre of mass

Therefore2 2 22 1 1 1

2 2 2 12 2 2p p m pm m m m

⎛ ⎞⎛ ⎞= = ⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠2 2

21 2vib

1 2

221 1

2 1

12 2 2

112 2

s

s

p pE k sm m

m p k sm m

∴ = + +

⎛ ⎞⎛ ⎞= + +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

118

Rotational energy in a diatomic gas molecule

A diatomic molecule can rotate around the x- or the y-

axis. We say there are two

rotational degrees of freedom.

Rotational kinetic energy levels of a diatomic molecule

22

rot 2 2yx IIE

ωω= +

22rot ,rot ,

2 2yx LL

I I= +

119

Energy in a diatomic gas molecule

( )2 2 21trans 2

1 332 2x y zK M v v v kT kT⎛ ⎞= + + = =⎜ ⎟

⎝ ⎠

221 1

vib2 1

112 2 s

m pE k sm m

⎛ ⎞⎛ ⎞= + +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

22rot ,rot ,

rot 2 2yx LL

EI I

= +

At high temperature,

At high temperature,

vib122

E kT kT⎛ ⎞= =⎜ ⎟⎝ ⎠

rot122

E kT kT⎛ ⎞= =⎜ ⎟⎝ ⎠

120

Energy levels of a diatomic molecule

121

Specific heat capacity cv (at constant volume) of a diatomic gas

222rot ,rot ,2 2 2 21 1

molecule cm2 1

1 1 1 112 2 2 2 2 2 2

yxx y z s

LLm pE Mv Mv Mv k s Mgym m I I

⎧ ⎫⎧ ⎫⎛ ⎞⎛ ⎞ ⎪ ⎪⎧ ⎫= + + + + + + + +⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎜ ⎟⎜ ⎟⎩ ⎭ ⎪ ⎪⎝ ⎠⎝ ⎠⎩ ⎭ ⎩ ⎭

molecule trans vib rot cmE K E E Mgy= + + +negligible

Contribution to cv

:12 k 1

2 k12 k 1

2 k 12 k 1

2 k12 k

122

Approximately what fraction of the sea-level air density is found at the top of Mount Everest, a height of 8848 m above sea level?k

= 1.38 ×

10-23

J K-1

1. 0.052. 0.13. 0.34. 0.55. 0.7

1 2 3 4 5

123

What is the rms

speed of a nitrogen molecule in this room?

1. 50 m s-1

2. 100 m s-1

3. 300 m s-1

4. 500 m s-1

5. 700 m s-1

1 2 3 4 5

124

In a vacuum, how high would an object go if thrown upward with initial speed 500 m s-1

?

1. 100 m2. 1200 m3. 2600 m4. 13000 m5. 25000 m

1 2 3 4 5