1 Pure droplet formation (Chapter 7)The basic question is what makes the existence of a droplet thermodynamically preferable to theexistence only of water vapor. We have already derived an expression for the saturation vaporpressure over water. But what is the vapor pressure over a solution droplet? Atmospheric liquidclouds nucleate on tiny aerosol particles with dry diameters less that 1 µm diameter. These aerosolparticles are the “seeds” of clouds. Without them, clouds could not exist in anything like the formwe observe.

Below we discuss some of the physics that is relevant to cloud formation

Step 1

Figure 1: Cluster formation (Lamb and Verlinde)

Gas molecules must collide and stick, to form a nucleus of a few molecules that is thermody-namically stable and sufficiently large to spontaneously grow. It turns out that the stability of sucha nucleus is favored by low temperatures (low particle energies). Thirteen molecules seems to bethe magic number required to get things going.

How do we reach a stable nucleus of size r?? What is the rate at which such embryos are

created? Suppose that:

• rn is the number concentration of the vapor phase with thermal velocity vth (T )

• C (n) is the concentration of embryos containing n molecules (m�3)

• r (n) is the embryo radius

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• R(n) is the rate of addition of molecules to an n-molecule cluster (# s�1 m�3 ) with stickingprobability a (T )

From dimensional and physical reasoning we might suppose that the rate of addition is:

R(n)⇠ rnpr2 (n)vth (T )a (T )

In equilibrium, we could derive from principles in statistical mechanics the concentration ofembryos of a given size. Since it requires thermodynamic work to create an embryo of sizensmaller than the critical size, the concentration of embryos of critical size n

? is

C (n?) =C (1)exp✓�DG

?

kT

◆

where DG increases with n and C (1) = rn.

Figure 2: Critical energy barrier (Lamb and Verlinde)

However, a complication is introduced by nucleation being an explosive non-equilibrium pro-cess where embryos do not in fact grow to a certain size as fast as they grow out of a certain size.The way this has been approached is to introduce something called the Zeldovich factor (⇠ 0.1) toaccount for non-equilibrium effects so that the nucleation rate of critical embryos is

J (n?)⇠ R(n?)C (n?)Z?

In this case the nucleation rate is

J ⇠ r2n

exp(�DG?/kT )pr

?2vth (T )a (T )Z

?

2

Step 2Once, a nucleus is formed, a certain supersaturation is required to maintain the droplet in equilib-rium with its environment, and a higher supersaturation to enable it to grow. We have shown thatthat Dµ = µv � µl = 0 if e = e

sat (T ). This is strictly true only over a flat surface of water. Overa curved surface, it turns out that e > e

sat (T ) for equilibrium. Qualitatively, the reason is that thetotal energy of the liquid surface includes not just the energy of water molecules themselves, butin addition the potential energy associated with the surface tension due to stretching the surface ofthe water into a curve. Consider that if the droplet were to burst, the surface tension would provideavailable energy to do dynamic work on the droplet. Therefore, at the same temperature, a curveddroplet is at a higher potential with respect to water vapor than is a flat surface of water. Thismeans that water vapor that ordinarily would be at higher potential than a flat surface water, anddrive condensation, is now at a lower potential and is associated with evaporation.

Starting our prior expression

dG

dt=�S

✓∂T

∂ t

◆

S

+V

✓∂ p

∂ t

◆

V

+µ✓

∂N

∂ t

◆

T,p

and noting thatdG

dt= N

✓dµdt

◆N +µ (T, p)

✓dN

dt

◆µ(T,p)

We obtain the Gibbs-Duhem expression

N

✓dµdt

◆N =�S

✓∂T

∂ t

◆

S

+V

✓∂ p

∂ t

◆

V

It can be show that �SDT =�DH|N,p. Thus:

�DHN,p +V Dp = NDµH,p

We showed previously that, if there is no enthalpy barrier and DH = 0, then this can be expressedas Dg = Dp/r , or, per molecule

Dµ =1n

Dp =⇣

µv �µsat

v,l (T )⌘= kT D lne = kT ln

e

esat (T )

where esat (T ) is determined by the Clausius-Clapeyron Equation. If e > e

sat then the current isfrom vapor to the droplet. If e = e

sat , then the system is in equilibrium since Dµ = 0.We must now consider that there is an energy barrier to droplet formation due to surface tension

forces. Suppose that enthalpy of surface tension is given by

H = sA

where A is the surface area of the droplet and s is the surface tension (units Newtons per meter orJoules per meter squared) which is about

s = 7.5⇥10�2 Nm�1 = 7.5⇥10�2 Jm�2

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Then the magnitude of the energy barrier produced, per molecule of droplet growth isdH/dt

dNl/dt=

dH

dNl

If the molecular enthalpy and molecular number are proportional such that dH/dNl = H/Nl =Dhcurvature, and nl = dNl/dVd is the molecular number density in a droplet of volume Vd and radiusr:

Dhcurvature =dH

dN=

snl

dA

dVd

=8prdr

4pr2dr

snl

=2snlr

(1)

Thus, from the Gibbs-Duhem expression (which we divide by N) we have a new expression forDµ that includes the energy barrier from the surface tension for a droplet of radius r

Dµtot = µv �µsat

l,v (T,r)

so�Dhcurvature +

1n

Dp = Dµtot

Substituting our expressions for Dh (Eq. 1) and Dp, we now have

�2snlr

+ kT lne

esat

l,v (T )= Dµtot (2)

Evaporation requires Dµtot < 0 and µv < µsat

l,v , allowing molecules to “fall down”, away from theliquid surface. Similarly, surface tension represents an energy barrier that always forces moleculesaway from the liquid surface. Thus, for Dµtot to be positive, which would allow for condensationonto a curved droplet, the vapor pressure e must be greater than it would be over a flat surface. Infact what is required is

lne

esat

l,v (T )>

2snlkTr

For a curved droplet to be in equilibrium with its environment, Dµtot = 0 and we get the KelvinEq., which expresses the saturation vapor pressure over a curved surface with radius r

esat

l,v (T,r) = esat

l,v (T )exp✓

Dh

kT

◆= e

sat

l,v (T )exp✓

2snlkT

1r

◆(3)

or, perhaps more familiarly in terms of per mass

esat

l,v (T,r) = esat

l,v (T )exp✓

2srlRvT

1r

◆(4)

A couple of things to notice about this equation. First as r approaches infinity, the curvatureeffect on the saturation vapor pressure becomes insignificant. In other words, even if the dropletis a sphere, from a thermodynamic standpoint, it can basically be considered to be a flat surface.For atmospheric applications, the Kelvin effect can be neglected, and we can consider droplets tobe essentially flat if they are larger than about 1 µm. Second, at the other extreme, for extremelysmall droplets, huge supersaturations are required for the droplet to be in equilibrium with its envi-ronment. For recently nucleated particles, say 0.1 µm across, the supersaturation at equilibrium isin excess of 100% (i.e. 200% relative humidity). We never get relative humidities this high in theatmosphere, and since any pure droplet forming from vapor must start from an embryo of a fewmolecules, clearly pure water droplets cannot form in the atmosphere.

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Figure 3: Dependence of system free energy on cluster size and saturation ratio in a homogenousgas-phase medium. (Lamb and Verlinde)

2 Solution droplet formation

2.1 Kohler equationThe Kelvin Equation (Eq. 3) has the form

esat

l,v (T,r) = esat

l,v (T )exp✓

2snlrkT

◆

which was obtained by adding an energy barrier due to surface tension so that the total potential is

Dµtot = kT lne

esat (T )� 2s

nlr

Thus we can interpret the Kelvin Eq. as being the equilibrium solution where Dµtot = 0. We foundthat maintain equilibrium requires extraordinarily high values of supersaturation S = e/e

sat � 1in order to compensate for the surface tension potential when the droplets are just nucleated andsmall.

Somehow, we need to increase Dµtot so that the vapor is at a higher or equal potential thanthe potential level of the droplet molecules µv,l(T,r), so that being lazy, the droplet molecules arehappiest to stay in the liquid phase.

An effective way to do this is to add solute to the liquid. What this does effectively is to reducethe density of water molecules at the interface between the liquid and vapor phase, since the water

5

10−4

10−3

10−2

10−1

100

101

10−2

10−1

100

101

102

103

104

105

106

R (µm)

Supers

atu

ratio

n (

%)

13−mer

T = −5 °C

Figure 4: Dependence of the supersaturation of water vapor at equilibrium over a pure waterdroplet. The size of a 13 molecule nucleus is shown for reference.

molecules are separated by solute ions. Consequently the density of water vapor needs to be lessin order for there to be equilibrium.

Let’s say we define a solution “activity” such that

µl (T,aw) = µsat

l(T,aw = 1)+ kT lnaw (5)

where aw is bounded by 0 and 1. If aw < 1 then kT lnaw < 0 and the chemical potential of liq-uid is depressed compared to what it would be at equilibrium over a flat surface of pure water,i.e. µl (T,aw) < µsat

l(T,aw = 1). If aw = 1 then the chemical potential of the liquid is what we

normally assume for pure water, and

µl (T,aw = 1) = µsat

l(T ) (6)

Thus, since, right at the surface of the water, vapor is at equilibrium with the underlying liquidsurface, and

µsat

v(T,aw)⌘ µl (T,aw) = µsat

l(T )+ kT lnaw (7)

so if aw < 1 then µsatv

(T,aw)< µsatv

(T ). We can alter our expression for Dµ from before to includethe energy per molecule (or work) required to take the step upwards from the solution potential tothe pure liquid potential, i.e.

Dµ = µsat

l(T )�µl (T,aw)⌘ µsat

v(T )�µl (T,aw) =�kT lnaw (8)

6

Figure 5: Illustration of three phases and the effects of a solution (Lamb and Verlinde)

We showed earlier that the energy barrier that must be overcome to bring water vapor into equilib-rium with a flat surface of pure water is Dµ = µsat

v�µv =�kT lnRH. Thus, the water vapor is in

equilibrium with the solution when the two energy barriers are equal and

kT lnRH = kT lnaw (9)

where, aw corresponds to the relative humidity over a flat solution that brings the potential µv intoequilibrium with µl (T,aw). Expressed as an energy barrier, with respect to µsat

lcurvature is a

positive energy barrier, but because µl (T,aw) < µsat

l(T ), the solution effect is a negative energy

barrier since it lowers µ .Dhsolution = kT lnaw

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Thus, the combined energy barrier for a curved solution droplet is

ÂDh = Dhcurvature +Dhsolution =2snlr

+ kT lnaw

From the Gibbs-Duhem relation�ÂDh+

1n

Dp = Dµtot

the combined potential difference per molecule between the vapor and the liquid phase, for acurved solution droplet, is now

Dµtot = µv �µl (T,r,aw) = kT lne

esat (T )� 2s

nlr� kT lnaw (10)

The second and third terms represent the sum of the energy barriers to diffusional flows Dh. Ifthe activity of the solution can be lowered to aw < 1 then Dµ increases and the existence of theliquid phase more stable and likely, despite the curvature effect. There is a bigger “wall” for theliquid molecules to jump over to get into the vapor phase, or a smaller wall required to initiatecondensation. Our revised equilibrium expression then for the vapor pressure over a solutiondroplet is obtained by solving for Dµtot = 0

esat (r,aw,T ) = e

sat (T )exp�ÂDh/kT

�= e

sat (T )exp✓

2snlr

+ kT lnaw

�/kT

◆

esat (r,aw,T ) = e

sat (T )aw exp✓

2snlrkT

◆(11)

esat (r,aw,T ) = e

sat (T )aw exp✓

2srlRvTr

◆(12)

This is known as the Kohler equation.

2.2 Activity of solution dropletsFor an “ideal” solution (usually true if very dilute)

aw =n0

n+n0(13)

where n0 is the molecules of water and n molecules of solute. This is known as Raoult’s Law. Fordilute solutions

aw = 1�n/n0 (14)

But we must remember that salts dissociate, i.e. NaCl dissociates to Na+ and Cl�. We representthis using what is really a bit of a kluge, but popular nonetheless - the so called Van’t Hoff factor i.

n ! in (15)

8

where, for NaCl, and NH4HSO4 (ammonium bisulfate) i = 2, and for (NH4)2SO4 (ammoniumsulfate), i = 3. i is essentially the number of solute ions that the molecule dissolves into. Theproblem with this approach is that the number i stops meaning anything physical when one startstalking about organic non-ionic species, which nonetheless are soluble.

What do we see from this?

1. The higher the concentration of the solute, the lower the equilibrium relative humidity

2. The more dissociation, the higher the solution effect.

Effectively what is happening is that, while curvature makes it easier for molecules to escape aliquid droplet, dissolving a solute makes it harder for them to escape. Effectively there are fewerliquid molecules per unit surface area for them to escape.

The Kohler curves are shown below. These show the relative humidity relative to saturationover a flat surface of pure liquid water that would be in equilibrium with a droplet of radius r

containing dissolved solute. Effectively, it is esat (r,aw,T )/e

sat (T ).

10−1

100

101

0.99

0.992

0.994

0.996

0.998

1

1.002

1.004

1.006

1.008

1.01

S

r (µm)

(NH4)2SO

4 (r

s=0.02 µm)

(NH4)2SO

4 (r

s=0.03 µm)

(NH4)2SO

4 (r

s=0.04 µm)

(NH4)2SO

4 (r

s=0.06 µm)

(NH4)2SO

4 (r

s=0.08 µm)

(NH4)2SO

4 (r

s=0.1 µm)

(NH4)2SO

4 (r

s=0.15 µm)

(NH4)2SO

4 (r

s=0.3 µm)

Figure 6: Kohler curves for ammonium sulfate with dry radius rs.

The location of peak saturation ratio S⇤ is at r⇤, the critical radius. Below this size the solution

effect dominates the Kelvin effect, and solution droplets are in equilibrium with their environment(i.e Dµ = 0). If the particle grows spontaneously, it is then too large for the ambient equilibriumsaturation for the droplet, so it shrinks to its original size. Likewise, if it shrinks it grow again toits equilibrium size. These are haze particles, which are in stable equilibrium. Above this size thehaze particles are said to be activated. The Kelvin effect dominates and Dµ < 0 if the particles

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grow spontaneously. A particle that grows spontaneously will be have an equilibrium saturationlower than the ambient saturation, so it keeps growing spontaneously by vapor diffusion. The totalGibbs free energy of the system will continue to decrease. Once the droplets get big enough, theKelvin effect stops mattering, and the droplet acts effectively like a plane surface of water.

Figure 7: Stability of the Kohler curve (Lamb and Verlinde)

Therefore, the function of a solute in droplet formation is to lower the amount of chemicalpotential energy of the water vapor (i.e. supersaturation) required to get the droplet to the sizer⇤ where it can start to grow spontaneously of its own accord. The supersaturations required are

typically less than 1%, which is quite manageable in clouds.One consideration is that not all solutions are ideal, particularly if they are concentrated. An

ideal solution has linear variation between aw = 1 for pure water to aw = 0 for pure solute. Anon-ideal solution satisfies these extremes but otherwise has aw < aw (linear). Solutes in solutiondecrease the concentration of water, but they may also affect the activity of water itself throughstrong molecular interactions.

Main points

• While surface tension increases the energy barrier required to maintain a curved surface,presence of a solute in the water decreases the energy barrier. For a salt, the magnitude ofthe effect is proportional to the number of solute ions

• Solution droplets can have an equilibrium relative humidity less than 100%. The equilibriumis stable.

• A droplet is in unstable equilibrium where the curvature effect and the solute effect are inbalance. Any further growth is unstable and leads to spontaneous condensation.

2.3 Cloud condensation nucleiIf a haze aerosols is not activated then, neglecting curvature effects, from Eq. 13, the equilib-rium water concentration of water can be determined nw from the equilibrium requirement that atequilibrium the relative humidity and activity are equal. A complication is that aerosols exhibitphenomena known as deliquescence and effloresence with hysteresis in between. For an initiallydry aerosol, the relative of mass of liquid to dry aerosol mass increases with increasing relative

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Figure 8: Figures 7.11 and 7.13 in Lamb and Verlinde. Dependence of particle mass on saturationratio, based on data obtained from a single particle levitated electrodynamically under conditionsof controlled relative humidity. The particle is large, so the effect of curvature on water uptakecan be ignored and the response of the particle to changing humidity is the same as that of a bulksolution. The deliquescence point is a function of the chemical composition.

humidity only once it passes a critical relative humidity that is a function of the composition. Thesize then decreases with decreasing relative humidity according to the aerosol activity down to theefflorescence point, which is typically much less than the deliquescence point. Deliquescence isseen in your salt shaker and efflorescence in concrete floors with too much moisture.

Figure 9: Aerosol size distributions from continental and maritime sources. Dependence of thecritical supersaturation on the dry-particle diameter. Calculation made using Kohler theory withvan’t Hoff factor i=2.5

Cloud condensation nuclei (or CCN) represent the subset of atmospheric aerosol particles withdry radii that are sufficiently large to activate at a particular supersaturation. It is often assumed thatthe number of salt solution droplets that “activate” at supersaturation S is of form n(S) ⇠ kCS

k�1

(m�3 %�1) where C and k are constants. That is, we assume that the size distribution of saltparticles is such that for every S there are n(S)dS (m�3) particles for which the critical super-saturation Scrit lies in the interval (S, S+ dS). For a given chemical composition, large particlesactivate at lower critical supersaturations S

?.

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