Aji

Hence the activity of water decreases as electrolyte 2 is added to the system, until the solution becomes saturated in that electrolyte, too. The aerosol is exposed to the atmosphere and therefore its DRH also decreases.

The preceding analysis can be extended to aerosols containing more than two salts. Thus one can prove that water activity reaches a minimum at the deliquescence point of the aerosol. Another consequence of this analysis is that the DRH of a mixed salt is always lower than the DRH of the individual salts in the particle. Wexler and Seinfeld (1991) solved (10.76) for the case of the system containing NH4NO3 and NH4C1. Their calculations at 303 K are depicted in Figure 10.8.

When there is only NH4C1 in the particle (xnh4no3 = 0) the DRH of the particle is 77.4%. If the RH is below 77.4%, the particle is solid; if it is above 77.4% it is c o

^-DRH(NH4C1)

a

1 1

m -——

b

S\DRH(NH4N03K

c

DRH*

1

1

ANH4N03

FIGURE 10.8 Water activity at saturation for an aqueous solution of NH4NO3 and NH4C1 at 303 K. (Reprinted from Atmos. Environ., 25A, Wexler, A. S., and Seinfeld, J. H., Second-generation inorganic aerosol model, 2731-2748. Copyright 1991, with kind permission from Elsevier Science Ltd., The Boulevard, Langford Lane, Kidlington OX5 1GB, UK.)

an aqueous solution of NH4 and CI . (Note that the calculations in Figure 10.8 are at 303 K.)

There are seven different RH composition regimes in the figure:

(a) RH > DRH(NH4C1) > DRH(NH4N03). When the RH exceeds the DRH of both compounds, the aerosol is an aqueous solution of NH4 , N03, and Cl~.

(b,) DRH(NH4C1) > RH > DRH(NH4N03)—left of [1], In this regime the aerosol consists of solid NH4C1 in equilibrium with an aqueous solution of NH4 , NO J, and CI".

(b2) DRH(NH4C1) > RH > DRH(NH4N03)—right of [1], If the aerosol contains enough NH4N03 so that its composition is to the right of line [1], the aerosol is an aqueous solution of NH4, NO J, and CI . In this regime addition of NH4NO3 results in complete dissolution of NH4C1 even if the RH is higher than its DRH.

(c0 DRH(NH4C1) > DRH(NH4N03) > RH > DRH*—left of [1], The aerosol consists of solid NH4C1 in equilibrium with a solution of NH|, N03, and CI".

(c2) Right of [1], left of [2]. In this regime there is no solid phase and the aerosol consists exclusively of an aqueous solution of NH4, N03, and Cl~.

(c3) Right of [2]. Here the aerosol consists of solid NH4N03 in equilibrium with an aqueous solution of NH4, NO J, and CI . (d) RH < DRH*. If the relative humidity is below DRH* = 51%, the aerosol consists of solid NH4N03 and solid NH4C1.

In conclusion, for this mixture of salts, if both are present, the aerosol will contain some liquid water for RH >51%. However, if we are below the solid line in Figure 10.8, the aerosol will also contain a solid phase in equilibrium with the solution. Only above the solid line is the particle completely liquid.

It is instructive to follow the changes in a particle of a given composition as RH increases. For example, consider a particle consisting of 40% NH4N03 and 60% NH4C1 as RH increases from 40 to 90%, assuming that there is no evaporation or condensation of these salts. At the beginning (RH = 40%) the particle is solid and it remains solid until RH reaches 51.4% ( = DRH*). At this point, the particle deliquesces and consists of two phases—a solid phase consisting of solid NH4C1 and an aqueous solution with the composition corresponding to the eutonicpoint (xnh4no3 = 0.811,xnh4ci = 0.189). As the RH increases further, more NH4C1 dissolves in the solution and the composition of the aqueous solution follows line [1], For example, at RH = 60% the aerosol consists of solid NH4CI and a solution of composition xnh4no3 = 0.73, xnh4ci = 0.27. At 70% RH, most of the NH4C1 has dissolved and the composition of the solution is close to the net particle composition xnh4no3 = 0.42, xnh4c:i = 0.58. Finally, when the RH reaches 71%, all the NH4C1 dissolves and the particle consists of only one phase (the aqueous one) with composition equal to the particle composition. Note that there is only one step change in the mass of the particle corresponding to the mutual DRH* (e.g., see Figure 10.7). Further increases in the RH cause continuous changes of the aerosol mass and not step changes. On the contrary, step changes are observed during particle evaporation and will be discussed subsequently.

This discussion of phase transitions has been extended to mixtures of more than three salts by Potukuchi and Wexler (1995a,b). The phase diagrams become now three

FIGURE 10.9 Deliquescence relative humidity contours (solid lines) for aqueous solutions of H+ — NH4 — HSO4 — SO4 - NOJ shown together with the lines (dashed) describing aqueous-phase composition with relative humidity. Labels on the contours represent the deliquescence relative humidity values. Total hydrogen is total moles of protons and bisulfate ions. Total sulfate is number of moles of sulfate and bisulfate ions. X = ammonium (ammonium + total hydrogen); Y = total sulfate (total sulfate + nitrate). (Reprinted from Atmos. Environ. 29, Potukuchi, S., and Wexler, A. S., Identifying solid-aqueous phase transitions in atmospheric aerosols. II Acidic solutions, 3357-3364. Copyright 1995, with kind permission from Elsevier Science Ltd., The Boulevard, Langford Lane, Kidlington 0X5 1GB, UK.)

FIGURE 10.9 Deliquescence relative humidity contours (solid lines) for aqueous solutions of H+ — NH4 — HSO4 — SO4 - NOJ shown together with the lines (dashed) describing aqueous-phase composition with relative humidity. Labels on the contours represent the deliquescence relative humidity values. Total hydrogen is total moles of protons and bisulfate ions. Total sulfate is number of moles of sulfate and bisulfate ions. X = ammonium (ammonium + total hydrogen); Y = total sulfate (total sulfate + nitrate). (Reprinted from Atmos. Environ. 29, Potukuchi, S., and Wexler, A. S., Identifying solid-aqueous phase transitions in atmospheric aerosols. II Acidic solutions, 3357-3364. Copyright 1995, with kind permission from Elsevier Science Ltd., The Boulevard, Langford Lane, Kidlington 0X5 1GB, UK.)

dimensional (Figure 10.9). Note that the solid phases that appear are (NH4)2S04, NH4HSO4, letovicite ((NH4)3H(S04)2), NH4N03, and the double salt (NH4)2S04 ■ 2NH4N03. The labels on the contours show relative humidities at deliquescence. The bold lines in this figure are the phase boundaries separating the various solid phases. On these lines both phases coexist. For example, let us assume that there is enough ammonia present so X = 1 in Figure 10.9. If sulfate dominates then Y — 1 and the system is in the upper right-hand corner of the diagram with a DRH of 80% corresponding to (NH4)2S04. If there is enough nitrate available so that Y = 0.3 then the DRH is 68% corresponding to (NH4)2S04 • 2 NH4NO3.

The mutual deliquescence points of a series of pairs are given in Table 10.4.

Next Page

TABLE 10.4 Deliquescence RH (DRH*) at Mutual Solubility Point at 303 K

Compound 1

Compound 2

DRH*

DRHi

DRH2

NH4NO3

NaCl

42.2

59.4

75.2

NH4NO3

NaN03

46.3

59.4

72.4

NH4NO3

NH4CI

51.4

59.4

77.2

NaN03

NH4CI

51.9

72.4

77.2

NH4NO3

(NH4)2S04

52.3

59.4

79.2

NaN03

NaCl

67.6

72.4

75.2

NaCl

NH4CI

68.8

75.2

77.2

NH4CI

(NH4)2S04

71.3

77.2

79.2

Source: Wexler and Seinfeld (1991).

Source: Wexler and Seinfeld (1991).

10.2.4 Crystallization of Single and Multicomponent Salts

The behavior of inorganic salts when RH is decreased is different from that discussed in Sections 10.2.2 and 10.2.3. For example, for (NH4)2S04, as RH decreases below 80% (the DRH of (NH4)2S04), the particle water evaporates, but not completely. The particle remains liquid until a RH of approximately 35%, where crystallization finally occurs (Figure 10.4). The RH at which the particle becomes dry is often called the efflorescence RH (ERH). This hysteresis phenomenon is characteristic of most salts. For such salts, knowledge of the RH alone is insufficient for determining the state of the aerosol in the RH between the efflorescence and deliquescence RH. One needs to know the RH history of the particle.

The ERH of a salt cannot be calculated from thermodynamic principles such as the DRH; rather, it must be measured in the laboratory. These measurements are challenging because of the potential effects of impurities, observation time, and size of the particle. Measured ERH values at ambient temperatures are shown in Table 10.5 (Martin 2000). Pure NH4HS04, NH4NO3, AND NaN03 aqueous particles do not readily crystallize even at RH values close to zero.

Particles consisting of multiple salts exhibit a more complicated behavior (Figure 10.7). For example, the evaporation of a KCl-NaCl particle is characterized by two step changes: the first at 65% with the formation of KC1 crystals and the second at 62% with the

TABLE 10.5 Efflorescence Relative Humidity at 298 K

Salt

erh (%)

(NH4)2S04

35 ±2

NH4HSO4

Not observed

(NH4)H(S04)2

35

nh4no3

Not observed

Na2S04

56 ± 1

NaCl

43 ±3

NaNOs

Not observed

NH4CI

45

KCl

59

Source: Martin (2000).

Source: Martin (2000).

I I i evious I age equilibrium vapor pressure over a curved surface: the kelvin effect 461

complete drying of the particles and the crystallization of the remaining NaCl. The ERH of a particle consisting of multiple salts depends on the composition of the particle. For example, as H2S04 is added to an (NH4).,S04 particle, its ERH decreases from approximately 35% to around 20% for a molar ratio of NH4 : S04" = 1.5 to approximately zero for NH4 : S04~ = 1 (ammonium bisulfate) (Martin et al. 2003). For a particle consisting of approximately 1:1 (NH4)2S04:NH4N03, the ERH is around 20%, while for a 1:2 molar ratio it decreases to around 10%. As a result, particles that are either very acidic (containing ammonium bisulfate) or contain significant amounts of ammonium nitrate tend to remain liquid during their atmospheric lifetime even when the ambient RH is quite low (Shaw and Rood 1990).

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).

10.3 EQUILIBRIUM VAPOR PRESSURE OVER A CURVED SURFACE: THE KELVIN EFFECT

In our discussion so far of aerosol thermodynamics, we have assumed that the aqueous aerosol solution has a flat surface. However, the key aspect that distinguishes the thermodynamics of atmospheric particles and drops is their curved interface. In this section we investigate the effect of curvature on the vapor pressure of a species A over the particle surface. Our approach will be to relate this vapor pressure to that over a flat surface. To do so we begin by considering the change of Gibbs free energy accompanying the formation of a single drop of pure A of radius Rp containing n molecules of the substance:

AG = Gdr0piet ^pure vapor

Let us assume that the total number of molecules of vapor initially is Nr\ after the drop forms the number of vapor molecules remaining is N\ = NT — n. Then, if gv and g/ are the Gibbs free energies of a molecule in the vapor and liquid phases, respectively

AG = Nlgv + ng, + 4k R2po - NTgv where 4nRpa is the free energy associated with an interface with radius of curvature Rp and surface tension a. Surface tension is the amount of energy required to increase the area of a surface by 1 unit. This equation can be rewritten as

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

4tiRl where vi is the volume occupied by a molecule in the liquid phase. Thus, combining (10.78) and (10.79),

4ti Rl

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 = 0, dg = vdp or gi — gv = (vi — v,,)dp . Since vv vi for all conditions of interest to us, we can neglect vi relative to vv in this equation, giving gt — gv = ~vv dp. The vapor phase is assumed to be ideal so vv = kT/p. Thus, integrating, we obtain fPA dp g,-gv = -kT (10.81)

Jpl P

where p°A is the vapor pressure of pure A over a flat surface and pA is the actual equilibrium partial pressure over the liquid. Then gl-gv = -kT ln^ (10.82)

We can define the ratio Pa/p°a as the saturation ratio S. Substituting (10.82) into (10.80), we obtain the following expression for the Gibbs free energy change:

4 kT

Figure 10.10 shows a sketch of the behavior of AG as a function of Rp. We see that if S < 1, both terms in (10.83) are positive and AG increases monotonically with Rp. On the

FIGURE 10.10 Gibbs free energy change for formation of a droplet of radius Rp from a vapor with saturation ratio S.

EQUILIBRIUM VAPOR PRESSURE OVER A CURVED SURFACE: THE KELVIN EFFECT

other hand, if S > 1, AG contains both positive and negative contributions. At small values of Rp the surface tension term dominates and the behavior of AG is similar to the S < 1 case. As Rp increases, the first term on the RHS of (10.83) becomes more important so that AG reaches a maximum value AG* at Rp = Rp and then decreases. The radius corresponding to the maximum can be calculated by setting 0AG/dRp = 0 in (10.83) to get

Since AG is a maximum, as opposed to a minimum, at Rp = Rp, the equilibrium at this point is metastable. Equation (10.84) relates the equilibrium radius of a droplet of a pure substance to the physical properties of the substance, a and vh and to the saturation ratio S of its environment. Equation (10.84) can be rearranged recalling the definition of the saturation ratio as

Expressed in this form, the equation is frequently referred to as the Kelvin equation. The Kelvin equation can also be expressed in terms of molar units as where M is the molecular weight of the substance, and p/ is the liquid-phase density.

The Kelvin equation tells us that the vapor pressure over a curved interface always exceeds that of the same substance over aflat surface. A rough physical interpretation of this so-called Kelvin effect is depicted in Figure 10.11. The vapor pressure of a liquid is determined by the energy necessary to separate a molecule from the attractive forces exerted by its neighbors and bring it to the gas phase. When a curved interface exists, as in a small droplet, there are fewer molecules immediately adjacent to a molecule on the surface than when the surface is flat. Consequently, it is easier for the molecules on the surface of a small drop to escape into the vapor phase and the vapor pressure over a curved interface is always greater than that over a plane surface.

Table 10.6 contains surface tensions for water and a series of organic molecules. At 298 K the organic surface tensions are about one-third that of water and their molecular volumes (M/p/) range from three to six times that of water.

FIGURE 10.11 Effect of radius of curvature of a drop on its vapor pressure.
TABLE 10.6 Surface Tensions and Densities of Selected Compounds at 298 Ka

Compound

Molecular Weight

Density, g cm 3

Surface Tension, dyn cm 1

Water

18

1.0

72

Benzene

78.11

0.879

28.21

Acetone

58.08

0.787

23.04

Ethanol

46.07

0.789

22.14

Styrene

104.2

0.906

31.49

Carbon tetrachloride 153.82

1.594

26.34

"a has units 10~3kgs~2.

in the SI system of N m 1 ( = kg s

~2). In the cgs system a

has units dyncnT1. ldyncm"1^

FIGURE 10.12 Saturation ratio versus droplet size for pure water and an organic liquid (dioctyl phthalate DOP) at 20°C.

Droplet Diameter, (im

FIGURE 10.12 Saturation ratio versus droplet size for pure water and an organic liquid (dioctyl phthalate DOP) at 20°C.

Figure 10.12 gives the magnitude of the Kelvin effect Pa/p°a for a pure water droplet and a typical organic compound (DOP) as a function of droplet diameter. We see that for water the vapor pressure is increased by 2.1% for a 0.1-pm and 23% for a 0.01-pm drop over that for a flat interface. Roughly, we may consider 50 nm diameter as the point at which the Kelvin effect begins to become important for aqueous particles. For higher-molecular-weight organics, the Kelvin effect is significant for even larger particles and should be included in calculations for particles with diameters smaller than about 200 nm.

10.4 THERMODYNAMICS OF ATMOSPHERIC AEROSOL SYSTEMS 10.4.1 The H2S04-H30 System

The importance of sulfuric acid in aerosol formation in a humid environment has been emphasized in a number of studies (Kiang et al. 1973; Mirabel and Katz 1974;

Jaecker-Voirol and Mirabel 1989; Kulmata and Laaksonen 1990; Laaksonen et al. 1995). It is instructive (o study first this simplified binary system before starting the discussion of more complicated atmospheric aerosol systems.

Sulfuric acid is very hygroscopic, absorbing significant amounts of water even at extremely low relative humidities. In Figure 10.13, sulfuric acid aqueous solution properties are shown as a function of the mass fraction of H2SO4 in solution. Included in Figure 10.13 are the equilibrium relative humidity over a flat solution surface, RH, the solution density, p, the boiling point, B.P., and the surface tension, o. In addition, the solution normality (in N or equivalents per liter), the mass concentration, ah,so4P-the particle growth factor (the ratio of the actual diameter of the droplet Dr to that if the water associated with that were completely evaporated, Dpq), Dp/Dp0, and the Kelvin parameter D^-, In (pH^SQi/pifeSO-i) are s^own ir! the same figure.

Calculation of the Properties of a H2SO4 Droplet (Liu and Levi 1980) Consider a 1-jim H2SO4-H2O droplet in equilibrium al 50% RH. Using Figure 10.13, calculate the following:

1. The H2S04 concentration in the solution

2. The density of the solution

3. The boiling point of the solution

4. The droplet surface tension

5. The solution normality

6. The mass concentration of H2S04 in the solution

7. The size of the droplet if all the water were removed

8. The Kelvin effect parameter

9. The size of this droplet at 90% RH.

Using the curve labeled RH and the corresponding scale at the right of the graph, we find that the solution concentration is 42.5% H2SO4 or thai _v!iiSOj = 0.425 g H2S04 per g of solution. The rest of the solution properties can be calculated using this solution concentration, the appropriate curve, and the appropriate scale. At this concentration the solution has a density, p, of i .32 g cm-3, a boiling point of 115°C, a surface tension, cr, of 76dyncm_l, a normality of 1 I N, and a mass concentration, iH.so.P- Of 0.55 g H2S04 cm 3 of solution. The particle size growth factor, Dp/Dpo, is 1.48. Thus the H2S04—H20 solution droplet would become a droplet of pure H2SO4 of 1/1.48 = 0.68 (im if all the water were removed. The Kelvin effect parameter DlA) In (ph2so,/Ph,soJ 1S 1 J -3 x 10_4pm, indicating that y^soj 0.68

Thus (/JiijSOj/Ph,so4) = 10017, and in this case the Kelvin effect is negligible; the increase in water vapor pressure due to the droplet curvature is only 0.i66% above that of a flat surface. The panicle size growth factor at 90% relative humidity is 2.12. Thus the 1 pm diameter drop at 50% RH will grow to become a drop of (2.12/1.48) (1) = 1.43 pm at 90% RH.

Kelvin Effect Parameter, D^q

10"4|am

Solution Normality, N s M

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