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Droplet Formation Gibbs Free Energy of a Droplet As 1 Droplet Formation How do spherical cloud droplets form? What are the necessary physical conditions? To start with, we need to intro duce the concept of surface tension. Weintro duce the Gibbs Free Energy of a droplet as G = U TS + pV Where U is the total internal energy,Sisentropy. Recall that U can also b e expressed as U = TS p V which is the rst law of thermo dynamics. So wenowhave G = U TS + ST +p V + VP and some of the terms obviously vanish leaving G = ST + VP Consider nowa water droplet immersed in water vap or which has partial pressure e. The droplet and vap or are not in equilibrium meaning e 6= e s We let the Gibbs free energy p er unit mass of the vap or b e G and of the droplet G . V d Next we make use of a standard 'trick' in lots of areas of physics and we makea very small p ertubation to a quantity. In this case we will vary the partial pressure by an amount e in sucha way so as the temp erature remains constant. Hence G = V e ; G = V e v v d d Now clearly 2 V >> V v d so that G G =V V e V e v d v d v We assume that the vap or satis es the ideal gas law RT V = v e so that e G = V e = RT v v e Integrating at constant T then gives G G = RT l ne+constant v d Implicit throughout this is the understanding that G is a function of temp erature and pressure and therefore the constantwould change with Temp erature. At equilibrium on the vap or pressure phase diagram, e = e and G = G Therfore, s v d our constantmust equal RT l ne s so that e G G = RT l n v d e s We next intro duce a mass M of water vap or at partial pressure e and temp erature o T. With no droplet presentwewould have G = G M o v o 3 b ecause by construction G is the free energy p er unit mass. Next we assume that v 2 a spherical droplet of radius a condenses. Its area is A =4a and its mass is M = d 3 4a =3. The total mass is now d M = M + M o v d and the total free energy of the system is now G = G M + G M + A v v d d where is the surface tension p er unit area. So in forming a droplet we necessarily pick up a new term in the total free energy of the system. we can now write G G = G M + G M + A G M + M o v v d d v v d or G G =G G M + A o d v d using the previously derived expression for G G yields v d e G G = RT l n M + A o d e s or in terms of the radius of the cloud droplet e 4 3 2 a RT l n +4a G G = d o 3 e s So now its clearly seen that the total Gibbs free energy dep ends on the radius of the droplet of the form 3 2 Ga=G a + a o Now refer to gure 4.10 in Wallace and Hobbs: e dep ends on so when e<e the logarithim is negative and Ga is always p ositive s e s and would clearly increase with increasing a. But for e>e wehave a critical p oint: s 4 a = a d at this p oint Ga =0 a so 2 3 a =2 a d d or 2 2 a = = d 3 RT l ne=e d s It is physically more reasonable to write the ab ove in terms of the partial pressure so that 2 e = e exp s RT a d d This shows that the smaller the droplet the b etter. For instance, for a = 0.01 microns RH = 112.5 meaning the air is sup eraturated by 12.5. For d a = 0.1 microns, RH = 101 d So what happ ens? Well if a droplet is in equilibrium at radius a then a condensation of a small quantityof d vap or around the droplet will increase a and it continues to grow. If, however, evap oration o ccurs then a shrinks and will continue to do so. If the cloud droplet is to survive than it must attain a radius greater than a . d Since the RH of clouds is usually not greater than 101 then a must b e greater than d 0.1 microns. A droplet of this size is unlikely to form via collisions and sticking of many smaller droplets so in real life, a condensation nucleus ice, dirt, p ollution, etc really needs to exist..
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