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1 Partition functions for artificial systems Consider three systems, each with a single particle at temperature 105 K. System A has two states, one with energy 0 eV and the other with 10 eV. System B has three states, one with energy 0 eV and the other two each with 10 eV. System C has four states, two each with energy 0 eV and the other two each with 10 eV. Let ZA be the partition function for system A, etc,. . . . a) Which of the following is true? Explain your answer. i) ZA = ZB = ZC. ii) ZA = ZB != ZC. iii) ZA = ZC != ZB. iv) ZB = ZC != ZA. v) None of the partition functions are the same. Now suppose that system A has a single particle and two states, one with energy 0 eV and the other with 10 eV. System B has two distinguishable particles and each could be in one of the two states of system A. b) Is ZA = ZB in this case? Explain your answer. 2 Interstellar heat bath The molecule CN is often found in interstellar molecular clouds. This molecule has many states associated with rotational motion. Observations indicate that about 10% of all such molecules are in any single one of the first three excited states, each of which has the same energy, 4.7 × 10−4 eV, above the ground state. The remaining 70% of the molecules are in the ground state. Assuming that the molecules are in thermal equilibrium with a heat bath, determine the temperature of the heat bath. This addresses a famous issue in cosmology and is discussed in detail in P. Thaddeus, Annual Review of Astronomy and Astrophysics, vol. 10, p. 305 (1972). 3 Two level system Consider a system consisting of a single particle that could be in one of two possible states. The energies of the states are −! and !. a) Determine an expression for the partition function of the system. b) Determine an expression for the mean energy of the system. 1 c) Determine an expression for the heat capacity of the system. d) Suppose that the energies of each state are each changed by adding the same constant energy, E0. Show that this will change the partition function and the mean energy of the system according to E → E + E0. Show that it will not change the probabilities with which the system will be in either state. 4 One-dimensional harmonic oscillators: canonical ensemble Consider an ensemble of N identical distinguishable quantum harmonic oscillators. The states of a single oscillator are labeled by n = 0, 1, 2,… and these have energy E = !ω ! n + 1 2 ” . These are all in equilibrium with a bath at temperature T. a) Use the canonical ensemble formalism (partition function, etc., . . . ) to show that the mean energy for the ensemble is E = N!ω !1 2 + e−!ωβ 1 − e−!ωβ ” = N!ω !1 2 + 1 e!ωβ − 1 ” b) Determine the specific heat capacity (heat capacity per particle) of the ensemble. c) Determine the mean energy and heat capacity in the high temperature limit kT % !ω. d) Show that the mean energy in the low temperature limit (kT & !ω) is E = N!ω !1 2 + e−!ω/kT ” . e) Determine the specific heat capacity in the low temperature limit and verify that C → 0 as T → 0. This behavior of the heat capacity is also required in all cases by the third law of thermodynamics. 5 Third law of thermodynamics Suppose that a system has states, labeled s = 1, 2, 3,…, and that these all have distinct energies that satisfy E1 < E2 < E3 < …. We would like to consider whether it is possible that the lowest energy state is occupied with certainty and none of the other states are occupied with certainty. To do this we require that the probabilities satisfy p2 p1 = p3 p1 = p4 p1 = … = 0. (1) a) Suppose that Eq. (1) is true. Find the temperature of the system. b) Show that if T → 0 then Eq. (1) is true. c) Determine the entropy of the system as T → 0. 2 The result lim T→0 S = (correct answer to part c)) is the third law of thermodynamics. The consequences of this for heat capacities is discussed in section 2.20 of the text. d) Consider the thermal expansion coefficient at constant pressure of any system !∂V ∂T ” P Use one of the Maxwell relations to relate this to a derivative of entropy and then use the third law to show that the thermal expansion coefficient must approach zero as T → 0. 6 Semi-classical gas The semi-classical “particle in a box” model of a gas of distinguishable particles results in the Helmholtz free energy F = −NkT # ln V + 3 2 ln !mkT 2π!2 “$ where m is the mass of a gas molecule. a) Determine the entropy of this gas. b) Determine the chemical potential of this gas. c) Consider two ideal gases that are initially isolated. Gas A consists of molecules with a smaller mass, gas B of molecules with a larger mass. Initially each has the same volume, number of particles and temperature. The two gases are allowed to interact and can exchange particles and energy. In which direction will particles flow as the gases reach equilibrium? Explain your answer. 7 Mean values of position and momentum for a particle in one dimension Consider a particle that can move in one dimension. Let x and p denote the position and momentum of the particle respectively. Suppose that the energy of the particle is E = p2 2m + U(x) where U(x) is any potential energy. a) Show that the partition function takes the form Z = ZxZp where Zx := αx % e−U(x)βdx and Zp := αp % e−p2β/2mdp where αx and αp are constants that are independent of temperature. 3 The constants αx and αp are irrelevant for the thermodynamics that follows and can both be set equal to 1. b) The probability density for the position of the particle, regardless of its momentum is p(x) = 1 Z % e−Eβdp. Show that p(x) = 1 Zx e−U(x)β. c) The mean value of the position, regardless of momentum, is x := % x p(x)dx. Show that x = 1 Zx % x e−U(x)βdx. d) Evaluate x for a one dimensional classical harmonic oscillator. e) Consider a particle that is trapped in a vertical region 0 < y ! ∞ and which is subject to potential u(y) = mgy where m > 0. Sketch the potential and indicate its minimum. If a particle were released from rest at any y > 0, where would you expect to eventually find it? Now suppose that the particle is in contact with a heat bath at temperature T. Determine an expression for y. How does this compare to your previous expectation? Comment on this as T → 0 and T → ∞. 4 2020 HW 17 QI Z= & e-EsB slates B= kT= /8.625 ev a) So ZA= e-OR te-loevB Za= 1+e-10в ZB = e-Of +e-t1ов +e-10B Zg = 1+2e-108 Zc = е-ове-ав b =2 +2e-10B. So ZBZA ZCZA Also Zc7ZB =D ZAC ZB Cъс =DV Hee states f B ave 1st pawhicle 2″4 parhice E ZB= 1+e’01s +e-1cs te-гов bwer lower 이 =ZA2 lower upper loev upper lo we 10ev SO ZB≠ ZA upper upper 2aev 2020 Hw 17Q2 About 10% we in each of the three excited states =0 70% grand stale The fracion in each state is to= Ps = e-eB/ Then fraction in graurd state is fo= a7 1 . one exciled is fi= 0,1 So f fo 2/2-3 = – (E,-Eo)B e e-EoB/2 =- (E1-E) 1] – ΔΕ = [Ert) EA/T =) =07 kin -4.7x 10-4 D T= 8.61×105 ev/ T=2.8K a) Z= Ze-EsB S +e-GB = e ER+e-EB =D Z= e GB +e-EB = 2 cash (EB) =0 Phys 362 2020 HW17 4 2 lnz ap =ie EBte -ce-6B E =_& [e+EB_e-6B] eEBte-EB -E EB -GB Sinh EB Cosh EB DE ав де c) C = = ат ат ав and B =ET ae But aB эт кт C= KT² JB Now 2€2 gede Sinh EB coshEBJ こ E”[cosh²B-sinh²eB] E coshEB COshEB-E sinhEB sinhEB CoshEB cosh²&B But cosh²x – sinh²x = (e*+e-*)²- (ex-e-x)² 4 4 = 1 =0 2E 62 аB CoshEB =D C = + 6²1 62 4 = kT2 Cosh²eB k72 [e+6Bte-6B] 2 Here Z = e-(#+ Eo)B+ e- (E+Ea)8 = e EB e- EoB + e-EB-EuS = e- EoB [eEAte-6B] previous partion furctian So Znew = e-EoB Zold. Then Enew=- hrew = – In [e-Eostad] 2 2 B 일을 [proutgg-] = Eo- hZold = Eo +Eld 2020 HW1794 a) Z= (Zsingle) Zsingle= Σe-EB single oscillator states The single oscillator states we labeled n=0,1,2,3,. and have erergy En= ћw (n+’½). So Thus Esigle = Σe-Enp Se-t-hw (n+k)B 10 0=0 8 -hwBn = e -hwplz Σ e 0=0 geometric senes r=e-hwB -hwB/2 e 1-e-hw/B since e-hwB <1 N Z= e-kupm/e (T) Now z E- B 20 ) 2 ( [ = E = E 二 SB Snle-hw BNA] + In[(+e-mo@)-N]} hw N Σ Nhw 2 Nhw 2 + 2 GwBN/₂ – N in (i-e-ho + 2 tw e-t-hwB l-e-tu e -hw B 1-e-hwB Ite-twB l-e-twß 2E Nhw 2 f+ b) c= et 2 2 =D E= Nhw Nhw and eđ ат ав ат 2 ( KT²) JB e hwB-1 N tw ehwB =+ 2KT² (ehwB1)² C Nk () (ehmop-1)”2 2020 HW 17 04 ehwp. ehwf+1 ehwB B=T 27 к c) kT>ħw =0 hwB<< ħwß =0 e 2 1+ ħwß So N E 2thwß ~ 신 쓸KT (24tws) NKT kT>>ħw m =NKT 2020 HW 17Q4 Then C Nk C Nk KT (1+thwB-1)2 E= Ntw 2 Here tiw B»1 {1+2e Nho 1+ E = Nhws I+ze-tiwp = Nhw (+ e-hwp) TA e ym$>1 and e-twß <1 2-twB 1-e-twB aE = 2 2E ат ат ав C=NK ()* e-hw leT Nhw = LT² [e-hwB]-thw = Nhw² [e-twß KT2 e) as T- Ο C0 a) Ps P こ Z い Phys 362 2020 Hw17 05 e-Esß e EiB = e-(ES-E)3 =0 This is only possible if (Es-Ei)B =00 B=0 Thisis only possible if T-D0 b) Reversing the above, T704 B-0 00 =0 @ -(Es-E)B So if TO O PSDO P c) In gereral S = +klhz + E Now E=E S= k h{Ze-BB+ = k in je-B (ite(E-ER (2) +5/ = K{he-EB + In (. )] +E 二 K(-E1B) +kln[1 ] +/ +E [ T -DO. appraacheso P Using 6= E-TS + PV dG= dE-TdS-SdT+ PdV + vPp = (Tas-pdv) -Tds-SdT +vdP +Pай = dG= – SdT+vdp ・())d 226 226 Этар “是個小 T Thus (一) But if s-00 as T-D0 ther (-D0 as T.DO os 아이곳) D0 as T-D0 a) F=E-TS dF= dE-Tas-sdT 2020 HW 17 Q6 = Ids-PdV -Tas-sdT+ MaN =0 dF= – PaV-sdT +HdN s= NK&{T [nV +&h N AM 中N+[+ OF STVN kinnIn/MKT b) M=+()v M=-KT [InV +n =0 た c) The gas with mdlecues with smaller mass (A) has larger chemicad potential (fiom part b). So in order to reack equilibrium particles must leave this yas and ester the other, So flow from A to B. a) Zo Se-E(K )F daxdp = e-PPom e-ullt dude = Se-PPhm dp Se-ullрх Phys 362 2020 HW 17 Q7 Zp Zx b) PXx) = Z e- E(xp)Pdp こ e -u)B dp e-U(x)B Z Je-piplem dp Zp But Z= ZxZp gives e-u(x)B P(x) = Zx X= fx Plw) plx) dx xe-umß dx Zx 2020 Hw17 97 d) Here U(x) = {Mw²x² and -00xXE 0 gives: e) Zx =0 If released from rest partiche would go to y=o since Fy is in negative y direction. 1 y= Sy e-mgyPdy with Zy= Jerimgy 0 dy s so y = de May ET dyKTMg-4e-MayBo+e-Mgyß dydu MgBJ dv kT =< MgB y= ma Not the same 70 AS TO ўDO (only y=0 likzly) 7200 I -00 (all ouistances equally likely) mgB
