Solutions to exercises week 45 FYS2160

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1 Solution to exercie week 45 FYS2160 Kritian Bjørke, Knut Oddvar Høie Vadla November 29, 2017 Schroeder 5.23 a) Writing Φ = U T S µn in term of infiniteimal change of the quantitie involved: dφ = du T ds SdT µdn Ndµ. Inerting the thermodynamic identity for du give dφ = P dv SdT Ndµ. The partial derivative of Φ wrt. T, V or µ i obtained by holding the two other quantitie fixed: = S, T b) V,µ = P, V T,µ µ T,V = N. A wa done for F and G in chapter 5.2 in Schroeder, expre an infiniteimal change in the total entropy S total a a um of change in the entropy S of the 1

2 ytem and the entropy S R of the reervoir it i in contact with: ds total = ds + ds R. Auming V fixed, apply the thermodynamic identity to ds R (olve the expreion for du R in term of ds R ), and ue the fact that du R = du and dn R = dn to write ds R in term of ytem variable. The total entropy of the ytem plu reervoir (expreed in term of ytem variable) i then ds total = 1 T (du T ds µdn) = 1 T dφ. The tendency of the total entropy to increae therefore mean that the grand free energy tend to decreae (ince ds total i poitive, dφ mut be negative in order to make dφ/t poitive). c) Φ = U T S µn = U T S + P V P V µn = G P V µn = P V, where we in the lat tep ue that G = µn when T and P are contant (from the thermodynamic identity of G, dg = SdT + V dp + µdn with dn = 1). d) For the unoccupied tate: U = S = N = 0 Φ = 0. For the occupied tate: U = U 0 = 13.6 ev, S = 0, N = 1 2

3 Φ occupied = U 0 µ = U 0 + kt ln V N ( ) 3/2 2πmkT, h 2 where eq in Schroeder i ued for the chemical potential of an ideal ga. Here m i the electron ma, and the electron concentration i given a N/V = m 3. Putting in number give u (the logarithm i calculated to be equal to 17.8) Φ occupied = 13.6 ev ev/k 5800 K ev < Φ unoccupied = 0 ev. To find the temperature at which the occupied and unoccupied tate are equally likely, we et their grand free energie to be equal and olve for T (neglecting the ln T 3/2 dependence, which i inignificant compared to the factor T outide the logarithm, when T i varied): Schroeder 7.1 Φ occupied = Φ unoccupied U kT = 0 Probability of heme ite being occupied: P = 1 e (ɛ µ)/kt +1 T = U k 8800 K Blood in equilibrium with air, that we can view a an ideal ga, giving the chemical potential: µ = kt ln( V Z int Nv Q ) = kt ln( kt Z int P v Q ), where P i the partial preure. 3

4 From thi we get P = 1 e ɛ/kt kt Z int = 1, P +1 0 P v Q P +1 where P 0 = e ɛ/kt kt Z int v Q. T = 310K, kt = ev, ɛ = -0.7 ev, P 0 = bar Schroeder 7.7 Grand potential/free energy: Φ = U T S µn Grand partition function: Z = From exercie 5.23 we have = N. µ T,V For a Φ = kt ln Z we get ( ) Φ = N. µ T,V So Φ and Φ obey the ame differential equation. e [E() µn()]/kt We et an initial value µ = 0, then Φ = U T S = F and Z = Z (the ordinary partition function). So for µ = 0, Φ = kt ln Z = kt ln Z = F = Φ, where we ue F = kt ln Z from ection

5 Φ and Φ obey the ame differential equation and and are the ame at the initial condition µ = 0, therefore they mut be the ame function, giving u: Φ = Φ = kt ln Z. Schroeder 7.14 From Figure 7.7 the ditribution more or le coincide for ɛ µ a few time greater than kt. Thi give: n BE n FD = e(ɛ µ)/kt +1 e (ɛ µ)/kt 1 = 1+e (ɛ µ)/kt 1 e (ɛ µ)/kt 1 + 2e (ɛ µ)/kt, for (ɛ µ)/kt 1 For a 1% difference between n BE and n FD (Boltzmann ditribution lie between BE and FD ditribution (Boltzmann ditribution lie between BE and FD ditribution) we require: 2e (ɛ µ)/kt < 0.01 (ɛ µ) kt > ln(200) 5.3 For particle at room temperature T = 300K we get: (ɛ µ) > ev. For an ordinary ga of particle in a box, energy level meaured in the uual way ɛ i poitive and µ/kt i ln(v Z int /Nv Q ). The condition above i then: V Z int Nv Q = kt Z int P v Q > 200 For nitrogen at room temperature and atmopheric preure: kt Z int P v Q = , which i eaily atified. Schroeder 7.15 Total number of particle: N = n Boltzmann = = e (ɛ µ)/kt = e µ/kt Single-particle partition function: e ɛ/kt 5

6 Z 1 = = e ɛ/kt Thi give: N = Z 1 e µ/kt µ = kt ln( Z 1 N ) Schroeder 7.20 For an electron ga at the centre of the un we have the Fermi temperature: ( ) T F = ɛ 2/3 F k = h2 3N 8mk πv = K Thi temperature i cloe to the actual temperature in the un, T = 10 7 K, o we can not threat the electron ga a a degenerate Fermi ga (T 0) or an ordinary claical ideal ga (T T F ). Schroeder 7.26 a) The Fermi energy i: ( ) 2/3 ɛ F = h2 3N 8m πv J ev, with N/V = N A /(37cm 3 ) where N A i the Avogadro number. b) From equation 7.48 we have: C V NkT = π2 k 2ɛ F = π2 2T F 1.0K 1 The predicted heat capacity i maller by almot a factor of 3. c) Entropy of liquid 3 He: S = T 0 C V T dt = (2.8K 1 )Nk T 0 dt = (2.8K 1 )NkT Entropy of olid 3 He: S = k ln(2 N ) = Nk ln 2, 6

7 thi applie for temperature down to low value before it freeze out and goe to 0. Same entropy for: Nk ln 2 = (2.8K 1 )NkT T = ln 2 2.8K 1 = 0.25K By Clauiu-Clapeyron relation the lope of the olid-liquid phae boundary on the PT graph i proportional to the entropy difference S liquid Solid. We predict the lope to be negative at temperature lower than 0.25 K and poitive for temperature greater than 0.25 K. We ee thi behaviour in Figure