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Physical Chemistry II (2013 2nd semaster) 1. Subject code: 458.202 2. Lecturer : Prof. Jihwa Lee 3. Textbook: “Physical Chemistry” 9th Ed. by Atkins 4. Lecture covered: Ch. 20- 23 5. 4 Exams: Ch. 20, Ch. 21, Ch. 22, Ch. 23 Friday 7:00-9:00 pm Credit:3 10/5, 10/25, 11/22, 12/14 6. 7. Grade: Exam (80%) + Homework (20%) All lecture-related information are uploaded in the departmental homepage. http://cheme.snu.ac.kr (lecture board) 8. Presence: ≥ ¾ required, < ¾ → F grade 9. Teaching assistants: 이재한 (880-8941, 010-7337-2345, Rm. 302-513) [email protected] 윤홍식 (880-8941, 010-3629-3643, Rm. 302-513) [email protected] 1 Introduction Two fundamental principles governing natural processes • Thermodynamics * Energy conservation law: dU = dq + dw, ΔU = q + w. * Law of Entropy: dS = dqrev /T. * Direction of change, Driving force of a change, Equilibrium point. • Kinetics * Rate of physical and chemical changes. * Understanding and formulation of the rate in terms of molecular motions. 2 Physical Chemistry II Change 2012. 8. 23 Department of Chemical & Biological Engineering, Seoul national university Ch.20 Molecules in motion • Molecular motion in gases 1. Kinetic model of gases 2. Collisions with walls and surfaces 3. Rate of effusion 4. Transport properties of perfect gas • Molecular motion in liquids 1. Conductivity of electrolyte solutions 2. Mobility of ion • Diffusion 1. Thermodynamic view 2. Diffusion equation 3. Diffusion probability 4. Statistical view 4 Physical & Chemical Changes in gas 1. Unequal pressure at constant T 2. Unequal temperature at constant P 3. Unequal concentrations of species • Equilibrium is attained via gaseous motions involving collisions. • A change can be quantitatively understood (how fast does a change occur?) by modeling the gaseous motions: gas kinetic theory. 5 Gaseous state 1) molecule has a small volume ( d = 2 ~ 4 Å ) 2) random thermal motion 3) collision among themselves and with the container wall. 4) weak attraction at small intermolecular distance 1. V(r) 4 2. 3. 4. attraction stable r 2 1 3 repulsion 6 Maxwellian speed distribution • Energy exchange between gas molecules by ceaseless collisions. • At thermal equilibrium, a time-invariant speed distribution function exists. • f(v) is determined only by a single parameter T. T1 • f(v) dv = N v2 exp(-½ mv2/ kBT) dv, where the normalization constant N = 1/ ∫ ∞ f(v) dv • f(v) is a probability function 0 • Why does the nature adopt such a distribution ? • Is there any simple hidden principle ? T4 >T1 Energy Partitioning NA E = ∑ ½ mvi2 : fixed N,V,T constant i =1 Internal energy U 1 mole Ar Q: How is the energy distributed among NA Ar atoms ? Q: How is the thermal energy absorbed by each mode of N2 motions (translation, rotation, vibration)? 1 mole N2 q Motions of gas molecules • Etotal = Etrans + Erot + Evib+ Eelec translation • Energy quantization E rotation x 0.1 x 0.01 Evib Eelec 0 Etran Erot vibration 9 Ludwig Boltzmann (1844-1906) Vienna, Austria 10 Boltzmann energy distribution law • Pi = Ni /N= exp (-εi/kT)/ ∑ exp (-εi/kT); discrete ε state q =∑ exp (-εi /kT); molecular partition function • P = exp (-ε/kT) dε / ∫ exp(-ε/kT) dε ; continuous ε state q = ∫ exp (-ε/kT); degenerate states • Degenerate states: states with the same energy but with different motions. gi = degenercy • Pi = ni / N = gi exp (-εi/kT)/q , gi = degeneracy • The propobability for a molecule with energy of ε~ ε+ dε dP = g(ε) exp(-ε /kT) dε / q 11 • pi = ni / N = gi exp (-εi/kT)/q = gi exp (- ½ mvi2 /kT)/q • gi = degeneracy • The translational energy εi =½ mvi2 is almost continuous because the energy gap is so small, there ε (v also) can be treated as a continuous variable • f(v)dv = N 4πv2 dv exp (- ½ mv2 /kT) • For a given system f(v)dv is determined by a single parameter T What is temperature? • A parameter governing energy partitioning in a system consisting of many molecules (atoms). …… Maxwellian speed distribution Another way to obtain f(v)dv • Consider 1D velocity distribution fx(vx)dvx • ε = ½ mvx2 is non-degenerate, so g =1 for all vx • According to the Boltzmann distribution law, fx(vx)dvx = (m/ 2πkT)1/2 exp(-½ mvx2 /kT) dvx cf: ∫ exp(-½ mvx2 /kT) dvx = (2πkT/m)1/2 • <vx> = (2kT/πm)½ θ φ • fx(vx), fy(vy), and fz(vz) are independent with one another and have the same function. • fx(vx) fy(vy) fz(vz) dvxdvydvz = (m/2πkT)3/2 exp {-½ m(vx2 +vy2 +vz2 )/kT} dvxdvydvz velocity space • Expressing it in polar coordinate (v,θ,φ), f (v,θ,φ) dvdθdφ = (m/2πkT)3/2 exp(-½ mv2)/kT) v2sinθ dvdθ dφ • Integrating over dθ and dφ, f (v) dv = 4π (m/ 2πkT)3/2 exp(-½ mv2)/kT) v2 dv = 4π (M /2πRT)3/2 exp(-½ mv2)/kT) v2 dv ; text eq. (20.4) • Mean speed = <v> = ∫0,∞ v f(v) dv = (8RT/πM) 1/2 • Most probable speed c* = (2RT/πM) ½ ; df /dv│v=c* = 0 • Root mean square speed crms= (<v2>)½ = (3RT/M) ½ 13 Relative speed vrel • In treating gas collisions, the relevant speed is the relative speed vrel. • rrel = r2 - r1 • C is the center of mass (CM): • R = r1 + (m2/ M) rrel, where M = m1+m2 • r1 and r2 scan be expressed in terms of R and rrel ; r1 = R - (m2/ M) rrel , r2 = R + (m/ M) rrel • Differenciating with time, v1 = V - (m2/ M) vrel , v2 = V + (m1/ M) vrel • E = E1 + E2 = ½ m1v12+ ½ m2v22 = ½ (m1+ m2) V2 + ½ {(m1m22+ m2m12)/ M2} vrel 2 = ½ M V2 + ½ (m1m2/ M) vrel 2 = ECM + Erel The reduced (effective) mass of the relative motion is μ ≡ m1m2/ M • p = p1 + p2 = m1v1+ m2v2 = m1 {V - (m2/ M) vrel } + m2 {V + (m1/ M) vrel } = MV • Motions of two particles can be considered the combination of the CM motion and the relative motion, which are independent. 14 How to get f(vrel) and <vrel> ? 1. Separation of motions: CM motion + relative motion • The two motions are independent of each other. • Then, the relative motion is simply that of a single particle with a mass μ. • f (vr) dvr = 4π (μ/ 2πkT)3/2 exp(-½ mvr2)/kT) vr2 dvr • <vr> = ∫0,∞ vrel f(vrel) dvrel = (8kT/πμ)1/2 • For collisions between the same molecules (m1= m2 =m), μ = m2/2m = m/2. Thus, <vrel> = √2 <v> = √2 2. Analytical method (Fig.1) • WhileV1 is fixed, -V2 can have any orientation in space. • Since Vrel = V1 -V2 • The speed │Vrel │ = {2v2 -2v2 cos (π-θ)}1/2 = √2 v (1+cos θ) • Averaging over the all orientations, <vrel> = √2 <v>[∫ 0, 2π dφ ∫0, π (1+cos θ) sinθ dθ] / 4π = 2-1/2 <v> ∫0, π sinθ dθ = √2 <v> [-cos θ]0, π = √2 <v> = √2 • This is only a special case in which v1 = v2 = v. • A general is more difficult to average but should give the same result. Fig.1 15 Concepts & terms related to gaseous collisions Number density N ≡ # of gas molecules (atoms) per unit volume • For a ideal gas pV =nRT =NkT, so N = N/V = p/kT d Collision diameter d & collision cross section σ (Fig. 2) • Collision is a relative motion between molecules, so one can consider that one is fixed while the other is moving with. • Molecule can be effectively considered a sphere of radius d. • Collision occurs when the center of the projectile (approaching molecule) lies between the two lines parallel to vrel. Collision diameter d = 2(r1+r2) = d1+d2 Collision cross-section σ = πd2 • Effectively speaking, collision is made between a moving sphere (projectile) of diameter d and a stationary point molecule (target). •The efficiency of collision is determined by the cross-section(area) σ. Collision frequency z ≡ # of collisions made by one molecule per unit time (1sec) • A molecule with a collision cross-section σ travels a distance <crel> in 1 sec. • The volume swept by the moving projectile is that of a curved circular cylinder of a length <crel>. • All the point molecules within the volume are hit by the projectile. • Then, the collision frequency z = <crel> σN = <crel> σ p/kT 16 Collision number Z ≡ # of collisions within a unit volume • Every and each molecules in unit volume experiences z collisions. • Then, Z = z N /2 = <crel> σ (p/kT)2/2 • The factor ½ is needed because each collision is counted twice. Mean free path λ ≡ average distance traveled between two successive collisions l2 • λ = lim(n→∞) Σ1,n (li /n) n l1 • λ = <crel> /z = <c> / <crel> σ p/kT = kT/√2σ p l3 l4 • Calculation of λ: examples N2 at T = 298K and p = 1atm → λ = (1.38x10-23J/K)(298K)/(0.43x10-18m2)(133 Pa/Torr)(760 Torr/atm) = 9.5x10-8 m = 95 nm Probabilistic interpretation of λ l5 most recent collision • Let p(x) ≡ probability to experiencing a collision after traveling a distance x without experiencing a collision. • Then, p(x+dx) = p(x) (1- αdx), where α is the constant of proportionality. • (1-αdx) is the probability of not experiencing a collision within dx. • dp(x) = p(x+dx) – p(x) = - αdx p(x) → dp(x)/ p(x) = d ln p(x) = - α dx. • Upon integrating, [ ln p(x)]1,p(x) = [- αx/λ]0,x → ln p(x) = - αx/ → p(x) = e –αx • Since p(x) is a probability function N ∫0,∞ e –αx = - 1/α[e –αx] 0,∞ = - N/α (0-1) = N/α =1 • N = 1/ α and p(x) = αe –αx • mean free path λ = <x> = α ∫0,∞ xe –αx = 1/α • p(x) = (1/λ) e –αx/ λ 17 Wall (surface) collision rate Zw • Zw plays an important role in thermal conduction by gas and in gas adsorption on a surface. • Zw ≡ # of molecules striking a wall /cm2 s • Zw = 1/2N │<vx>│∆t A/ A ∆t = 1/2N <vx> = 1/4N <v> = 1/2 (p/kT)(2kT/πm)½ = p/(2πmkT)½ • Zw √m, p, √T Rate of effusion • Effusion: a phenomenon in which gas molecules exiting into vacuum through a small orifice under a condition λ > d (diameter). • The exiting molecules do not experience any collision in going out through the hole. • The exiting molecules have a speed distribution which is the same as that in the interior. Such a gas flow is called molecular or Knudsen flow (collisionless). • Effusion rate = Zw A0 = p A0 /(2πmkT)½ √m • It can be used in gas separation. λ p, T 18 Angular distribution Rate of effusion • Zw = 1/2N │<vx>│A =1/2N <vx> =1/2N<v cosθ> • Zw (θ) cosθ Thin film deposition cosθ distribution of an effusing gas • E-beam evaporation • Thickness uniformity filament heater schematic diagram of an E-beam evaporator • Film thickness ∝r-2 cos2 θ = (2r0 cos θ) -2 cos2 θ = 1/4r02, which is constant on every point on the surface of the sphere. • So, thickness uniformity can be obtained by placing the wafers as shown in the figure. 19 Transport properties of perfect gas 1. Non-uniform N → ∂N(z)/ ∂z ≠ 0 → mass transport ; diffusion 2. Non-uniform T → ∂T(z)/ ∂z ≠ 0 → energy transport ; thermal conduction 3. Non-uniform vx → ∂vx(z)/ ∂z ≠ 0 → momentum transport ; viscosity Flux (J) ≡ quantity transported /unit area and time • Transport has a direction, so J is a vector quantity. • Empirical observation: J ∝gradient Hard sphere model (simple approximation) • Atoms (molecules) are treated as an impenetrable (hard) sphere of diameter d. • This is equivalent to assuming an intermolecular potential shown below. • Intermolecular force is neglected. V(r) = 0: r ≥ d = ∞: r< d σ = 4πd2 real potential hard sphere model 20 Diffusion • Jz = - D ∂N(z)/ ∂z :1D , J = - DN(r) :3D Fick’s 1st law • Modeling: consider the net flux across a plane at z= 0 D =½ λ : diffudion coefficient • The model above is only an approximation because it assumes that all the molecules impinge upon the plane at z = 0 perpendicularly. • In reality, molecules strike the surface at an angle θ. • Correcting for this effect, Jz = - ½ λ (dN/ ∂z)0 <cos2θ> • <cos2θ> = 2/3. Therefore, • D = (1/3) λ θ 21 Thermal conduction Energy ε = νkT : thermal conductivity Viscosity η: viscosity = kg m-1 s-1 = [(kg ms-2)/m2] m/s • 1 P (Poise) = 0.1 kg m-1 s-1 • 1 cP = (1/100) P: used for liquid • 1μP = 10-6 P: used for gas 22 23 cP 24 25 v p dependence of η • λ = = kT/√2σ p ∝1/p ∝and N = p/kT ∝p • η ∝ λN is independent of p • However, η↑ with ↑p at high p and ∝η↓ with ↑p at low p. • Deviation at high p is due to attractive intermolecular force. • Deviation at low p is due to that λ < l (characteristic system dimension), in which λ becomes independent of p. • Then, η ∝N ∝p only. T dependence of η • N ∝ 1/T, λ ∝T, ∝T1/2, so η ∝ T1/2. • However, experimental data (red curve) shows that the T-dependence of η is sharper than T1/2. • As T ↑, also increases. • A faster molecule can approach more closely to the center of the collision partner, which effectively makes σ smaller, hence λ larger. • As a result, η ↑ more sharply than T1/2 predicted by the hard sphere model. εk = ½ mv2 26 Fluid flow down a circular tube • Δp = (p2 - p1) causes the flow. • In a Laminar flow, there is a friction force (viscous drag) between two adjacent layers given by • F/A = η (dvz/dr), where A is the boundary area. (kg m-1-s-1)(ms-1) = (kg m-s-1)(ms-1)m-2 = force/ unit area • Ffriction = A(Fin – Fout) = 2πl η [r(dvz/dr)r – (r+dr)(dvz/dr)r+dr] = 2πl η [rvz’- (r+dr) {vz’(r) + vz’’(r)dr}] = - 2πl η (vz’+ rvz’’) dr • In the steady-state, this force is balanced by the pressure force Fpressure = Δp 2πrdr • - 2πrl η (vz’+ rvz’’) dr = Δp 2πr dr • Integrating over r, - rvz’ = (Δp/2ηl ) r2 + C1 • The boundary condition: vz’ = 0 at r =0, where vz is maximum → C1 = 0. • Thus, - vz’ = (Δp/2ηl ) r. • Integrating once more, - vz = (Δp/4l η) r2 + C2 • The boundary condition: vz = 0 at r = R. Molecules sticks at the wall, where vz = 0 → C2 = - (Δp/4l η) R2 • vz = (Δp/4l η)(R2 - r2); Poiseuille equation • parabolic velocity profile a) b) a) Knudsen (molecular ) flow, b) b) Laminar (viscous) flow Cross section 27 Volume flow rate dV/dt • dV/dt = ∫0,R 2πrvz dr = ∫0,R 2πr dr = Δp (R2 - r2) /4l η dr = πΔp R4/8l η Throughput Q • Q = p1 dV1/dt = p2 dV2/dt = p(z) dV(z)/dt, where 1, 2, and z are the inlet, outlet, and any point in between. • At constant T, PV = NkT→ N = PV/kT • Since p(z) does not varies with t, Q = d(pV)/dt = kT (dN/dt) = constant along the tube. dp Pressure drop along the tube • p decreases with ↑z along the tube. • Since Q= p dV/dt is constant, dV/dt increases with z. • Considering a small volume element as shown in the figure, dQ = p dV/dt = (πR4/8l η) p(-dp) • Integrating the above, Q = -(πR4/8l η) ∫p1, p2 pdp = (πR4/16l η) (p12 – p22) = (πR4/8l η) pavg(p1–p2), where pavg = (p1+ p2)/2 • Alternatively, Q = - (πR4/8l η) ∫p1, p pdp = (πR4/16zη) (p12- p2) • (p12- p22)/ l = (p12- p2)/z • p(z) = [{p12l -z (p12- p22)} /l ]1/2 R z Z= 0 dz p p1 p2 0 l z 28 Viscosity of liquids • In liquid, molecules are surrounded by the Sulfuric acid 27 neighboring ones, which exert attractive force. • Molecular hopping requires an activation energy Ea. • Consider a liquid contained between two plates, one is moving and the other is fixed. • When the plate is moving with the 1st layer, the 2nd layer molecules (eg. blue circle) feels tangential force // to v by the attractive interaction with the top molecules. • As a result, the blue molecule acquires extra kinetic energy in the direction of v. • Thus, the energy barrier for hopping to the right is lowered, while that to the left is increased . • The probability P(E > Ea) ∝e- Ea/RT∝ mobility η∝e-Ea/RT 100 K η of liquid Ar at 35 and 190 atm. • η ∝e Ea/RT • Experimentally, this is widely observed. H2O v fixed r0 29 Molecular motion in liquids Conductivity of electrolyte solutions • Ohm’s law, conduction & conductivity R = ρl /A; ρ = resistivity (Ω m) V = IR = I ρl /A → V/l = σ(I/A), E = σJ • σ = 1/ρ : conductivity (Ω-1m-1) SI unit = Sm-1 • S (siemens) = Ω-1 • G = 1/R : conductance (Ω-1) strong electrolyte Molar Conductivity Λm Λm ≡ κ /c , where (κ = σ) • Ionic conduction • Charge carrier: (+) and (-) ions • Degree ion ionization α • α ~1 for strong electrolyte and α <<1 for weak electrolytes. • Empirical relation for strong electrolytes weak electrolyte Λm = Λm0 – K c1/2 : Kohlausch’s law • Λm0 = limc→0 Λm : limiting conductivity Λm0 = ν+ λ++ ν- λ- ; Law of independent migration of ions • MgCl2: ν+ = 1, ν- = 2 30 Weak electrolytes • Weak electrolytes are not fully ionized in solution: ex: CH3COOH, NH3 data Measurement of Λm0 ; y =(a + bx) type • 1/ Λm vs. c Λm plot • Intercept = 1/ Λmo • Slope = Ka(Λmo)2 → Ka is obtained. • At high c the data deviates from the linear dependence because of ion-ion interactions. 31 The mobilities of ions a) Drift speed, mobility • Felec = qE = zeE = ze (V/l ) • Ffric = fv; f = friction coefficient • At steady-state, the two forces are balanced, namely Felec + Ffric = 0. • zeE = f v • f = 6πηa : Stokes law a = hydrodynamic radius of solvated ion • vd = zeE/ f = zeE/ 6πηa ; drift speed • Mobility u = vd/E = ze/ 6πηa b) Mobility and conductivity • J(ion) = (A vd Δt) ν cNA / A Δt = vd ν cNA • J(charge) = zevd νcNA = zvd νc F(arad) 1F = NAe = 96416 C(oulomb) • J(charge) = zuEνcF • J = κ E → κ = J/ E = zuνcF • Applying this equation to the cations(+) and anions(-) in the limit of c→0, Λm0 = ν+λ+ + ν-λ- = (z+ u+ ν++ z- u- ν-) cF • For a symmetrical z:z electrolyte (e.g. CuSO4 with z=2), Λm0 = ν+λ+ + ν-λ- = z (u+ ν++ u- ν-) cF 32 Ionic atmosphere Conductivity and Ion-ion interactions • Λm = Λm0 – K c1/2 : Kohlausch’s law • K = A + B Λm0 : Debye-Hückel- Onsager theory A∝ z2 /ηT1/2, B ∝ z3/T3/2 Debye-Hückel theory • Ionic atmosphere of radius rD • Shielded Coulomb potential : φi = (zie/4πεr) exp (– r/rD), • The free energy of an ion is lowered due to the presence of the ionic atmosphere of the opposite charge. • activity a = γ b ~ γ c (γ = activity coefficient) • For ions in solution γ± = (γ+ γ-)1/2 : mean activity coefficient • log γ± = - A│z+ z-│ I1/2 , where A = 0.529(in H2O at 25o C) I = ½ ∑ zi2 (bi/bo) : Ionic strength → I ∝ c1/2 Coulomb potential ∝ 1/r • Ion-ion interactions affect the conductivity in two ways; 1) Relaxation effect due to a asymmetrical ionic atmosphere. 2) Electrophoretic effect due to viscous drag by oppositely moving ionic atmosphere • Felec + (Ffric + Frelax + Fep) = 0 E=0 E≠0 33 Diffusion • J = - D ∇c (3D), Jx = - D ∂c /dx (1D); Fick’s 1st law of diffusion • Mobility of ion u = vd /E = ze/ 6πηa in E field • u and D are not independent and inter-related. Thermodynamic view of diffusion • In a mechanical system the force acting on a particle is F = - ∇V • In an electrical system E = -∇Vel (V = electrical potential: volt) • In analogy to these systems, diffusion can be viewed as material transport driven by a thermodynamic force given by Fther = - ∇μ. μ = μo + RT ln c → Fther = - RT (dc/dx)/c • Ffric = NA f v (1 mole) • Under a steady-state condition, Fther+Ffric = 0 RT (dc/dx)/c = NAf vd and vd = RT (dc/dx)/c fNA • Then, J = cvd = - RT (dc/dx)/ f ………..(1) • Fick’s 1st law: J = - D (dc/dx) ………………….(2) • Comparing the two equations above, D = RT/ f (1 mole), D = kT/ f (1 particle) ; Stokes- Einstein relation • Then, D = kT/6πηa • qE = zFE (1mole)= f vd and mobility u = vd /E = zF/f • Therefore, D = uRT/ zF ; Einstein relation 34 Diffusion Equation • Flux (J) ≡ quantity transported /unit area and time • Jx = - D ∂c /dx (1D); Fick’s 1st law of diffusion • dJx = Jx+dx - Jx = - D [(∂) x+dx - (∂c/∂x) x] = - D (∂2c /∂x2) dx • ∂Jx /∂x = - D (∂2c /∂x2) • Influx – outflux = # molecules increase (decrease) in unit volume/sec = concentration /sec • ∂c/∂t = D (∂2c /∂x2); Fick’s 2nd law of diffusion • J = - D∇c (3D) •∇ ∙ J = -∇∙(D∇c) = - D∇2c • ∂c/∂t = D∇2c (r,t) (3D) Jx Jx+dx x x+dx Convection • Transport of material due to motion of fluid is called convection. • Jxconv = cvx , (3D: Jconv = cv ) • Assuming vx is constant , ∂Jxconv /∂x = - vx (cx+dx - cx) = - vx (∂cx/∂x) • ∂c/∂t = - vx (∂c/∂x (3D: ∂c/∂t = - v∙∇c • Combining diffusion and convection, ∂c/∂t = D∇2c - v∙∇c: generalized diffusion equation • If a reaction occurs, it has also to be included. For a 1st-order reaction - kc term is included. • This is the key equation in designing a chemical reactor. • Finite element method in carrying out computer calculation. 35 δ(x) = ∞; x = 0 = 0; x≠ 0 ∫-∞,∞ δ(x) dx =1 Solutions of Diffusion Equation • ∂c/∂t = D (∂2c /∂x2) • Solution c = c(x,t) 1-D Delta-function (point) source • c(x,0) = (n0/A)δ(x), where n0 is the total # of moles present. • The solution is a Gaussian function c(x,t) = (n0/A)(4πDt)-1/2 exp(-x2/4Dt) • The normalized function p(x) = c(x,t)/(n0/A) is the probability density function with σ = (2Dt)1/2. This is the same type as the normal distribution function (see Figure) • Standard deviation σ ≡ <(x- <x>)2>1/2 • The peak width increases with time t. • <x> = 0 because c(x,t) is an even function. • <│x│> = 2 ∫ 0,∞ x p(x) dx = 2(Dt/π)1/2 • xrms = <x2>1/2 = (2Dt)1/2 : diffusion distance • In 3-D diffusion rrms = <r2>1/2 = <x2 +y2 +z2>1/2 = (6Dt)1/2 3-D Delta-function (point) source • c(x,0) = n0 δ(r) • Solution is c(x,t) = n0(4πDt)-3/2 exp(-r2/4Dt) • rrms = <r2>1/2 = <x2 +y2 +z2>1/2 = (2Dt)1/2 normal distribution function 36 Statistical view of diffusion model of 1D random walk • N = total # of walks N = m(→) + n(←) Falling table tennis balls array of tacks • m = # of the walks to the right (→) • n = # of the walks to the left (←) • p = probability of walking to the right(→) = ½ • q = probability of walking to the left(→) = ½ • step distance (보폭) = λ • step time = τ → → ← → ← ← → → → ← → ← ← ….. • x = (m-n)λ after time t = Nτ • # of the possible event = W(m,m) = N! / m! n! (p+q)N = Σ m=0,N NCm pm qN-m = NC0 qN +NC1 p qN-1 +NC2 p2 qN-2 + ……∙NCN-1 pN-1 q +NCN pN = 1 • W is a symmetric function with a maximum at m* = N/2. • p(m) = W(m) /Σm W(m) = (1/√2π) exp[-(m-m*)2 /m*] • Stirling’s approximation ln N! = ln(2π)/2 + (N+1) lnN- N is used. • This treatment leads to p(x,t) = (2τ/πt)1/2 exp(-x2 τ/2tλ2) • By comparing exp(-x2 τ/2tλ2) with that of the solution of the point source exp(-x2/4Dt), 1/4D = τ /2λ2. Therefore, D = λ2 /2τ : Einstein –Smoluchowski equation 37