PHYS427fa08_HW07_Solution - UTUClPhynes427 Homework 7...

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Unformatted text preview: UTUClPhynes427 Homework 7 Solutions Philip Powell October 16, sees 1. Prohlern 1 {KS-EH 5.2} Here we consider the Earth‘s atmosphere, which we take to be eoropoeed of molecules of mass M. and to be in thermal equilibrium at a temperature T = rfitg. First, we recall that the chemical potential of an ideal gas [without a gravitational potential} is [K351i 5.12a} ii = Tln [1} The definition of chemical potential is p = {fiFfflN}T__V, and the effect of the gravitational field is to decrease the energyr per molecule by an amount. Gme-r {recall that the gravitational potential is negative}, where m in the mass of the Earth. Thus, in the prmenee of a gravitational field the chemical potential beoornoa nErj) _ GmM m 1i|[r}=1'ln(nq T We can express the product Gm in terms of gravitational acceleration at the earth’s surface {g} by noting: Gm. g 2 EE- => Gm = QR: l3} 'I‘hua1 we have 2 to] = Tia — M33 (4} Now, if the atmuapherc is in oqujl1'.l:n1'iu.rrlI the chemical. potential will he the same at all i", and will equal a constant on: 111:?) JHQRE ___ Til] ( “Q j r — tan [5} We can evaluate it; by setting 1" = R, the Earth’s radius1 so that we obtain 2 via a M33 = m (Tl—ml) — Mali {a} “Q 1” no Solving this equation for nEr] gives _ m+flafi r r, [1r], 1“er = willie The total number of moleculm in the atmosphrne may he obtained by integrating the density over the atmosphere’s vohnne: _ mT'n-r W2 an}; a {a {a} l Sutetituting our expression for nfir} gives ‘2 W N = Mamie—fig j dr raefl-Lfl— {9} R We can see that this!r integral divergm by melting the substitution u = M933!” so that we have MIR N=41m[R}Ie' “5“ (Mg‘qufa ' an (i) {10] T In the limit a —r [i the exponential approaehee l, ao the integral diverges thanlw to the a". . Problem 2 [KSaK 5.3} Here we eonaider a eolnmn of atoms of mam M at temperature T in a uniform gravitational field 9. As in the last problem: the ehenn'ral potential is guru by ,1: 2m +ng an Note that in this problem we nee expression U = ng for the gravitational potential, mauring the potential enemas,r from ground level {and warming the gravitetirmal fielrl is constant This stands in eontraet to the last problem, where we chose the zero of potential energr to be infinity. In equilibrium the eheraieal potential must be constant throughout the eolumn. Thea, we may eat this expression equal toita veloeaty=fland write Tin + my = T In [12] Solving for “(til we obtain nit} = flue—£1” {131' Now that we have the height diatrihution functionr we can empress the average height as _ it” dy wit!) {it} — m [14} The numerator can be evaluated as follows: m: an ._ M f demo} = 1mu day-5 "i oE—g o o T a m 2 “WE dye—“i If I a sale0 .r-‘_"\. QII— Nu—a’ = 0—2 {15} hileanwhihe1 the denominator is simply do :Icl f dentin = fie] drew” a e _ fl . — I, [15) Thus, we find that the average height '15 1 'r {a} — E = E {1?} Thus, the thermal average potential energy per atom is to = May} =— no The thermal average of the total energy [kinetic + potential} per atom is therefore 37' 57 " =_ =_-u 1 {hi , +r 2 (a where the average kinetic energy per particle follows from the equipartition theorem. The heat capacity per atom is thus _ 3E 0-5? => o=g gee] . Problem 3 {KS-5K 5.8] {3} Consider a model tor carbon monoxide poisoning in which we have N fixed Heme sites, each of which can either be empty, oeenpied by an 0; molecule, or occupied by a CO molecule. The energy of these states are taken to be l], 5,4 and s}; respectively. We first ignore the CD, and consider a system only with 03. In this case the grand partition flinetion is Z = Zea—err HEN 1 +eiitog-E‘slr’f = 1 + nope-WT {21) The probability of a site being eecupied by an ()3 molecule is therefore ., closet—"IT Flog} _ 1+A[02]e-EAE? If we tales MO?) = lfl’a, T = 3?“, and 13(02} = .9 we find a, = —e.sr c‘y' [232! (b) Now we allow CC! to be present as well so that the grand partition function beeomes z = l + nose-W7 + nook-E's” (24] [223 The probability of a site now being oocupied by so {)2 molecule is )LEDZ leg—=sz Ploil = 1 + Mflgle—“l’ + stoop-see (25] Using Alfi'fl} = 10—7 and setting this probability equal to 10%, solving for 53 yields 83 = —e.ss or“ (as) . Problem 4 {KSsK 5.14) Consider a hemoglobin molecule, which is capable of binding up to four 0;; molecules. We denote the energy of a bound ()2, relative to free 02, as e and write the absolute activity of the free oxygen molecule as A = el‘f’. The grand partition function of the hemoglobin is a = 1 + any” + eerie” + axle—3W + X‘s—4"” (27} where the coeflicionts {l,d,fi,d,l} give the mnnber of ways which a given number of sites can be ocnlpierl. The probability of only one D: molecule being adsorbed on a hemoglobin molecule is therefore rile—E” = its} Pl (1.4 Similarly, tho probability of mostly four polygon molecules are adsorbed is {2‘9} These probobilitics are plotted as a function of A on the next page1 with the “typical” value E m Ill EV. AS we would expect, as A de- oc {or ru —r on, which mmaponds to a large density of 02 molecules] the probability of four {)2 molecules being adsorbed approoohm 1. [LB _ as 11.4 112 .-..I. 5. Probiem E- (KflaK 6.1] Here wo studs.r the Fermi—Dirac distribution: 1 He} = m [an] Difl'ETEntifiting! we find {if 1 ate—xsirr E = 7m {31} Evaluating this expression at a" = ,u wo obtain df 1 Te ‘32] 6. Problem 6 {K351i 5.2} Looking again at the Formi—Dimc distribution and writing E 2 p + 6 we can write 1 to + a: = W + 1 (as: Letting :5 —r *6 we 3150 have 1 HF — 5] = m [34.] Adding those two expressions gives 1 l :5 - = .— fflu+ i+ftu L5} di_}.;l,,+l+fl_“f+1 = {ta—WT +1]+{3“T + 1] (Eif'r + 11(3—53'1" + _ 2"” + 3‘15” + 2 _ Ear“? + E—fiy’r + 2 = 1 {353 Thus, we can write f{#+fl'II=I-fl.u-r5} {35] ...
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