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Gas/Particle Partitioning. Why is gas/particle partitioning important? Dispersion of Pollutants Introduced into the Atmosphere as Determined by Residence.

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Presentation on theme: "Gas/Particle Partitioning. Why is gas/particle partitioning important? Dispersion of Pollutants Introduced into the Atmosphere as Determined by Residence."— Presentation transcript:

1 Gas/Particle Partitioning

2 Why is gas/particle partitioning important? Dispersion of Pollutants Introduced into the Atmosphere as Determined by Residence Times (Husar and Patterson, 1980)

3 Washout by rain: Washout consists of capture of gas and particle phases Gases equilibrate with rain via Henry’s Law (reversible air-water exchange) Particles are irreversibly captured For PCBs: W g ~ , W p ~ 10 4 – 10 5 For PAHs: W g ~ , W p ~ 10 3 – 10 6

4 Atmospheric reactions Many (most?) organic species are reactive in the atmosphere via direct or indirect photolysis Half-lives of PAHs: Gas phase = minutes-hours Particle phase = hours-days How does this contradict figure 1? Health effects: Particles can deposit in lungs. What role does chemical makeup of these particles play?

5 Background of gas/particle distribution Parameters affecting G/P distribution: 1.Chemical’s properties (liquid VP, Koa) 2.particle characteristics: size distribution, carbon content, composition 3.Atmospheric conditions: T, RH

6 How do we measure G/P partitioning? Traditional hi-vol air sampler uses a filter to capture particle phase and then an adsorbent to capture gas phase Filters: GFF, QFF, PTFE Adsorbents: PUF, polymer resin, tenax

7 Denuder under laminar flow, gases diffuse faster, so adsorb to surface best for hi conc and lo MW gases (NO x, etc)

8 Impactor MOUDI micro-orifice uniform deposit impactor nozzle size and thus air velocity changes in each chamber 8-10 stages 0.2 – 0.3 um cutoff filter gets the rest

9 Cyclone

10 sampling artifacts? Sorption of gases to filters or particles during sampling Volatile losses from filter, particles, impactor stages especially at high T, high flow (I.e. large pressure drop Effect on partition coefficient?

11 Theory: adsorption vs. absorption

12 Particle phase concentration dynamics? Adsorption (Pankow, 1987) Absorption (Pankow, 1994) K p = mass normalized partition coefficient C g, C p = gas and particle phase concentrations N S = # of adsorption sites Q l, Q v = enthalpy of desorption and vaporization a TSP = aerosol surface area TSP = total suspended particle mass/volume f om = fraction of organic matter MW om = molecular wt of NOM  om = activity coeff. of SOC in OM

13 If absorption is important, K oa may be a better predictor of K p those previous equations used p as the descriptor variable. we could also use K oa : K oa = octanol-air partition coefficient f om = fraction of organic matter MW om, MW oct = molecular wt of NOM or octanol  om,  oct = activity coeff. of SOC in NOM or octanol  oct = density of octanol

14 Adsorption Junge, 1977 (based on linear Langmuir isotherm) A limited number of sites, S, of which S 1 are occupied and S 0 = S – S 1 are unoccupied Rate of evaporation = k 1 S 1 Rate of condensation = k 2 C g S 0 At eqbm: k 1 S 1 = k 2 C g S 0 = k 2 C g (S- S 1 ) S 1 /S =  and b = k 2 /k 1 S 1 (C s ) (C g )

15 Surface area of particles  amount sorbed/ total  (cm 2 /cm 3 ) Curved lines are constant VP (mm Hg) c = energetics term (strength of interaction)

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17 Temperature dependence Yamasaki (1982) semi-volatile compounds have significant conc in both the gas and particle phases

18 Adsorption: Pankow 1987 K p = mass normalized partition coefficient C g, C p = gas and particle phase concentrations N S = # of adsorption sites Q l, Q v = enthalpy of desorption and vaporization a TSP = aerosol surface area Linearize by taking log of both sides: Pankow suggests that at eqbm, m = -1, assuming: N s does not vary within a compound class Q l -Q v does not vary within a compound class

19 Absorption: Pankow 1994 TSP = total suspended particle mass/volume f om = fraction of organic matter MW om = molecular wt of NOM  om = activity coeff. of SOC in OM Linearize by taking log of both sides: Pankow suggests that at eqbm, m = -1, assuming:  om does not vary within a compound class THUS Pankow suggest the slope should be –1 regardless of whether absorption or adsorption dominates

20 What is observed in the field?

21 K p alone often gives slopes of -1 Mader and Pankow, 2002 normalizing K p to om, OC, or EC improves R 2, but slopes still -1 accounting for OC and EC usually gives best R 2, slopes still -1

22 difference between PAHs and PCBs in same air mass sampled twice July 22-23, 1994 Simcik et al, 1998 Slopes: Chicago PAH = PCB = LM PAH = PCB = Despite a transport time of ~3.5 hrs and reduction in concentration to 1/3, Kp and slopes have not changed. Slopes not necessarily –1 at eqbm (?)


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