HistoricallyDiffusion Chapter 18General process of flowHeat as the exampleFick’s First lawUse of diffusion Coefficientspersonal passive samplers,SO2 accommodation coef.ParticlesProbability distribution using random movementDiffusion distancesFick’s Second LawDiffusion in a GC columnDiffusion in a sphereEstimating Diffusion Coefficientsgaseswater; liquidsTurbulent DiffusionLake systemAtmospheric System1HeatTemplow2 Temphigh1xThe heat that flows thru a slab of material is proportional to the cross sectional area, A, of the slab and the time, t, for a given TempHeat flow is also  to Temp/x for a given A and time if Temp/x is smallΔxΔTAΔtΔq if we think about really small thickness of xdqdtkAdTdxdq/dt = the rate of heat transfer with respectto time dT/dx = the temperature gradientk = thermal conductivity2dqdtkAdTdxk has the units of meter/Cmetersec/Kcalo2material kAl 4.9x10-2Steel 1.1x10-2Pb 8.3x10-3air 5.7x10-6glass 2.0x10-4Temphigh1Templow2 Lat steady state for a const. temp gradient across the rodLTTkAtΔq213Diffusion1b. cool 1a. hot2b. low voltage 2a. high voltage3b. low mass 3a. high mass4b. low pressure 4a. high hydrostatic pressureflux = flow area-1 time-1a gradient drives the flow4page 184 table 9.1Page 184 Table 9.15Let us consider a gas diffusing in into a zone where it is constantly collected or removedO3O3O3O3[O3]O3O3x inlet x xODFluxO33 = areatimeOmoles-)()(3;( )( )[ ]moles Otime areaDOx3 3-If we measure the # of moles of O3 collected over a period of time; know the diffusion coef. for O3 in air, [O3] can be calculated 6Diffusion and sticking coefficientsthe average speed of gas molecules is given bycRTMw8The rate of collisions per unit time with a wall of surface A in a given volume is rate = 1/4 c x area x concgasrateRTMwA Cgas -2 rate/A = # molecules time-1 area-1 = fluxif we think about the # of effective collisions, i.e. the ones that actually stick to the wall, a factor is introduced called “sticking” factorsurface recombinationaccommodation coefficient removal rateRTMwA Cgas - -2 Judeikis et al. were interested in the effectiveness of coal surfaces in the uptake of SO2 gas.7rate/A = # molecules time-1 area-1 = fluxflux = radial velocity Cgas D C/rr SO2coal soot coatingmeasure SO2on surface-D d [SO2]/d r = rad. vel x [SO2] ln {[SO2]/[SO2,o]} = rad. vel x  r /D ln {[SO2]/SO2,o]} = krate t8Removal of SO2 along a tube reactor coated withfly ash. The accommodation coef. = 4.4x10-4. Thetotal pressure was 55 torr, with O2 = 6 torr andSO2 = 9m torr; % RH= 0 9The Stokes-Einstein Equation (particles)Let us now think of diffusion in terms of chemical potentialWe can think of the driving force of diffusion as the negative gradient of the chemical potential, i.e.d/dx = free energy/mol /dxThe frictional force resisting the flow, due to an imbalance in chemical potential, is the frictional coef. f (force/velocity) on each molecule x the velocity, v, of the flow, for a mole this force is f vNof v No= -d/dxFlux has the units of moles, molecules or mass per area per time Flux = moles/(cm2 time)Conc x velocity = moles/cm3- x cm/time = moles/(cm2 time)We can define a diffusion velocity caused by a driving force orchemical potential, or the concentration gradient such that:C v = Fluxrecall 10dRTPdp substituting for P from PV=nRT, and n/V = CP= CRT dRTCdC f  Nov = -d/dxvRTN CdC dx- -f0/C v = Flux = -RTNdC dxf0/11FluxRTNdC dx-f0/diffusion coef D= RT/(f No)Stokes (including Cunningham’s slip factor) showed that for unit spheres and nonturbulent viscous flow that the resisting force on a particle flowing through a fluid isf = 6r/Cc where = is the viscosity of the medium (poise)air(20oC) = 1.83x10-4 g/(cm sec) r is the radius of the particles and Cc = 1+/r(A+ Qe-rb/) where  is the mean free path of air = 0.067 mparticle Ccsize (m)0.01 22.20.05 4.970.1 2.870.25 1.690.5 1.331 1.165 1.0312The Randomness of DiffusionConsider 17 boxes arranged in a row PAGE 185 FIGURE 9.213page 186 figure 9.3fit to random walk distribution Normal Gaussian Distribution p mnmnn( ) exp///2221 22p xx( ) exp/12 21 222  14Where we would like to go with this is relate the sigma, , which is a basic feature of the normal distribution to the diffusion coefficient D p xx( ) exp/12 21 222   p mnmnn( ) exp///2221 2222= n/22/nif we multiply this by x, the distance across a box,  can be related to an actual distanceWe will then calculate the flux across from one box to anotherThe concentration gradient which caused the fluxSubstitute into Flux = -D dC/dt15BOX # 3 4 5 6x40 0 8 5t 60 0 24 0 4 6t 0 42 0 14 7t0 28 0 7 8tIf we look at 5th box in seventh time step (7t), 12 particles are entering box #5 from the left and2 from the rightat the eighth (8t) step, 7 particles leave box #5 and go back to box #4this gives a net flux of 5 particles between the 7th and 8th steps or Flux = 5/2tWe now define an average change in concentration per length between adjacent boxes because for every step one box is emptied and the adjacent one filledC = N/xthe spatial conc. gradient dC/dx is dC/dx = C/x = N/x216If we look at time step 6, the gradient dC/dx driving the diffusion between boxes for time step 7 is dC/dx = -N/x2 = -(24-4)/x2recalling that F = -D dC/dx and F= 5/(2t)txxtD8)()/(20)2/(522t= nt and x = n1/2/2 xx = (2Dt)1/2for three dimensional movements = (4/Dt)1/217Example How long does it take for a gas molecule of biphenyl and 0.25 m particle to diffuse from the center of a 5 cm sphere to the a walls of the sphere? Assume a diffusion coef of 0.06 cm2/sec for biphenyl and 1.6x10-6 cm2/sec for the particle.2.5 cmfor biphenyl s = (4/Dt)1/21sec2cm4x0.062cm2.53.14xDSt42t = 82 secondsfor the aerosol, D= 1.62x10-6 cm2/sec(d= 0.25 m) t = 35.5 daysin water biphenyl diffusion is much slower than in airD10-6 sot = tens of days18Fick’s