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Aerosol Self Nucleation SO 2 nucleation Experiment (over the web) at the UNC smog chamber facility in Pittsboro Homogeneous Nucleation Theory Some examples.

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Presentation on theme: "Aerosol Self Nucleation SO 2 nucleation Experiment (over the web) at the UNC smog chamber facility in Pittsboro Homogeneous Nucleation Theory Some examples."— Presentation transcript:

1 Aerosol Self Nucleation SO 2 nucleation Experiment (over the web) at the UNC smog chamber facility in Pittsboro Homogeneous Nucleation Theory Some examples Global Calculation of nucleated particles from SO 2 going into the atmosphere/year???

2 Aerosol Self Nucleation Why are we interested?  Contribute to natural aerosol concentrations  global warming implications  health implications  serve as sites for the sorption of other gas phase compounds-toxic Usually they are very small  pyrene (gas).0007  m  viruses.002 -.06  m  if condensation nuclei start as clusters of 3-8.001-.005  m molecules

3 If gases are coming together to form particles or clusters free energy of the system surface tension level of gas saturation amount of cluster growth vapor pressure of the gas molecules

4 Free energy and surface tension What is surface tension  if a liquid has a meniscus surface we could define a force per unit length,  t, that the liquid surface moves from the flat surface of the liquid  t x l = force  force x distance = work  if the distance is dy  work =  t x l x dy  dy x l has the units of area  work/area =  t = surface tension  the free energy of the meniscus moving from position a to b or dy:   G =  H -T  S ;  H = work + heat   G =  t x  A + heat -T  S

5 Free energy and surface tension G =  t x dA + heat -T  S often the free energy of just the surface is given as:   G S =  t x  A for a spherical liquid nuclei or small cluster  G S = 4  r 2 x  t for gas molecules forming a small cluster where N l gas molecules -> o o o o the change in total free energy is the change in going from a pure vapor to a system that contains particle embryos  G T = G embryo system - G gas vapor

6 Free energy and surface tension  G T = G embryo system - G gas vapor let  g = chemical potential of the remaining gas,  l the liquid or embryo system; N T will be the total number of starting gas molecules; after embryo formation the N g = # of gas molecules, so, N g = N T - N l where N l the number of liquid embryo molecules   G T =  g N g +  l N l + 4  r 2 x  t - N T  g  Substituting N T = N g + N l   G T = N l {  l -  g } + 4  r 2 x  t

7 Free energy and surface tension  G T = N l {  l -  g } + 4  r 2 x  t the number of molecules in a liquid cluster, N l, is the volume of the cluster divided by the volume of one molecule, v l where N l = 4/3  r 3 / v l   G T = 4/3  r 3 / v l {  l -  g } + 4  r 2 x  t  the Gibbs Duhem equation describes the change in chemical potential with vapor pressure  d  = v dp ; since v g >>> v l  d {  l -  g } = v g d P  {  l -  g } = - kT ln P/P o

8 dU = SdT+ TdS -Vdp-pdV+ For a closed system which only does pressure volume work subtracting 0 = SdT -Vdp+ At constant temperature, one obtains the Gibbs-Duhem Equation for gases d {  2 -  1 } = v g d P

9 Free energy and and saturation  {  l -  g } = - kT ln P/P o  define P/P o as the saturation ratio S   G T = 4/3  r 3 / v l {  l -  g } + 4  r 2 x  t   G T = - 4/3  r 3 / v l { kT ln S } + 4  r 2  t  A plot of  G T vs particle diameter for different saturation ratios >1,shows it to go thru a maximum and then fall; this max is called the critical diameter (or radius r c )  differentiating and solving for r c  r c = 2  t v l /(kT ln S);  ln S = 2  t Mw/(RT  r c ); molar units (Kelvin equation) what happens to vapor pressure over a particle as r decreases and why??  ln P/P o = 2  t Mw/(RT  r c );

10 An expression for cluster #, N l  If we go back to  G T = - 4/3  r 3 c / v l { kT ln S } + 4  r 2  t  and take the derivative with respect to r again, and set this equal to zero, one gets:  4  r 2 c / v l { kT ln S }= 8  r  t  mulyiplying both sides by r/3 we get something that looks like the cluster # N l where N l = 4/3  r 3 / v l since r c = 2  t v l /(kT ln S)  substituting we obtain a valve for N l, the number of molecules in a cluster with a radius of r c and as function of saturation

11 Estimate cluster r c and the cluster #, N l  substituting molar values in the N l expression one obtains:  r c = 2  t Mw/(RT  ln S );  critical #s (N l ) and r c for 3 organics  saturation ratio 2345  acetone (# N l )265673321 (r c in nm) 2.01.31.0.8  benzene (# N l )7061778856 (r c in nm) 3.0 1.81.51.3  styrene (# N l )1647413202132 (r c in nm) 4.2 2.7 2.1 1.8

12 A Flow reactor study  -pinene + O 3  nucleated particles

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14 1. Assume the number of nuclei/cc formed is proportional to 1/N l N l =number of molecules in a cluster  -pinene seed products that form are proportional to their saturation ratio

15 Plot of 1/N l vs S

16 Do “high” saturation ratios of the products really occur??

17 CHO O O CH 3 O O O Criegee2 Criegee1 O O O  -pinene O 3 COOH pinic acid + other products O pinonic acid CHO O COOH + CO, HO 2, OH COOH O norpinonaldehyde norpinonic acid Mechanism

18 Gas phase pinonaldehdye O O mg/m 3 Time in hours EST

19 Particle phase pinonic acid model pinonic acid data norpinonic acid

20 If we can calculate the saturated vapor pressure, P o ?? We can compare it to the partial pressure of compounds predicted or measured in the gas phase and estimate a P/P o Where P o is in atmospheres

21 Boiling points can be estimated based on chemical structure (Joback, 1984) T b = 198 +  T b  T ( o K) -CH 3 = 23.58 K -Cl = 38.13 -NH 2 = 73.23 C=O= 76.75 C benz H-= 26.73 Joback obs (K)(K) acetonitrile347355 acetone322329 benzene358 353 amino benzene435457 benzoic acid 532522 toluene386384 pentane314309 methyl amine295267 trichlorethylene361360 phenanthrene598613

22 Time (min)Sat ratio

23 A a remote–web nucleation experiment at the smog chamber Do rural concentrations of SO2 nucleate in sunlight??

24 The Chamber had two sides Or Darkness Formaldehyde SO2 300 m 3 chamber Teflon Film walls NO &NO 2

25 Adding SO 2 to the Chamber SO2 ppm tank x volume from tank = SO2 ppm chamber x chamber volume Tank flow is 5 l/min; so how long do we leave the tank on to get 0.02 ppm SO2 in the chamber, if the tank is 1000 ppm????


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