Equilibrium Vapor Pressure over a Curved Surface The Kelvin Effect

Spann and Richardson (1985) have shown that for particles of composition between NH4HSO4 and (NH4)2S04 the crystallization RH ranges from 10 to 40%, indicating that for certain compositions the solution cannot be dried in the atmosphere. Particles of this composition would likely be present even at low relative humidities in the atmosphere as supersaturated salts and will not undergo deliquescent behavior. The same is true for NaCl with a crystallization RH of around 42% (Shaw and Rood, 1990). 

3 EQUILIBRIUM VAPOR PRESSURE OVER A CURVED SURFACE  

Let US assume that the total number of molecules of vapor initially is Nj after the drop forms the number of vapor molecules remaining h N = Nr — . Then, if g, and gi are the Gibbs free energies of a molecule in the vapor and liquid phases, respectively, 

Note that the number of molecules in the drop, n, and the drop radius Rp are related by 

We can define the ratio Pa/Pa e saturation ratio S. Substituting (9.82) into (9.80), we obtain the following expression for the Gibbs free energy change  

Inclusions of insoluble dust minerals (CaC03, Fe203, etc.) can increase the efflorescence RH of salts (Martin 2000). For example, for (NH4)2S04, the ERH can increase from 35% to 49% when CaC03 inclusions are present. These mineral inclusions provide well ordered atomic arrays and thus assist in the formation of crystals at higher RH and lower solution supersaturations. Soot, on the other hand, appears not to be an effective nucleus for crystallization of salts because it does not contain a regular array of atoms that can induce order at least locally in the aqueous medium (Martin 2000). 

We now need to evaluate gi — gv, the difference in the Gibbs free energy per molecule of the liquid and vapor states. Using (10.13) at constant temperature and because dn, = 0, dg = vdp or gi — gv = (vi — vv)dp. Since vv v/ for all conditions of interest to us, we can neglect v/ relative to v, in this equation, giving gi — gv = —vv dp. The vapor phase is assumed to be ideal so vv = kT/p. Thus, integrating, we obtain 

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