metal, then a pore or pores will form in the metal. In Figure 14.4a, a single pore is formed in the center and when the oxidation is complete, all that will remain is the shell of oxide with an empty center.
Figure 14.4b illustrates a more likely case in which pores form throughout the metal. However, if the oxidation is taken to completion, the net result is the same: the pores will coalesce into one single large pore leaving only the empty oxide shell. Sometimes this porosity is referred to as the “Kirkendall porosity.” The Kirkendall porosity, discussed later, is caused by differences in diffusion coefficients in alloys. Naming the porosity that is the result of oxidation the same as that produced by interdiffusion is confusing at a minimum. The mechanisms of the porosity formation are really quite different even though they are both produced by differences in diffusion coefficients of two different atomic or ionic species. Calling this oxidation-induced porosity the “Kirkendall porosity” seems ill-advised.
Another way to form porosity is at the metal–oxide interface as shown in Figure 14.4c. An alternative way of tracing the disappearance of the nickel metal is that the metal–oxide interface simply moves with the disappearing metal. However, this requires that, for a finite-size piece of metal, the oxide must plastically deform to keep up with the shrinking nickel. In some cases, the compressive stresses on a plane surface may become large enough to cause the oxide layer to buckle and debond from the nickel and form a pore at the interface. At a corner, where even greater defor- mation of the oxide is necessary to take care of shrinking Ni–NiO interfaces, debonding and pores are even more likely to form. In both of these situations, pores form and oxidation stops at these points unless cracks actually form in the oxide layer.
14.3 OSMOSIS
14.3 Osmosis 487 molecule is differentiated from a suspended body solely (his italics) by its dimensions, and it is not
apparent why a number of suspended particles should not produce the same osmotic pressure as the same number of molecules” (Einstein 1956, 3).
14.3.2 d
efInItIonTo borrow a definition from the literature (Chang 2000, 999),
Osmosis is the net movement of solvent molecules through a semipermeable membrane from a pure solvent or from a dilute solution to a more concentrated solution.
14.3.3 S
emIpermeablem
embraneWhat is a semipermeable membrane? It is a thin sheet (usually) of material that has a very definite pore size, as shown in Figure 14.5, such that molecules and ions smaller than the pore size can eas- ily pass through the membrane and anything larger than the pore size cannot. So it is permeable to some chemical species and not others, hence, semipermeable. Semipermeable membranes are commercially available with effective pore sizes from 0.1 nm to much larger. The permeability of a membrane is specified by its molecular weight cutoff, MWCO, expressed in daltons (Haney et al.
2013) where 1 dalton is the unified atomic mass unit equal to one-twelfth of a carbon-12 atom (Atomic mass unit). Frequently, these membranes are made from cellulose fibers (see Figure 14.6) and can be made with a wide range of MWCO from 1 to 20,000 daltons and have many applications including the dialysis of blood.
Cellulose O
CH2OH
H H OH
H H
OH
O CH2OH H OH H
H OH O
O
n H
H H
FIGURE 14.6 The structure of cellulose that is frequently the main chemical constituent of semipermeable membranes.
Semipermeable
membrane Pore size
0.2 nm Small molecule
0.3 nm Large
molecule 3 nm
Water molecules and small ions 0.1 nm
Net flow
FIGURE 14.5 Schematic showing the operation of a semipermeable membrane in which atoms, ions, molecules, or particles greater than the pore size cannot go through the membrane while those smaller than the pore size can.
14.3.4 t
hermodynamIcSFigure 14.7 illustrates how the principle of osmosis and osmotic pressure, π, can be observed. Consider a system containing components 1 and 2.
Pure solvent, X1 = 1,* is separated from a solution of composition X1 < 1 by a semipermeable membrane that is permeable to the solvent that can freely flow through the membrane in both directions but the solute in the solution cannot. In terms of diffusion, the solvent has a finite diffu- sion coefficient through the membrane while the diffusion coefficient of the solute through the membrane is zero. As a result, there is a Gibbs energy difference produced by this difference in concentration,
∆
∆
G G G G RT X G
G RT X
solution solvent solvento
solvent
= − = + − o
=
ln ln .
1 1
(14.10)
With this Gibbs energy difference, solvent flows from the pure solvent to the solution increasing its volume. If the apparatus is constructed with a tube for the increase in volume of the solution, as shown in Figure 14.7, the increase in height of the solution in the tube will be such that the pressure that it exerts, π, will just balance the energy dif- ference produced by the solution–solvent interaction. That is, because dG Vdp SdT= − and at constant temperature, where V = molar vol- ume of the pure solvent. Now, ΔG must be negative since X1 < 1, so
∆G V dp V
P P
= − = −
∫
+π π. (14.11)Equating Equations 14.10 and 14.11, gives
π = −RT X V ln 1.
The mole fraction of the solute is X2 and X1+X2=1 so lnX1=ln
(
1−X2)
≅ −X2 because X2 is usually small, soπ = =
+
RTX
V RT
V n n n
2 2
1 2
π ≅RTn = Vn2 RTC
1
2 (14.12)
where n1 and n2 are the number of moles of the solute and solvent, respectively, and n2n1 and V n V≅ 1 is the total volume and C2 is the concentration of solute in mol/m3 or something similar.
Equation 14.12 is known as the van’t Hoff equation for osmotic pressure (Laidler and Meiser 1995).
14.3.5 e
xampleSuppose that C2 = 10−3 mol/L = 1 mol/m3. Then, at 300 K, π
π
= =
( )( )( )
= × ≅ × −
RTC
Pa bar
2
3 2
8 314 300 1 2 49 10 2 49 10
.
. .
* Xi is the mole fraction of the component i, Chapter 1.
Solvent Solution h
Solute
Semipermeable membrane
P
P
π Imagined
closure
FIGURE 14.7 An apparatus for the measurement of osmotic pressure. Pure solvent is separated from the solution by a semipermeable membrane: permeable to the solvent but not the solute. As a result, solvent diffuses through the membrane into the solution until an osmotic pressure, π, builds up on the solution side of the mem- brane countering the entropic Gibbs energy between the solvent and the solution. If the vessel were sealed at the bottom of the vertical tubes as indicated, then the pres- sure would build up on the solution side. If the membrane could move, then it would move toward the faster-dif- fusing solvent side, a perfect example of marker motion during interdiffusion: the marker moves opposite to the direction of the flux of the faster diffusing species.
14.3 Osmosis 489 Because 1 atm = 760 mm Hg, if the solvent were water, which has a density of ρ = 1/g/cm3
while that of mercury is ρ = 13.6 g/cm3, 1 atm = 1.01325 bar = 1.03 × 103 cm H2O. Or the height to which the water would rise in the right-hand vertical tube in Figure 14.7 would be h=1 03 10. × 3
(
cm atm)
×1 01325.(
atm bar) (
2 49 10. × −2bar)
=25 99. cm! Note that p= ρgh so for 1 atm p= ×(
1 103kg m3)(
9 81. m s2) (
10 3. m)
=1 01 10. × 5Pa so that 1.03 × 103 cm of water is correct.On the other hand, if the membrane was able to physically move laterally under the applied pres- sure, the osmotic pressure would cause the membrane in Figure 14.7 to move relative to the two ends of the water column, and keep moving until the pure solvent was gone. This membrane is acting as an inert marker relative to the diffusion fluxes of the solvent and the solute where one diffuses much faster than the other: the marker moves toward the faster diffusion species! How fast this marker moves depends on how fast the solvent diffuses through the membrane and into the solution.
14.3.6 m
eaSurIngm
olecularW
eIghtSEquation 14.12 could be written as
π = =
RTn
Vn RT
M m
V
2 1
2 (14.13)
where:
m2 is the mass of component two so m2/V is the concentration in kg/m3 M is the molecular weight.
So the osmotic pressure can be used to measure the molecular weight of a particular substance, including polymers. For example, suppose 1 g of poly(ethylene glycol) is dissolved in 1 L of water, so the concentration is 1 kg/m3. If the water is raised to a height of 7.26 cm at equilibrium, the molecu- lar weight of the polymer can be calculated from
M RT m
V atm Pa atm
=
=
( )( )
×
(
×)
π
2
3
5
8 314 300 7 26
1 03 10 1 01 10 .
.
. .
(
1kg m/ 3)
M=3 503. kg mole/ .
14.3.7 d
eSalInIzatIonofS
eaWaterNow consider the case where a pressure, P, is applied to the solute side of Figure 14.7. From Section 14.3.5, if a pressure in excess 2.49 × 10–2 atm pressure is applied to the solution side of the membrane, then the pure solvent, water, is forced through the membrane from the solution side to the pure solvent side of the membrane. This process is not too surprisingly called reverse osmosis, and it is used to desalinate seawater and provide drinking water for regions near saltwater. However, seawater contains about 3.5 w/o soluble salts, with NaCl as the main constituent at 2.6 w/o (Hecht 1967). The concentration that must be entered into Equation 14.12 is the concentration of all of the ionic species in solution. So in a liter of solution there is roughly 35 g of salt. Because most of it is NaCl with M = 58.443 g/mol,
C g l
g mole l m
C mole m
2 3 3
2 3 3
35
58 443 2 10
1 20 10
= ×
×
= ×
/
. /
.
so from Equation 14.12, neglecting small changes in molar volume because of the solution, π
π
= =
( )( ) (
×)
= ×
RTC Pa
2 3
6
8 314 298 1 20 10 2 97 10
. .
.
or 2.97 × 106 Pa/1.013 × 105 Pa/atm × 14.7 psi/atm = 431 psi or 29.3 atm!
The practical implication of this result is that desalination by reverse osmosis is an energy intensive process, and in general, industrial-scale desalination plants are almost always located close to a source of electri- cal generation. In addition, the high pressures necessary to purify sea- water by reverse osmosis requires that the membrane materials be quite strong. One of the materials used is made from aramid—poly(aromatic imide)—(see Figure 14.8), fibers, the same family of polymer fibers as the Kevlar™ fibers used in bullet-proof vests. Of course, as the water is removed from the seawater the concentration of dissolved salts increases and the osmotic pressure increases as well. As a result, not only does the reverse osmosis equipment need to be robust enough to withstand the pressures, but also a significant amount of pressure–volume work must be expended to produce fresh water from seawater. Nevertheless, the process is energy-competitive with other desaliniza- tion processes such as evaporation and condensation of the water from the sea water salt solution and is widely used throughout the world where freshwater is scarce.*
14.3.8 n
oteWorthyAs mentioned above, the main reason for introducing osmotic pressure and its effects is that some of the phenomena associated with osmosis are very similar—if not identical—to processes that occur during interdiffusion in solids, particularly where the diffusion coefficients of the interdiffusing spe- cies are different. For osmosis, the model is that that solvent can diffuse quite readily through the semipermeable membrane while the molecules, ions, or particles (Einstein 1956) cannot: so their diffusion coefficients through the membrane are zero. As a result, the faster (only) diffusing spe- cies, the solvent, diffuses through the membrane and increases the volume of the solution phase and depletes the volume of the solvent on the pure solvent side of the membrane, Figure 14.7. Eventually equilibrium will be reached when the osmotic pressure calculated in Equation 14.12 is reached. By the way, if the two chambers on either side of the membrane in Figure 14.7 are on the order of cen- timeters in dimensions, because the diffusion coefficient of water and salts in water are on the order of 10–5 cm2 s and x2≅4Dt, times for equilibration will be on the order of 3–8 h. Nevertheless, when equilibrium is reached, the volume on the solution side will have increased considerably while that on the solvent side will have decreased.
Consider the following thought experiment. Suppose both chambers on either side of the mem- brane were sealed at the intersection of the vertical tubes and chambers and not open to the atmo- sphere (imagined closure in Figure 14.7). Then, the solvent would move into the solution until there would be a positive pressure in the solution chamber and equal and negative pressure on the pure solvent phase. The sum of the absolute values of the pressures would equal the osmotic pressure for the open system. Stated another way, if the vessel were an elastomer—or some other material that would allow free expansion and contraction—then the solution side would expand and the pure solvent side would shrink.
As discussed above, “Suppose that the membrane could move along the tube”? In that case, the membrane—a marker indicating macroscopic changes is volume—would move toward the pure solvent side under the increased pressure on the solution side because of the faster diffusion of the
* Climate change appears to be generating severe drought conditions throughout many parts of the world and societies may be forced to produce fresh water through desalination of sea water—an energy intensive process regardless of the technology used. This creates a serious dilemma! Unless the desalinization plant is powered with renewable energy, any other source of power will increase atmospheric levels of greenhouse gases that will exacerbate climate change.
N H
N H O
O n Aramid (aromatic polyamide)
FIGURE 14.8 Typical structure of aramid polymers that are used because of their high strengths for the semipermeable membranes in seawater desalinization plants.
14.4 Interdiffusion in Metals and the Kirkendall Effect 491 solvent into the solution. These same processes occur in solids in which the diffusion coefficients of
the interdiffusing species are different.