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 considering 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 uniform 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 population 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 temperatures it is necessary to devise strategies that populate the low energy levels of a system 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 demagnetization is based on the fact that, in the absence of a magnetic field, the unpaired electrons 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 thermodynamic 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 illustration). 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 corresponds 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 increasing 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 typical 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 introduces 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 independent 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