Thermodynamic Laws/Definition of Entropy

Transcription

1 This is a review of Thermodyamics ad Statistical Mechaics. Thermodyamic Laws/Defiitio of Etropy st law of thermodyamics, the coservatio of eergy: du = dq dw = Q W, ( where dq is heat eterig the system ad dw is work doe by the system. Note the covetio: (+ for eergy eterig the system ad (- for eergy leavig the system. d law of thermodyamics, etropy: I ay spotaeous trasitio, the etropy of the uiverse icreases. There are may equivalet statemets of the secod law, such as: heat ever flows from cold to hot, or, there is o such thig as a perpetual motio machie. I a reversible trasformatio, the etropy of the uiverse does ot chage. Note: this does ot mea that the etropy of some sub-uiversal system will ot icrease or decrease. It is importat to cosider all parts of your system. To fid the chage i etropy of a system betwee state A ad state B, coect A ad B by a reversible path. The S(B S(A = B A dq T. ( Usig Equatio, we ca illustrate the poit about reversible trasformatios above: Cosider a reversible isothermal expasio of a ideal gas i cotact with a thermal reservoir at temperature T. For a ideal gas, U = 3 kt, for a isothermal expasio du = 0, ad dq = P dv, so Q dq P dv T = T = S = T. (3 P = RT, so V P VB T dv = R V A V dv = R l V B > 0, (4 V A but heat leaves the reservoir, thus S res = Q T ad S uiverse = 0. Carot Cycle The Carot Cycle is usually discussed with a ideal gas as the workig substace, but i reality ay thermodyamic system (a paramaget, a electrochemical cell, etc. ca be used. A Carot Cycle is a cycle ivolvig two reversible isothermal trasitios ad two reversible adiabatic trasitios. If the workig substace is a ideal gas, the P V diagram of the cycle looks like Fig., ad the S T diagram looks like Fig., The efficiecy of the Carot Cycle is give by η = W Q = Q Q Q = T T T = T T, (5

2 Figure : Carot Cycle: P V diagram Figure : Carot Cycle: S T diagram which is work doe by the system divided by the heat absorbed by the system. For a Carot Refrigerator, ru the cycle backwards: η ref = Q W = Q Q Q = Q Q = T T = T, (6 T T heat extracted from the system divided by work required to do so. Note that as T T, η ref. I geeral, the efficiecy is η = Typical Phase Diagrams what you get out what you eed to put i. (7 Figs. 3 ad 4 show the phase diagrams for a sigle compoet system.

3 Figure 3: P v diagram: Pressure vs. specific volume Thermodyamic Potetials Let s begi with Eq. : du = dq dw. (8 If we coect states by reversible processes we get T ds = dq ad dw = P dv for a gas system. So, du = T ds P dv, (9 ad settig oe of s or v equal to zero yields, ( ( U U = T, S V Defie F U T S, } {{ } Helmholtz free eergy V df = SdT P dv, S = P. (0 ( F = ( F = S. T V V T ( Defie Defie H U + P V, } {{ } ethalpy G H T S, } {{ } Gibbs free eergy dh = T ds + V dρ, dg = SdT + V dρ, ( ( H S ρ = T, H = V. ρ S ( G T ρ = S, ( G ρ T = V ( (3 3

4 Figure 4: P T diagram: Pressure vs. Temperature. The dotted lie characterizes the aamolous behavior of water (ad ay substace that expads whe froze. The differetial form tells us the atural variables to use with each fuctio. A, for example (Fig., is best expressed i terms of T ad V : A(T, V. I physics we ofte do problems at costat (V, T ad work with A. Chemists are proe to a costat (T, ρ ad work with G. Note that we ca get Maxwell s relatios from the fact that each of these is a exact differetial. For example, thus but so df = ( F T V dt + ( F V ( F = S ad ( F T V V T F T V = ( S V T F T dv, (4 = ρ, (5 V T, (6 ( ρ = T, (7 which is oe of Maxwell s relatios. Note from Dr. Collis: I have ot see Maxwell s relatios o ay GRE example test (yet!. Also, all of the above assume N is fixed. If N ca vary each gets aother term ad we must worry about chemical potetials: du = T ds P dv + µdn where ( U N S,V V = µ = chemical potetial (8 The rest of the potetials also get this term, ad we get a ew potetial, the grad potetial - Φ: Φ = A µn where dφ = SdT ρdv NdU (9 4

5 Heat Capacity Heat capacity is specific heat if you divide by volume or mass c V = Q holdig volume costat (0 ( T U = where U is iteral eergy ( T V c P = U holdig pressure costat ( T c P > c V because the volume expads at costat pressure ad the system does work which meas it stores less eergy. Note that for o P V T systems aalogous heat capacities ca be defied. I a magetic system, P H ad V M ad we ca defie c H ad c M. A statistical approach ca be used to obtai the classical heat capacity of commo systems. If the Hamiltoia of a particle i the system ca be writte as N N H = A j Pj + B j Q j (3 where P j ad Q j are geeralized mometa of coordiates, the j j U = NkT f Equipartitio Theorem (4 where N is the total umber of particles ad f = N + N, the umber of quadratic coordiates ad mometa. Thus, c V = fnk (5 ad the specific heat is fk. You get k per degree of freedom (mometum of coordiate degrees of freedom cout equally. For this expressio to work, the temperature must be large compared to the quatum of eergy associated with a particular motio. For example, if a diatomic molecule has a vibratioal frequecy ω, the the quatum of eergy is ω (E = ( + ω. So kt hω to get NkT cotributios to U ( NλT for x ad NkT for ρ. Cosider the example of a diatomic molecule like N. See Fig. 5. We ca write the Hamiltoia as H = Mv x cm + Mv y cm + Mv z cm + µṙ + kr + Iω z + Iω x (6 Note that I y = 0 for quatum mechaical reasos, so there is o rotatioal term about the y-axis. We immediately kow that at high temperatures H = 7kT N, c V = 7Nk, ad c V = c V N = 7 k. (7 Oe fial thig: For ad ideal gas, c V = 3Nk = 3 R where N is the umber of molecules, is the umber of moles, k is Boltzma s costat, ad R is the uiversal gas costat. Also c P = ( H = 5 Nk. I geeral, for gases with ideal-like behavior, but iteral degrees T P of freedom (rotatios, stretches, c P = c V + Nk. 5

6 Figure 5: Diatomic molecule Statistical Mechaics of systems at costat T ad V The probability a state of the system is occupied is proportioal to e Es/kT where E is the eergy of the state. The expectatio value of the eergy (which gives us U, the iteral eergy, is s E se Es/kT s e Es/kT If the system is made up of idetical particles or a set of idetical smaller systems (a buch of harmoic oscillators, we ca just fid the expected eergy of the particle of a smaller system, ad multiply by N. A couple of examples. Cosider the two level system, a buch of particles ca have eergy O or E. The expectatio value of the eergy of N particles is < E >= N ( Oe O/kT + Ee E/kT e O/kT + e E/kT U = NEe E/kT ; +e E/kT as T, U N E. The average particle eergy is halfway betwee the groud state ad excited state. If U N > E, (8 the e E/kT + e E/kT >, (9 ad e E/kT + > ee/kt < ad T < 0 (30 6

7 Figure 6: umber microstates vs. U The two level problem is a classic problem which leads to egative temperatures. To see why, let s look at it a differet way. Let s plot the umber of cofiguratios of the system that will give a particular eergy. If the total eergy is 0, there is oly oe cofiguratio, ad i the groud state. Same for U = NE so the plot looks like Fig. 6. Here, S = k l(ω ad du = T ds P dv, so S =, where Ω is the multiplicity. U T Refer to Fig. 6. Whe S is icreasig, T is positive. Whe S is decreasig, T is egative. This a characteristic of ay system with a maximum total eergy: the plot of the umber of microstates vs. eergy will have a maximum, ad thus egative temperatures. Eistei model of a solid The Eistei model is simply a buch of atoms held together by sprigs. See Fig 7. We approximate it as N harmoic oscillators with agular frequecy ω. They are idetical, so we ca just cosider oe. E = ( + ωe (+ ω/kt e (+ ω/kt (3 We immediately see a trick: defie β = kt. The, ( + ωe (+ ωβ = β ( e (+ ωβ. (3 7

8 Figure 7: Eistei s model of a solid This is the same sum as i the deomiator of Eq. 3, so let s focus o it: e (+ ωβ = e ( ωβ/ e ωβ } {{ } =x (33 = e ωβ/ x (34 ad So, β =0 = e ωβ/ x = e ωβ/ (35 e ωβ (36 = e ωβ/ e ωβ/ (37 = sih ( ωβ (38 sih ( ωβ ( ωβ cosh ω (39 E = sih( ωβ cosh ( ωβ sih( ωβ ω ( e ωβ/ + e ωβ/ = ω e ωβ/ e ωβ/ = ω tah ( ωβ (40 as T ad β 0. (4 8

9 We ca use series expasios: e ωβ/ + ω β ad e ωβ/ ω β, E ω ( = ωβ β = kt. (4 Lo ad behold, we recover the equipartitio result! kt per degree of freedom ad degrees of freedom: mv + kx. 9