《物理化学》课程教材讲义（英文）16 Statistical thermodynamics（1）the concepts

Statistical16thermodynamics 1:the conceptsThedistributionofmolecularStatisticalthermodynamicsprovidesthelinkbetweenthemicroscopicpropertiesofmatterstatesanditsbulkproperties.Twokeyideasareintroducedinthischapter.ThefirstistheBoltzmanndistribution,which is used to predict the populations of states in systems at thermal16.1Configurationsandweightsequilibrium. In this chapter we see its derivation in terms of the distribution of particles over16.2 The molecular partitionavailablestates.Thederivationleadsnaturallytotheintroductionofthepartitionfunction,functionwhichisthecentralmathematicalconceptofthisandthenextchapter.Weseehowtointerpret thepartitionfunctionandhowtocalculateit inanumberof simplecases.Wethen116.1Impactonbiochemistryseehowtoextractthermodynamicinformationfromthepartitionfunction.InthefinalpartThe helix-coil transition inpolypeptidesof the chapter, we generalize the discussion to include systems that are composed ofassembliesofinteractingparticles.Verysimilarequations aredevelopedtothose inthefirstThe internal energy andpart of the chapter, but they are much more widely applicable.theentropy16.3 Theinternal energyThe preceding chapters of thispart of thetext have shown howthe energy levelsof molecules can be calculated,determined spectroscopically,and related to their16.4 The statistical entropystructures.The nextmajor step is to see how a knowledge of these energylevels canbeusedto accountfortheproperties ofmatterinbulk.Todo so,wenowintroduceThecanonical partitionfunctionthe concepts of statistical thermodynamics, the link between individual molecular16.5Thecanonicalensembleproperties and bulk thermodynamic properties.The crucial step in going from the quantum mechanics of individual molecules16.6 The thermodynamictothethermodynamics ofbulk samples is torecognizethat thelatter deals with theinformation inthe partitionaveragebehaviour of large numbers of molecules.For example, the pressure of a gasfunctiondepends on the average force exerted by its molecules, and there is no need to specify16.7Independentmoleculeswhich molecules happen to be striking the wall at any instant. Nor is it necessary toconsiderthefluctuations inthepressure as different numbers of molecules collideChecklist of key ideaswith the wall at different moments. The fluctuations in pressure are very small com-Further readingpared with the steady pressure: it is highly improbable that there will be a sudden lullFurther information 16.1:in the number of collisions,or a sudden surge.Fluctuations in other thermodynamicTheBoltzmanndistributionproperties also occur,butforlargenumbersofparticles theyarenegligiblecomparedFurther information 16.2to the mean values.TheBoltzmannformulaThis chapter introduces statistical thermodynamics in two stages.The first,theFurther information 16.3:derivation of the Boltzmann distribution for individual particles,is of restrictedTemperatures below zeroapplicability,butithastheadvantageoftakingusdirectlytoaresultofcentralimport-ance in a straightforward and elementary way.We can use statistical thermodynamicsDiscussionquestionsoncewehavededuced theBoltzmanndistribution.Then(in Section16.5)weextendExercisesthe arguments to systems composed of interacting particles.Problems

Statistical thermodynamics 1: the concepts Statistical thermodynamics provides the link between the microscopic properties of matter and its bulk properties. Two key ideas are introduced in this chapter. The first is the Boltzmann distribution, which is used to predict the populations of states in systems at thermal equilibrium. In this chapter we see its derivation in terms of the distribution of particles over available states. The derivation leads naturally to the introduction of the partition function, which is the central mathematical concept of this and the next chapter. We see how to interpret the partition function and how to calculate it in a number of simple cases. We then see how to extract thermodynamic information from the partition function. In the final part of the chapter, we generalize the discussion to include systems that are composed of assemblies of interacting particles. Very similar equations are developed to those in the first part of the chapter, but they are much more widely applicable. The preceding chapters of this part of the text have shown how the energy levels of molecules can be calculated, determined spectroscopically, and related to their structures. The next major step is to see how a knowledge of these energy levels can be used to account for the properties of matter in bulk. To do so, we now introduce the concepts of statistical thermodynamics, the link between individual molecular properties and bulk thermodynamic properties. The crucial step in going from the quantum mechanics of individual molecules to the thermodynamics of bulk samples is to recognize that the latter deals with the average behaviour of large numbers of molecules. For example, the pressure of a gas depends on the average force exerted by its molecules, and there is no need to specify which molecules happen to be striking the wall at any instant. Nor is it necessary to consider the fluctuations in the pressure as different numbers of molecules collide with the wall at different moments. The fluctuations in pressure are very small com￾pared with the steady pressure: it is highly improbable that there will be a sudden lull in the number of collisions, or a sudden surge. Fluctuations in other thermodynamic properties also occur, but for large numbers of particles they are negligible compared to the mean values. This chapter introduces statistical thermodynamics in two stages. The first, the derivation of the Boltzmann distribution for individual particles, is of restricted applicability, but it has the advantage of taking us directly to a result of central import￾ance in a straightforward and elementary way. We can usestatistical thermodynamics once we have deduced the Boltzmann distribution. Then (in Section 16.5) we extend the arguments to systems composed of interacting particles. The distribution of molecular states 16.1 Configurations and weights 16.2 The molecular partition function I16.1 Impact on biochemistry: The helix–coil transition in polypeptides The internal energy and the entropy 16.3 The internal energy 16.4 The statistical entropy The canonical partition function 16.5 The canonical ensemble 16.6 The thermodynamic information in the partition function 16.7 Independent molecules Checklist of key ideas Further reading Further information 16.1: The Boltzmann distribution Further information 16.2: The Boltzmann formula Further information 16.3: Temperatures below zero Discussion questions Exercises Problems 16

56116.1CONFIGURATIONSANDWEIGHTSThedistributionofmolecularstatesWe consider a closed system composed of N molecules. Although the total energy isconstantatE,it isnotpossibletobedefiniteabout howthat energyis shared betweenthe molecules. Collisions result in the ceaseless redistribution of energy not onlybetween themolecules but also amongtheirdifferentmodes of motion.The closestwe can cometoa descriptionofthedistribution ofenergyisto reportthepopulationofa state, the average number ofmolecules that occupy it, and to say that on averagethere are n, molecules in a state of energy e, The populations of the states remainalmost constant, but the precise identities of the molecules in each state may changeateverycollision.Theproblem weaddress inthis section is thecalculationofthepopulations ofstatesforanytypeofmolecule inanymode of motion at any temperature.Theonlyrestric-tion is that the molecules should be independent, in the sense that the total energyof the system is a sum oftheir individual energies.Weare discounting (at this stage)the possibility that in a real system a contribution to the total energy may arise frominteractions between molecules. We also adopt the principle of equal a priori prob-abilities,theassumption that all possibilitiesforthedistribution ofenergyare equallyprobable.Apriorimeans in this contextloosely'as far as oneknows.Wehavenoreasonto presume otherwise than that,for a collection of molecules at thermal equilibrium,vibrational states of a certain energy,for instance, are as likely to be populated asrotational states of the same energy.One very important conclusion that will emerge from the following analysis is thatthepopulationsofstatesdependona singleparameter,the'temperature.Thatis,statist-ical thermodynamics provides a molecular justification for the concept of temperature and some insight into this crucially important quantity.16.1ConfigurationsandweightsAny individual molecule may exist in states with energies Eo, E, .... We shall alwaystake eg, the lowest state, as the zero ofenergy (e =O),and measure all other energiesrelative to that state.To obtain the actual internal energy, U, we may have to add aconstant to the calculated energy of the system.For example, if we are considering thevibrational contributiontotheinternal energy,then wemustadd thetotal zero-pointenergyof anyoscillators in the sample(a)InstantaneousconfigurationsAt any instant there willbe ngmolecules in the state with energyEg,n, with ,and soon.Thespecification ofthesetof populationsno,np..in theform (ng,n,.Jisastatement of the instantaneous configuration of the system. The instantaneou业业站configurationfluctuateswithtimebecausethepopulations change.Wecan picturealargenumberofdifferentinstantaneousconfigurations.One,forexample,mightbe(N,o,o,...J,corresponding to every moleculebeing in its ground state.Anothermight be (N -- 2,2,0,0, ... J, in which two molecules are in the first excited state.The latter configuration is intrinsically more likely to be found than the formerbecause it can be achieved in more ways: (N,o,0,...J can be achieved in only oneFig.16.1Whereasa configurationway, but (N- 2,2,0, .. J can be achieved in N(N-1) different ways (Fig. 16.1; see{5,0,0, ... J can be achieved in only oneJustification 16.1).At this stage in the argument, we are ignoring the requirementway,a configuration [3,2,0,...f can bethat thetotal energy of the system should be constant (the second configuration hasachieved in the ten different ways showna higher energy than the first). The constraint of total energy is imposed later in thishere, where the tinted blocks representsection.differentmolecules

16.1 CONFIGURATIONS AND WEIGHTS 561 The distribution of molecular states We consider a closed system composed of N molecules. Although the total energy is constant at E, it is not possible to be definite about how that energy is shared between the molecules. Collisions result in the ceaseless redistribution of energy not only between the molecules but also among their different modes of motion. The closest we can come to a description of the distribution of energy is to report the population of a state, the average number of molecules that occupy it, and to say that on average there are ni molecules in a state of energy εi . The populations of the states remain almost constant, but the precise identities of the molecules in each state may change at every collision. The problem we address in this section is the calculation of the populations of states for any type of molecule in any mode of motion at any temperature. The only restric￾tion is that the molecules should be independent, in the sense that the total energy of the system is a sum of their individual energies. We are discounting (at this stage) the possibility that in a real system a contribution to the total energy may arise from interactions between molecules. We also adopt the principle of equal a priori prob￾abilities, the assumption that all possibilities for the distribution of energy are equally probable. A priori means in this context loosely ‘as far as one knows’. We have no reason to presume otherwise than that, for a collection of molecules at thermal equilibrium, vibrational states of a certain energy, for instance, are as likely to be populated as rotational states of the same energy. One very important conclusion that will emerge from the following analysis is that the populations of states depend on a single parameter, the ‘temperature’. That is, statist￾ical thermodynamics provides a molecular justification for the concept of tempera￾ture and some insight into this crucially important quantity. 16.1 Configurations and weights Any individual molecule may exist in states with energies ε0, ε1, . . . . We shall always take ε0, the lowest state, as the zero of energy (ε0 = 0), and measure all other energies relative to that state. To obtain the actual internal energy, U, we may have to add a constant to the calculated energy of the system. For example, if we are considering the vibrational contribution to the internal energy, then we must add the total zero-point energy of any oscillators in the sample. (a) Instantaneous configurations At any instant there will be n0 molecules in the state with energy ε0, n1 with ε1, and so on. The specification of the set of populations n0, n1, . . . in the form {n0, n1, . . . } is a statement of the instantaneous configuration of the system. The instantaneous configuration fluctuates with time because the populations change. We can picture a large number of different instantaneous configurations. One, for example, might be {N,0,0, . . . }, corresponding to every molecule being in its ground state. Another might be {N − 2,2,0,0, . . . }, in which two molecules are in the first excited state. The latter configuration is intrinsically more likely to be found than the former because it can be achieved in more ways: {N,0,0, . . . } can be achieved in only one way, but {N − 2,2,0, . . . } can be achieved in 1 –2N(N − 1) different ways (Fig. 16.1; see Justification 16.1). At this stage in the argument, we are ignoring the requirement that the total energy of the system should be constant (the second configuration has a higher energy than the first). The constraint of total energy is imposed later in this section. Fig. 16.1 Whereas a configuration {5,0,0, . . . } can be achieved in only one way, a configuration {3,2,0, . . . } can be achieved in the ten different ways shown here, where the tinted blocks represent different molecules

56216STATISTICALTHERMODYNAMICS1:THECONCEPTSFig.16.2 The 18 molecules shown here canN=18be distributed into four receptacles(distinguished by the three vertical lines)in 18! different ways.However,3! of theselections that put three molecules in thefirst receptacle are equivalent, 6! that putsix molecules into the second receptacle are3!6!514!equivalent, and so on. Hencethe numberof distinguishable arrangements is181/31615!4!.If,as a result ofcollisions,the system were to fluctuate between the configurations(N,0,0, ...] and (N - 2,2,0, .. 1, it would almost always be found in the second,more likely state (especially if N were large). In other words, a system free to switchbetween the two configurations would show properties characteristic almost exclus-ively of the second configuration.A general configuration (ng,n,...I can be achievedin W different ways, where W is called the weight of the configuration. The weight oftheconfiguration (no.n,...J is given bythe expressionComment16.1N!Moreformally,Wis called theW=(16.1)multinomialcoefficient (seeAppendix2)ng!n,!n,!...Ineqn 16.1,xl,xfactorial,denotesEquation 16.1 isa generalization of theformula W=N(N-1),and reduces to it forx(x-1)(x-2)...1,andby definition0!=1.the configuration (N-2,2,0,...].Justification16.1Theweight ofa configurationFirst, consider theweightofthe configuration (N-2,2,0,0,...J.One candidateforpromotion to an upper state can be selected in N ways.There are N-1 candidatesfor the second choice, so the total number of choices is N(N-1).However, weshouldnotdistinguishthechoice (Jack,Jill)fromthechoice(Jill,Jack)becausetheyleadtothesameconfigurations.Therefore,onlyhalfthechoicesleadtodistinguish-able configurations,and thetotal number ofdistinguishable choices is N(N-1).Now we generalize this remark.Consider the number of ways of distributingNballs into bins.The first ball can be selected in N different ways, the next ballin N -1 different ways for the balls remaining, and so on. Therefore, there areN(N-1)...1 =N! ways of selecting the balls for distribution over the bins.However,iftherearen.balls in thebinlabelled e,there wouldben!different waysin which the same balls could have been chosen (Fig. 16.2). Similarly, there aren,! ways in which the n, balls in the bin labelled e, can be chosen, and so on.Therefore,thetotal number of distinguishable ways ofdistributing theballs so thatthere are ninbin Eo,n,inbin Ey,etc.regardless ofthe order in which theballs werechosen is N!/n,!n,!...,which is the content ofeqn 16.1.Illustration16.1Calculating theweightofa distributionTo calculate the number of ways of distributing 20 identical objects withthearrangement 1, 0, 3, 5, 10, 1, we note that the configuration is [1,0,3,5,10,1] withN= 20; therefore the weight is20!9.31×108W=110!3!5!10!1!Self-test 16.1 Calculate the weight of the configuration in which 20 objects are[4.19×1010]distributed in the arrangement 0, 1, 5, 0, 8, 0, 3, 2, 0, 1

562 16 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS Comment 16.1 More formally, W is called the multinomial coefficient (see Appendix 2). In eqn 16.1, x!, x factorial, denotes x(x − 1)(x − 2) . . . 1, and by definition 0! = 1. 3! 6! 5! 4! Fig. 16.2 The 18 molecules shown here can N = 18 be distributed into four receptacles (distinguished by the three vertical lines) in 18! different ways. However, 3! of the selections that put three molecules in the first receptacle are equivalent, 6! that put six molecules into the second receptacle are equivalent, and so on. Hence the number of distinguishable arrangements is 18!/3!6!5!4!. If, as a result of collisions, the system were to fluctuate between the configurations {N,0,0, . . . } and {N − 2,2,0, . . . }, it would almost always be found in the second, more likely state (especially if N were large). In other words, a system free to switch between the two configurations would show properties characteristic almost exclus￾ively of the second configuration. A general configuration {n0,n1, . . . } can be achieved in W different ways, where W is called the weight of the configuration. The weight of the configuration {n0,n1, . . . } is given by the expression W = (16.1) Equation 16.1 is a generalization of the formula W = 1 –2N(N − 1), and reduces to it for the configuration {N − 2,2,0, . . . }. Justification 16.1 The weight of a configuration First, consider the weight of the configuration {N − 2,2,0,0, . . . }. One candidate for promotion to an upper state can be selected in N ways. There are N − 1 candidates for the second choice, so the total number of choices is N(N − 1). However, we should not distinguish the choice (Jack, Jill) from the choice (Jill, Jack) because they lead to the same configurations. Therefore, only half the choices lead to distinguish￾able configurations, and the total number of distinguishable choices is 1 –2N(N − 1). Now we generalize this remark. Consider the number of ways of distributing N balls into bins. The first ball can be selected in N different ways, the next ball in N − 1 different ways for the balls remaining, and so on. Therefore, there are N(N − 1) . . . 1 = N! ways of selecting the balls for distribution over the bins. However, if there are n0 balls in the bin labelled ε0, there would be n0! different ways in which the same balls could have been chosen (Fig. 16.2). Similarly, there are n1! ways in which the n1 balls in the bin labelled ε1 can be chosen, and so on. Therefore, the total number of distinguishable ways of distributing the balls so that there are n0 in bin ε 0, n1 in bin ε1, etc. regardless of the order in which the balls were chosen is N!/n0!n1! . . . , which is the content of eqn 16.1. Illustration 16.1 Calculating the weight of a distribution To calculate the number of ways of distributing 20 identical objects with the arrangement 1, 0, 3, 5, 10, 1, we note that the configuration is {1,0,3,5,10,1} with N = 20; therefore the weight is W = = 9.31 × 108 Self-test 16.1 Calculate the weight of the configuration in which 20 objects are distributed in the arrangement 0, 1, 5, 0, 8, 0, 3, 2, 0, 1. [4.19 × 1010] 20! 1!0!3!5!10!1! N! n0!n1!n2!

56316.1CONFIGURATIONSANDWEIGHTSIt will turn out to be more convenient to deal with the natural logarithm of theweight, ln W,rather than with the weight itself. We shall therefore need the expressionN!In W=ln=In N!-In(no!n,!n! -)no!n,in!...=In N! (ln no! + In n,!+ln n?!+...)= In NIIn n!where in the first line we have used In(x/y) = n x-In y and in the second ln xy =ln x+ln y.One reason for introducing ln Wis that it is easier to make approximations. InComment16.2particular,wecan simplify thefactorials byusing Stirling'sapproximation in theformAmore accurateform of Stirling'sIn x!=xlnx-x(16.2)approximationisThen the approximate expression for the weight isxl（2元)12x+++e-In W=(NInN-N)-Z(n,ln n;-n)=NlnN-Zn,ln n;(16.3)and is in error byless than Ipercentwhen x is greater than about 10. We dealThe final form of eqn 16.3 is derived by noting that the sum of n,is equal to N, so thewithfarlargervaluesofx,andthesecond andfourth terms inthe second expression cancel.simplified version in eqn16.2 isadequate.(b)The Boltzmann distributionWehave seen that the configuration (N-2,2,0, .. dominates (N,0,0,..., and itshould be easy to believe that there may be other configurations that have a muchgreater weight than both. We shall see, in fact, that there is a configuration with sogreat a weight that it overwhelms allthe rest in importance to such an extent that thesystem will almost always befound in it.The properties ofthe system will thereforebecharacteristic of that particular dominating configuration. This dominating config-uration can befound bylookingfor thevalues ofn,that leadto a maximum value of W.Because W is a function of allthe n, we can do this search by varying the n; and look-ingforthevaluesthat correspondtodW=o(justas in the searchfor themaximum ofany function), or equivalently a maximum value of ln W. However, there are twodifficultieswith this procedure.Thefirst difficultyisthat the onlypermitted configurations arethose corresponding to the specified, constant, total energy of the system. This requirement rules outmany configurations; (N,o,0, ...) and (N-2,2,0, ... J, for instance, have differentenergies, so both cannot occur in the same isolated system. It follows that, in lookingfor the configuration with the greatest weight, we mustensure that the configurationalso satisfies the conditionZn;s=E(16.4)Constant total energy:where E is the total energy of the system.The second constraint is that, because the total number of molecules present isalso fixed (at N),we cannot arbitrarily vary all the populations simultaneously.Thus,increasing the population of one stateby1demands thatthe population of anotherstate must be reduced by 1. Therefore, the search for the maximum value of Wis alsosubjecttothe conditionEn;=NConstant total number of molecules:(16.5)We show in Further information 16.1 that the populations in the configuration ofgreatest weight, subject to the two constraints in eqns 16.4 and 16.5, depend on theenergyofthestateaccordingtotheBoltzmanndistribution:

16.1 CONFIGURATIONS AND WEIGHTS 563 It will turn out to be more convenient to deal with the natural logarithm of the weight, ln W, rather than with the weight itself. We shall therefore need the expression ln W = ln = ln N! − ln(n0!n1!n2! · · · ) = ln N! − (ln n0! + ln n1! + ln n2! + · · · ) = ln N! −∑ i ln ni ! where in the first line we have used ln(x/y) = ln x − ln y and in the second ln xy = ln x + ln y. One reason for introducing ln W is that it is easier to make approximations. In particular, we can simplify the factorials by using Stirling’s approximation in the form ln x! ≈ x ln x − x (16.2) Then the approximate expression for the weight is ln W = (N ln N − N) −∑ i (ni ln ni − ni ) = N ln N −∑ i ni ln ni (16.3) The final form of eqn 16.3 is derived by noting that the sum of ni is equal to N, so the second and fourth terms in the second expression cancel. (b) The Boltzmann distribution We have seen that the configuration {N − 2,2,0, . . . } dominates {N,0,0, . . . }, and it should be easy to believe that there may be other configurations that have a much greater weight than both. We shall see, in fact, that there is a configuration with so great a weight that it overwhelms all the rest in importance to such an extent that the system will almost always be found in it. The properties of the system will therefore be characteristic of that particular dominating configuration. This dominating config￾uration can be found by looking for the values of ni that lead to a maximum value of W. Because W is a function of all the ni , we can do this search by varying the ni and look￾ing for the values that correspond to dW = 0 (just as in the search for the maximum of any function), or equivalently a maximum value of ln W. However, there are two difficulties with this procedure. The first difficulty is that the only permitted configurations are those correspond￾ing to the specified, constant, total energy of the system. This requirement rules out many configurations; {N,0,0, . . . } and {N − 2,2,0, . . . }, for instance, have different energies, so both cannot occur in the same isolated system. It follows that, in looking for the configuration with the greatest weight, we must ensure that the configuration also satisfies the condition Constant total energy: ∑ i ni εi = E (16.4) where E is the total energy of the system. The second constraint is that, because the total number of molecules present is also fixed (at N), we cannot arbitrarily vary all the populations simultaneously. Thus, increasing the population of one state by 1 demands that the population of another state must be reduced by 1. Therefore, the search for the maximum value of W is also subject to the condition Constant total number of molecules: ∑ i ni = N (16.5) We show in Further information 16.1 that the populations in the configuration of greatest weight, subject to the two constraints in eqns 16.4 and 16.5, depend on the energy of the state according to the Boltzmann distribution: N! n0!n1!n2!. Comment 16.2 A more accurate form of Stirling’s approximation is x! ≈ (2π) 1/2xx+ 1 – 2 e−x and is in error by less than 1 per cent when x is greater than about 10. We deal with far larger values of x, and the simplified version in eqn 16.2 is adequate

56416STATISTICALTHERMODYNAMICS1:THECONCEPTSe-pe,n(16.6a)NEe-Pewhere o≤e,≤e. .... Equation 16.6a is the justification of the remark that a singleparameter,heredenotedβ,determines themostprobablepopulationsofthestates ofthe system.We shall see in Section 16.3b that1β=(16.6b)KTwhereTis thethermodynamictemperature and k is Boltzmann's constant.In otherwords,thethermodynamictemperature is theunique parameterthatgovernsthe mostprobablepopulationsofstates ofa system atthermal equilibrium.InFurtherinformation16.3,moreover, we see thatβis a more natural measure oftemperaturethan Titself.16.2ThemolecularpartitionfunctionFromnowonwewritetheBoltzmanndistributionase-Re,(16.7)qwhere p,is thefraction ofmolecules in the statei, p,= n,/N, and q is the molecularpartitionfunction:q=e-ke[16.8] The sum in qis sometimesexpressed slightlydifferently.Itmayhappenthat several stateshave the same energy,and so give the same contribution to the sum.If,for example,g, stateshavethe same energy , (so thelevel isg-folddegenerate),we could writeZg;e-Re(16.9)-levelsiwhere the sum is now over energy levels (sets of states with the same energy), notindividual states.Example16.1Writinga partition functionWrite an expression for the partition function ofa linearmolecule (such as HCl)treated as arigid rotor.Method To use eqn 16.9 we need to know (a) the energies of the levels, (b) thedegeneracies,thenumberofstatesthatbelong to each level.Whenevercalculatingapartitionfunction,theenergies ofthelevelsareexpressed relativetoOforthestateoflowest energy.The energy levels ofarigid linear rotor were derived in Section 13.5c.Answer From eqn 13.31, the energy levels of a linear rotor are hcBJ(J+ 1), withJ = O, l, 2,.... The state oflowest energy has zero energy, so no adjustment needbe made to the energies given by this expression. Each level consists of 2J +1degenerate states.Therefore,S1eZ(2]+1)e-83hcB10+1)J=0The sum can be evaluated numericallyby supplying the value of B(from spectro-scopyorcalculation)andthetemperature.ForreasonsexplainedinSection17.2b

564 16 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS = (16.6a) where ε0 ≤ ε1 ≤ ε2 . Equation 16.6a is the justification of the remark that a single parameter, here denoted β, determines the most probable populations of the states of the system. We shall see in Section 16.3b that β = (16.6b) where T is the thermodynamic temperature and k is Boltzmann’s constant. In other words, the thermodynamic temperature is the unique parameter that governs the most probable populations of states of a system at thermal equilibrium. In Further information 16.3, moreover, we see that β is a more natural measure of temperature than T itself. 16.2 The molecular partition function From now on we write the Boltzmann distribution as pi = (16.7) where pi is the fraction of molecules in the state i, pi = ni/N, and q is the molecular partition function: q =∑ i e−βεi [16.8] The sum in q is sometimes expressed slightly differently. It may happen that several states have the same energy, and so give the same contribution to the sum. If, for example, gi states have the same energy εi (so the level is gi -fold degenerate), we could write q = ∑ levels i gie−βεi (16.9) where the sum is now over energy levels (sets of states with the same energy), not individual states. Example 16.1 Writing a partition function Write an expression for the partition function of a linear molecule (such as HCl) treated as a rigid rotor. Method To use eqn 16.9 we need to know (a) the energies of the levels, (b) the degeneracies, the number of states that belong to each level. Whenever calculating a partition function, the energies of the levels are expressed relative to 0 for the state of lowest energy. The energy levels of a rigid linear rotor were derived in Section 13.5c. Answer From eqn 13.31, the energy levels of a linear rotor are hcBJ(J + 1), with J = 0, 1, 2, . . . . The state of lowest energy has zero energy, so no adjustment need be made to the energies given by this expression. Each level consists of 2J + 1 degenerate states. Therefore, gJ εJ q = ∞ ∑ J=0 (2J + 1)e−βhcBJ(J+1) The sum can be evaluated numerically by supplying the value of B (from spectro￾scopy or calculation) and the temperature. For reasons explained in Section 17.2b, 5 6 7 5 6 7 e−βεi q 1 kT e−βεi ∑ i e−βεi ni N

56516.2THEMOLECULARPARTITIONFUNCTIONthis expression applies only to unsymmetrical linear rotors (for instance, HCl,not CO,).Self-test 16.2 Write the partition function for a two-level system, the lower state18(at energy O) being nondegenerate,and the upper state (at an energy e) doubly[q=1 + 2e-Be]degenerate.36(a)An interpretation of thepartitionfunction2eSome insight into the significance of a partition function can be obtained by considering how q depends on the temperature.When T is close to zero, the parameterβ=1/kTis close to infinity.Then everyterm except one in the sum definingq is zero0because each one has the form e*with x→ oo,The exception is the term with ,= 0(or thegotermsatzeroenergyif theground state isgo-fold degenerate),becausethenFig.16.3 The equallyspaced infinite arrayofe,/kT = 0 whatever the temperature, including zero. As there is only one survivingenergy levels used in the calculation of theterm when T=0, and its value is go, it follows thatpartition function.Aharmonic oscillator(16.10)lim q=80has the same spectrum oflevels.That is, at T= O, the partition function is equal to the degeneracy of the ground state.Comment16.3Now consider thecasewhen Tis sohighthat foreachterm inthe sume./kT0.Thesumofthe infinite series S=1+x+Becausee--1when x=0,each term inthesumnowcontributes1.Itfollowsthatthex?+..is obtained by multiplyingbothsum is equal tothenumber ofmolecular states,which in general isinfinite:sidesbyx,whichgivesxS=x+x+x+..=S-1andhence S=1/(1-x).(16.11)limg=In some idealized cases,the molecule mayhave onlyafinite numberofstates; then theupper limit ofq isequal to the numberofstates.For example, ifwewereconsideringonly the spin energylevels of a radical in a magnetic field, then there would be only10two states (m, = ). The partition function for such a system can therefore beexpected to rise towards 2 as T is increased towards infinity.Weseethatthemolecularpartitionfunctiongivesan indication ofthe numberofstatesthatarethermally accessibletoa moleculeat thetemperature ofthesystem.AtT-O,onlythe ground level is accessible and q=go. At very high temperatures, virtually all statesareaccessible,andqis correspondinglylarge.1PExample16.2EvaluatingthepartitionfunctionforauniformladderofenergylevelsEvaluatethe partition function for a molecule with an infinite number of equallyspaced nondegenerateenergylevels(Fig.16.3).Theselevelscanbethoughtofas thevibrational energylevels ofa diatomic molecule in theharmonicapproximation.8Method Weexpectthepartitionfunction to increasefrom IatT=Oandapproach0510infinityasTtoco.Toevaluateeqn16.8explicitly,notethatkTle11+x+x2+Fig.16.4 The partition function for thesystem shown in Fig.16.3 (a harmonicAnswer If the separation of neighbouring levels is g,the partition function isoscillator)asafunctionoftemperature.Exploration Plotthe partitionq=1+e-+e-2pe+.=1+e-e+(e-e)2+function ofa harmonicoscillator-e-Beagainst temperature for several values ofThisexpression isplotted in Fig.16.4:notice that,as anticipated,q rises fromI tothe energy separation How does q varyinfinity as the temperature is raised.with temperature when Tis high, in thesense thatkT(orβe<1)?

16.2 THE MOLECULAR PARTITION FUNCTION 565 0 e 2 3 e e e Fig. 16.3 The equally spaced infinite array of energy levels used in the calculation of the partition function. A harmonic oscillator has the same spectrum of levels. 0 5 10 10 5 0 q kT/ Fig. 16.4 The partition function for the system shown in Fig.16.3 (a harmonic oscillator) as a function of temperature. Exploration Plot the partition function of a harmonic oscillator against temperature for several values of the energy separation ε. How does q vary with temperature when T is high, in the sense that kT >> ε (or βε << 1)? Comment 16.3 The sum of the infinite series S = 1 + x + x2 + · · · is obtained by multiplying both sides by x, which gives xS = x + x2 + x3 + · · · = S − 1 and hence S = 1/(1 − x). this expression applies only to unsymmetrical linear rotors (for instance, HCl, not CO2). Self-test 16.2 Write the partition function for a two-level system, the lower state (at energy 0) being nondegenerate, and the upper state (at an energy ε) doubly degenerate. [q = 1 + 2e−βε ] (a) An interpretation of the partition function Some insight into the significance of a partition function can be obtained by con￾sidering how q depends on the temperature. When T is close to zero, the parameter β = 1/kT is close to infinity. Then every term except one in the sum defining q is zero because each one has the form e−x with x → ∞. The exception is the term with ε0 ≡ 0 (or the g0 terms at zero energy if the ground state is g0-fold degenerate), because then ε0 /kT ≡ 0 whatever the temperature, including zero. As there is only one surviving term when T = 0, and its value is g0, it follows that lim T→0 q = g0 (16.10) That is, at T = 0, the partition function is equal to the degeneracy of the ground state. Now consider the case when T is so high that for each term in the sum εj /kT ≈ 0. Because e−x = 1 when x = 0, each term in the sum now contributes 1. It follows that the sum is equal to the number of molecular states, which in general is infinite: lim T→∞ q = ∞ (16.11) In some idealized cases, the molecule may have only a finite number of states; then the upper limit of q is equal to the number of states. For example, if we were considering only the spin energy levels of a radical in a magnetic field, then there would be only two states (ms = ± 1 –2 ). The partition function for such a system can therefore be expected to rise towards 2 as T is increased towards infinity. We see that the molecular partition function gives an indication of the number of states that are thermally accessible to a molecule at the temperature of the system. At T = 0, only the ground level is accessible and q = g0. At very high temperatures, virtually all states are accessible, and q is correspondingly large. Example 16.2 Evaluating the partition function for a uniform ladder of energy levels Evaluate the partition function for a molecule with an infinite number of equally spaced nondegenerate energy levels (Fig. 16.3). These levels can be thought of as the vibrational energy levels of a diatomic molecule in the harmonic approximation. Method We expect the partition function to increase from 1 at T = 0 and approach infinity as T to ∞. To evaluate eqn 16.8 explicitly, note that 1 + x + x2 + ··· = Answer If the separation of neighbouring levels is ε, the partition function is q = 1 + e−βε + e−2βε + ··· = 1 + e−βε + (e−βε ) 2 + ··· = This expression is plotted in Fig. 16.4: notice that, as anticipated, q rises from 1 to infinity as the temperature is raised. 1 1 − e−βε 1 1 − x

56616STATISTICALTHERMODYNAMICS1:THECONCEPTS1.4Oq1.21.500.5110510kTIkTleFig.16.5 The partition function for a two-level system as afunction oftemperature.Thetwographs differ in thescaleofthetemperature axisto showthe approachtolasT→O and theslowapproach to2as T→ o.Exploration Consider a three-level system withlevels O,g, and 2e.Plot thepartitionfunction againstkTle.HighLowtemperaturetemperatureSelf-test 16.3 Find and plot an expression for the partition function of a systemwith one state at zero energy and another state at the energy e.[q= 1 +e-pe, Fig, 16.5]It follows from eqn16.8 and theexpressionforqderived in Example16.2 for a uni-formladder of states of spacing ,1(16.12)q=1-e-βethat the fraction of molecules in the state with energy &, ise-Pe,=(1 -e-βe)e-βe(16.13)P;=aFigure 16.6 shows how P;varies with temperature. At very low temperatures, whereqis close to 1, only the lowest state is significantly populated. As the temperature israised, the population breaks out of thelowest state,and the upper states become1.00.7βe:3.00.3progressively more highly populated. At the same time, the partition function risesq1.051.581.993.86from I and its value gives an indication of the range of states populated.The name‘partition function' reflects the sense in which q measures how the total number ofFig.16.6The populations ofthe energymolecules is distributed-partitioned-over the available states.levels of the system shown in Fig.16.3Thecorresponding expressionsforatwo-level systemderived in Self-test16.3areat different temperatures, and thee~Be1correspondingvaluesofthepartition(16.14)Po=Pi=functioncalculatedinExample16.2.1 +e-Pe1 +e-βeNotethatβ=1/kT.These functions are plotted in Fig. 16.7.Notice how the populations tend towardsExploration To visualizethe contentEequality(Po=,Pi=)asT→coAcommon erroristo supposethatallthemoleculesofFig,16.6inadifferentway,plotthefunctions Po PuPa,and psagainstkT/einthesystem will befound intheupper energy statewhen T=oo; however,we see

566 16 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS q 2 10 5 10 1.5 1.4 1.2 1 0 0.5 1 q kT/ kT/   Fig. 16.5 The partition function for a two-level system as a function of temperature. The two graphs differ in the scale of the temperature axis to show the approach to 1 as T → 0 and the slow approach to 2 as T → ∞. Exploration Consider a three-level system with levels 0, ε, and 2ε. Plot the partition function against kT/ε. Low temperature High temperature  3.0 1.0 0.7 0.3 q: 1.05 1.58 1.99 3.86 5 Fig. 16.6 The populations of the energy levels of the system shown in Fig.16.3 at different temperatures, and the corresponding values of the partition function calculated in Example 16.2. Note that β = 1/kT. Exploration To visualize the content of Fig. 16.6 in a different way, plot the functions p0, p1, p2, and p3 against kT/ε. Self-test 16.3 Find and plot an expression for the partition function of a system with one state at zero energy and another state at the energy ε. [q = 1 + e−βε , Fig. 16.5] It follows from eqn 16.8 and the expression for q derived in Example 16.2 for a uni￾form ladder of states of spacing ε, q = (16.12) that the fraction of molecules in the state with energy εi is pi = = (1 − e−βε )e−βεi (16.13) Figure 16.6 shows how pi varies with temperature. At very low temperatures, where q is close to 1, only the lowest state is significantly populated. As the temperature is raised, the population breaks out of the lowest state, and the upper states become progressively more highly populated. At the same time, the partition function rises from 1 and its value gives an indication of the range of states populated. The name ‘partition function’ reflects the sense in which q measures how the total number of molecules is distributed—partitioned—over the available states. The corresponding expressions for a two-level system derived in Self-test 16.3 are p0 = p1 = (16.14) These functions are plotted in Fig. 16.7. Notice how the populations tend towards equality (p0 = 1 –2 , p1 = 1 –2 ) as T → ∞. A common error is to suppose that all the molecules in the system will be found in the upper energy state when T = ∞; however, we see e−βε 1 + e−βε 1 1 + e−βε e−βεi q 1 1 − e−βε

56716.2THEMOLECULARPARTITIONFUNCTIONPe0.50.5p00010050.5kTIekTIeFig.16.7 The fraction of populations of the two states ofa two-level system as a function oftemperature (eqn 16.14).Notethat,as thetemperature approaches infinity,the populationsofthetwostatesbecome equal (andthefractionsbothapproach0.5).ExplorationConsider a three-level system withlevels O,,and2.PlotthefunctionsPo国Pr,andpz against kTle.from eqn 16.14 that, as T-oo, the populations of states become equal. The sameconclusion is true of multi-level systems too: as T→ oo, all states become equallypopulated.Example16.3UsingthepartitionfunctiontocalculateapopulationCalculatetheproportionof ,moleculesintheirground,firstexcited,and secondexcitedvibrational states at 25°C.Thevibrational wavenumber is214.6cm-!Method Vibrational energy levels have a constant separation (in the harmonicapproximation,Section 13.9),so thepartition function isgiven byeqn 16.12 andthe populations byeqn 16.13.To use the latter equation, weidentify the indexi with the quantum number u, and calculate P, for u = 0, 1, and 2. At 298.15 K,kT/hc=207.226cm-lAnswerFirst,wenotethathev214.6cmBe==1.036kT207.226cm-lThen it follows from eqn 16.13that thepopulations areP,=(1-e-be)e-pe=0.645e-1.036u)Therefore,Po=0.645,P1=0.229,P2=0.081.TheI-Ibond is not stiffand the atomsare heavy: as a result, the vibrational energy separations are small and at roomtemperature several vibrational levels are significantly populated.The value of thepartitionfunction,q=1.55,reflectsthis small but significantspread ofpopulations.Self-test16.4Atwhattemperature wouldthe=1levelofl,have(a)halfthepopu-lation oftheground state,(b)thesame population as theground state?[(a)445K,(b)infinite]

16.2 THE MOLECULAR PARTITION FUNCTION 567 0 0 0.5 0.5 1 1 p p0 p1 0 0.5 1 0 5 10 p p0 p1 kT/ kT/ Fig. 16.7 The fraction of populations of the two states of a two-level system as a function of temperature (eqn 16.14). Note that, as the temperature approaches infinity, the populations of the two states become equal (and the fractions both approach 0.5). Exploration Consider a three-level system with levels 0, ε, and 2ε. Plot the functions p0, p1, and p2 against kT/ε. from eqn 16.14 that, as T → ∞, the populations of states become equal. The same conclusion is true of multi-level systems too: as T → ∞, all states become equally populated. Example 16.3 Using the partition function to calculate a population Calculate the proportion of I2 molecules in their ground, first excited, and second excited vibrational states at 25°C. The vibrational wavenumber is 214.6 cm−1 . Method Vibrational energy levels have a constant separation (in the harmonic approximation, Section 13.9), so the partition function is given by eqn 16.12 and the populations by eqn 16.13. To use the latter equation, we identify the index i with the quantum number v, and calculate pv for v = 0, 1, and 2. At 298.15 K, kT/hc = 207.226 cm−1 . Answer First, we note that βε == = 1.036 Then it follows from eqn 16.13 that the populations are pv = (1 − e−βε )e−vβε = 0.645e−1.036v Therefore, p0 = 0.645, p1 = 0.229, p2 = 0.081. The I-I bond is not stiff and the atoms are heavy: as a result, the vibrational energy separations are small and at room temperature several vibrational levels are significantly populated. The value of the partition function, q = 1.55, reflects this small but significant spread of populations. Self-test 16.4 At what temperature would the v = 1 level of I2 have (a) half the popu￾lation of the ground state, (b) the same population as the ground state? [(a) 445 K, (b) infinite] 214.6 cm−1 207.226 cm−1 hc# kT

56816STATISTICALTHERMODYNAMICS1:THECONCEPTSItfollowsfrom ourdiscussion ofthepartitionfunctionthattoreachlowtempera-Magnetictures it is necessary to devise strategies that populate the low energy levels of a sys-fieldofftem at the expense of high energy levels.Common methods used to reach very lowtemperatures include optical trapping and adiabatic demagnetization.In opticaluNMuNMtrapping,atoms inthegasphaseare cooled byinelasticcollisions withphotons fromintense laser beams,which act as walls ofa very small container.Adiabatic demagne-tization isbased on the factthat, inthe absenceofa magneticfield,theunpaired elec-trons of aparamagnetic material are orientated at random, but in thepresence of aMMmagnetic field there are more β spins (m,=-) than α spins (m,=+). In thermo-Cdynamicterms,theapplicationofamagneticfieldlowerstheentropyofasampleand,atagiven temperature,theentropyofa sampleislower when thefield is onthan whenMit is off.Evenlowertemperatures can be reached ifnuclearspins (which alsobehavelike small magnets) are used instead of electron spins in the technique of adiabaticMagneticfieldonnucleardemagnetization,whichhasbeenusedtocoolasampleofsilvertoabout0Temperature,T280 pK.In certain circumstances it is possible toachieve negative temperatures,andthe equations derived later in this chapter can be extended to T< O with interestingFig.16.8 The technique ofadiabaticconsequences (see Further information 16.3).demagnetizationisusedtoattainverylowtemperatures.The upper curve shows thatlllustration16.2Coolingasamplebyadiabaticdemagnetizationvariation oftheentropyofaparamagneticsystem in the absenceofanappliedfieldConsider the situation summarized by Fig. 16.8. A sample of paramagneticThe lower curve shows that variation inentropy when a field is applied and hasmaterial, such asa d-or f-metal complex with several unpaired electrons, is cooledmade the electron magnets more orderly.to about I K by using helium. The sample is then exposed to a strong magneticThe isothermal magnetization step isfromfield while it is surrounded by helium, which provides thermal contact with theA toB;the adiabaticdemagnetization stepcold reservoir.This magnetization step is isothermal,and energy leaves the system(at constantentropy)is from Bto C.as heat whiletheelectron spins adopt thelowerenergy state (AB inthe illustration). Thermal contact between the sample and the surroundings is now brokenby pumping away the helium and the magnetic field is reduced to zero. Thisstepisadiabaticand effectivelyreversible,so the state ofthesample changes fromB to C. At the end of this step the sample is the same as it was at A except that itnow has a lower entropy.Thatlower entropyin the absence ofa magneticfield cor-responds to a lower temperature. That is, adiabatic demagnetization has cooledthe sample.(b)ApproximationsandfactorizationsIn general, exact analytical expressions for partition functions cannot be obtained.However, closed approximate expressions can often be found and prove to be veryimportant in a number of chemical and biochemical applications (Impact 16.1).Forinstance,the expressionforthepartition function fora particleofmass mfree tomovein a one-dimensional container oflengthXcanbeevaluated by making useofthefactthat the separation of energy levels is very small and that large numbers of states areaccessibleatnormal temperatures.As shownintheJustificationbelow,inthiscase2元m+(16.15)qx=h2βThis expression shows that the partition function for translational motion increaseswith thelengthof theboxand themass oftheparticle,for in each casethe separationoftheenergylevelsbecomessmallerandmorelevelsbecomethermallyaccessible.Fora given mass and length ofthebox, the partition function also increases with increas-ingtemperature(decreasingβ),becausemore states becomeaccessible

568 16 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS Magnetic field on Magnetic field off Entropy, S A C B 0 Temperature, T Fig. 16.8 The technique of adiabatic demagnetization is used to attain very low temperatures. The upper curve shows that variation of the entropy of a paramagnetic system in the absence of an applied field. The lower curve shows that variation in entropy when a field is applied and has made the electron magnets more orderly. The isothermal magnetization step is from A to B; the adiabatic demagnetization step (at constant entropy) is from B to C. It follows from our discussion of the partition function that to reach low tempera￾tures it is necessary to devise strategies that populate the low energy levels of a sys￾tem at the expense of high energy levels. Common methods used to reach very low temperatures include optical trapping and adiabatic demagnetization. In optical trapping, atoms in the gas phase are cooled by inelastic collisions with photons from intense laser beams, which act as walls of a very small container. Adiabatic demagne￾tization is based on the fact that, in the absence of a magnetic field, the unpaired elec￾trons of a paramagnetic material are orientated at random, but in the presence of a magnetic field there are more β spins (ms = − 1 –2) than α spins (ms = + 1 –2). In thermo￾dynamic terms, the application of a magnetic field lowers the entropy of a sample and, at a given temperature, the entropy of a sample is lower when the field is on than when it is off. Even lower temperatures can be reached if nuclear spins (which also behave like small magnets) are used instead of electron spins in the technique of adiabatic nuclear demagnetization, which has been used to cool a sample of silver to about 280 pK. In certain circumstances it is possible to achieve negative temperatures, and the equations derived later in this chapter can be extended to T < 0 with interesting consequences (see Further information 16.3). Illustration 16.2 Cooling a sample by adiabatic demagnetization Consider the situation summarized by Fig. 16.8. A sample of paramagnetic material, such as a d- or f-metal complex with several unpaired electrons, is cooled to about 1 K by using helium. The sample is then exposed to a strong magnetic field while it is surrounded by helium, which provides thermal contact with the cold reservoir. This magnetization step is isothermal, and energy leaves the system as heat while the electron spins adopt the lower energy state (AB in the illustra￾tion). Thermal contact between the sample and the surroundings is now broken by pumping away the helium and the magnetic field is reduced to zero. This step is adiabatic and effectively reversible, so the state of the sample changes from B to C. At the end of this step the sample is the same as it was at A except that it now has a lower entropy. That lower entropy in the absence of a magnetic field cor￾responds to a lower temperature. That is, adiabatic demagnetization has cooled the sample. (b) Approximations and factorizations In general, exact analytical expressions for partition functions cannot be obtained. However, closed approximate expressions can often be found and prove to be very important in a number of chemical and biochemical applications (Impact 16.1). For instance, the expression for the partition function for a particle of mass m free to move in a one-dimensional container of length X can be evaluated by making use of the fact that the separation of energy levels is very small and that large numbers of states are accessible at normal temperatures. As shown in the Justification below, in this case qX = 1/2 X (16.15) This expression shows that the partition function for translational motion increases with the length of the box and the mass of the particle, for in each case the separation of the energy levels becomes smaller and more levels become thermally accessible. For a given mass and length of the box, the partition function also increases with increas￾ing temperature (decreasing β), because more states become accessible. D F 2πm h2 β A C

56916.2THEMOLECULARPARTITIONFUNCTIONJustification16.2Thepartition functionforaparticle inaone-dimensionalboxThe energy levels of a molecule of mass m in a container of length X are given byeqn9.4a with L=X:nhE.n=1,2,..8mx2The lowest level (n=1) has energy h2/8mX2, so the energies relative to that level are,=(n2-1)e=h2/8mx2The sum to evaluate is therefore29x=e-(n-1)pe台The translational energylevels are very closetogether in a containerthe size ofatyp-ical laboratory vessel; therefore, the sum can be approximated by an integral:-1)edne-n'pedrTheextension of thelower limitton=O and the replacement of n?-1 by n? intro-duces negligible error but turns the integral into standard form. We make thesubstitution x= nβe, implying dn=dx/(βe)/2,and therefore thatn12axBBAnother usefulfeature of partition functions is used to derive expressions when theenergy ofa molecule arises from several different, independent sources: if the energyis a sum of contributions from independent modes of motion, then the partitionfunction is a product of partitionfunctions for each modeof motion.For instance,suppose themolecule we are consideringis freetomove inthree dimensions.Wetakethe length of the container in the y-direction to be Y and that in the z-direction to beZ.The total energy ofa molecule is the sum of its translational energies in all threedirections:En,nn,=e(X)+e()+e(2)(16.16)wheren,n,andn,arethequantumnumbersformotion in thex-y-,and z-directions,respectively.Therefore, because ea+b+c= e"ee, the partition function factorizes asfollows:q-Ze-Re-pe-pen-Ee-Beme-Reme-BegallnallnZe-pe(16.17)=qxq9zIt is generally true that, if the energy ofa molecule can be written as the sum ofindependent terms, then the partition function is the corresponding product ofindividualcontributions

16.2 THE MOLECULAR PARTITION FUNCTION 569 Justification 16.2 The partition function for a particle in a one-dimensional box The energy levels of a molecule of mass m in a container of length X are given by eqn 9.4a with L = X: En = n = 1, 2, . . . The lowest level (n = 1) has energy h2 /8mX2 , so the energies relative to that level are εn = (n2 − 1)ε ε = h2 /8mX2 The sum to evaluate is therefore qX = ∞ ∑n=1 e−(n2 −1)βε The translational energy levels are very close together in a container the size of a typ￾ical laboratory vessel; therefore, the sum can be approximated by an integral: qX = ∞ 1 e−(n2 −1)βεdn ≈ ∞ 0 e−n2 βεdn The extension of the lower limit to n = 0 and the replacement of n2 − 1 by n2 intro￾duces negligible error but turns the integral into standard form. We make the substitution x 2 = n2 βε, implying dn = dx/(βε) 1/2, and therefore that π1/2/2 qX = 1/2 ∞ 0 e−x2 dx = 1/2 = 1/2 X Another useful feature of partition functions is used to derive expressions when the energy of a molecule arises from several different, independent sources: if the energy is a sum of contributions from independent modes of motion, then the partition function is a product of partition functions for each mode of motion. For instance, suppose the molecule we are considering is free to move in three dimensions. We take the length of the container in the y-direction to be Y and that in the z-direction to be Z. The total energy of a molecule ε is the sum of its translational energies in all three directions: εn1n2n3 = ε n1 (X) + ε n2 (Y) + ε n3 (Z) (16.16) where n1, n2, and n3 are the quantum numbers for motion in the x-, y-, and z-directions, respectively. Therefore, because ea+b+c = ea eb ec , the partition function factorizes as follows: q =∑ all n e−βε n1 (X) −βε n2 (Y) −βε n3 (Z ) =∑ all n e−βε n1 (X) e−βε n2 (Y) e−βε n3 (Z) = ∑ n1 e−βε n1 (X) ∑ n2 e−βε n2 (Y) ∑ n3 e−βε n3 (Z) (16.17) = qXqYqZ It is generally true that, if the energy of a molecule can be written as the sum of inde￾pendent terms, then the partition function is the corresponding product of individual contributions. D F A C D F A C D F A C D E F 2πm h2 β A B C D E F π1/2 2 A B C D E F 1 βε A B C D E F 1 βε A B C 5 6 7 n2 h2 8mX2