Step 1. Step 2. g l = g v. dg = 0 We have shown that over a plane surface of water. g v g l = ρ v R v T ln e/e sat. this can be rewritten

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1 The basic question is what makes the existence of a droplet thermodynamically preferable to the existence only of water vapor. We have already derived an expression for the saturation vapor pressure over water. But what is the vapor pressure over a solution droplet? Atmospheric liquid clouds nucleate on tiny aerosol particles with dry diameters less that 1 µm diameter. These aerosol particles are the seeds of clouds. Without them, clouds could not exist in anything like the form we observe. Below we discuss some of the physics that is relevant to cloud formation Step 1 Gas molecules must collide and stick, to form a nucleus of a few molecules that is thermodynamically stable. It turns out that the stability of such a nucleus is favored by low temperatures (low particle energies). Thirteen molecules seems to be the magic number required to get things going. The physics of nucleation formation is beyond the level of this course. Step 2 Once, a nucleus is formed, a certain supersaturation is required to maintain the droplet in equilibrium with its environment, and a higher supersaturation to enable it to grow. We have shown that that g = g v g l = 0 if e = e sat (T ). This is strictly true only over a flat surface of water. Over a curved surface, it turns out that e > e sat (T ) for equilibrium. Qualitatively, the reason is that the total energy of the liquid surface includes not just the energy of water molecules themselves, but in addition the potential energy associated with the surface tension due to stretching the surface of the water into a curve. Consider that if the droplet were to burst, the surface tension would provide available energy to do dynamic work on the droplet. We showed that in equilibrium, at constant temperature and pressure, the difference in the Gibbs free energy at the interface between two phases is equal. For liquid and vapor: g l = g v dg = 0 We have shown that over a plane surface of water this can be rewritten g v g l = ρ v R v T ln e/e sat µ v µ l = kt ln e/e sat In some ways, physically, this expression is a bit easier to understand since is easy to understand saturation on a molecular level. What this equation states is that e = e s (i.e. saturated) if the chemical potential energy of each vapor molecule is the same as that of each water molecule. Let imagine a box of fixed volume with nothing but water vapor in it, and out of this box forms a droplet. What we want to ask is what is the change in Gibbs free energy associated with such a transformation, and under what conditions would the change in Gibbs free energy be negative (i.e. we would have a spontaneous process that requires no external work). If we can answer this 1

2 question and show these conditions regularly occur in the atmosphere, then we ve successfully explained why cloud droplet form from thin air. For the empty box, the total energy of the system is just the chemical potential energy of the gas G i = n v V µ v where, n v is the initial number density of the vapor molecules, and V is the volume. For the box with the droplet in it, the total energy of the system is the sum of the chemical potential energy of the droplet, the chemical potential energy of the remaining vapor, and the surface tension energy associated with the curvature of the droplet G f = n l V µ l + (n v n l ) V µ v + σa where, n l is the number density of water molecules in the box, µ l is the chemical potential energy of liquid water molecules, A is the surface area of the droplet, and σ is the surface tension of the droplet, which is about σ = N m 1 = J m 2 Therefore, the change in Gibbs free energy associated with droplet formation is G = G f G i = Aσ + n l V (µ l µ v ) which is kind of intuitive if you think about it. We need to form surface tension energy, and whatever added energy is associated with the chemical potential of the liquid. But we can go further. Remembering we can substitute to get which we can show is equivalent to µ v µ l = kt ln e/e sat G = Aσ n l V kt ln ( e/e sat) G = Aσ mr v T ln ( e/e sat) G = 4πr 2 σ 4 3 πρ lr 3 R v T ln ( e/e sat) To analyze this equation, we examine two conditions Condition I: e < e sat In this case, both terms are positive, and G increases exponentially with increasing r. Since we require G < 0 for an irreversible transformation, the droplet cannot form without external work, for example by lifting the parcel so that it cools. 2

3 Condition II: e > e sat Now the second term is negative, and if it dominates, then G < 0. Plotting the equation, we would see that initially G increases with r, it reaches a maximum, and then decreases with r. The question is what is the critical radius r required for just a tiny increase extra to send the droplet on its way to spontaneous irreversible growth? To calculate, r, we simply take the differential of G with respect to r and equate to zero. d G dr ] r=r = 8πr σ 4πρ l r 2 R v T ln ( e/e sat) = 0 r = 2σ ρ l R v T ln (e/e sat ) Solving we get the saturation vaporation over a pure droplet, taking into account its surface tension. e sat (r) = e sat exp ( 2σ ) ρ l R v T r This is known as the Kelvin equation. A couple of things to notice about this equation. First as r approaches infinity, the curvature effect on the saturation vapor pressure becomes insignificant. In other words, even if the droplet is 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 to be essentially flat if they are larger than about 1 µm. Second, at the other extreme, for extremely small droplets, huge supersaturations are required for the droplet to be in equilibirum with its environment. For recently nucleated particles, say 0.1 µm across, the supersaturation at equilibrium is in excess of 100% (i.e. 200% relative humidity). We never get relative humidities this high in the atmosphere, and since any pure droplet forming from vapor must start from an embryo of a few molecules, clearly pure water droplets cannot form in the atmosphere. To understand this further, we could also right the Kelvin Equation in the form ( σ e sat (r) = e sat exp rkt which is similar to the Boltzmann distribution expression we derived earlier. Remember that dg = n µ 1 + µ 2 dn = 0 d ln n = µ 1 µ 2 which conveniently leads to the Boltzmann distribution n = n 0 exp ( µ 1 µ 2 For the case of evaporation off a flat surface µ = l v and µ v = kt, such that n = n 0 exp( l v kt ) 3 )

4 T = 5 C 10 4 Supersaturation (%) mer R (µm) Figure 1: Dependence of the supersaturation of water vapor at equilibrium over a pure water droplet. The size of a 13 molecule nucleus is shown for reference. 4

5 We ve already incorporated this physics in our calculation of e s. Here then we ve added the extra component of droplet surface tension which acts to reduce the energy barrier between the liquid and vapor phase. In other words, now we have µ = µ v µ l = l v 2σ/r It takes less energy for a water molecule to break free from a curved surface than from a flat surface. The molecules are less tightly bound by each other in the liquid phase. So this makes it very hard to form a droplet composed of liquid alone. In fact in the atmosphere it is impossible. Somehow, we need to increase µ so that it is harder for a droplet to evaporate, such that it is happy to stay in the liquid phase. The way the atmosphere does this is by including aerosol matter in the droplet. The solution effect and the Kohler curve An effective way to lower the gibbs free energy density of the liquid water is to dilute it with a solute. What this does effectively is reduce the density of water molecules at the interface between the liquid and vapor phase, since the water molecules are separated by solute ions. Consequently the density of water vapor needs to be less in order for there to be equilibrium. Let s say we define a solution activity such that the activity equals the relative humidity that is required for the liquid and vapor phase to be in equilibrium. In other words, at equilibrium Since µ sat v (T, a w ) = µ l (T, a w ) µ sat v (T, a w ) = µ l (T, a w = 1) + kt ln RH µ l (T, a w ) = µ l (T, a w = 1) + kt ln a w where a w is bounded by 0 and 1. If a w < 1 then the chemical potential of liquid is depressed compared to what it would be at equilbrium over a flat surface of pure water. So we can alter our express for µ from before so that we have µ = µ v µ l = l v 2σ r kt ln a w If the activity of the solution can be lowered then the last term is positive and this increases the energy barrier between the liquid and vapor phase µ, making existence of the liquid phase more stable. There is a bigger wall for the liquid molecules to jump over to get into the vapor phase. Our revised expression then for the vapor pressure over a solution droplet is e sat (r, a w, T ) = e sat (T ) a w exp ( ) ( ) 2σ 2σ = e sat (T ) a w exp ρ l R v T r rkt This is known as the Kohler equation. For an ideal solution (usually true if very dilute) a w = n 0 n + n 0 5

6 where n 0 is the molecules of water and n molecules of solute. This is known as Raoult s Law. For dilute solutions a w = 1 n/n 0 But we must remember that salts dissociate, i.e. NaCl dissociates to Na + and Cl. We represent this using what is really a bit of a kluge, but popular nonetheless - the so called Van t Hoff factor i. n in where, for NaCl, and NH 4 H (ammonium bisulfate) i = 2, and for (ammonium sulfate), i = 3. i is essentially the number of solute ions that the molecule dissolves into. The problem with this approach is that the number i stops meaning anything physical when one starts talking 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 a liquid droplet, dissolving a solute makes it harder for them to escape. Effectively there are fewer liquid molecules per unit surface area for them to escape. The Kohler curves are shown below. These show the relative humidity relative to saturation over a flat surface of pure liquid water that would be in equilibrium with a droplet of radius r containing dissolved solute. Effectively, it is e sat (r, a w, T ) /e sat (T ). 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 dg = 0). If the particle grows spontaneously, it is then too large for the ambient equilibrium saturation for the droplet, so it shrinks to its original size. Likewise, if it shrinks it grow again to its equilibrium size. These are haze particles, which are in stable equilibrium. Above this size the haze particles are said to be activated. The Kelvin effect dominates and dg < 0 if the particles grow spontaneously. A particle that grows spontaneously will be have an equilibrium saturation lower than the ambient saturation, so it keeps growing spontaneously by vapor diffusion. The total Gibbs free energy of the system will continue to decrease. Once the droplets get big enough, the Kelvin effect stops mattering, and the droplet acts effectively like a plane surface of water. Therefore, the function of a solute in droplet formation is to lower the amount of chemical potential energy of the water vapor (i.e. supersaturation) required to get the droplet to the size r 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. 6

7 =0.02 µm) =0.03 µm) =0.04 µm) =0.06 µm) =0.08 µm) =0.1 µm) =0.15 µm) =0.3 µm) S r (µm) Figure 2: Kohler curves for ammonium sulfate. 7

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