Chapte 5 Page 5.1 CHAPTER 5 Poblem 5.1: 1 (a) u () 4 0.90.93 3.0 (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 = 4 1 7 Now in dimensionless fom: 13 7 f() 4 3 u () 1 f() 0 1 3 0 1 3 4 (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
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 1 1 13 7 = o min min 1.1 4 Initialize to find oot 1 esult oot.5.001 esult 3.984 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.
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 = 11.85 DT ( ) 3.849 1 0.0081T 1 1 T 00 400 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. 3.5 3.4 DT ( ) 3.3 3. 3.1 0 100 00 300 400 Poblem 5.4: Fom Table A5.5 fo wate k 1.38010 1 eg debye 10 18 egcm 3 K T 3 0.5 o 3.3910 8 cm ε o 09.1kK μ 1.85debye α 1.49 10 8 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 () 0.90.95 3 800 Note that the potentials plotted ae in units of 10-1 egs u LJ () u D () u P () 400 0 u tot () 400 Poblem 5.5: 800 0 1 3
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 3.505 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 11.85 3.49 150.8 48.1 K 174.8 3.84 09.4 54.3 CH 4 1.78 3.78 190. 45.4 Xe 4.83 4.07 89.7 57. C H 74.48 4.1 305.4 48. 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:
Chapte 5 Page 5.5 1.013510 conv 10 4 cm 3 conv 7.337 10 3 K atm eg cm 3 atm 1 1.38110 1 eg K FLUID A=/(kTc) B=(conv)( 3 )/T c A 0.808 0.094 K 0.834 0.095 CH 4 0.875 0.087 Xe 0.77 0.098 C H 0.898 0.087 Thus, ε = k 0.84Tc Aveage 0.84 0.09 3 = 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
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 1.0810 18 Q 3.810 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) 3 4 1 5 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 1 0 1.5 f π π 0 1.591 f π π 0 1.591 f π π 3 0 4 4 4 4 4 4 4 4 u 5. 10 3.53 (c) μ u d 3 f 1 π π 0 4 4 μq d 4 f π π 0 4 4 μq d 4 f π π 0 4 4 f 1 ( 00 0 ) f ( 00 0 ) 3 f 3 ( 00 0 ) Q d 5 f 3 π π 0 4 4 u 1.35 10 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 1.4 10 15 Poblem 5.9: μ 1.310 18 Q 1.710 fo SO sinθ A f 1 θ A θ B ϕ sinθ B cos( ϕ) cos θ A cosθ B
Chapte 5 Page 5.7 f θ A θ B ϕ f 3 θ A θ B ϕ π π (a) f 1 0 3 cos θ A3cosθ B cosθ B 1 sinθ A sinθ B cosθ B cos( ϕ) 3 4 1 5 cos θ A 5cosθ B 15 cosθ A cosθ B 4cosθ A cosθ B sinθ A sinθ B cos( ϕ) 1 π π f 0.755 10 1 π π f 0.755 10 1 π π f 3 0.5 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 0.49 10f 1 3π π 0 4.59 10 1 π 3π f 3 0.5 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 510 8 10 8 310 7 310 14 10 14 u p ( d) u a ( d) 110 14 0 110 14 10 14 0 110 7 10 7 310 7 410 7 d Fo example, at 10 A sepaation: u p 10 7.7 10 15 u a 10 7.59 10 15 Poblem 5.10: 10 8 μ 1.310 18 α 3.710 4 Q 1.710 3.89510 8 k 1.380110 1 ε 39.3k T 300 dispesion
Chapte 5 Page 5.8 u d 1 4ε u d 1.41 10 14 chage-chage u qq 0 dipole-dipole 1 μ 4 u μμ 3k T u μμ 1.18 10 15 dipole-polaization μ α u αμ u αμ 4.37 10 1 dipole-quadupole u μq kt quadupole-quadupole μq 4 u μq.08 10 1 14 Q 4 u QQ 5k T 10 u QQ 9.341 10 18 total u u d u qq u μμ u αμ u μq u QQ u 1.599 10 14 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 3.89510 8 ε o 39.3k
Chapte 5 Page 5.9 1 μ 4 μ α 3 kt F 1 F 1.108 4ε o o ε ε o F ε k 1 ε o F ε k 453.047 K k 3.89 10 cm 8 u eff () 4ε 1 u LJ () 4ε o o 1 o 3.10 8 3.710 8 810 8 10 13 u eff () 110 13 u LJ () 0 110 310 8 410 8 510 8 10 8 710 8 810 8 Poblem 5.1: Fo ammonia.9310 8 μ 1.4710 18 α.10 4 Q 110 T 300 k 1.380110 1 ε 37.9k 810 8 Using the angle-aveaged equations: ε 1 F dis 4 F dis 3.74 10 8 μ 4 F μμ kt 7 F μμ 1.07 10 8 1μ α F αμ 7 F αμ.794 10 9
Chapte 5 Page 5.10 F μq 1 μ Q kt 9 F μq.1 10 10 F QQ 8 Q 4 kt 11 F QQ 7.873 10 1 F F dis F μμ F αμ F μq F QQ F 5.18 10 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 710 7 dyne 1.5310 8 cm τ F τ 1.071 10 14 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 0.393 d 0.153 u() 4ε 1 109.47 θ π s 0. 180 Now find vaious distances and plug into LJ functionality to get pai inteactions. a : aα 0. u aα u aα u aα.83 10 15 a : aβ s d u aβ u aβ u aβ.404 10 15
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 1.594 10 15 Finally, U 3u aα 4u aβ u aγ U.139 10 14 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. 1 0.393 1.51 0.588 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 /. 1 1 1 0.19 0.94
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