Lecture 2 – The First Law (Ch. 1)Lecture 2 – The First Law (Ch. 1)Wednesday January 9Wednesday January 9thth

•Statistical mechanics

•What will we cover (cont...)

•Chapter 1

•Equilibrium

•The zeroth law

•Temperature and equilibrium

•Temperature scales and thermometers

Reading: Reading: All of chapter 1 (pages 1 - 23)All of chapter 1 (pages 1 - 23)1st homework set due next Friday 1st homework set due next Friday

(18th).(18th).Homework assignment available on web Homework assignment available on web

page.page.Assigned problems: 2, 6, 8, 10, 12Assigned problems: 2, 6, 8, 10, 12

Statistical MechanicsWhat will we cover?

Probability and StatisticsProbability and Statistics

PHY 3513 (Fall 2006)

50 55 60 65 70 75 80 85 90 95 1000

1

2

3

4 Mean 78%Median 81%

D

C+

BB+

A

Num

ber

of s

tude

nts

Score (%)

Probability and StatisticsProbability and StatisticsProbability distribution functionProbability distribution function

0

2

4

6

8

10

12

14

16

18

0 20 40 60 80 1000

Num

ber

of s

tude

nts

Final score (%)

PHY2048 - Fall 2002458 studentsMean 63±0.5

Sigma 27.5±1.5Area 470±33

Input parameters: Quality of teacher and level of difficultyAbilities and study habits of the students

2

2

1exp

22

x xf x

Gaussian statistics:

0 1 2 3 4 5 6 7 80

20

40

60

80

100

120

PHY2048 - Fall 2002 (test 2)

Mean 3.03±0.09Sigma 3.41±0.32Area 561±75

522 students

Num

ber

of s

tude

nts

Score (out of 8) 2

2

1exp

22

x xf x

Input parameters: Quality of teacher and level of difficultyAbilities and study habits of the students

Probability and StatisticsProbability and StatisticsProbability distribution functionProbability distribution function

Gaussian statistics:

The connection to thermodynamicsThe connection to thermodynamics

1/ 2

1/ 2

1/ 2

2

8

3

m

rms

kTv

m

kTv

m

kTv

m

3/ 2

2 24 exp / 22

mf v v mv kT

kT

Maxwell-Boltzmann Maxwell-Boltzmann speedspeed distribution function distribution function

2 21 2 1 2

3 3 2 3PV Nmv N mv N NkT

Equation of state:

Input parameters:Temperature and mass (T/m)

Probability and EntropyProbability and Entropy

Suppose you toss 4 coins. There are 16 (24) possible outcomes. Each one is equally probably, i.e. probability of each result is 1/16. Let W be the number of configurations, i.e. 16 in this case, then:

1 1

1; 1tot i

i

p P p W pW

Boltzmann’s hypothesis concerning entropy:

lnBS k Wwhere kB = 1.38 × 1023 J/K is Boltzmann’s constant.

The bridge to thermodynamics The bridge to thermodynamics through through ZZ exp / ;j

j

Z E kT js represent different configurations

1/ kT

Quantum statistics and identical Quantum statistics and identical particlesparticlesIndistinguishable events

Heisenberguncertaintyprinciple

The indistinguishability of identical particles has a profound effect on statistics. Furthermore, there are two fundamentally different types of particle in nature: bosons and fermions. The statistical rules for each type of particle differ!

The connection to thermodynamicsThe connection to thermodynamics

1/ 2

1/ 2

1/ 2

2

8

3

m

rms

kTv

m

kTv

m

kTv

m

3/ 2

2 24 exp / 22

mf v v mv kT

kT

Maxwell-Boltzmann Maxwell-Boltzmann speedspeed distribution function distribution function

Input parameters:Temperature and mass (T/m)

ConsiderConsider T T 00

Energy

# of

bos

ons

1110987654321

0

Bose particles (bosons)Bose particles (bosons)

Internal energy = 0Entropy = 0

Energy

# of

fer

mio

ns

1

0

Fermi-Dirac particles (fermions)Fermi-Dirac particles (fermions)

Pauli exclusion principle

EF

Internal energy ≠ 0Free energy = 0Entropy = 0

Particles are indistinguishable

ApplicationsApplications

Insulating solid Diatomic molecular gas

Specific heats:

Fermi and Bose gases

The zeroth & first LawsChapter 1

Thermal equilibriumThermal equilibrium

System 1

System 2Heat

Pi, Vi Pe, Ve

If If PPii = = PPee and and VVii = = VVee, then system 1 and systems 2 are already in , then system 1 and systems 2 are already in

thermal equilibrium. thermal equilibrium.

Different aspects of equilibriumDifferent aspects of equilibrium

1 kg

1 kg

Mechanical equilibrium:

Pe, Ve

Piston

gas

Already in thermalequilibrium

When Pe and Ve no longer change (static) mechanical equilibrium

P, nl, Vl

P, nv, Vv

Different aspects of equilibriumDifferent aspects of equilibrium

Chemical equilibrium:

Already in thermal and mechanical equilibrium

liquid

vapor

nl ↔ nv nl + nv = const.

When nl, nv, Vl & Vv no longer change (static) chemical equilibrium

A, B & AB

Different aspects of equilibriumDifferent aspects of equilibrium

Chemical reaction:

A + B ↔ AB # molecules ≠ const.

Already in thermal and mechanical equilibrium

When nA, nB & nAB no longer change (static) chemical equilibrium

Different aspects of equilibriumDifferent aspects of equilibrium

If all three conditions are met:

•Thermal•Mechanical•Chemical

Then we talk about a system being thermodynamic equilibrium.

Question:Question:

How do we characterize the equilibrium state of a system?How do we characterize the equilibrium state of a system?

In particular, thermal equilibrium.....In particular, thermal equilibrium.....

The Zeroth LawThe Zeroth Law

AA CC CCBBa)a) b)b)

VVAA, , PPAA VVCC, , PPCC VVCC, , PPCCVVBB, , PPBB

“If two systems are separately in thermal equilibrium with a third system, they are in equilibrium with each other.”

AA BBc)c)

VVAA, , PPAA VVBB, , PPBB

The Zeroth LawThe Zeroth Law

AA CC CCBBa)a) b)b)

VVAA, , PPAA VVCC, , PPCC VVCC, , PPCCVVBB, , PPBB

“If two systems are separately in thermal equilibrium with a third system, they are in equilibrium with each other.”

•This leads to an equation of state, f(P,V ), where the parameter, (temperature), characterizes the equilibrium.

•Even more useful is the fact that this same value of also characterizes any other system which is in thermal equilibrium with the first system, regardless of its state.

More on thermal equilibriumMore on thermal equilibrium

characterizes (is a measure of) the equilibrium.

•Continuum of different mechanical equilibria (P,V) for each thermal equilibrium, .

•Experimental fact: for an ideal, non-interacting gas, PV = constant (Boyle’s law).

•Why not have PV proportional to ; Kelvin scale.

Each equilibrium is unique. Erases all information on history.

Equations of stateEquations of state•An equation of state is a mathematical relation between state variables, e.g. P, V & .

•This reduces the number of independent variables to two.

General form:General form: ff ((PP,,VV,,) = 0 or ) = 0 or = = ff ((PP,,VV))

Example:Example: PVPV = = nRnR (ideal gas law)(ideal gas law)

•Defines a 2D surface in P-V- state space.

•Each point on this surface represents a unique equilibrium state of the system.

ff ((PP,,VV,,) = 0) = 0

Equilibrium state

Temperature ScalesTemperature Scales

-300 -200 -100 0 100 200

-273.15 oCPr

essu

re

Temperature (oC)

P = aP = a[[TT((ooC)C) + + 273.15]273.15]

Gas Pressure ThermometerGas Pressure Thermometer

Celsius scale

Steam point

Ice point

LN2

0

10

20

-300 -250 -200 -150 -100 -50 0 50 100

T = aP + bValue Error

b -267.2 2.8a 19.5 0.2 = 2.2

Data Linear fit

Temperature (oC)

Pres

sure

(ar

b. u

nits

)

P T17.7 7913.8 03.63 -195.97

An experiment that I did in PHY3513An experiment that I did in PHY3513

0 100 200 300 400

Pres

sure

Temperature (K)

TT(K)(K) = T = T((ooC) + 273.15C) + 273.15

The ‘absolute’ kelvin scaleThe ‘absolute’ kelvin scale

Triple pointof water:273.16 K

Other Types of ThermometerOther Types of Thermometer

•Thermocouple: = aT + bT2

•Metal resistor: R = aT + b

•Semiconductor: logR = a blogT

How stuff worksHow stuff works

Low Temperature ThermometryLow Temperature Thermometry