PHYS 352 Homework 2 Solutions

Transcription

1 PHYS 352 Homework 2 Solutions Aaron Mowitz (, 2, and 3) and Nachi Stern (4 and 5) Problem The purpose of doing a Legendre transform is to change a function of one or more variables into a function of variables conjugate to the original ones. In other words, we want to trade the dependence on one set of variables for another. The entropy and its differential as a function of energy and volume are written as S(E, V ) = E T + pv T ( ) ( ) ds = de + dv = E V V E T de + p T dv To find the Legendre transform with respect to energy J, we subtract ( ) E V E = E T entropy: from the J S E T = F T where we recall that F E T S. To find dj, we simply take the differential of the above expression ( ) E dj = ds d = p T T dv + E T 2 dt So we see that J is a function of T and V. This isn t surprising, since the Helmholtz free energy is a function of temperature and volume. Now to find the Legendre transform with respect to both energy and volume Y, we subtract ( ) V E V = pv T from the entropy, in addition to what we subtracted before. Equivalently, we can just subtract this from J: Y J pv T + pv = F T = G T where we recall G E T S + pv = F + pv. To find dy, we again just take the differential, which yields ( ) pv dy = dj d = E + pv T T 2 dt V T dp = H T 2 dt V T dp where we use the definition of the enthalpy, H E + pv. So we see that Y is a function of T and p. Again, this makes sense, since the Gibbs free energy is a function of temperature and pressure. Problem 2 The distribution of velocities of a classical ideal gas is given by the Maxwell-Boltzmann distribution: ( ) m 3/2 f (v) = exp [ m v 2 2πk B T 2k B T

2 This can be found in many ways (e.g. using symmetry considerations or the canonical ensemble), and the derivation is in most standard textbooks, so we will not reproduce it here. We can use this distribution to calculate the average speed. Since the M-B distribution only depends on the magnitude of velocity, the integration is most easily done in spherical coordinates: v = v f (v) d v dω v ( ) m 3/2 = 2πk B T 0 8kB T = πm v 3 exp [ mv2 dv 2k B T Note that this is different from both the root-mean-square speed, v rms v 2 3k = B T m, and the 2k most probable speed, v p = B T m, the speed at which the M-B distribution is a maximum. Problem 3 (i) Since the Hamiltonian of a classical spin is H = µ z B = µ 0 B cos θ, the Boltzmann factor is given by e µ 0Bβ cos θ. To find the canonical partition function, we integrate this over the sphere: = e µ 0Bβ cos θ dω = 2π 0 = 2π µ 0 Bβ e µ 0Bβ cos θ d (cos θ) dφ ( e µ 0Bβ e µ 0Bβ ) = µ 0 Bβ sinh (µ 0Bβ) (ii) The average magnetization is given by µ z = µ 0 cos θe µ 0Bβ cos θ dω We want to show this equals T ln B = β B. We see that B = B e µ 0Bβ cos θ dω = µ 0 β cos θe µ 0Bβ cos θ dω If we divide this by β, we get exactly what we wrote down for the average magnetization. (iii) To find the magnetic susceptibility, we first compute the derivative of the average magnetization with respect to B: µ z B = ( ) µ 0 cos θe µ 0Bβ cos θ dω B = µ 2 0β cos 2 θe µ 0Bβ cos θ dω 2 µ 0 cos θe µ 0Bβ cos θ dω B = β µ 2 0 cos 2 θe µ 0Bβ cos θ dω β ( ) 2 2 µ 0 cos θe µ 0Bβ cos θ dω 2

3 We now evaluate this at zero magnetic field. We note that B=0 =, which can be seen by either setting B = 0 in the original definition of the partition function, or by using the fact that =. Using this, we find lim x 0 sinh x x χ = µ z B [ = β B=0 µ 2 0 cos 2 θdω ( 6π 2 ) 2 µ 0 cos θdω which is inversely proportional to T, since the quantity in brackets is independent of temperature. The quantity in brackets above is the proportionality factor χ 0. The second term vanishes, since the average of cosine over the sphere is zero. So we are left with χ 0 = µ 2 0 cos 2 θdω = µ

4 4) Fluctuations i) Show that, in the canonical ensemble (ΔE) 3 = T 4 ( C V T ) + 2T 3 C V V Let us first consider the LHS: (ΔE) 3 (E E ) 3 = E 3 3E 2 E + 3E E 2 E 3 = E 3 3 E 2 E + 2 E 3 Now, we also have the relation between moments of the energy and the partition function E n = ( ) n n β n (ΔE) 3 = 3 β β 2 β 2 3 ( β ) 3 This looks quite complicated, but consider the following 2 ln β 2 And very similarly: 3 ln β 3 ln β = β = E = ( β β ) = 2 ( 2 β ) + 2 β 2 = E 2 + E 2 = (ΔE) 2 = β [ 2 ( β ) β 2 = 3 β β 2 β ( β ) 3 = (ΔE) 3 That is rather useful! Let us further note that C V = ( E T ) V. One may change variables (k B ): Now we can write (ΔE) 3 = 3 ln β 3 β = T β T = ( T 2) T = T2 T = 2 E E = T2 β2 { T2 T T } = T 2 T {T2 C V } = T 2 {2TC V + T 2 ( C V T )} = T4 ( C V T ) + 2T3 C V Which is what we wanted to show. 4

5 ii) Show that, in the grand canonical ensemble (ΔN) 2 = T ( N μ ) Although this proof can be done using a trick similar to the one shown on part (i), let us do it in brute force, which is quite easy in this case. Using the relations of the partition function and the average particle number N = T μ ; N2 = T 2 2 μ 2 We may write T ( N μ ) = T ( T μ μ ) = T 2 ( 2 μ 2 ) T 2 2 ( 2 μ ) = N 2 N 2 = (ΔN) 2 And we got the expected relation. ii) Compute (ΔN) 2 for an ideal gas. The grand canonical partition function of an ideal gas is Using this we calculate the average particle number Taking another derivative will give us the fluctuation = exp [Ve βμ ( 2πm βh 3 ).5 N = β ln = β μ μ = Veβμ ( 2πm.5 βh 3 ) (ΔN) 2 = β ( N μ ) = Ve βμ ( 2πm.5 βh 3 ) = N (ΔN) 2 = N β,v This result tells us that we should expect no fluctuations compared to the equilibrium particle number in the thermodynamic limit: (ΔN) 2 N = N /2 0 5

6 5) Adsorption Centers i) Find fraction of occupied adsorption sites. The partition function of one adsorption site is = e β(e μ)n = + e β(e μ) Different adsorption sites are independent, so the total partition function is n=0 = ( ) N = ( + e β(e μ) ) N We can relate the average adsorbed particle number to this partition function,as we did in the previous question: N = β μ = Nβe β(e μ) ( + e β(e μ) N ) β( + e β(e μ) ) N = N e β(e μ) + e β(e μ) The fraction of adsorbed sites is thus f = N N = e β(e μ) + e β(e μ) = + e β(e μ) ii) Binding energy of O 2 molecules. Apparently this part gave you guys a lot of trouble. To solve it efficiently it is useful to relate the chemical potential to the given quantities in the problem. We are given that f = 0.5, which allows us to relate E and μ: 2 = e β(e μ) E = μ + e β(e μ) The chemical potential of an ideal gas can be derived from the Sackur-Tetrode relation μ = T ( N ) E,V S = kn ln [ V 3/2 N (m E 3h 2 N ) kn = kt ln [ V 3/2 N (m E 3h 2 N ) = kt ln [ N V ( 3h2 m Note that for an ideal gas N/V = P/kT. Further, in room temperature, O 2 molecules have translational + rotational degrees of freedom (linear molecule), so = 0.5(3 + 2)NkT = 2.5NkT. Using these facts we may write N E ) 3/2 6

7 h 2 E = μ = kt ln [ P kt ( 3 0π mkt ) The mass of a O 2 molecule is m = kg. In STP conditions kt = k(273[k) /43 ev, and the pressures is P = Atm 0 5 Pa. Plugging these numbers, the binding energy is E 43 ln [ ( ) ln[ ev 0.38eV 43 3/2 7