Problem Set 5 Solutions - McQuarrie Problems 3.20 MIT Dr. Anton Van Der Ven

Transcription

1 Prblem Set 5 Slutns - McQuarre Prblems 3.0 MIT Dr. Antn Van Der Ven Fall Fall Prblem 3-4 We have t derve the thermdynamc prpertes f an deal mnatmc gas frm the fllwng: = e q 3 m = e and q = V s the parttn functn fr the grand canncal ensemble, where T, V, are fxed. The characterstc ptental fr the grand canncal ensemble s the grand canncal ptental (snce E = T S pv + ) The thermdynamc prpertes fr the grand canncal ensemble are: S = p = V = but = e q s, 3 3 m 5 m = q = V e = () V exp h h h = E T S = pv d = SdT pdv d The grand canncal ptental s related t accrdng t Startng wth, =,V T, T,V ln (see table 3-1) and p 3 m = = V exp T,V 3 m p = V = exp T, Puttng thse tgether we can get the deal gas equatn f state, namely p = V w S, m 5 m S = = k ( ) Ve +( ) V e h h,v 1 h h ((3-8-1))

2 But frm (3-8-1) we can get m m S = k Ve Ve h T h [ 3 ] 5 m S = k Ve T h }{{} 5 S = k ((3-8-)) T e = V m 3 [ 3 ] m V = ln h Puttng ths nt (3-8-) we get [ 3 ]) { [ 3 ]} 5 m V m V S = k + k ln = k ln e 5 + ln h h [ 3 ] m V 5 S = k ln e G S = G V = G = p,t G s related t the sthermal-sbarc parttn functn accrdng t (see Table 3-1): [ ] (m) 3 () 5 G = ln = ln ph 3 1 Ths s the same expressn as that btaned n the canncal ensemble (see Chapter 5). Ths s due t the equvalence f ensembles when s very large. Prblem 3-10 h We are dealng wth the sthermal-sbarc ensemble ths tme, wth the parttn functn fr an deal mnatmc gas gven t us n the prblem as [ ] 3 5 (m) () = ph 3 The sthermal-sbarc s fr fxed (, T, P ). The characterstc ptental fr ths ensemble s the Gbbs free energy The thermdynamc prpertes are: h G = E T S + pv dg = SdT + V dp + d p, p T,

3 Startng wth V, G = ln ( m) ln ( ) ln h G V = = p whch s the deal gas equatn f state. w fr : [ ] 3 G (m) () 5 = = ln ph3 And nw fr S, [ 3 ] m = ln + ln p }{{} ln p [ ] 3 5 G (m) () 5 S = = k ln + k ph 3 p, [ 3 ] m V 5 S = k ln + k ln e h [ 3 ] m 5 V S = k ln e When lkng at fluctuatns, we derved [ ] 1 E E P (E) exp C v C v Fr an deal mnatmc gas (derved explctly n chapter 5) 3 E = E 3 C v = = k 4 We are nw asked what s the prbablty that the -partcle system wll sample an energy that dffers by 10 % frm the average energy, E = 3? (We can let = A = M) E = E = 10 6 E = (E) k T = = 10 C 3 v k 4 S nw we can g back t P (E), P (E) exp p,t p h T, h These are the same expressns as btaned n the grand canncal ensemble (Prblem 3-8) and n the canncal ensemble (see chapter 5). Ths s due t the equvalence f ensembles n the thermdynamc lmt,.e s very large such that fluctuatns are neglgble. Prblem 3-1 3

4 1 11 P (E) exp Ths s an extremely small prbablty whch valdates ur earler assumptns. Prblem 3-18 Derve an expressn fr the fluctuatn f the pressure n the canncal ensemble.. We knw the pressure n state s E p = V and by defntn E E exp V p = E exp Fluctuatn s defned as =(p p) = p p p. Usng the methds develped n class: Step 1: Multply bth sdes by the parttn functn E pq = p exp Step : Get dervatve wth respect t mechancal varable s cnugate. ) E (pq) = p exp V V ) Q p E p + Q = p exp V V V ) ) E p E p exp + Q = p exp V V V E E 1 p p E E E 1 p exp + Q = exp + p exp V V V V Step 3: Dvde thrugh by the parttn functn [ ] exp E E p E 1 p 1 Q V exp p exp E E V p V V + = + Q Q Q Q p p p p + = + V V Rearrangng a lttle... [ ] p = p p = p p V V p te: V has n mmedate macrscpc nterpretatn, t must be calculate n any specfc case and depends E n the partcular spectrum f V. Ths cnclusn hlds fr all generalzed frces n the frm A = E a, where a s an extensve dsplacement cnugate t A. Hence, we cannt make an unqualfed assertn that fluctuatns n all knds f external frces wll be small. 4

5 Cmpare ths wth fluctuatns n extensve quanttes such as E, H, r whch can be expressed n terms f thermdynamc respnse varables suchs as heat capactes r cmpressbltes. Specfc calculatns f the fluctuatns n p f a perfect gas by Fwler s estmated as (p p p ) fr a cubc centmeter f gas under standard cndtns. Ths s apprxmately mlecules n the gas. 1938) 1 n 3, where n s the number f (Surce: The Prncples f Statstcal Mechancs, Rchard C Tlman, Oxfrd Unversty Press, frst edtn Prblem 3-4 Shw that H H = C p n an, p, T ensemble., P, T fxed means we are wrkng n the sthermal-sbarc ensemble.. The parttn functn n ths ensemble s = V H = (E E pv e Where the E s are the energes f the system when t has vlume V. We als remember that H = E + pv. Usng the methds develped n class: r Step 1: Multply bth sdes by the parttn functn V, E pv + pv ) e Step : Get the temperature dervatve at cnstant (, P ) (The cnugate varable t H n ths case) H 1 E pv +H (E + pv ) e = 1 (E + pv ) E pv e,p V, V, Step 3: Dvde thrugh by the parttn functn H 1 V, (E + pv ) e E pv 1 V, (E + pv ) E pv e + H =,P } {{} }{{ } H but we knw,p = C. p S, H H H = H = C p H H 5,P H

6 Prblem 3-6 V p Shw that =.. V,T Frm Gbbs-Duhem we have At cnstant T, we then get V We can use the chan rule and get:,t SdT V dp + d = 0 d V = dp d d d = dp d V,T dp V,T Usng partal dervatve manpulatn d d dv = dp V,T dv P,T dp,t dv But fr a sngle cmpnent system = the mlar vlume = V and we get d p,t d dv = dp V,T V dp,t Puttng ths all tgether we get d V dp = d dv V,T,T Prblem h Shw that gven n table 4-1 s very large fr electrns n metals at T = 300 K. V 1m Take a-metal havng the fllwng prpertes 10 - stable n the bcc crystal structure wth lattce cnstant a = m - tw a atms per bcc unt cell - number f valance electrns per a atm = 1 - valance electrns n a can be cnsdered nearly free S we can get the fllwng values t substtute nt the rgnal equatns: = V ( ) 3 h = J s J k = K m e = kg Puttng thse all tgether we get 6 h 3 = 154 >> 1 V 1m Therefre Bltzmann statstcs cannt be appled t electrns n metals. Must use Ferm-Drac statstcs. 6

7 Prblem 4-6 n 1 n S = x 1 x wth n 1 and n =0, 1, and. The means wth the restrctn that n 1 + n =. {n } n1 n n 1 n Let s cnsder x x fr several values f. {n } 1 =0 {n } =0 pssble cmbnatns f n 1 and n are 0 and 0. = = 1 =1 pssble cmbnatns f n 1 and n are {n } n 1 n n 1 n = x x = x 1 + x {n } = pssble cmbnatns f n 1 and n are n 1 n = x1 x = x + x1x + x 1 {n } n 1 n = x x = x x + x x n 1 n = x x = x x n 1 n Fr >4 x 1 x = 0 because n 1 + n 4 {n } Puttng everythng tgether we get n 1 1 n =3 pssble cmbnatns f n 1 and n are n 1 n {n } n 1 n x 1 x = 1+ x 1 + x + x + x1x + x 1 + x1 x + x1x + x1 x =4 pssble cmbnatns f n 1 and n are n 1 n 4 {n } {n } 1 1 7

8 w lets cnsder S = 1+ x k + x 1 + x + x x 1 + x 1 x + x S = 1+ x 1 + x 1+ x + x = 1+ x + x 1x + x 1 n1 Ths expressn cntans the exact same terms as that btaned wth x x k=1 k {n } 1 n. 1 + x 1x Prblem 4-8 We need t shw ths ( remember the upper (lwer) sgns s fr Ferm-Drac (Bse-Ensten) : S = k [n ln n (1 n ) ln (1 n )] Start wth the parttn functn and g frm there. Becuase f the equvalence f ensembles n the thermdynamc lmt, we can calculate the entrpy usng the ensemble that ffers the mst mathematcal cnvenence. Fr Fermns ln r Bsns, ths s the grand canncal ensemble. S = k ln+ ε [ ] e ε ( ε ) ln exp = ε = ε 1 e 1 e ε e ε ε S = k ln 1e + 1 e ε S = k { ln u n ln v} = k { ln u n ln n n ln u} 1 S = k {1 ( n ) ln u n ln n } = k ( 1 n ) ln n ln n u Rememberng that u v = 1 S = k { ( 1 n ) ln (1 n ) n ln n } = 1+ e V, ε 1 ε ln = ln 1 + e T make ths a lttle easer t manpulate we can wrte ths shrthand by makng the fllwng substtutns. Let: S we nw have ε u = 1e and v = e u = 1 v and u v = 1 v n = u ε S = k {n ln n (1 n ) ln (1 n )} 8

9 Prblem 4-1 -dstngushable ndependent partcles, each f whch can be n state +ε r -ε. + = number f partcles wth energy +ε = number f partcles wth energy ε wth + + = The ttal energy s gven as: E = + ε ε = + ε ε We nw have t evaluate the parttn functn Q by summng exp E ver levels and cmpare t t the result Q = q.. We knw E Q = exp where labels a state n whch the system can resde. Instead f summng ver the states that the system can be n, we can als sum ver the pssble energy levels, makng sure we take accunt f the degeneracy f each energy level,.e. E Q = ( E) exp where (E) s the number f states wth energy E. Fr ths system we have already stated the allwed energy levels are where + can vary frm 0. Q = e ε! ε + e + ( + )! f we let x = e ε and rewrte usng the bnmal expansn we have Q = e ε 1 + e ε r Q = e ε + e ε E E = + ε ε Fr each allwed energy level E, there are (E) pssble states cmpatble wth ths energy. Snce the partcles are dstngushable,!! (E) = = +!! + ( + )! Ths represents the number f ways that + partcles ut f can be n the + state. S nw we can wrte! Q = ( E) exp ( E ) = exp [ ( + ε ε )] Snce E s unquely a functn f + + ( + )! E + Ths can be rewrtten n a ncer frm f we remember the bnmal expansn S we have n n! x n1 (1 + x) n = n1!( n n 1 )! + w we need t cmpare ths t Q = q, where q s the sngle partcle parttn functn defned as n 1 =0 9

10 q = e l ε l where l labels the sngle partcle states. In ths example there are tw sngle partcle states, wth energy -ε and + ε. and = q = e ε + e ε ε Q = q = e + e ε E The last part f ths prblem asks us t calculate and plt the heat capacty fr ths system. We knw C v = and E s gven as and C v s then ε ln Q ln e ε + e E = = ε e ε ε ε ε e e e ε E = ε = ε e + e ε eε + e ε E = ε tanh ( ε ) Plttng Cv vs. k ε C v = E ε ε = k sech 10

11 Prblem 5-4 Calculate the entrpy f e at 300K and 1 atm. The entrpy f an deal gas (eqn 5-0): [ 3 ] 5 m Ve S = k ln h te ths s neglectng electrnc exctatns (see Chapter 5). Sme data: m = kg k = J k h = J s p = 1 atm = P a Puttng that tgether we can get and V = = p 3 m = h S = k ln [ ] S = k ln where = the number f e atms. We are nw asked t estmate the translatnal degeneracy -Frm ur study f fluctuatn thery, we fund that the fluctuatn n energy f a thermdynamc system (wth very large) s exceedngly small. -Therefre, the energy f the gas s essentally always very clse t E (see dscussn n page 63 f McQuarre) and we can use the expressn fr the entrpy n the mcrcanncal ensemble. S = k ln where s the degeneracy at fxed energy E. Cmpare ths wth [ ] S = k ln and we get 7 = whch makes sense snce t shuld be large because s n the rder f 10. Prblem 5-9 What s the DeBrgle wavelength f Arÿ at 98K? h = m Use: m = kg k = J k h = J s 11 1

12 and we get = m w cmpare ths wth the nter-atmc dstance The vlume per Ar atm s V = p wth p = Pa. S V = V 3 m. The nteratmc dstance = >>. (See page 83 n the relevance f ths result.) 1

13 Prblem Set 5 Addtnal Prblems Slutns 3.0 MIT Dr. Antn Van Der Ven Fall Fall Prblem 1 (a) A system f nn-nteractng partcles wth tw pssble states ether 0 r ε. A gd rule s t assume partcles (e.g. atms, electrns, etc) are ndstngushable unless they are lcalzed n a crystal r n a surface. The number f atms n the excted state can be determned usng Bltzmann statstcs under the assumptn that we are wrkng at hgh temperature and/r lw densty:: ε = ε where ε s the prbablty an atm wll be n state ε. Ths prbablty s determned usng the sngle partcle parttn functn and can be wrtten as ε exp ε = ε; (McQuarre 4-14) exp But ur system can be n nly tw states, s the sum n the denmnatr can be fund explctly: ε exp ε = ε exp [0] + exp S ε can be wrtten as ε exp ε = ε = = ε ε 1 + exp 1 + exp ε = ε 1+exp[ ] (b) The ttal energy s smply U = ε = ε (McQuarre 4-1 and 4-13) U = ε = ( 1ε 1 + ε ) = ) ε exp U = ( ε ) = ε ε 1 + exp U = ε ε 1+exp[ ] [(1 ) 0 + ε] te: Snce many partcles wll ccupy the same state (ether 0 r ε) these partcles must be Bsns. At lwer temperatures we wuld have t use Bse-Ensten statstcs (McQuarre 4-6) whch wuld lead t a much mre cmplcated prblem snce we wuld have t determne the chemcal ptental. 1

14 Prblem The magnetzatn s gven as M = n Snce the partcles are nn-nteractng, the energy at, = cnstant and H = 0 s cnstant. E s ndependent f the number r arrangement f up versus dwn spns. Snce the abslute scale f energy s nt mprtant fr thermdynamcs, we can arbtrarly set the cnstant energy equal t zer gvng us [ ] = exp [ M state H] = exp n H = exp [ n H ] states n 1,n,...n =1 n 1,n,...n =1 We can evaluate the sum snce n = 1 s, =1 Ths s bascally sayng that we have atms lcalzed n a crystal and the magnetc mment at each ste can be ether up r dwn. The prblem asks us t determne the thermdynamc prpertes as a functn f T,, H. Let us als assume fr smplcty that we can wrk at cnstant vlume. Therefre, ur cntrllng varables are T,, V, H. We need t make the apprprate Legendre transfrm t the entrpy. Remember the entrpy can be wrtten startng frm E: E = TS pv + HM + rearrangng t get thngs n terms f S and we get S = E + pv HM k Legendre transfrm such that ur cntrllng varables are V,, T, H S E + HM = ( TS E + HM ) = = ln k where s the characterstc ptental fr ths ensemble wth V,,T,H cnstant and s the parttn functn. can be wrtten as t = states +1 = exp [ n H ] = exp [ n H ] n 1,n,...n =1 =1 n = 1 S = V,,H M = exp [ ( E state M state H)] where we sum ver all pssble energy states and magnetzatns M state. = (exp [ H ] + exp [ H ]) We knw the characterstc ptental f an ensemble s related t the parttn functn fr that ensemble accrdng Furthermre, we knw frm therm that = ln = ln d = SdT pdv + d MdH whch gves us the fllwng relatnshps fr the prpertes f the system: H V,,T

15 and frm stat mech we have frm abve Lets get the entrpy, S S = k ln (exp [ = p = V,T,H V S = T,,H = ln [exp [ H ] + exp [ H ]] H (exp [ H ] exp [ H ]) H ] + exp [ H ]) + exp [ H ] + exp [ H ] w fr the magnetzatn exp [ H ] exp [ H ] M = = H exp [ H ] + exp [ H ] The energy E E = + TS + HM = 0 V,,H S = k {ln (exp [ H ] + exp [ H ]) tanh ( H )} V,,T M = tanh ( = E TS HM H ) = tanh ( H ) Hwever f yu defne a quantty called the nternal magnetc energy (whch s a quantty analgus t the enthalpy n the T, p, ensemble) yu can get E H = E H = E HM H tanh ( The last part f ths prblem asks yu t determne the behavr f the energy and entrpy as T 0. E H (T 0) = H lm S = lm k {ln (exp [ H ] + exp [ H ]) tanh ( H )} = 0 H ) ST ( 0) = 0 s n accrdance wth the thrd law f thermdynamcs. Prblem 3 (a) M = (see slutn t Prblem - Methd 1) exp [ H ] exp [ H ] M = = H exp [ H ] + exp [ H ] =1 n V,,T M = tanh ( 3 H ) = tanh ( H )

16 (b) The parttn functn fr the, V, H, T fxed ensemble s where s = states = exp [ E s M H )] s ( s We want t determne the fluctuatns n the extensve quantty M. Use the 3-step prcedure develped n class. Step 1: Multply bth sdes by the parttn functn M = M s exp [ (E s M states Because E s s always cnstant (see cmment n Prblem ) we can therefre arbtrarly set t t zer and wrte H )] ) [ )]) M = n exp H n n 1,n,...n =1 =1 Step : Get dervatve wth respect t mechancal varable s cnugate. s ) [ )]) M + M = n exp H n H H n 1,n,...n =1 =1 { ) [ )])} ) [ )]) M +M n exp H n = n exp H n H n 1,n,...n =1 =1 n 1,n,...n =1 =1 Step 3: Dvde thrugh by the parttn functn M + M = M H r we can wrte t lke ths wth: (M) = M M = 1 M H M = 1 tanh (H) H, (c) As, tanh ( H ) 1. Therefre M T 0 = wrds, the grund state wth all the spns algned has n fluctuatns. and (M ) = {1 1} = 0. In ther T 0 4

17 Prblem 4 Frm the frst and secnd law we have de = T ds pdv + HdM + d E = T S pv HM + (Euler Frm) We are wrkng wth fxed, T, V, M n ths prblem. We need t make a Legendre transfrm because T s a cntrllng varable nstead f E. Frm the dfferental frm f F, namely we get that We are wrkng under cnstant magnetzatn s S E = F = ln Q k Q = exp [ E states F = F H = M = n = (n + n ) =1 where n + = number f up spns and n = number f dwn spns. That means that the sum n the expressn f Q must be perfrmed ver nly thse states wth fxed M (.e. fxed n + and n ) Als the atms n the system d nt nteract, meanng that the energy s ndependent f the number and arrangement f up/dwn spns and s therefre cnstant = E. Q = exp [ E states] = exp [ E ] = exp [ E ] states wth magnetzatn M ln Q state] df = SdT pdv + HdM + d M T,V, states wth magnetzatn M where s the number f states that are cnsstent wth a magnetzatn M = (n + n ).! = n +!( n + )!! Q = exp [ E ] n +!( n + )!!! F = ln exp [ E ] = E ln n +!( n + )! n +!( n + )! usng Strlng s apprxmatn, F = E {( ln ) n + ln n + + n + ( n + ) ln ( n + ) + ( n + )} but we knw frm abve that M + M = (n + n )= ( n + ) = n + = we can substtute ths back nt ur expressn fr F t get F as a functn f M M + M + M M F = E ln ln ln 5

18 F Rememberng that H = M T,V, F 1 M M 1 H = = ln + ln + M T,V, M + M H = ln ln M + H = ln M and t s easy t shw that M = tanh ( H ) whch s the same result as n the ther ensemble f Prblems and 3. Ths s due t the equvalence f ensembles n the thermdynamc lmt. Prblem 5 (a) Grand canncal ensemble: = exp [ E ] exp [ ] = Q (, V, T ) exp [ ] =0 =0 }{{} Ths sum extends ver allstates wth ttal # f partcles = q exp [ E ] = Q (,V,T ) =! fr a ne cmpnent gas f nn-nteractng partcles. We can wrte q n clsed frm as we have dne befre as 3 m q = V S nw can be wrtten as: q qe = exp [ ] =!! =0 =0 If we nw substtute n fr q we can wrte as = e zv wth z = e m h h But f we let a = qe then we can wrte ths sum n a frm fr whch the slutn s knw, namely =0 a! a = e (b) The characterstc ptental fr the grand canncal ensemble s pv = = ln = E T S d = SdT pdv d 6 3

19 (c) ( zv ) p = = = z V T, V z = = V = zv ( ) () = (because V fxed) ) q q + e = e!! =0 =0 + = Step 3: Dvde thrugh by the parttn functn + = = 3 e m frm (b) we have that = zv wth z = h. z = V = zv T,V puttng thse tw parts tgether and elmnatng z gves us the famlar deal gas equatn p =. V We need t get () =. Fllw the 3-step prcedure. Step 1: Multply bth sdes by the parttn functn q == e! =0 q e = q e!! =0 =0 Step : Get dervatve wth respect t mechancal varable s cnugate. dvde by V t get the denstes = zv = z = V V = 1 = = pv 7

20 (d) T secnd rder n 1 ln P( ) ln P( ) = ln P( ) + () = assumng a Gaussan dstrbutn arund wth varance we get ( ) 10 1 P 7 10 = exp = e whch s very unlkely. P( ) 8