Lecture 7: quick review

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Hilary Fisher
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1 Lecture 7: quick review We discussed the meaning of the critical r* and ΔG*: since it s unstable, if we can form a drop with r > r*, the system will keep going falling down the energy curve, trying to get to a minimum energy. For water, the trouble is that we compute pretty large r* for the kinds of water saturation ratios we expect in the true atmosphere How can we get this large of a critical embryo to form spontaneously?? And when we invoke classical nucleation theory and compute RATES of nucleation (rates of pushing embryos over the critical energy barrier), the rates only get reasonable if S > 3 that s a lot of water vapor! We have to search for some alternative explanation (other than homogeneous nucleation ) as to why clouds form where they do (which is observed to be fairly close to S=1) Heterogeneous nucleation? This means another substance / surface is involved.

Heterogeneous Nucleation on Insoluble Surfaces We ll consider formation of a small water droplet on a flat surface first (not so realistic, but we can extend to a curved surface modeling an aerosol

2 Heterogeneous Nucleation on Insoluble Surfaces We ll consider formation of a small water droplet on a flat surface first (not so realistic, but we can extend to a curved surface modeling an aerosol particle later) Furthermore, we assume the surface is insoluble, but wettable (hydrophilic) A hydrophobic surface is insoluble and not wettable (water beads up on it) V = vapor phase L = liquid phase C = substrate phase (solid) r = radius of spherical cap Model this r θ θ

The contact angle θ The contact angle is the angle between the substrate, and the tangent to the droplet (or cap) surface If the surface is completely wettable, θ = 0 If the surface is completely

3 The contact angle θ The contact angle is the angle between the substrate, and the tangent to the droplet (or cap) surface If the surface is completely wettable, θ = 0 If the surface is completely hydrophobic, θ = 180 r How are these variables related? Realize that, at equilibrium, a force balance must exist between the surface tensions (force / unit length) Note the directions they are given in sketch θ σ CL ±σ LV component in x - dir = σ VC σ CL +σ LV cosθ = σ VC cosθ = σ VC σ CL σ LV = m

4 Aside: Pendant Drop Shape Analysis ( The shape of a drop of liquid hanging from a syringe tip is determined from the balance of forces which include the surface tension of that liquid. The surface or interfacial tension at the liquid interface can be related to the drop shape through the following equation: γ = Δρ g R 02 /β where γ = surface tension Δρ = difference in density between fluids at interface g = gravitational constant R 0 = radius of drop curvature at apex β = shape factor β, the shape factor can be defined through the Young-Laplace equation expressed as 3 dimensionless first order equations as shown in the figure below. Modern computational methods using iterative approximations allow solution of the Young-Laplace equation for β to be performed. Thus for any pendant drop where the densities of the two fluids in contact are known, the surface tension may be measured based upon the Young-Laplace equation.

5 Free energy (change) analysis 4 Recall for pure water drop we found, ΔG = (n l 3 πr 3 )( µ l µ v )+ 4πR 2 σ We have an analogous expression for the cap, except: We don t have the full spherical volume (so we need to find V L ) We have two surface energies to compute: between liquid and vapor, as before, and between substrate and liquid (need to find the surface areas they go with, too) (net along liquid substrate boundary) Bulk thermo term; energy change with the bulk phases. Surface tension terms Use this geometry to figure out V L, A LV, A CL : 2 A CL = πx 0 = πr 2 (1 m 2 ) m cosθ

6 Spherical cap (Wikipedia) A = 2πr 2 (1 m) We use contact angle to define a and h, namely: (r h) cosθ = r r cosθ = r h h = r(1 cosθ) = r(1 m)

7 Spherical cap (continued) V = π 6 r(1 m) [ 3r 2 (1 m 2 )+ r 2 (1 m) 2 ] = π 6 r(1 m) [ 3r 2 3r 2 m 2 + r 2 (1 2m + m 2 )] = π 6 r(1 m) [ 3r 2 3r 2 m 2 + r 2 2mr 2 + r 2 m 2 ] = π 6 r(1 m) [ 4r 2 2r 2 m 2 2mr 2 ] = π 6 r(1 m)2r 2 [ 2r 2 m 2 m] = π 3 r 3 (1 m)(1 m)(2+ m) = π 3 r 3 (1 m) 2 (2+ m) We use contact angle to define a and h, namely: (r h) cosθ = r r cosθ = r h h = r(1 cosθ) = r(1 m) The area touching the substrate is A CL = πa 2 : r 2 = a 2 + (r h) 2 a 2 = r 2 (r h) 2 = r 2 r 2 (cosθ) 2 a 2 = r 2 (1 m 2 ) A CL = πr 2 (1 m 2 )

8 Back to free energy calculation: Now we need to substitute in the values for volume and areas we derived. After doing this and cleaning up the equation, we get Recall that ΔG = n L V L Δµ VL + A LV σ LV + A CL (σ CL σ VC ) = n L V L Δµ VL + A LV σ LV + A CL ( mσ LV ) = n L V L Δµ VL +σ LV (A LV ma CL ) ΔG = π(2 3m + m 3 ) n LΔµ VL r 3 +σ LV r 2 3 Δµ VL = RT ln S As before, we ll look to see if ΔG(r) has a minimum anywhere. To do that, again take the derivative dδg/dr and set it equal to zero when we evaluate the point at which this is true, we ll call the (r, ΔG) values there r* and ΔG* (or, critical values). We find that r * = 2σ LV n L RT ln S This is exactly the same as for homogeneous nucleation! Why?

9 Critical free energy The free energy at the critical point is given by: ΔG * = ΔG(r * ) = ΔG * = 3 16πσ LV (2+ m)(1 m)2 2 3(n L RT ln S) πσ LV 3(n L RT ln S) f (m) 2 Same as barrier height in homogeneous nucleation case Geometrical factor, f(m) < 1 (turns out to be the ratio of the volume of the cap to the volume of the whole sphere) The presence of the insoluble surface aids the nucleation of the liquid by lowering the free energy to form liquid, for any given S. The amount of lowering is the same for all radii, so the critical radius is the same As a liquid cap forms, it essentially replaces part of the large solid-vapor surface free energy with a smaller liquidsolid surface free energy

10 Consider effect of f(m) The range of values of m can be 0 θ m 1 When θ = 180, then m = -1 and f(m) = 1 Hydrophobic surface Same as homogeneous nucleation case Drop simply rests on the surface and does no interact with it When θ = 0, then m = 1 and f(m) = 0 Completely wettable, hydrophilic surface ΔG* = 0 (no barrier to drop formation) Wettable surface supports condensation at RH=100%! (remember so far, the solid substrate is assumed flat)

11 Effect of contact angle on nucleation rate The rate of nucleation on a flat substrate depends on the number of critical embryos (here, spherical caps of critical size), and the rate with which those embryos acquire new vapor molecules to grow past the barrier As before, we compute the number density of critical embryos on the surface using a Boltzmann distribution: n(r * ) = n s1 exp( ΔG * /kt) Here n s1 is the surface density of single adsorbed water molecules The rate is J s J s = J 1 n(r * ) = Bπr *2 n(r * ) p n 2πmkT πr*2 s1 exp( ΔG * /kt) ( cm 2 s 1 )exp( ΔG * /kt) Notice that now we have rate of nucleation per unit area (not volume) Figure chooses threshold ~1 cm -2 s -1

Nucleation on an insoluble, curved surface Heterogeneous pathway has the potential to allow water condensation at more realistic S values, but to really evaluate, we have to fix the assumption that

12 Nucleation on an insoluble, curved surface Heterogeneous pathway has the potential to allow water condensation at more realistic S values, but to really evaluate, we have to fix the assumption that the surface is an infinite, flat plane We let the drop of radius r form on a particle of radius r N (see figure) and repeat a similar derivation to get the free energy change We get a similar finding: ΔG is equal to the homogeneous expression, modified by f But now, f depends on both contact angle and size We express f in terms of m and the size ratio, x = r N / r f(m,x) 1 for r 0 ; f(m,x) f(m) for r (flat surface)

13 Nucleation rates Notice that now, to find a threshold saturation ratio with a measurable nucleation rate, since we have a finite surface, we only need a single cap of the critical size to form anywhere on the particle surface particle acts as a nucleus So the new criterion is (with J s from before): J S A N =1 s -1 = 4πr N 2J S Decreasing wettability When the particle is very large, results similar to flat surface For small particles, even highly wettable, harder to nucleate because of the curvature effect Don t expect particles smaller than ~ 0.01 µm to participate in nucleation process spherical, insoluble particle must have a radius greater than 0.01 µm and be (nearly) perfectly wettable for it to serve as an effective nucleus, at supersaturations available in clouds. What does this curve represent?

14 poor nucleators of water; but (possibly) great for ice nucleation. Pruppacher & Klett (1978)

15 Heterogeneous nucleation on soluble particles Up to now we have studied homogeneous nucleation and heterogeneous nucleation involving insoluble substrates or particles. We have found that both homogeneous nucleation and heterogeneous nucleation (with exception of large, perfectly wettable particles) don t explain well atmospheric observations of cloud formation, due to the large supersaturations required to initiate the phase change. We now examine the formation of a droplet by condensation on a small soluble particle (we are going to identify these particles, that can participate readily in cloud formation, as cloud condensation nuclei, CCN). An example of a highly soluble species is a salt (ubiquitous in the atmosphere), which quickly dissociates in solution into positive & negative ions. But, please keep in mind that solubility is a specific chemical concept and the way it gets used frequently in discussing CCN is not quite correct. I prefer the term hygroscopic, by which I mean any species that can form aqueous solutions at ambient relative humidities. It includes salts, acids, bases, and many organic compounds (not all of which necessarily dissociate into ions). To understand how hygroscopic compounds can enhance vapor liquid phase transitions, we have to go back to the Kelvin-type equation

Criterion for water vapor equilibrium Defini&ons: p w o vapor pressure of pure water, over a flat surface p w vapor pressure of water in the environment p sol vapor pressure of water over a solution

16 Criterion for water vapor equilibrium Defini&ons: p w o vapor pressure of pure water, over a flat surface p w vapor pressure of water in the environment p sol vapor pressure of water over a solution RH relative humidity in the environment RH = p w p w o 1 a w "water activity" of a solution a w = p sol p w o 1 The criterion for equilibrium between the par&cle (that consists of an aqueous solu&on) and the environment is p w = p sol or RH = a w or, small particle : RH = a w exp 4σM s ρ s RTD p a w is a func&on of the solu4on composi4on. Can we calculate a w for a given composi&on? 16

$Ideal solu4ons (Raoult s Law) Raoult s Law: p i = x i p i o x i = mole fraction of i in solution p A O Applying this to water, as we had in previous slide: p B O p sol x w = x w p w o = mole fraction$

17 Ideal solu4ons (Raoult s Law) Raoult s Law: p i = x i p i o x i = mole fraction of i in solution p A O Applying this to water, as we had in previous slide: p B O p sol x w = x w p w o = mole fraction of water in solution a w = x w (ideal behavior!) REAL solutions (solid curves) typically deviate (positively or negatively) from Raoult s Law (dashed lines) Generally, we need lab measurements to trace out these real activity curves The modified water activity equation is a w = γ w x w γ w = f (x w ) γ is called an activity coefficient Note that as x i 1, Raoult s Law applies and as x i 0, Henry s Law applies (linear, but different slope) 17

18 Representa4ons of water ac4vity We saw that if solu&on behaves ideally, a w = x w Write the mole frac&on of water in terms of the number of moles of water n w and of solute, n s : x w = a w = n w n s + n w 1 a w = n s n w +1 If the solute dissociates (many atmospheric aerosol components are salts, so this applies), then the number of moles of the solute in solu&on INCREASES. If there are n s total moles of undissociated solute, and each salt is composed of ν ions, then a w = νn s + n w 1 a w n w =1+ νn s n w However, many salts do not fully dissociate or, put more correctly, the ions in solu&on do not act as if there were completely independent of one another. The molal osmo&c coefficient Φ is used to make correc&ons for nonideality (more typical for ionic species, rather than γ) ln a w = νmm w 1000 Φ m is the molality (moles of A per kg water) 18

19 Representa4ons of water ac4vity (cont d) Write the defining equa&on for Φ as follows, and then do a series expansion: n s Expand on defini&on of n s (similarly for n w ): 1 a w 1 a w = exp νmm Φ w 1000 =1+ νmm Φ w νmm w Φ 1000 Since = mm w we are now in a posi&on to compare this expression to the defini4on of n w 1000 the van t Hoff factor, i: 1 =1+i n s a w n w From which we no&ce that Typically used to describe solutions in the Köhler equation i = νφ+ mm w 1000 i νφ ( νφ) n s = V sρ s M s V s is the volume of the solute (dry particle) so 1 a w 1 a w =1+i V sρ s M w V w ρ w M s =1+ V s V w κ This is the definition of κ, the hygroscopicity parameter (applies at any RH that supports a solution) 19

20 The κ parameter If we ignore the Kelvin effect for a moment and assume equilibrium, such that a w =RH, 1 RH Convert into wet and dry diameters: V w V s 1 RH =1+κ V s V w 1 = 1 RH RH = κ V s V w = V w V dry = κ RH 1 RH 3 D wet = V wet 3 D dry V dry 3 D wet D dry = V w +V s V s =1+κ RH 3 1 RH =1+ V w V s Think about: what does a small κ mean? What about κ~o(1)? volume additivity has been applied Suppose the par&cle is composed of several solutes, each with its own κ i. What is the total water content? The ZSR assump&on says that each solute brings its own water with it. V w = V w,i = a w i 1 a w V TOT = V w,i + V s,i i i κ i V s,i i = V w +V s V w = V TOT V s = a w V s ε i κ i 1 a i w ε i = V s,i V s κ mix = ε i κ i i volume fraction in the dry particle 20

21 How much water does this correspond to? Kreidenweis et al., ERL,

22 Note that assuming κ=constant is not accurate below ~90% RH 22

23 Solubility Hygroscopicity Solubility: Think about puwng salt into pure water in a beaker. Is there a limit as to how much salt we can dissolve? Salt in Aside: if aw,sat = xw (ideal), then aw,sat = 1-xs,sat (mole fraction salt) or rather aw,sat = 1-νxs,sat The amount of water here is determined by the composition of the saturated solution RH DRH See Table, previous slide The composition of the saturated solution is associated with a particular aw of that solution. Under the criterion of equilibrium, RH = aw A saturated solution So aw,sat = RHD, the relative humidity of deliquescence 23

24 Metastable solu4ons It is oxen observed that a dry par&cle deliquesces at the expected DRH; but if the same solu&on par&cle is dried out, the wet par&cle can persist to RH well below the DRH. What s going on? This corresponds to the formation of a metastable solution The thermodynamics for the metastable solutions for many common salts have been measured and documented so that e.g. AIM can predict water contents even in this regime Similar to the situa&on for gas- to- par&cle conversion, there is a barrier to nuclea&on. Interes&ngly, aerosol lab studies oxen find a generally reproducible efflorescence RH at which the par&cle finally completely dries out (e.g., for ammonium sulfate ERH ~ 40%) Argue that most atmospheric par&cles have some water on them (water persists to low RH) 24

25 diameter hygroscopic growth factors GF (RH) = diameter at RH / dry diameter 25

26 Example experiments on inorganic aerosol: potassium chloride (KCl) humidified diameter dry diameter Parameterization: V w V s RH = κ (1 RH) Deliquescence (formation of saturated solution) Relative Humidity, RH (%) 26

27 Extra stuff

28 derivation of spherical cap area and volume from Lamb & Verlinde Figure 7.7

30 (Atmospheric Research, 2000)

Köhler Curve. Covers Reading Material in Chapter 10.3 Atmospheric Sciences 5200 Physical Meteorology III: Cloud Physics

Köhler Curve. Covers Reading Material in Chapter 10.3 Atmospheric Sciences 5200 Physical Meteorology III: Cloud Physics Köhler Curve Covers Reading Material in Chapter 10.3 Atmospheric Sciences 5200 Physical Meteorology III: Cloud Physics Review of Kelvin Effect Gibbs Energy for formation of a drop G = G &'()*+, G ).'+