Chapter 5 Page 5.1 CHAPTER 5. r Force times distance has units of energy. Therefore, fxr=u, or f / is dimensionless.

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1 Chapte 5 Page 5.1 CHAPTER 5 Poblem 5.1: 1 (a) u () (b) Foce times distance has units of enegy. Theefoe, fx=u, o f/ is dimensionless. d f = d u 1 d f 4ε 1 = f = 4 ε1 d 13 f = 4 ε 1 f ε = Now in dimensionless fom: 13 7 f() 4 3 u () 1 f() (c) f d = x Newton's nd law in dimensionless fom. Using dimensional analysis we find dt that dimensionless t is given by (m/. Theefoe, the dimensionless foce can be witten as shown. d x = dt 4 x 1 x x IC1: t = 0 x =.5 IC: t = 0 d dt x = 0 initial conditions

2 Chapte 5 Page 5. Rough phase-space sketch of what the solution would look like. Poblem 5.: 1 u() 4 d d u Thus, u min 13 7 = 48 4= 0 at the minimum in the potential = o min min Initialize to find oot 1 esult oot esult Poblem 5.3: Let D be the distance at which the potential enegy becomes equal to the kinetic enegy and the diection of the colliding atoms ae evesed. Let x = (/D). 4ε x x = kt x x We can solve this with the quadatic fomula: kt = 0 4ε 1 1 x = kt ε We have chosen the + sign because x must be positive and the squae oot tem is always seen to be geate than unity.

3 Chapte 5 Page 5.3 Thus, kt 1 1 ε = x = o D 1 1 = D 1 1 kt and D 1 = 1 Fo A, 3.49 ε kt ε ε k 1 = DT ( ) T 1 1 T The vaiation of the diamete with tempeatue looks like the figue shown at the ight. Notice that at highe tempeatues, thee is lage penetation coesponding to smalle had-sphee diametes. The value is the had sphee value only at 0 K whee no penetation into the epulsive coe occus DT ( ) Poblem 5.4: Fom Table A5.5 fo wate k eg debye egcm 3 K T o cm ε o 09.1kK μ 1.85debye α cm T 300K 1 4ε o 1 μ 4 1 αμ 1 u LJ () 10 1 u D () 3k T eg o 10 1 u P () eg o 10 1 eg u tot () u LJ () u D () u P () Note that the potentials plotted ae in units of 10-1 egs u LJ () u D () u P () u tot () 400 Poblem 5.5:

4 Chapte 5 Page 5.4 Fom Appendix 5 we have the following values fo methane: LJ Kihaa Sutheland SW ε LJ 1.78k d 0.77 S 3.94 SW 3.40 LJ 3.78 K ε S 434.8k ε SW 88.8k ε K 3.k g 1.85 Poblem 5.: ε k = ATc 3 = Bk Tc Pc FLUID (A) Tc (K) Pc (atm) A K CH Xe C H ε 3 Pc A = B = Note: A and B ae both dimensionless. To make units in ktc ktc equation fo B compatible, use the following convesion facto:

5 Chapte 5 Page conv 10 4 cm 3 conv K atm eg cm 3 atm eg K FLUID A=/(kTc) B=(conv)( 3 )/T c A K CH Xe C H Thus, ε = k 0.84Tc Aveage = 0.09 ktc Pc Poblem 5.7: 1 PV NkT Fi = 3 i Note: we'll use an oveba to indicate expectation value i a d u () = f() = d u () a o f() = Fi = F cosθ but because these ae centic foces, = 0 and cos = 1 i Fi a = i Fo lage N, we may eplace the sum with an integal. i i i Now, the numbe of molecules located between and + d fom any given molecule is: N () = 4π ρ() but we have N - 1 molecules that we can choose as the efeence. Thus, pais = ( N 1) 4π ρ() Howeve, this double counts each pai. We divide by to get the numbe of unique pais. Again, in the limit as N becomes lage N - 1 = N and the summation in the viial can be eplaced by an integal. We then find the expectation value in the usual way: a = i i 0 a N π exp β a V πa N βa d = exp V 0 d

6 Chapte 5 Page 5. a πa N 1 π( Nk T) = i V i ( βa) = Va πnkt Finally, PV = NkT1 3 av Poblem 5.8: μ Q fo HCl sinθ A f 1 θ A θ B ϕ sinθ B cos( ϕ) cos θ A cosθ B f θ A θ B ϕ f 3 θ A θ B ϕ d 10 7 cm 3 cos θ A3cosθ B cosθ B 1 sinθ A sinθ B cosθ B cos( ϕ) cos θ A 5cosθ B 15 cosθ A cosθ B 4cosθ A cosθ B sinθ A sinθ B cos( ϕ) (a) f 1 ( 0π 0 ) f ( 0π 0 ) 3 f ( π0 0 ) 3 f 3 ( 0π 0 ) μ u d 3 f μq 1( 0π 0 ) d 4 f μq ( 0π 0 ) d 4 f Q ( π0 0 ) d 5 f 3( 0π 0 ) (b) f π π f π π f π π f π π u (c) μ u d 3 f 1 π π μq d 4 f π π μq d 4 f π π f 1 ( 00 0 ) f ( 00 0 ) 3 f 3 ( 00 0 ) Q d 5 f 3 π π u u μ d 3 f 1( ππ 0 ) μq d 4 f ( ππ 0 ) μq d 4 f ( ππ 0 ) Q d 5 f 3( ππ 0 ) u Poblem 5.9: μ Q fo SO sinθ A f 1 θ A θ B ϕ sinθ B cos( ϕ) cos θ A cosθ B

7 Chapte 5 Page 5.7 f θ A θ B ϕ f 3 θ A θ B ϕ π π (a) f cos θ A3cosθ B cosθ B 1 sinθ A sinθ B cosθ B cos( ϕ) cos θ A 5cosθ B 15 cosθ A cosθ B 4cosθ A cosθ B sinθ A sinθ B cos( ϕ) 1 π π f π π f π π f u p ( d) μ d 3 f 1 π π 0 μq d 4 f π π 0 μq d 4 f π π 0 Q d 5 f 3 π π 0 π 3π (b) f 1 0 π 3π 1 f f 1 3π π π 3π f u a ( d) μ d 3 f 1 π 3π 0 μq d 4 f π 3π 0 μq d 4 f 3π π 0 Q d 5 f 3 π 3π 0 d u p ( d) u a ( d) d Fo example, at 10 A sepaation: u p u a Poblem 5.10: 10 8 μ α Q k ε 39.3k T 300 dispesion

8 Chapte 5 Page 5.8 u d 1 4ε u d chage-chage u qq 0 dipole-dipole 1 μ 4 u μμ 3k T u μμ dipole-polaization μ α u αμ u αμ dipole-quadupole u μq kt quadupole-quadupole μq 4 u μq Q 4 u QQ 5k T 10 u QQ total u u d u qq u μμ u αμ u μq u QQ u Poblem 5.11: We use the same paametes as in poblem 5.10, but we will neglect the quadupola pat of the potential. The quadupola potion of the potential does not have the same dependence that would allow factoization into effective LJ paametes. Fom the equations on pages 19 and 17 of the text we have: o ε o 39.3k

9 Chapte 5 Page μ 4 μ α 3 kt F 1 F ε o o ε ε o F ε k 1 ε o F ε k K k cm 8 u eff () 4ε 1 u LJ () 4ε o o 1 o u eff () u LJ () Poblem 5.1: Fo ammonia μ α.10 4 Q 110 T 300 k ε 37.9k Using the angle-aveaged equations: ε 1 F dis 4 F dis μ 4 F μμ kt 7 F μμ μ α F αμ 7 F αμ

10 Chapte 5 Page 5.10 F μq 1 μ Q kt 9 F μq F QQ 8 Q 4 kt 11 F QQ F F dis F μμ F αμ F μq F QQ F dynes 8 Poblem 5.13: τ = o F τ = F sinθ whee is the angle between the vectos (90 degees in this case) and the new toque vecto is in the positive y diection. F dyne cm τ F τ dynecm Poblem.14: U = u aα u aβ u aγ u bα u bβ u bγ u cα u cβ u cγ But, all sites ae equivalent in tems of and, so: u aα = u bβ u cγ = u aβ = u cβ = u bα = u bγ u aγ = u cα U = 3u aα 4u aβ u aγ Now set up constants and LJ functionality. ε 7k d u() 4ε θ π s Now find vaious distances and plug into LJ functionality to get pai inteactions. a : aα 0. u aα u aα u aα a : aβ s d u aβ u aβ u aβ

11 Chapte 5 Page 5.11 a : by Law of Cosines, ac d d d dcos( θ) ac 0.89 aγ ac s aγ 0. u aγ u aγ u aγ Finally, U 3u aα 4u aβ u aγ U egs Poblem 5.14: The potential depends upon the two sites inteacting. As not enough infomation is given in the poblem to completely specify the oientation of the othe inteacting molecule, we will take the easiest case whee an identical molecule is oiented exactly above it so that we have like-like site inteactions The potential is zeo when =. Thus, the line of constant potential fo u = 0 should be dawn as a cicle aound each site of adius /

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