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???

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  m  if condensation nuclei start as clusters of  m molecules

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

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

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

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

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

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

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 );

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

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 ) (r c in nm)  benzene (# N l ) (r c in nm)  styrene (# N l ) (r c in nm)

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

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

Plot of 1/N l vs S

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

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

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

Particle phase pinonic acid model pinonic acid data norpinonic acid

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

Boiling points can be estimated based on chemical structure (Joback, 1984) T b =  T b  T ( o K) -CH 3 = K -Cl = NH 2 = C=O= C benz H-= Joback obs (K)(K) acetonitrile acetone benzene amino benzene benzoic acid toluene pentane methyl amine trichlorethylene phenanthrene598613

Time (min)Sat ratio

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

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

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????