Statistical Mechanics of a System with Equally Spaced Non-degenerate Energy Levels, Assignments of Chemistry

Download Statistical Mechanics of a System with Equally Spaced Non-degenerate Energy Levels and more Assignments Chemistry in PDF only on Docsity! Page 1 of 10 CHEM 156 Winter 2002 Problem Set 6 ANSWER KEY 1. The California SuperLotto PlusSM drawing selects six numbers. The first five numbers are chosen by sampling without replacement from a set of numbers 1-47. The sixth number, the MEGA number, is chosen from a separate pool of numbers 1-27. To win the jackpot prize, a ticket must match all six numbers, but does not have to match the order of the first five numbers. Purchasing a single set of numbers for a SuperLotto PlusSM drawing costs $1. A. Calculate the total number of possible outcomes for a SuperLotto PlusSM drawing. ( ) 353,416,41 27 12345 4344454647 27 !5!5-47 47! 27 5 47 number. MEGA for the outcomes ofnumber the timesnumbers fivefirst for the outcomes ofnumber theofproduct the toequal is outcomes possible ofnumber totalThe number.MEGA theof drawing theoft independen is numbers fivefirst theof drawing thenumber,MEGA than the numbers of pool separate a fromchosen are numbers fivefirst theSince MEGA##5 =Ω ⋅ ⋅⋅⋅⋅ ⋅⋅⋅⋅ =⋅ ⋅ =⋅      =Ω⋅Ω=Ω first B. Calculate the probability of winning a SuperLotto PlusSM jackpot by playing a single set of numbers. 353,416,41 1 drawing.per possible outcome winningoneonly is thereprize,jackpot the win tomatched bemust numbers 6 all Since outcomes. possible ofnumber total by the divided outcomes winningofnumber the toequal isjackpot thewinning ofy probabilit thelikely,equally is drawing theof outcomeeach that Assuming = Ω Ω = total winningp C. Calculate the cost of purchasing one of every possible set of numbers for a SuperLotto PlusSM drawing. Is it possible to make a profit by purchasing every possible set of numbers? (This has been attempted several times for other state lotteries by Australian Peter Mandral by feeding entry forms through laser printers.) ers.other winn split withnot isjackpot the thatassuming amount, than thismore worth isjackpot theifprofit a make topossible isIt 3.$41,416,35or ($1), price purchase the timesA)part (from outcomes possible of number total the toequal is numbers ofset possibleevery of one purchasing ofcost The Page 2 of 10 2. Consider a canonical ensemble with N subsystems that has a total energy of E. A. What is the probability that subsystem 1 has energy ε1 and that subsystem 2 has energy ε2? ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) ( )( ) ( )( ) ( )( ) ( )!1E !3-E !-E !E 21 !1E !3-E !-E !E !3 !1 !1!E !1E !3!-E !3-E :asgiven be y willprobabilit they,probabilit equal an occur with willstate possibleeach that Assuming systems. 2-Nother for the remaining energy 2)1(-E is there2,energy has subsystem2 and 1energy has subsystem1After t.independennot are iesprobabilit the2, subsystem toavailableenergy of pool thefrom depletes 1 subsystemin usedenergy thesince However, P(B). * P(A)event each of iesprobabilit theof product the toequal is B)&P(A occuringboth eventst independen twoofy probabilit The 21 21 21 21 21 21 , 2),( 21 −+ −++ ⋅ + ⋅−−= −+ −++ ⋅ + ⋅ − − = − −+ −+ −++ = Ω Ω = + −+− N N NNp N N N N N N N N p NE NE εε εε εε εε εε εε εεεε εε B. Calculate the probability that subsystem 1 has energy ε1 and that subsystem 2 has energy ε2 if the number of subsystems, N, is extremely large and if the total energy, E, is much, much greater than ε1 and ε2. ( )( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) 2121 21 21 21 21 21 11 1 11 1 11 1 ppp .subsystem2 to)-(E availableenergy ofamount affect the tlysignificannot doesenergy of poolcommon thefrom missingenergy 2). subsystem toavailableenergy effect thenot does subsystem1in (t independen considered are subsystems two theify probabilit theequalsresult theion,approximat Under this 11 1 EE :obtain wereader, class thefrom 311 page similar to tionssimplifica Using !1E !3-E !-E !E 21 2 21 1 1 1 2 2 2 2 2 21 21 εεεε εε εε εε εε εε ε ε εε ε εε ε ε εε ε ε ε ε ε ε εε εε + + + + ++ +       + ⋅      + =      + ⋅      + ⋅      + ⋅      + ≈⋅≈       + ⋅      + ≈           + ⋅           + =      + ⋅      + = + ⋅≈ −+ −++ ⋅ + ⋅−−= p N NE N E N N N N NE E NE N N E Np N N NNp Page 5 of 10 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) kT kT A kT kT kTkT kT V kTkT kT V kT V kT kT V kT kT kT V kT kT e e nNE e e NE kT ee e NkTE T kTee e NkTE T e T e e NkTE T e e e NkT T e e NkTE ε ε ε ε εε ε εε ε ε ε ε ε ε ε ε ε εε ε ε − − − − −−− −− −−− −− − −− −− − −− −− −− −− − =⋅= − =     −     −−      − =         ∂ −∂     −     −−      − =                       ∂     ∂ −      ∂ ∂      −−      − =           ∂      −∂      −−      − =             ∂      −∂      − = 1 moles ofnumber theisn whereNnN with or, 1 111 1 1 111 1 1 1 11 1 1 1 11 1 11 1 1 A 2 2 1 2 2 1 2 2 1 2 2 1 2 1 1 2 ( ) ( )      −− − =       − +         − = +=+=+= = += − − − − − − kT kT kT kT kT kT N eNk e e T N S e Nk T e e N S qNk T E qk T E Qk T E S Qk T E S ε ε ε ε ε ε ε ε 1ln 1 1 1 ln 1 lnlnln hable.distinguis are and movecannot molecules that theassume will wespaced)unevenly be ld(which wou levelsenergy onal translatino are thereSince .#6)question (see N!by dividing bycorrect must wegas, ain assuch molecules, ishableindistinguFor crystal. in theposition by their labelled bemay which solid ain molecules assuch molecules, habledistinguisfor trueholds This .qQ and levels,energy ofset own its has molecules N theofEach q.function partition MOLECULAR not the Q,function partition SYSTEM theusemust weS, gcalculatinWhen ln N ( )      −=       − −= −=−=−= − − kT kT N eNkTA e NkTA qNkTqkTQkTA ε ε 1ln 1 1 ln lnlnln Page 6 of 10 B. Calculate the energy, entropy, and Helmholtz free energy of a system of 1 mole of molecules described in part A, when the temperature is equal to the energy level spacing divided by Boltzmann’s constant (T=ε/k). ( )( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )[ ] ( )( ) ( )( ) ( ) ( )( ) ( ) ( )[ ] [ ] εε ε ε ε ε εε ε ε ε ε ε εεε ε ε ε ε ε ε ε 23 103807.1 652.8 2323 23 23 1231 1 1 1 1 1 23 2323 1 1 123 10762.210505.3652.810505.3 :TS-EAh answer wit theChecking 10762.2 10762.2 1ln10022.61ln 1ln1ln 652.8 0407.11314.80407.11 4587.05820.011ln 1 1 1 1ln1 1 1 1ln 1 10505.3 5820.010022.6 1 1 10022.6 1 10022.61 1 23 ×−=      −×=     −×= = ×−= ×−= −×=−= −     =     −= =       ⋅ == +=      −−      − = −−      − =      −−         − = ×= ×= − ×= − ×= − = −× −− −− − − − − − − − − − − − − K J K J kT A A kT kT kT kT kT A K J k A A A eeNA e k NkeNkTA K J S mole Kmol J moleRS moleRe e moleRS ekNmole e e k Nmole S eNk e e T NS E ee e molemoleE e e nNE C. Calculate the energy, entropy, and Helmholtz free energy of a system of 2 moles of molecules described in part A, when the temperature is equal to the energy level spacing divided by Boltzmann’s constant (T=ε/k). ( )( ) ( ) ( )( ) ε εεε ε ε ε 23 2424 1 1 123 10009.7 5820.0102044.1 1 1102044.1 1 10022.62 1 ×= ×= − ×= − ×= − = − − − − − E ee emolemoleE e e nNE kT kT A Page 7 of 10 ( ) ( ) ( ) ( ) ( ) ( )[ ] ( )( ) ( )( ) ( ) ( )( ) ( ) ( )[ ] [ ] εε ε ε ε ε εε ε ε ε ε ε ε ε ε 23 103807.1 3048.17 2323 23 23 1241 1 1 1 1 1 10524.510009.73048.1710009.7 :TS-EAh answer wit theChecking 10524.5 10524.5 1ln102044.11ln 1ln1ln 30.173048.17 0407.12314.80407.12 4587.05820.021ln 1 1 2 1ln2 1 2 1ln 1 23 ×−=      −×=     −×= = ×−= ×−= −×=−= −     =     −= ==       ⋅ == +=      −−      − = −−      − =      −−         − = −× −− −− − − − − − − − K J K J kT A A kT kT kT K J k A A A eeNA e k NkeNkTA K J K J S mole Kmol J moleRS moleRe e moleRS ekNmole e e k Nmole S eNk e e T N S D. Calculate the energy, entropy, and Helmholtz free energy of a system of 1 mole of molecules described in part A, when the temperature is twice the energy level spacing divided by Boltzmann’s constant (T=2ε/k). ( )( ) ( ) ( )( ) ( ) ( ) ( ) ( )[ ] ( )( ) ( )( ) K J K J mole Kmol J moleRS moleRe e moleRS ekNmole e e k Nmole S eNk e e T N S E ee e molemoleE e e nNE A A kT kT kT kT kT A 16.141629.147035.11314.8703499.11 932752.0770747.011ln 1 1 2 1 1 1ln1 12 1 1ln 1 10283.91028288.9 541.110022.6 1 1 10022.6 1 10022.61 1 2 1 2 1 2 1 2 1 2 1 2323 23 2 1 23 2 1 2 1 123 ==      ⋅ == +=              −−      − =      −−         − =      −−        − = ×=×= ×= − ×= − ×= − = − − − − − − − − − − − − ε ε ε εε εεε ε ε ε ε ε ε