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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
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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
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