Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Lecture 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 thermometersReading: 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, 12Statistical MechanicsWhat will we cover?Probability and StatisticsProbability and StatisticsPHY 3513 (Fall 2006)50 55 60 65 70 75 80 85 90 95 10001234Mean 78%Median 81%DC+BB+ANumber of studentsScore (%)Probability and StatisticsProbability and StatisticsProbability distribution functionProbability distribution function0246810121416180 20 40 60 80 1000Number of studentsFinal score (%)PHY2048 - Fall 2002458 studentsMean 63±0.5Sigma 27.5±1.5Area 470±33Input parameters: Quality of teacher and level of difficultyAbilities and study habits of the students( )( )221exp22x xf xss p� �-= -� �� �� �Gaussian statistics:0 1 2 3 4 5 6 7 8020406080100120PHY2048 - Fall 2002 (test 2)Mean 3.03±0.09Sigma 3.41±0.32Area 561±75522 studentsNumber of studentsScore (out of 8)( )( )221exp22x xf xss p� �-= -� �� �� �Input parameters: Quality of teacher and level of difficultyAbilities and study habits of the studentsProbability and StatisticsProbability and StatisticsProbability distribution functionProbability distribution functionGaussian statistics:The connection to thermodynamicsThe connection to thermodynamics1/ 21/ 21/ 2283mrmskTvmkTvmkTvmp� �=� �� �� �=� �� �� �=� �� �( )( )3/ 22 24 exp / 22mf v v mv kTkTpp� �= -� �� �Maxwell-Boltzmann Maxwell-Boltzmann speedspeed distribution function distribution function2 21 2 1 23 3 2 3PV Nmv N mv N NkTe� �= = � = =� �� �Equation of state:Input parameters:Temperature and mass (T/m)Probability and EntropyProbability and EntropySuppose 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 11; 1tot iip P p W pW= = = � =�Boltzmann’s hypothesis concerning entropy:lnBS k W=where kB = 1.38 × 1023 J/K is Boltzmann’s constant.The bridge to thermodynamics The bridge to thermodynamics through through ZZ( )exp / ;jjZ E kT= -�js represent different configurations1/ kTb =Quantum statistics and identical Quantum statistics and identical particlesparticlesIndistinguishable eventsHeisenberguncertaintyprincipleThe 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 thermodynamics1/ 21/ 21/ 2283mrmskTvmkTvmkTvmp� �=� �� �� �=� �� �� �=� �� �( )( )3/ 22 24 exp / 22mf v v mv kTkTpp� �= -� �� �Maxwell-Boltzmann Maxwell-Boltzmann speedspeed distribution function distribution functionInput parameters:Temperature and mass (T/m)ConsiderConsider T T  00Energy# of bosons11109876543210Bose particles (bosons)Bose particles (bosons)Internal energy = 0Entropy = 0Energy# of fermions10Fermi-Dirac particles (fermions)Fermi-Dirac particles (fermions)Pauli exclusion principleEFInternal energy ≠ 0Free energy = 0Entropy = 0Particles are indistinguishableApplicationsApplicationsInsulating solid Diatomic molecular gasSpecific heats:Fermi and Bose gasesThe zeroth & first LawsChapter 1Thermal equilibriumThermal equilibriumSystem 1System 2HeatPi, ViPe, VeIf 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 equilibrium1 kg1 kgMechanical equilibrium:Pe, VePistongasAlready in thermalequilibriumWhen Pe and Ve no longer change (static)  mechanical equilibriumP, nl, VlP, nv, VvDifferent aspects of equilibriumDifferent aspects of equilibriumChemical equilibrium:Already in thermal and mechanical equilibriumliquidvapornl ↔ nv nl + nv = const.When nl, nv, Vl & Vv no longer change (static)  chemical equilibriumA, B & ABDifferent aspects of equilibriumDifferent aspects of equilibriumChemical reaction:A + B ↔ AB # molecules ≠ const.Already in thermal and mechanical equilibriumWhen nA, nB & nAB no longer change (static)  chemical equilibriumDifferent aspects of equilibriumDifferent aspects of equilibriumIf all three conditions are met:•Thermal•Mechanical•ChemicalThen 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 LawAACCCCBBa)a)b)b)VVAA, , PPAAVVCC, , PPCCVVCC, , PPCCVVBB, , PPBB“If two systems are separately in thermal equilibrium with a third system, they are in equilibrium with each other.”AABBc)c)VVAA, , PPAAVVBB, , PPBBThe Zeroth LawThe Zeroth LawAACCCCBBa)a)b)b)VVAA, , PPAAVVCC, , PPCCVVCC, , 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 ;