1. Recall: Criterion for water vapor equilibrium
Definitions:
The criterion for equilibrium between the particle (that consists of an aqueous solution) and the environment is
aw is a function of the solution composition. MORE DILUTE (adding water)  aw closer to 1

2. Ideal solutions (Raoult’s Law)
Applying this to water, as we had in previous slide:
(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
g is called an “activity coefficient”
Note that as xi1, Raoult’s Law applies
and as xi0, Henry’s Law applies (linear, but different slope)

3. Representations of water activity
We saw that if solution behaves ideally, aw = xw
Write the mole fraction of water in terms of the number of moles of water nw and of solute, ns:
If the solute dissociates (many atmospheric aerosol components are salts, so this applies), then the number of moles of the solute in solution INCREASES. If there are ns total moles of undissociated solute, and each salt is composed of n ions, then
Sometimes (usually) the salt does not fully break apart into its ions, or the ions interact with each other and don’t act completely independent. We replace n with the van’t Hoff factor i. It plays the same role but can have a non-integer value. Think of it as the “effective” number of moles of ions in solution.
The van’t Hoff factor was used in the “classical” derivation of the Köhler equation, so we’ll use it for today’s development. (largely replaced in modern treatments…)
Examples: NaCl  Na+ Cl- n=2
(NH4)2SO4  NH4+ NH4+ SO42- n=3

4. Expressing solution water activity in terms of i
We start with the assumption that Raoult’s Law applies, so aw = xw.
Using the van’t Hoff factor, (1)
How many moles of salt do we have in the solution drop? Depends on size (and composition) of the initial dry particle. We will represent the mass of dry salt particle as m, and the molecular weight of the salt as M, so the moles of undissociated salt (ns) are given by (dissociation handled in Eq. 1, by multiplying by i)
How many moles of water do we have in the solution drop? Depends on the radius r that we grow the initial dry particle to. We will represent the mass of water as the difference between the total droplet mass once it has grown to size r, and the mass of the initial dry particle. The total droplet mass depends on its volume and the density of the solution, r’:
Now we subtract the dry particle mass m from the total mass to get mass of water, and to get moles of water nw, we divide by the molecular weight of water, Mw. Plug into equation (1):
ns = moles of (undissociated) salt
nw = moles of water

5. Deriving the “classical” Köhler equation
More familiar form:
Köhler equation
We noted that xw ( ≈ aw) decreases if the dissolved substance dissociates, and the larger the number of ions released, the lower is xw , for the same amount of water assumed to be associated with the solution (i.e., same size drop)
From the Köhler equation, we immediately see that this means the S required for equilibrium is lower
We just saw how to express the mole fraction of water as follows,
where m is the mass of solute, M is the molecular weight of solute, and r’ refers to the density of the solution (at the composition represented by the choice of m and r)
Now we’ll develop the “classic” Köhler equation by making some approximations, primarily to the treatment of the water activity term but also using an expansion in place of the exponential term

6. Using a series expansion for ex of the form
Previous equation can be examined through a more physical viewpoint by rewriting in a slightly different form, for the case of a DILUTE SOLUTION.
For a dilute solution,
This is starting equation
This is approximation
Note we have the form
Define
We have
Using a series expansion for ex of the form
and applying the binominal expansion
(neglecting higher-order terms) gives

7. Hence, curvature term (+) solute term (-)
Note that supersaturation, s, is defined as S – 1 and is often expressed as a percent:
s, % = (S - 1) x 100
curvature
term
(+)
solute
term
(-)
Think about scanning r and finding the S value for which the solution droplet is in equilibrium. This results in curves shaped like this:
The solute term is initially important in reducing the required saturation ratio. As the solution becomes more dilute, corresponding with r increasing, the importance of the –b/r3 term diminishes. Eventually when r becomes even larger the a/r term (curvature) itself is no longer a factor and the drop behaves as a plane surface of pure water.
Take dS/dr and see if it is positive or negative…

8. curvature
term
(+)
solute
term
(-)
Or dry particle size, rd
The peak in the Köhler Curve (for a given choice of m) can be found by setting
The solution droplet radius at which this derivative is zero is called the critical radius, r*
And the corresponding saturation ratio is called the critical saturation ratio, or more commonly, the critical supersaturation, sc. We can show, for the simplified (classic) Köhler equation, the numerical values of these quantities are given by:

9. Take the viewpoint of the droplet: the curves are its vapor pressure over its surface. IF the environment is the SAME, then it’s in equilibrium with its environment.
Where on this curve do we find stable, unstable, and metastable equilibria?

10. Consider the meaning of the critical supersaturation:
What happens if we put this dry particle into an environment where S-1 = 0.3% ?

11. aerosol; ammonium sulfate. Dotted curve is for
Houghton (1985)
Köhler Curves
Kelvin Curve
Curves are for
aerosol; ammonium sulfate.
Dotted curve is for

12. Important points to glean from the Köhler curves (previous slide):
As the dry particle size increases, the critical supersaturation decreases (for a chosen composition, it is easier to “activate” larger dry particles).
For similar-sized dry particles, but with different composition, those creating the largest suppression of water vapor pressure over the solution are the easiest to activate. (This means solutes that dissociate are easier to activate than those that do not, and those that dissociate into more than two ions are easier to activate than those that dissociate into only 2) CAVEAT: The statement ignores the influence of the dry particle density. In other words, particles of the same dry size but different densities contribute different amounts of mass to the solution, which has to be considered.
The size of the droplet at activation decreases with decreasing dry particle size (since the wetted size is so small, the Kelvin effect is large and this leads to high required critical supersaturations – see next slide; cam also see it’s still concentrated and not very dilute)
For large dry particles, the size at activation is very large and the Kelvin (curvature) effect very small – it is very easy to activate these particles: they could even be (nearly) insoluble and still activate (“giant” CCN)

13. Notice how small it is even at its critical size!

14. Now, let’s go back to the r
Now, let’s go back to the r* and sc equations and see what else we can learn from them:
For the generally-dilute solutions at the critical point, we can assume the surface tension is close to that of pure water, and the molar volume of water in the solution is close to the molar volume of pure water
So we approximate a as a constant (unless T changes a lot):
Let’s substitute for the mass of the dry particle in terms of the dry particle radius, and we will also approximate the mass of the total solution as the mass of water only (OK if it’s very dilute!):

15. This results suggests that a plot of ln sc vs
This results suggests that a plot of ln sc vs. ln rd, for a chosen composition of dry particle, should yield straight lines, with slopes of -3/2
This line is for pure water (insoluble particle)
This line is for a “salt” similar to NaCl
dry diameter, cm

16. “Kappa-Köhler” (Petters and Kreidenweis, ACP, 2007)
Let’s go back to the first form of the Köhler equation:
We substituted xw for aw (assuming Raoult’s law applied), and immediately wrote an expression for xw for a “dissociating salt”
Let’s go back to this point and find an alternative parameterization:
Here the k term can account for the nature of the solute, its dissociation, plus even nonidealities of the solution that is formed. If we want to relate it strictly to the model we already had, we find
where

17. These curves were computed from the basic equation with aw parameterized using k, without making the approximations leading to the “classic” Köhler equation (so slopes can deviate from -3/2)
We also propose this as a “grid” for plotting experimental data to determine a best-fit k

18. Slopes deviate because the assumption of very dilute solutions (volume water >> volume dry particle) breaks down

19. Utility of k When k = 0, aw = 1  pure water
Can see that k “scales” the water content at a given aw (for a bulk solution, aw = RH)
k values for “real” atmospheric particles range from 0 to 1.2
Shaded area = reported range for ammonium sulfate

21. How would we apply this information to the atmosphere????
Range of instrument s usually achieved