Homework 6. Duygu Can & Neşe Aral 10 Dec 2010

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1 Homework 6 Duygu Can & Neşe Aral 10 Dec 2010

2 homework 6 solutions 1.1 Problem 5.1 A circular cylinder of radius R rotates about the long axis with angular velocity ω. The cylinder contains an ideal gas of atoms of mass M at temperature τ. Find an expression for the dependence of the concentration n(r) on the radial distance r from the axis, in terms of n(0) on the axis. Take µ as for an ideal gas. The relation between the concentration of the particles and total chemical energy of the system can be found by combining Eq. (5.12a) and Eq. (5.15) in the book as follows: µ = µ ext + µ int, where µ ext is the potential energy per particle in the external potential and µ int is given by ( ) n µ int = τln, with n Q = (Mτ/2π h 2 ) 3/2. n Q The particles are under the centrifugal force F (r) = Mω 2 r, so their potential energy can be calculated as V (r) = r 0 F (r )dr = r 0 Mω 2 r dr = Mω2 r 2. 2 Then the total chemical potential of the system becomes Solving this equation for n(r) we get ( ) n(r) µ = τln Mω2 r 2. 2 n Q n(r) = n Q e µ/τ e Mω2 r 2 /2τ. Since the chemical potential is constant at a given temperature, the term n Q e µ/τ is also constant. So we can write n(r) = n(0)e Mω2 r 2 /2τ. 2

3 Problem Problem 5.6 (a) Consider a system that may be unoccupied with energy zero or occupied by one particle in either of two states, one of energy zero and one of energy ε. Show that the Gibbs sum for this system is = 1 + λ + λ exp( ε/τ) (1.1) Our assumption excludes the possibility of one particle in each state at the same time. Notice that we include in the sum a term for N = 0 as a particular state of a system of a variable number of particles. (b) Show that the thermal average occupancy of the system is < N > λ + λ exp( ε/τ) (1.2) (c) Show that the thermal average occupancy of the state at energy ε is < N(ε) >= λ exp( ε/τ)/ (1.3) (d) Find an expression for the thermal average energy of the system (e) Allow the possibility that the orbital at 0 and ε may be occupied by one particle at the same time; show that = 1 + λ + λ exp( ε/τ) + λ 2 exp( ε/τ) = (1 + λ)[1 + λ exp( ε/τ)] (1.4) Because can be factored as shown, we have in effect two independent systems. (a) The Gibbs sum is given by Eq. (5.61) in the book as = ASN λ N e ɛs/τ. In our system we have three different states for which occupancy and the corresponding energy values are (0,0), (1,0) and (1,ɛ). Inserting these values into the Gibbs sum we get = 1 + λ + λe ɛ/τ. (b) Recalling Eq. (5.62) in the book, for the average number of particles we can write N = λ λ ln = λ λ = λ + λe ɛ/τ. 3

4 1. Homework 6 s (c) Total number of the particles in the system is N = N(ɛ n = 0) + N(ɛ n = ɛ). Comparing this equation with the one found in part (b) we see that N(ɛ) = λ e ɛ/τ. (d) Using Eq. (5.63) in the book, the internal energy can be calculated as U = ɛ = 1 ɛ s λ N e ɛs/τ ASN = 1 ( ɛλe ɛ/τ ) = λe ɛ/τ Z λ = N(ɛ) ɛ. (e) We have to add a new state to the system having occupancy N=2 and energy ɛ s = ɛ + 0 = ɛ. The contribution to the Gibbs sum of this new state is λ 2 e ɛ/τ, hence we get 1.3 Problem 5.8 new = + λ 2 e ɛ/τ = 1 + λ + λe ɛ/τ + λ 2 e ɛ/τ = (1 + λ)[1 + λe ɛ/τ ]. In carbon monoxide poisoning the CO replaces the O 2 adsorbed on hemoglobin(hb) molecules in the blood. To show the effect, consider a model for which each adsorption site on a heme may be vacant or may be occupied either with energy ε A by one molecule O 2 or with energy ε B by one molecule CO. Let N fixed heme sites be in equilibrium with O 2 and CO in the gas phases at concentrations such that the activities are λ(o 2 ) = and λ(co) = , all at body temperature 37 o C. Neglect any spin multiplicity factors. (a)first consider the system in the absence of CO. Evaluate ε A suc that 90 percent of Hb sites are occupied by O 2. Express the answer in ev per O 2. (b) Now admit the CO under specified conditions. Find ε B such that only 10 percent of Hb sites are occupied by O 2. 4 (a) In our model we have three different states with occupancy numbers and corresponding anergy values as (1, ɛ A ), (1, ɛ B ) and (0,0), where (0,0) corresponds to the vacant state. So the Gibbs sum for the system can be given

5 Problem 5.10 by = ASN λ N e ɛ/τ = 1 + λ A e ɛ A/τ + λ B e ɛ B/τ. The fraction of sites occupied by O 2 is f A = (λ A e ɛ A/τ )/. In the absence of CO 2, λ B = 0 and the Gibbs sum reduces to and we get = 1 + λ A e ɛ A/τ f A = λ Ae ɛ A/τ 1 + λ A e ɛ A/τ. where λ A = Setting f A = 0.9 we can solve this equation for ɛ A. Here τ = eV, calculated for body temperature of 310 K. Hence we get ɛ A = 0.366eV. (b) In the presence of CO 2 the fraction of sites occupied with O 2 becomes f A = λ A e ɛ A/τ 1 + λ A e ɛ A/τ + λ B e ɛ B/τ. We want that only 10% of the sites are occupied with O 2, so setting 0.1 for f A and using the result of part (a), ɛ A = 0.366eV we can solve the last equation for ɛ B, which gives ɛ B = 0.55eV. 1.4 Problem 5.10 The number of particles is not constant in a system in diffusive contact with a reservoir. We have seen that < N >= τ ( ) (1.5) µ τ,v from (59). (a) Show that < N 2 >= τ 2 2 µ 2 (1.6) The mean square deviation < ( N) 2 > of N from < N > is defined by < ( N) 2 > = < (N < N >) 2 >=< N 2 > < N > 2 (1.7) [ < ( N) 2 > = τ µ 2 1 ( ) ] 2 2 (1.8) µ 5

6 1. Homework 6 s (b) Show that this may be written as < ( N) 2 >= τ < N > / µ (1.9) In Chapter 6 we apply this result to the ideal gas to find that < ( N) 2 > < N > 2 = 1 N (1.10) is the mean square fractional fluctuation in the population of an ideal gas in diffusive contact with a reservoir. If < N > is order of atoms, then the fractional fluctuation is exceedingly small. In such a system the number of particles is well defined even though it cannot be rigorously constant because diffusive contact is allowed with the reservoir. When < N > is low, this relation can be used in experimental determination of the molecular weight of large molecules such as DNA of MW (a) Using Eq.(5.52) and (5.53) we can write N 2 = ASN N 2 P (N) = 1 N 2 e (Nµ ɛ/τ) = τ 2 2 e (Nµ ɛ)/τ µ 2 = τ 2 2 µ 2 ASN (b) Using Eq. (5.80) we can write ) N = τ µ τ ( 2 ) µ 2 τ ( ) 2 τ 2 = 1 µ τ τ ( N)2, where the last equality follows by comparison with Eq. (5.82). 6

Homework #8 Solutions

Homework #8 Solutions Question 1 K+K, Chapter 5, Problem 6 Gibbs sum for a 2-level system has the following list of states: 1 Un-occupied N = 0; ε = 0 2 Occupied with energy 0; N = 1, ε = 0 3 Occupied