C.J. Gommes, Chim0698

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CEDRICJ.GOMMES

PHYSICALCHEMISTRY OFINTERFACES

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Contents

Introduction 0-1

Surface energies and Laplace’s pressure I–1

Wetting, a short introduction II–1

Solved problems involving wetting III–1

Intermolecular forces IV–1

Gibbs-Thomson Effects V–1

Thermal ﬂuctuations VI–1

Metastability and nucleation VII–1

Adsorption on surfaces VIII–1

Introduction

Gold is known as a shiny, yellow noble metal that does not tarnish, has a face centred cubic structure, is non-magnetic and melts at 1336 K. However, a small sample of the same gold is quite different, providing it is tiny enough: 10 nm particles absorb green light and thus appear red. The melting temperature decreases dramatically as the size goes down. Moreover, gold ceases to be noble, and 2-3 nm nanoparticles are excellent catalysts which also exhibit considerable magnetism. At this size they are still metallic, but smaller ones turn into insulators. Their equilibrium structure changes to icosahedral symmetry, or they are even hollow or planar, depending on size.

Emil Roduner, Size matters: why nano materials are different.

Everything goes nano these days. Examples from the marketing world are myriads, from cars to watches, shoes and toy helicopters. I personally own a music player bearing that name, to which I am listening at the very moment where I am writing these lines. Poli- cymakers even seem to be anticipating a nano-revolution. So what exactly is this hype all about? Figure 0–1: Thomas Graham (1805- 1869)

Let us not fool ourselves: nanomaterials aren’t anything new. The oldest known example of a nanotechnology is probably the Lycurgus Cup, which is a 4th century roman cup currently exposed at the British Museum. The cup exhibits dichroism due to the presence of colloidal silver within the glass. Another spectacular example is that of so-called Damascus blades, which are made up of steel with specific inclusions, which some researchers clim to be carbon nanotubes. Closer to us, the entire field of colloids, born at the end of the nineteenth century under the influence of people such as Thomas Graham and Michael Faraday, among others, is about nanometer-sized objects. The fact is that nanometer-sized objects have properties that are distinctly different from chemically identical, Figure 0–2: Michael Faraday (1791-1867) but larger objects. The quote at the beginning of the chapter is the spectacular example of gold. 0-2 c.j. gommes, chim0698

A first difference between today and say, twenty years ago is the fantastic progress made in the field of electron microscopy, which enables us to visualise nanometer-sized objects with unprecedented accuracy. Advanced techniques such as electron tomography even Bottom-up and top-down methods enable one a obtain 3D reconstructions of structures with nanome- refer to whether the larger structures are created last or first. Typically, ter resolution. The second difference is the progress in the synthesis lithography is top-down. Chemical of nano structured materials by a variety of techniques, both phys- methods, by which molecules assemble into nano structures is bottom-up. ical and chemical, top-down and bottom-up. The fields of synthesis of nano materials, and that of fundamental understanding of nanometer-scale phenomena are clearly feeding on each other. It is because we understand better nanometer-scale phenomena that we can create structures on that scale in a more controlled way. In turn, those better-defined structures enable us to better understand nanometer-scale phenomena. This is incidentally the reason why scientific or technological progresses are exponential: because they feed on themselves. There are, however, still some problems on the horizon. For example, in the field of imaging of nanostructures one is often limited to studying materials in vacuum, which is a very different environment from that in which most nanomaterials are generally used. The development of so-called in situ imaging techniques at the nanometer- Figure 0–3: One of the colloidal gold suspension prepared by Faraday in the scale, to visualise nano structures in working conditions, is still a 1850’s, which is still kept by the Royal challenge. One therefore often relies on indirect methods, such as Insitution. Read more about it here small-angle X-ray scattering. In the field of synthesis too, our current skills are limited. Although the word nano fabrication is sometime used, our current abilities are far from the self-multiplying robots that terrorised Great Britain in 2001 under the ominous name of "Grey Goo". A positive way to look at this is to think that there are many things left to be developed by new scientists and engineers in that field. The word colloidal was coined by Graham to characterise "glue-like" substances. The official IUPAC definition When people express their enthusiasm about nanoma- of that word is as follows: "State of subdivision such that the molecules or terials, they often say that it is fantastic that we can control the polymolecular particles dispersed in a shape of materials on a scale as small as a few nanometers. Before medium have at least one dimension we comment on this, we should first emphasise that there is already between approximately 1 nm and 1 µm, or that in a system discontinuities an existing and successful science, the purpose of which is to control are found at distances of that order." matter over a still smaller scale, that of the Angstrœm. That science In other words, colloidal is almost synonymous with nano. is naturally chemistry. So size per se is not the issue. The real difficulty with nano structures is that they at the interstice of two worlds: It is important that you become familiar with a few orders of magnitude. For the microscopic and the macroscopic. Systems in the former world example: the number of molecules are generally made of a few molecules, interacting with very strong or atoms in a macroscopic amount of matter is 1023, the size of an atom is covalent bonds, and subject to thermal fluctuations. By contrast, in 1 Å, the energy involved in chemical the latter world systems comprise a huge number of molecules (say, bonding is 1 eV, the thermal energy at one mole) interacting through very weak forces, and thermal fluctu- room temperature is 25 meV, etc. introduction 0-3

ation is generally negligible. What is difﬁcult and interesting about naometer-sized systems is not really that they are small, it is that they are intermediate between two worlds. It is often unclear which type Because nano materials are squeezed of physics applies: would a 5 nm particle better be thought of as a between the macroscopic and the microscopic worlds, one often refers to small crystal or a large molecule? The answer is obvious "it depends". them as being mesoscopic. The greek The purpose of CHIM0698-1 is to guide you into mesoscopic physics. preﬁx "meso" means "in between".

Figure 0–4: When a given object (here a cube) is cut into objects of half the initial size, the total surface area exposed is multiplied by two.

At this stage, you may wonder why the title of the course is "Physical Chemistry of Interfaces" and not, say "Intro- duction to nano sciences". The reason is that all many properties of nanostructured materials result from their having a large surface-to- volume ratios. In other words, interfaces play an important role at the nanometer-scale. This is illustrated in Fig. 0–4 with the unfolding of a paper cube, compared to that of eight smaller cubes. The surface is two times larger in the latter case. To put this in a more quantitative way, try to estimate the specific surface area of a material made up of silica particles with a diameter of 10 nm. Obviously, if you double the mass of material, you double the surface. So the only sensible quantity to evaluate is expressed in m2/g, for example. The word "specific" means "per unit of mass". Now, when I wrote "try to estimate.." I really meant it: before reading what follows try to do it! A possible way to calculate this is to estimate first the area of a single nanoparticle, and then to calculate how many nanoparticles there are in one unit of mass. The area of one spherical particle of radius R is 4πR2 so that the total area of the N particles in the material is

A = N 4πR2 (0–1) × The mass of a single particle is 4/3πR3 ρ where ρ is the density × (the speciﬁc mass) of the considered material. The number of parti- It is useful to know the order of mag- cles in a mass m of material is therefore nitude of the density of a few common materials. The density of water, and of most ordinary solvents is 1 g/cm2, that 4 N = m/ πR3ρ (0–2) of noble metals is around 20 g/cm3. 3 The density of silica (glass) and of graphite are both close to 2 g/cm2. 0-4 c.j. gommes, chim0698

The specific surface area of the material is finally A 3 = (0–3) m ρR In the particular case of R = 5 nm, and ρ = 2 g/cm3, one finds A/m = 300 m2/g. This is approximately the area of a tennis court! Because This is yet again an order of magnitude. the density of packings of spheres are approximately 50 %, one gram The exact densities for the packing of spheres range from 0.6 for extremely of that powder occupies about one cubic centimeter. In other words, loose random packings to 0.7405 it it the total area of a tennis court that fits in a cubic centimetre. for dense packings. The order of magnitude is 0.5. The inverse relation that exists between the surface area and the size of the structures is very general. It is independent of whether the structures are spherical, laminar, or cylindrical. Exercises 1 and 2 should convince you of this. Going back to the tennis court analogy, imagine that you have covered a tennis court with, say, a paper sheet with infinitely small thickness so that you can fold it infinitely easily. You then try to fold or crease the entire sheet into a tiny cube with a size of 1 cm. The first fold will get you about half the size of the court, 10 m., etc. To have the entire sheet fit in one cubic centimetre, you will have to fold the sheet a very large number of times, with increasingly smaller folds. What Eq. 0–3 tells you is that the average distance between each fold once it fits in the cube is about 5-10 nm.

Why is the surface area relevant at all? For two reasons. For specific applications, it is often desirable to have materials with a Figure 0–5: Tranmission Electron Mi- croscopy (TEM) image of a Pt/Carbon large surface area. For example, in the case of heterogeneous cataly- heterogeneous catalyst. The black dots sis, the active sites are generally the surface atoms of metal particles. are the Pt nanoparticles. The scale bar is As a rule of thumb: the large the surface area, the more active the 20 nm. catalyst. Another example is the case of separation processes based on adsorption. In that case too, the larger the surface area the more useful the material, imply because there is more room on the surface to capture contaminant. Another, less direct, reason is that the specific surface area of a material is roughly proportional to the fraction of atoms or molecules that are on a surface, as opposed to the bulk of the material. Atoms or molecules in the bulk of the material have more neighbours than those close to the surface, so they are more tightly bounded to each other. This means that the overall cohesion of a material should decrease when it is divided into small particles, because the fraction of atoms at the surface is larger. If you back to the epigraph at the beginning of the chapter, you can already understand why gold nanoparticles have a lower melting temperature. The modification of thermodynamic equilibria at the nanometer scale as opposed to chemically identical macroscopic systems is a very general phenomenon know as Gibbs-Thomson effects.

Let us estimate the fraction of atoms at the surface of a introduction 0-5

material. The situation is sketched in Fig. 0–6 in the case of an object having a deliberately complex shape. The volume of that object is V and its surface area is A. The atoms that are at the surface occupy a layer of thickness d all over the surface, where d is an equivalent atomic diameter. The volume of that layer is therefore A d, so the × fraction of atoms on the surface is

Fraction of atoms A d A × = ρ d (0–4) at the surface ' V m

where the equality results from expressing the volume as m/ρ. The fraction in the Right-Hand Side (RHS) is nothing but the specific surface area of the material. This equation has a very intuitive meaning: V/A has the dimension of a length, and it is the order of magnitude d of the size L of the object (check for a sphere). What Eq. 0–4 tells you L is simply that the fraction of atoms at the surface is proportional to d/L. If the object if of a size comparable to that of the atoms, then there are may atoms at the surface. On the other hand, if the object is macrosopic, there is a negligible fraction of atoms at the surface. In practice, consider for example the fraction of atoms at the sur- Figure 0–6: A complex-shaped object, face of an ice cube. The ice cube is about 1 cm across so L 10 2 with characteristic size L. All atoms − in a layer a thickness d, comparable to 10 ' m, and the diameter of the atoms is about d 10− m. So the frac- a molecular diameter, can be consid- 8 ' ered to be at the surface. The fraction tion is d/L 10− ; only ten atoms out of a billion are at the surface, ' of molecules at the surface is approx- which isa negligible quantity. Note in passing that it does not matter imately equal to the fraction of the 8 8 whether the answer had been 20 10− or 0.2 10− . If the answer is volumes of the red and grey regions of 100 times larger or smaller, it is negligible anyway. This is the reason the drawing. why we do not have to care too much about the exact size of atoms and of the ice cube for this type of calculation. This is the essence of order-of-magnitude calculations. If the result had not been negligibly small, say if we had found a fraction of 0.01 then we would have needed to pay more attention to the details. Because it makes a lot of difference whether 1 % or 100 % of atoms are at the surface. As an exercise, let us calculate the fraction of water molecules that are at the surface of a 10 nm droplet. We first have to find a reasonable value for d. This can be done by expressing that the density of single molecules is the same as that of macroscopic water. Assuming a cubic packing of molecules, which is naturally a rough approximation, one has

M/N A = ρ (0–5) d3

where M is the molar mass, NA is Avogadro’s number, and ρ is the macroscopic density. Using the values M = 18 g/mol, NA = 6.02 1023, and ρ = 1 g/cm3, one ﬁnds d 3.1 10 8 cm. Using that ' − 0-6 c.j. gommes, chim0698

value, the fraction of molecules on the surface is found to be

8 A d 3 d 3 3.1 10− × = × × 7 0.2 (0–6) V R ' 5 10− ' In a 10 nm droplet of water, twenty percent of molecules are at the surface! This is a huge fraction. If you compare the local environment of atoms or molecules at the surface and in the bulk of a material, you will find that the molecules at the surface are necessarily lacking neighbours with which to link. The overall cohesion of water molecules in a 10 nm droplet must therefore be significantly lower than bulk water, which must have a dramatic influence on the equilibrium vapour pressure at any given temperature. Do you expect the equilibrium vapour pressure over a 10 nm water droplet to be larger or smaller than over a macroscopic water surface? If you prefer you can think of it in terms of a boiling point: what is the boiling point of a 10 nm water droplet? This is one of the many question of this type that we will be addressing in CHIM0698.

Exercises

The main take-away message from this chapter is the relation between size and surface area. It is important that you feel comfortable with this before studying the following chapters. The following exercises should help you.

1. We obtained Eq. 0–3 by assuming spherical particles. How is that relation changed if the nanomaterial consists of elongated Figure 0–7: A carbon nanotube sample, needles of radius R instead of spheres? The fact that the needles imaged by transmission electron mi- are elongated means that end surfacer of the needles is negligible croscopy (TEM). The thickest nanotubes in the ﬁgure have an approximate compared to the lateral surface. diameter of 10 nm. 2. Repeat the exercise in the case of a material made up of thin layers of thickness t.

3. In a later chapter, we shall discuss nitrogen adsorption, which is an experimental method to determine the speciﬁc surface areas of materials. Using that technique on a given carbon nanotube sample (see e.g. Fig. 0–7), one found A/m = 230 m2/g. Can you use this data to estimate the approximate radius of the tubes in that sample? Do the calculation twice, ﬁrst assuming that the tubes are closed so that the measured surface is the outer surface, and then by assuming that the tubes are open so that the area is the sum of the inner and outer surfaces. In both cases, you may assume that the inner radius is about half the outer radius. I–0 c.j. gommes, chim0698

Further reading

The present lecture notes are supposed to be self-contained. These further-reading suggestions are mostly meant to broaden your perspective on nano materials and nanotechnologies. Some links are with restricted access, so you need to connect via your University account to read them.

On the web

• The NANOYOU (Nano for Youth) project is funded by the Euro- pean Commission to increase young people’s understanding of nanotechnologies. You can ﬁnd many useful resources on their website here.

• The Royal Institution has a beautiful website on Faraday’s work, and in particular about his celebrated colloidal gold suspensions. You can ﬁnd it here.

• Here is a general overview to nano sciences by Paul Alivisatos, on Youtube, who is a prominent researcher in the ﬁeld and a professor at the University of California at Berkeley.

Regular papers

• Emil Roduner, Size matters: why nanomaterials are different. Chem. Soc. Rev., 2006, 35, 583-592 This tutorial review by Emil Roduner (cited as an epigraph of the chapter) is a nice overview. You should not be able to understand all of it at this stage. However, I encourage to try reading regularly Figure 0–8: Richard Feynman (1918- this paper along this semester to see your progress in that ﬁeld. 1988). • R. Feynman, There is plenty of room at the bottom This paper is a transcript of a historical conference that physics genius and Nobel Laureate Richard Feynman gave in Pasadena in 1959. This conference is considered by many as a kick-off event in the ﬁeld of nanotechnologies. Surface energies and Laplace’s pressure

With respect to the cohesion and capillary action of liquids, I have had the good fortune to anticipate Mr. Laplace in his late researches, and I have endeavoured to show that my assumptions are more universally applicable to the facts, than those which that justly celebrated mathematician has employed.

Thomas Young, A course of lectures on natural philosophy and the mechanical arts, preface.

We have highlighted in the introduction that a specificity of nanometer-sized systems is to have a large specific surface area, which results in a large fraction of their atoms or molecules being positioned at interfaces. Because the local environment of molecules at interfaces is different from the bulk, their binding energy is different, Figure I–1: Thomas Young (1773-1829) which results in an energy cost of the surface. This is sketched in Fig. I–8. A convenient way to think about this is in terms of missing bonds. Form the point of view of the total system, the extra energy is proportional to the number of molecules at the interface, i.e. to the area of the interface. The extra energy, per unit area of the interface, is the surface energy γ. The dimension of γ is an energy per unit area, e.g. J/m2, which can also be expressed as a force per unit length, e.g. N/m. This is the reason for which γ is also referred to as the surface tension, particularly when dealing with liquids. To understand why the equivalence of N/m with J/m2 is not merely dimensional but also physical, consider the situation in Fig. I–3. A thin film of water is held in a wire, one side of which is free Figure I–2: Pierre Simon de Laplace to slide in such a way as to increase or decrease the area of the film (1749-1827) A. The mechanical equilibrium of the system can be analysed by applying the principle of virtual work. Because the natural tendency for the film is to reduce its area as much as possible, you have to apply a force F in order to prevent the area from decreasing. If you moved the wire to the right by a small distance dx, the mechanical work would be

δW = Fdx (I–1) I–2 c.j. gommes, chim0698

On the other hand, the change in energy resulting from the displace- ment dx is dU = 2γLdx. The factor Ldx is the variation of the area enclosed by the wire; γ is the energy per unit area of the free surface; the factor 2 accounts for the fact that the film has two free surfaces, one on top and the other on the bottom. When equating δW with dU, one finds F/L = 2γ (I–2) top view which is indeed proportional to γ. The factor 2 proves that if γ is to be interpreted as a force per unit length, then the force has to be F considered as being exerted on the surface itself and not to the film A L as a whole. In the present case, that force applies on both surfaces of the film. In principle, the wired frame of Fig.I– 3 could provide a means to side view measure the surface tension of liquids. In practice, however, liquids F films are generally too fragile for this is to be possible. Another experimental configuration is used, which is referred to as Wilhelmy’s Figure I–3: Mechanical equilibrium of a 1 plate. It consists in dripping a plate of roughened platinum into the liquid film on a wire. liquid and in measuring the force required for pulling it out of the 1 You will have to wait until we have studied wetting phenomena to under- liquid. stand why this is important. The situation is sketched in Fig.I– 4; the thickness of the plate is t and its length (in the direction orthogonal to the figure) is L, and we call P0 the ambient (e.g. atmospheric) pressure. Let us write all the forces exerted on the plate. The forces oriented downwards are

Forces = mg + P L t + 2γL (I–3) down 0 × The ﬁrst term is the weight of the plate, the second is the pressure that is exerted on top of the plate by the ambient gas (applied on a total area L t) and the last contribution is the same 2γL as in Fig. × I–3. On the other hand, the forces upwards are

Forces = F + (P ρgz)L t (I–4) up 0 − × F where the ﬁrst term is the force exerted by the measuring device, which are eventually interested in, and the second term is the pressure in the liquid just underneath the plate. The latter is simply the z ambient pressure minus the hydrostatic head. When balancing the forces up and down, one ﬁnds

F = mg + ρLtzg + 2γL (I–5) Figure I–4: A sketch of Wilhelmy’s plate while it is being pulled out of the liquid. The ﬁrst term is a known constant (independent of the liquid), the second term is the weight of the small part of the liquid being pulled out under the plate, and the last term is the one we are interested in. Using a plate that is sufﬁciently thin, one can make the second term surface energies and laplace’s pressure I–3

as small as desired. In this case the force needed to pull the plate out of the liquid is proportional to γ. In practice, a common experimental setup is a so-called Wil- helmy’s ring (as opposed to plate), in which it is a horizontal ring made up of a thin Pt wire that is pulled out of the liquid. The analysis is similar to that in Eq.I– 5, only L has to be understood as the length of the ring. A few experimental values of γ obtained using this type of setup are gathered in Tab.I– 1.

2 Substance Surface energy (mJ m− ) Nitrogen (77°K) 8.85 Ethanol (25°C) 23.2 Acetone (25°C) 23.5 Cyclohexane (25°C) 25 Table I–1: The surface energies of a few Formamide (25°C) 57.0 usual and not so usual substances. Note that two substances in the table are Water (10°C) 74.2 solid. Water (25°C) 72.0 Water (50°C) 67.9 Mercury (25°C) 485.5 Aluminum (25 °C) 35 Nickel (300°C) 2450 Glass 2000-4000

With the values from Tab.I– 1 in mind, we are in a better situation to examine the relative importance of the second and third terms in Eq.I– 5. We shall for now keep the discussion as general as possible and consider a system subject to it surface energy and gravity. The total energy may be written as

U = γA + mgh (I–6) where A is the total area of the liquid free surface, and the second contribution is the gravitational energy, with m the mass, g the acceleration of gravity, and h the height of the centre of gravity. Calling l a characteristic size of the system, the order of magnitude of each contribution is the following. The surface contribution is You may consider that "a characteristic size" is synonymous to "the order of magnitude of the size". In the case of a U γl2 (I–7) sur f ace ∼ sphere, you may chose the radius R or the diameter 2R. The type of mathemat- and the the gravity contribution is ical relations involving characteristic sizes, orders of magnitude, and the like, 4 are referred to as "scaling relations". Ugravity ρgl (I–8) The symbol means that a dimen- ∼ ∼ sionless factor is omitted, which is the where ρ is the density. To obtain that estimation, we have taken ac- reason why we should not care too much about the accurate deﬁnition of count that the mass scales as m ρl3 and the height as h l. ∼ ∼ the characteristic length. I–4 c.j. gommes, chim0698

Comparing the two contributions to the energy, one sees that the surface is dominant whenever the length satisﬁes r γ l (I–9) ρg

The right-hand side is referred to as the capillary length lc. In the case of water, with γ 0.1 N/m, ρ = 103 kg/m3, and g 10 m.s 2, ' ' − one finds l 3 mm. This corresponds to the common observation c ' that objects smaller than that value, such as dew drops, tend indeed to adopt a spherical shape. By contrast, larger droplets have a flat- tened shape as a result of the gravitational energy. It is convenient to summarise the present analysis by introducing a dimensionless number. In the present case, a natural choice is the ratio of a system’s characteristic length and its capillary length. For historical reason, it is the square of that ratio that have received a Figure I–5: Loránd Eötvös (1848-1919) name ρgl2 Bo = Eo = (I–10) γ Actually, it has received two names: it is sometimes referred to as the Bond number, and sometimes as the Eötvös number. Going back to the initial question, about the ratio of the second and third terms in Eq.I– 5, it can be written as ρgtz (I–11) 2γ Figure I–6: Wilfrid Noel Bond (?-1937) Because the typical height of the liquid z is expected to be compa- Eötvös and Bond have independently rable to the thickness of the film, t z, the ratio is about one half given their names to (roughly) the same ' dimensionless number. The Bond num- of Bond’s number. Experimentally, in Wilhelmy’s experiment we ber is used mostly in English-speaking want to measure the surface tension so that any other effect would countries and the Eötvös number is used in continental Europe. Besides be parasitic. In otegr words, we want Bond’s number to be as small his work on capillarity, Eötvös was as possible, which can be achieved by making the wire be as thin as also a pioneer in the discussion of the possible. Assuming a wire that would be about half a mm thick, and equivalence of inertial and gravitational masses, which is considered now a that the liquid is water, one finds that the weight of the liquid con- foundation of general relativity. tributes just a few percent to the force in the Wilhemy’s experiment. Most of the force is due to surface tension, which is good.

Let us step back for a moment and consider the case of Fig.I– 3 again from a microscopic perspective. What opposes the extension of the liquid film is the fact that more molecules are pulled towards the surface if the area A of the film is increased. This may seem like a crazy explanation, and yet it is in quantitative agreement with the experimental values of γ in Tab.I– 1. Let us try to make quantitative the sketch in Fig.I– 8 and relate γ to some more familiar characteristic of a fluid. The surface energy γ is the energy required to bring a molecule/atom to the surface and

Figure I–7: Jožef Stefan (1835-1893) surface energies and laplace’s pressure I–5

break a few bonds. On the other hand, if you think of it, from the point of view of a molecule, evaporation consists in breaking all the bonds that maintain it in the liquid. One therefore expects the energy relevant to γ to be a fraction of the enthalpy of evaporation ∆he. Let us see how quantitative we can make that analogy, by assuming an oversimpliﬁed liquid structure where the molecules/atom would be positioned according to a cubic lattice, with spacing d.2 In 2 Remember, this is equivalent to what that case, the number of bonds of a molecules in the bulk is 6, and we did on page 0-5 when we were trying to guess molecular dimensions. only 5 at the surface. The surface energy would then be written as

1 ∆h γ e 12 2/3 (I– ) surface ' 6 vm

2 2/3 because the area occupied by a molecule at the surface is d = vm , bulk according to the cubic model where vm is the molar volume. To put real numbers on this, consider the case of cyclohexane at 25 °C. Looking up in tables of physicochemical properties, or even Figure I–8: The physical origin of surface energies: the local environment on Wikipedia, you find the following values. Evaporation enthalpy of the atoms/molecules is different in 3 the bulk and at interfaces. he = 30.5 kJ/mol, density (specific mass) ρ = 773 kg/m , and molar mass M = 84.16 g/mol. Using the latter two values, you can estimate the molecular diameter d = 5.56 Å. From that value, Eq.I– 12 leads to γ = 26 mJ/m2. Compare that value to the actual experimental value in Tab.I– 1. The agreement is excellent! Of course we must Josef Stefan was an Austrian physicist, have been particularly lucky with this calculation because no one in born in today’s Solvenia, who gave his name to Eq. (I–12), among many his own mind would accept that molecules in a liquid would pack others. To understand the importance in a cubic way. Moreover, we have assumed that molecules on the of Stefan’s contribution, you have to consider that the very existence of surface loose just one neighbour. Why not half of them? You see atoms and molecules was controversial my point: this is yet again an order of magnitude calculation. The at the time. The final demonstration clear message, however, is that the story of molecules preferring to for the existence of atoms is generally considered to be the work of Jean Perrin be in the bulk rather than at the surface is indeed responsible for on Brownian motion, for which he macroscopic manifestation of surface energies. received the Nobel prize in 1926. That is 33 years after the death of Stefan! So far we have talked mostly of liquids, but the concept of surface energy applies to solids as well. Of course, it is difficult to imagine a Wilhelmy-like experiment with a solid, but there are other manifes- tations of surface energies for solids. For example, if you think of it a mechanical fracture in a solid creates newly exposed surfaces, so fracture criteria must have something to to with surface energies. Solids with higher surface energies ought to be tougher is some sense. Ac- tually a fracture is a process by which elastic deformation energy is transformed into surface energy. This simple observation by Alan Griffith resulted in a major breakthrough in solid mechanics, but we will not have time to discuss this further. What shall keep us busy in the remainder of the chapter is an effect that is important for both solids and fluids, which bears the name of Laplace pressure. The discovery that Thomas Young is refer- Figure I–9: Alan Arnold Griffith (1893- 1963) I–6 c.j. gommes, chim0698

ring to in the epigraph of this chapter is the observation that there has to be a pressure jump across any curved interface. That pressure jump is Laplace’s pressure. To understand the reason for this, consider two phases with volumes V1 and V2 within a ﬁxed total volume V = V1 + V2, as sketch in Fig.I– 10. The area of the contact between the two phases is A. According to the general laws of thermodynamics, if you let the system evolve it will reach an equilibrium state where free energy G is minimal. In other words, the following differential should vanish identically dG = P dV P dV + γdA (I–13) − 1 1 − 2 2 This means that if you change the shape of the two phases in any A imaginable way, the volumes V1 and V2, and the interface area A

will obviously change but the free energy will be left unchanged. V1 To convert this statement into something more intuitive, about the pressure difference P1 P2, we need some mathematics. V − 2 Consider a volume like shown in Fig.I– 10. Each point of the surface has a normal vector ~n pointing towards phase 2. Any infinitesi- Figure I–10: Two phases with volumes mal deformation of the surface can be described by a function V1 and V2 fill space. At equilibrium, the free energy is left unchanged if the surface is deformed in any possible δ~r = δψ(x)~n (I–14) way. This will eventually lead us to Laplace’s law. where δψ(x) is a function defined over the surface. If δψ is positive at a given point, the the surface is deformed outwards; it is deformed inwards if δψ is negative. The total change in volume can be written in terms of δψ as Z dV1 = dΣ δψ(x) (I–15) A

where the integral is over the total interface. Naturally dV2 is simply the opposite of dV1. The change in surface area dA also depends on δψ(x). According to a famous theorem from differential geometry, the relation can be written in terms of the local curvature of the surface. The actual relation is To gain some intuitive understanding of Z Eq.I– 16 consider the case of a spherical dA = dΣ K(x)δψ(x) (I–16) A surface with radius R, with δψ = dR. In that case the total curvature is 2/R where K(x) is the local total curvature of the surface at point x. Ge- everywhere on the surface so that the ometrically, the total curvature is the sum of the inverse of the two integration is simply a multiplication by the area 4πR2. On the other hand the principal radii of curvature at point x change in surface area is dA = 8πR dR, so that Eq.I– 16 is indeed satisﬁed. 1 1 K(x) = + (I–17) R1(x) R2(x)

This is all the mathematics we need to put Eq.I– 13 is a more under- standable form. surface energies and laplace’s pressure I–7

With the help of Eq.I– 16, Gibbs’ equation can be put under the form Z dG = dΣ [ P1 + P2 + γK] δψ (I–18) A − Expressing that this quantity should vanish for any possible δψ leads us to the important result

P P = γK (I–19) 1 − 2 which is known as Laplace’s law. It relate the pressure difference across an interface to the curvature. Note that all quantities in the equation may depend on the point considered on the surface. The pressures P1 and P2 may vary from point to point, but they difference is always proportional to the local curvature of the interface. It is important to understand that the curvature can be either positive or negative, as illustrated in Fig.I– 11.

a b c Figure I–11: A few menisci conﬁg- uration to illustrate the concept of total curvature, in relation to the local radii of creature. In (a) the curvature is K = 2/R, in (b) it is K = 1/R because one radius is inﬁnite, in (c) it is K = 1/R 1/R because the two 1 − 2 curvatures have opposite signs.

As we already mentioned, Laplace’s pressure is not typical of liquids. It applies to solids as well. Fig.I– 12 illustrates that point. Let us see if we can ﬁnd back the right order of magnitude of the observed deformation from the known surface energy of aluminium in Tab.I– 1 and from the macroscopic compression modulus K = 76 GPa. The latter quantity relates the change of volume of the material to the pressure via dV P = (I–20) V − K In our case, the pressure P results from the surface tension, which in the case of a spherical particle is written as

2γ Figure I–12: Compressive strain e on P = (I–21) the crystalline lattice parameters of of R Ag and Al nanoparticles, determined The ﬁnal relation between the change in volume and the size should from the shift of scattering peaks of X-ray diffraction patters. Taken from therefore be J. Phys. Chem. B 105 (2001) 6275. This dV 2γ contraction is a direct consequence of = (I–22) V − RK the Laplace pressure. if our analysis is correct. Assuming a particle with radius R = 5 nm, the predicted shrinkage of the nanoparticles should be of the order of dV/V 1/5, i.e. e 1/15. ' − ' − I–8 c.j. gommes, chim0698

The agreement is not quantitative for several reasons. First the compression modulus need not be the same for nanometer particles and for macroscopic Aluminium, if only because of the Laplace pressure. Second, the nanoparticles are not spherical, and it is an approximation to model them as spheres. Anyway, our oversimplified analysis predicts correctly the overall inverse relation between e and the particle size, and the order of magnitude is correct. Figure I–13: A corrugated cylinder. We shall finish the present chapter with interesting consequences It is stable or unstable depending of surface energies, which are referred to capillary instabilities. These on whether the Laplace pressure is larger at point 1 than at point 2, or the instabilities result from the tendency of any system with a larger opposite. surface-to-volume ratio to try and minimise its surface area. Consider for example the case of a cylinder with radius R. You can clearly lower the surface area if you transform it into a single sphere with the same volume. The same hold if you decompose it into a series of spherical droplets, provided the droplets are sufficiently large3. 3 Try to prove that as an exercise. What So a column is in principle unstable. This is something you know by is the smallest size of the droplets for this to be the case? observing a jet breakup into smaller droplets in a sink. To see how this happens imagine that there some infinitesimal corrugation to the column, so that it can be described by

r(x) = R + δ sin(2πx/λ) (I–23)

where x is the position along the column, R is the average radius, δ is the amplitude of the corrugation, and λ is its periodicity. The question we ask is: "Do we expect δ to increase with time or to decrease?". The situation is sketched in Fig.I– 13. The column is unstable (and δ increases) if the Laplace pressures at points 1 and 2 satisfy P P . 2 ≥ 1 Such a pressure difference would indeed induce a ﬂow of matter from the thinner to the thicker parts of the columns, thereby increasing its corrugation. Let us calculate those pressures. For that purpose, it is important to note that there are two orthogonal radii of curvature: one along the axes of the column R , and in the other is R . The easiest to k ⊥ calculate is R : ⊥ ( R + δ at point 1 Figure I–14: Joseph Plateau (1801-1883). R = (I–24) ⊥ R δ at point 2 − To calculate R , you have to remember from your calculus classes k that the radius of curvature of a curve parameterised as y(x) is related to the second derivative of y via

1 d2y (I–25) R(x) ' dx2

where the sign means that the relation is valid provided the curva- ' surface energies and laplace’s pressure I–9

ture is small enough. Applying that formula to Eq.I– 23, one ﬁnds ( 1 (2π/λ)2δ at point 1 = (I–26) R (2π/λ)2δ at point 2 k − Note the minus sign in the second equation, which means that the center of the tangent circle points outside of the surface, so that its effect on the pressure via Laplace’s law is negative. Putting all that together, one ﬁnds the following pressures " # " # 1 2π 2 γ 2π 2 1 P = γ + δ + γ δ (I–27) 1 R + δ λ ' R λ − R2

" # " # 1 2π 2 γ 2π 2 1 P = γ δ γ δ (I–28) 2 R δ − λ ' R − λ − R2 − where the second equality results from a Taylor development to the Figure I–15: John William Strutt, 3rd Baron Rayleigh (1842-1919). ﬁrst order in δ/R, which is justiﬁed because we have assumed that δ was small. The condition for instability P P takes the simple form 1 ≤ 2

λ 2πR (I–29) ≥ Isn’t that a beautifully simple result? EquationI– 29 means that a column is naturally unstable towards corrugation with long periodicity. The periodicity that is actually observed results from a balance between thermodynamics and kinetics. Thermodynamics favours large values of λ, but the growth of such an instability would be extremely slow because its would require the transport of atoms/molecules over distances comparable to λ. The values observed in practice are close to λ = 2πR. This is illustrated in Fig.I– 16, which shows the decomposition of a Cu nanowire into a series of spherical nanoparticles. For the records, this type of instability is know as the Plateau- Rayleigh instability, after the famous british physicist Lord Rayeigh, and the belgian physicist Joseph Plateau.

Exercises

1. Calculate the capillary length of mercury, with γ 0.5 N/M and ' ρ 13 103 kg.m 3, as well as that of water on the moon, where ' − gravity is three times less than on Earth. Figure I–16: Example of nanometer- scale Rayleigh instability: spontaneous 2. Repeat the scaling analysis that we did for Bond’s number to fragmentation of a Cu nanowire. Taken compare the surface energy to the kinetic energy of a ﬂowing from Appl. Phys. Lett. 85 (2004) 6275. liquid. That number is referred to as the Weber number We: ﬁnd an expression for it in terms of a characteristic length l, as well as on any other relevant quantities. I–10 c.j. gommes, chim0698

3. Same question as before, but for the relative importance of the viscous and surface forces. That number is generally referred to as the capillary number Ca.

4. The list is longer than Ca and We: dimensionless numbers can be derived to assess the relative importance of any physical force to the capillary force. Can you think of other examples?

5. Analyze the deﬁnition of Bond’s number Eq.I– 10 in terms of Laplace’s pressure.

6. Check the validity of Eq.I– 25 in the particular case of the equation p y(x) = R2 x2 (I–30) − corresponding to a circle of radius R.

7. FigureI– 17 displays a string of air bubbles rising in a viscous liquid (shampoo). Because air is almost inviscid, the pressure in the bubbles can be considered as homogeneous. This is not the case for the pressure in the liquid, which varies from point to point. The variability of the liquid pressure controls the shape of the bubbles via Laplace’s relation. Considering the shape topmost bubble in the ﬁgure, determine where the pressure on its surface is the largest and where it is the smallest.

8. By a direct calculation of the surface area and of the volume corresponding to Eq.I– 23, show that the instability condition Eq. I–29 is equivalent to stating that the area should decrease when δ increases, at constant volume.

9. Looking at Fig.I– 17, would you say that the capillary number (Cf. Figure I–17: Air bubbles rising in a viscous liquid. point 3) is larger or smaller than one?

10. When pouring water on a hydrophobic surface, the water forms a layer with thickness h, as sketched in Fig.I– 18. Assuming that the lateral ends of the layer are circular, calculate the thickness h. h Based on experiments that you can do at home, do you think this is a good approximation? Do you expect the ends of the layer to be ﬂatter or shaper than a circle? Figure I–18: Water on a hydrophobic surface. This is related to a more general problem wittily referred to as the housewife problem: when you pour Further reading a bucketful of water on a ﬂoor, what surface does it cover. • The series of lectures given by Thomas Young at the Royal Institu- tion, from which the epigraph of the chapter is taken, are available on Google books here. Browsing the table of contents of that book will give you a hint of the type of universal genius that Thomas Young was. II–0 c.j. gommes, chim0698

• There are many good books on surface tension and wetting. An excellent one is the following

de Gennes, Brochard-Wyart, and Queré Capillarity and Wetting Phenomena: Drops, Bubbles, Pearls, Waves Springer, 2004

• The Nobel Lecture of Pierre-Gilles de Gennes is also interesting to read.

• Here is a paper from Nature (487, 2012, 463) about how nanometer- scale Plateau-Rayleigh instabilities can be exploited to produce nanoparticles with sophisticated structure.

Figure I–19: Pierre-Gilles de Gennes (1932-2007). Wetting, a short introduction

But words are things, and a small drop of ink, Falling like dew, upon a thought, produces That which makes thousands, perhaps millions, think.

Lord Byron

The situations in which liquids are put in contact with solid surfaces are too many to enumerate, and their technological importance can hardly be overestimated. The examples from either technology or the natural world that immediately come to mind involve the painting of a surface, printing, lubrication, wicking of porous media by Figure II–1: Leonardo da Vinci (1452 - a liquid, the creating of water-tight yet air-permeable fabrics, rain 1419) drops sliding down a window or rolling down a hydrophobic leaf, the walking of a insect on a pond, the capillary ascent of sap in the xylem vessels of a plant, etc. The topic of wetting is not a new one. The first documented study about capillarity seems to be the deed of Leonardo da Vinci, and it has been followed by a long series of scientists. A name you might be familiar with is that of James Jurin who gave his name to the law relating the height of capillary rise to the diameter of the capillary. It is also a poorly known fact that Albert Einstein’s first published paper was about capillarity, in year 1900. Capillarity is an extremely active field of research, and the work of Pierre-Gilles de Gennes (Fig. Figure II–2: James Jurin (1684-1750) I–19) was instrumental in the bringing that topic to the forefront of research.

Figure II–3: Examples of a liquid deposited on a clean surface, displaying a typical truncated sphere shape with a characteristic contact angle. This is the case of water on differently treated glasses, taken from Phys. Chem. Chem. Phys. 6 (2004) 604. II–2 c.j. gommes, chim0698

The simplest question that you may want to ask is whether a liquid will spread uniformly on a solid or not. This is a question that is loosely referred to as that of the hydrophilicity or hydrophobicity of the surface. It is central to the question of painting a surface, for example. If you are practically minded you may start by doing experiments and deposit drops of liquids on the surface and see what happens. What you will probably obverse at your ﬁrst trial is something like in Fig. II–3: the liquid will adopt a truncated spherical shape. If you repeat the experiment with a variety of liquids on the same surface, you will notice that the angle θ changes from one liquid to the next. After experimenting with a few liquids, you may start to ratio- nalise your ﬁndings in terms of sound physicochemical quantities. Because small values of θ go along with large areas of the free surface of the liquid, it seems natural to see whether the liquid surface tension γ explains the variability of θ. After playing a bit with the Figure II–4: William Zisman (1905- data you will probably end up plotting them as 1 cos(θ) against γ, 1986). Picture taken from here. and you would observe something like Fig. II–5: the points fall on a single line, which you write empirically as

1 γ = (II–1) cos(θ) γc

The reason why it makes sense to call the denominator γc is that it is a critical surface tension. When γ = γc, the contact angle reaches the value θ = 0. From this, you may infer that any liquid with γ γ ≤ c would spread perfectly on the solid. This type of characterisation of wetting, in particular II–1 is called Zisman’s law. The values of γc for a few solids are given in Tab. II–1. In the case of clean glass, the value is quite large (γ 150 mN/) which shows c ' that any liquid with surface tension lower that than will spread on it. Figure II–5: Zisman plot, by which a series of liquids are deposited on a This is notably the case for water and of most organic solvents (see given surface and the contact angle Tab.I– 1 on pageI– 3). On the contrary, polyethylene will be wetted is measured. The extrapolation to cos(θ) = 1 yields the critical surface by most organic solvents but not by water. This corresponds to the tension of the surface. In this particular common observation that polyethylene is hydrophobic. You may also case, the surface is terphene, and the note from the values in Tab. II–1 that surface treatment may have critical surface tension is found to be close to 25 mN/m. dramatic effect, as already suggested by Fig. II–3. Almost no liquid will spread on ﬂuorinated glass, except perhaps liquid nitrogen.

If you are more theoretically minded you may want to think thoroughly about the question of wetting before experimenting. You may come up to the following. In a situation like that of Fig. II–3 there are three different interfaces, between the solid and air, the liquid and air, as well as between the solid and the liquid. Each interface has its own surface energy. The condition for the spreading of wetting, a short introduction II–3

the liquid on the surface necessarily depends on these three surface Solid γc (mN/m) energies. Clean glass 150 Fluorinated glass 10 We use the notations γS, γL, γSL for the surface energies of the Polyethylene 31 solid, of the liquid, and of the solid-liquid interface. Assuming that PVF4 18

you know all the relevant surface energies, you can calculate the Table II–1: Critical surface tension of between the dry solid and the wetted solid. Because that quantity is a few solids. PVF4 is a ﬂuorinated directly related to the tendency of the liquid to spread, it is referred polymer. to as the spreading parameter S. It can be written in terms of surface energies as follows S = γ (γ + γ ) (II–2) S − L SL where the term between brackets is the energy of the wetted solid. If S > 0 it is energetically favourable for the liquid to spread, which happens for large values of γS and a lower value of γL. It may happen that I sometimes forgot The case S < 0 leads to partial wetting, in which the contact angle the subscript L in γL, in line with the notation of Chapter 1. θ takes a ﬁnite value. All examples of Fig. II–3 correspond to that situation. In order to relate the contact angle to the surface energies, let us consider the sketch of Fig. II–6, which shows a close-up view of a triple line between a solid, a liquid, and air. The total free energy of such a system has contributions proportional to the area of each interface, which can be written as

G = ... + γS AS + γL AL + γSL ASL (II–3)

In this equation AS, AL and ASL are the surface areas solid-air, liquid-air, and solid-liquid interfaces. The shape of the liquid on the solid, and in particular the value of θ, is such that the free energy is 휸L stationary, i.e. that G is left unchanged by slight deformations of the dx cos(휃) drop. The dots in the equation stand for the volume contributions, 휃 휸LS 휸S which are assumed to be constant and are irrelevant in the present dx context. If you have in mind how we obtained the expression of the Laplace’s Figure II–6: Close up view of a triple pressure, you may remember that the free energy has to be stationary line between a solid S, a liquid L and air, with definition of the contact angle with respect to any possible deformation. Let us see, however, if we θ as well as of the surface energies γS, can learn something useful by considering only stationarity with re- γL and γSL. spect to displacements of the triple line. If the triple line moves by an infinitesimal quantity dx as shown in Fig. II–6, the areas are modified according to dA = l dx SL × dA = l dx S − × dA = l dx cos(θ) (II–4) L × where l is the length of the triple line, and θ is the contact angle. Introducing these expressions in the variation of the free energy Eq. II–3 leads to the following relation γ γ = γ cos(θ) (II–5) S − SL L II–4 c.j. gommes, chim0698

which is usually known as the Young-Dupré equation. In the particular case where the spreading parameter is equal to zero the Young-Dupré equation predicts θ = 0, as it should. For S > 0, the equation has no solution because this would require cos(θ) > 1. As we shall see in the rest of the chapter and also in the next, the contact angle is often sufﬁcient to calculate many useful properties of a system involving wetting. In other words, it is often unnecessary to know γS, γSL and γL in more detail than though their combination (γ γ )/γ . S − SL L

Figure II–7: Example of nanometer- scale wetting: environmental TEM shows that Cu nanoparticles adopt different shapes in presence of different vapours. Figure taken from Science 295 (2002) 2053.

Examples of wetting with different contact angles are given in Fig. II–3 and in Fig. II–7. In the former case, It is the surface that is modified via plasma treatment, which resulted in varying γS and γSL and hence θ. The latter case is more interesting: it consists in the wetting of solids at the nanometer scale. For example, the difference between Fig. II–7A and II–7E is the presence of carbon monoxide (CO), which adsorbs on the Cu surface and reduces its surface energy. As a consequence, the nanoparticle is found to flatten, in qualitative agreement with the Young-Dupré relation. In the case of crystalline solids, the sur- It is interesting to look back on Zisman’s law in the light of the face energy is generally dependent on the particular facet considered. In that Young-Dupré equation; it can be re-written as case, the lower energy configuration is not strictly a sphere, but a faceted 1 γ = L (II–6) shape known as Wullf’s shape (see Fig. cos(θ) γS γSL II–7). Most results obtained with liquids − remain, however, qualitatively correct in which suggest that γ = γ γ . One has to be careful, however, the case of solids. c S − SL that γSL is liquid-dependent so this is not the same as Zisman’s law. The empirical observation that Zisman’s law is correct seems to suggest that γSL is almost liquid-independent. This might be reasonable in the case of homologous liquids, e.g. a series of oils with increasing molar mass or surfactant solutions with increasing concentrations. However, Zisman’s law is empirical: there is no reason to believe that the critical surface tension of a given solid would be the same for oils wetting, a short introduction II–5

and for aqueous solutions of surfactants. You will have to wait until we discuss the physical origin of intermolecular forces in Chapter IV to understand this better.

Before we investigate the consequences of the Young- Dupré equation in some specific situations, it might be useful Figure II–8: Self-explanatory illustra- to draw your attention to some limitations of that law, so that we can tion of contact angle hysteresis. The afterwards knowingly forget about them. When playing around with advancing contact angle (top) is generally arguer than the receding angle liquid droplets on a surface you immediately notice the phenomenon (bottom). Figure borrowed from here. of triple-line pinning, by which the triple line seems to stick to the surface. That phenomenon is illustrated in Fig.II–8. In general, for a given liquid on a given solid, the angle can take any value in an interval θ θ θ , where the subscripts a and r stand for advancing r ≤ ≤ a or receding. The triple line moves macroscopically only when these angles are exceeded. a) 휃2 Some observations as familiar as the sticking of a water droplet 휃1 on a vertical window pane would be impossible without triple-line pinning. It is because the contact angles at the top and at the bottom of the droplet are different that such a droplet can be in mechanical b) equilibrium. See also exercise XY for that matter. 휃 휃2 The origin of contact angle hysteresis can be found in the hetero- 1 geneity of the surface. This is expected notably if a surface is made up of microscopically small patches with slightly different contact angles θ and θ , as illustrated in Fig. II–9a. In that case, the contact Figure II–9: Two type of surface hetero- 1 2 geneities that may lead to contact angle angle passes from one value to the other when the triple line passes hysteresis: (a) chemical heterogeneity from patch to the next. If the characteristic sizes of the patches are and (b) structure heterogeneity. very small, from a macroscopic perspective, it seems that the triple line has not moved. Another situation in which contact angle hysteresis is expected is the type of surface roughness shown in Fig. II–9b. In that case, the microscopic contact angle of the liquid with the surface is uniquely deﬁned. However, the macroscopic contact angle is subject to hysteresis. Another interesting consequence of surface roughness, besides hysteresis, is the following. Surface roughness increases he actual Figure II–10: The Westinghouse Elec- contact area of the liquid and the solid, and this is expected to in- trical Company research labs in Pitts- burgh, where Robert N. Wenzel was ﬂuence the contact angle. To see how, repeat the derivation of the employed when he published his work Young-Dupré relation, by replacing the areas dAS and dASL in Eq. on the wetting of rough surfaces. The II–4 by r dA and r dA , where r is the roughness factor of the recognisable dome is a Van De Graaff × S × SL generator. surface. This leads to the following relation for the contact angle θ∗ on the rough surface, compared to the contact angle on the chemically identical smooth surface.

cos(θ∗) = r cos(θ) (II–7)

Because r is necessarily larger than one, contact angles larger than II–6 c.j. gommes, chim0698

90 °are increased by surface roughness, and contact angles smaller The roughness factor of the surface r - than 90 °are made even shallower. In other words, roughness en- also called the rugosity - is deﬁned by IUPAC as the ratio of the actual surface hances both hydrophilicity and hydrophobicity. Eq. II–7 is known as area to the apparent (projected) surface Wenzel’s law. area.

A common illustration of wetting phenomena is the wicking of a porous solid by a liquid, as illustrated in Fig. II–11 which is achtypica capillarity problem. It is an empirical observation by James Jurin (Fig. II–2) that a wetting liquid rises in a vertical capillary tube in inverse proportion to the diameter of the tube. This situation can be modelled by assuming that the pore space of the material (the sugar cube) consists in cylindrical tubes. For methodological reasons, we shall analyse that situation in two different manners: ﬁrst based on surface energies, and then on Laplace’s pressure. The energy of the liquid column has two contributions: the gravitational and surface energies. The gravitational energy can be written as " # d 2 h Figure II–11: The wicking of a sugar Ug = g π h (II–8) cube by coffee is a common illustration 2 2 of capillarity. where h is the height of the column, d is the diameter of the capillary, and ρ is the liquid density. The term between square brackets is the volume of the column and h/2 is the height of its center of mass. On the other hand, the surface contribution is

U = πdh [γ γ ] (II–9) s SL − S It is the difference of surface energies that have to be considered because the dry solid is replaced with wet solid when h increases. Looking for the value of h that minimises Ug + Us yields immediately the value 4(γ γ ) h = S − SL (II–10) ρgd which is the celebrated Jurin’s law. It states that the liquid rises in a capillary provided γS > γSL, and the thinner the capillary the larger the rise. Try to think of the molecular origin of Here is the second analysis, based on Laplace’s pressure. If you Jurin’s law in relation to the surface-to- volume ratio of the capillary. look close to the surface of the liquid at the top of the capillary, you will notice that the surface is curved. Provided the diameter of the capillary is much smaller than the capillary length4, the free surface 4 Be sure you understand this, and go will have a spherical shape with radius of curvature given by back to chapter I if necessary.

d R cos(θ) = (II–11) 2 As a consequence of that curvature, the pressure in the liquid just wetting, a short introduction II–7

h Figure II–12: Two different and equivalent approaches to analyse the rising of a liquid in a capillary, based on energy minimisation or on Laplace’s pressure. d

below the meniscus is at a pressure

2γ P = P L (II–12) 0 − R where P0 is the ambient pressure. On the other hand, that pressure can also be calculated from the height of the liquid as

P = P ρgh (II–13) 0 − Equating these to independent estimations of P brings you to

4γ cos(θ) h = L (II–14) ρgd which is exactly identical to Eq. II–10 is the Young-Dupré equation holds. In our discussion of Jurin’s law so far, and particularly in Fig. II– 12, we have implicitly assumed θ 0 or equivalently γ γ . All ≤ SL ≤ S the results that we have obtained are still valid in the case of θ > 90°, but the formulae predict a negative capillary height. This means that when a capillary is dipped into a non-wetting liquid, the latter does invade the capillary until the point where the hydrostatic pressure balances exactly the Laplace pressure, which is in this case positive. An interesting way to look at it is to consider that you have to exert a given pressure to push a non-wetting liquid in the capillary. In a Jurin-like experiment that pressure is provided by gravity, but you may also imagine immersing the capillary in liquid and increas- Figure II–13: Example of two mer- ing mechanically the pressure. The pressure that you would have to cury (intrusion) porosimetry curves measured on two porous materials reach to push the liquid in is with different pore sizes (courtesy of Nathalie Job). 4γ cos(θ) P = L (II–15) i − d where the subscript i stands for intrusion. Note that Pi can be directly obtained by applying Laplace’s law at the meniscus. It is interesting to note that the pressure that you have to exert to push a non-wetting liquid in a capillary is inversely proportional to its diameter. The II–8 c.j. gommes, chim0698

smaller the diameter, the larger the pressure. The deep reason for this is that wetting phenomena in pores scale like the surface-to-volume ratio of the pores. You may at first think of mercury That observation can be used practically measure the size of pores porosimetry as a technique from an older age. After all, why not simply do it porous materials, by a technique called mercury porosimetry, or electron microscopy to measure pore sometimes intrusion porosimetry. The technique consists in immersing sizes? The reason is that the amount of material that you actually characterise the unknown porous material in mercury, and increasing progres- when doing microscopy is extremely sively the pressure. At each particular pressure the volume of mer- small. A fragment 100 nm across, which cury that has entered the pore space is measured. Mercury is used is the largest you can analyse in a microscope with nanometer resolution, 15 for those experiments because it does not wet any known solid. All has a mass of 10− g. Compared to solids are mercuro-phobic, so to speak. the kg or even ton of material that is used in any industrial context, this is an A typical mercury intrusion curve is shown in Fig. II–13: the in- extremely poor sampling. The strength truded volume is plotted as a function of the pressure. One observes of techniques like mercury porosimetry that intrusion occurs at a specific pressure that can be related to the is that they enable you to characterise a large quantity of material, with a pore size via Eq. II–15. In practice, the accepted value for the contact nanometer-scale resolution. angle of mercury with any material is θ 142 °, which together with ∼ the value of γL given in Tab.I– 1 leads to the following quantitative relation 1500 d[nm] (II–16) ∼ Pi[MPa] between the intrusion pressure and the pore size. In the case of the figure, with intrusion pressures Pi = 20 MPa and Pi = 50 MPa, the pore sizes are found to be 75 nm and 30 nm, respectively. In typical mercury porosimeters, the pressure can reach 200 MPa, which means that pores as small as 7 nm can be detected. In case you do not remember the There is an interesting aspect to intrusion porosimetry that is not Hagen-Poiseuille, you can always find it back using dimensional analysis. In often discussed, which concerns the kinetics of fluid intrusion. When dimensional form: the volume flow the pressure in the fluid is suddenly brought to a value larger than Q [m3/s] depends on the pressure gradient dP/dx [Pa/m], on the viscosity the intrusion pressure Pi, how fast does intrusion take place? This η [Pa.s] and on the tube diameter d [m]. question can be answered by noting that when a liquid invades a Only a single dimensionless number 4 porous material, there is a pressure drop in the liquid dP per unit can be formed, e.g. d dP/dx/(ηQ), which can only be equal to a constant. length dx of a pore, which can be approximately described by the This is exactly Eq. II–17, except for the Hagen-Poiseuille law, namely factor 128.

dP 128ηQ = (II–17) dx − πd4 where η is the viscosity of the ﬂuid, Q is the volumetric rate, and d is the pore diameter. Imagine that at given time t, the liquid has entered the pore over a length L(t), from the outer surface of the material to the meniscus. Over that distance, the pressure drops from P at the outer surface to Pi just behind the meniscus. This enables us to write the Hagen-Poiseuille law again as

πd4 Q = (P P ) (II–18) 128ηL(t) − i wetting, a short introduction II–9

Because Q = π(d/2)2dL/dt, this can be expressed as a differential equation, the solution of which is s d t L(t) = [P P ] (II–19) a) 4 η − i

The intrusion is proportional to √t: it is initially very rapid, and then very slow. This is the typical dynamics of many wetting phenomena. b) Although the fundamentals of wetting seem to be relatively well understood, the way in which they apply in speciﬁc natural or technological situations can be quite subtle. For example, Figure II–14: Cassie-Baxter type of wetting, in the case of (a) hydrophobic consider again Wenzel’s law for the wetting of a rough surface (Eq. and (b) hydrophilic surfaces. II–7). When experiments are done on very rough surfaces, signiﬁcant deviation from Wenzel’s law is often detected. Wenzel’s law is valid only for surfaces with small rugosity. To understand why this happens, remember that when deriving Wenzel’s law we have assumed that the entire surface was either dry or wetted by the liquid. But this need not be the case. Based on what we said about Jurin’s law in the case of a non-wetting liquid (Cf. mercury porosimetry), you may expect something like Fig. II–14a for a drop deposited on a rough surface. In that case, the apparent energy of the wetted surface γSL∗ could be modelled like

γ∗ φγ + (1 φ)γ (II–20) SL ' SL − L where φ is the fraction of the surface where the two phases are actually in contact. On the remainder fraction of the surface, the interface energy is simply γL because the crevices of the surface are ﬁlled with air. You may try to incorporate that type of thinking into our demonstration of the Young-Dupré equation, and this will lead you to a different relation between surface roughness and contact angle. Similarly, in the case of a wetting liquid you may think of the crevices on the surface as of a porous material, which may be in- Figure II–15: The Cassie-Baxter-like super-hydrophobicity of the leafs truded by the liquid following the usual laws of capillarity. In that of some plants is often referred to case, on a microscopic scale things may look like in Fig. II–142, which as the lotus effect. Picture from from is also different from the Wenzel’s model. In that case too, you may Wikipedia. use an approach similar to Eq. II–20 to derive the apparent energy of the dry surface γS∗. The model corresponding to the wetting mechanism of Fig. II–14 is known as the Cassie-Baxter model. In the particular case of hydrophobic surfaces it is sometimes referred to as the Lotus effect. The surface of the lotus leaf is covered by a waxy substance that confers it with hydrophocity; in addition to that, the surface comprises numerous micrometer-sized posts that make it similar to Fig. II–14a, and confers super-hydrophobicity to the surface. II–10 c.j. gommes, chim0698

In myriads of biological contexts, natural selection has lead to structured surfaces having wetting properties desirable to the species. This is the case of super-hydrophobic leafs, which help keeping the leafs clean from any macroscopic impurity. This is also the case of the legs of insects walking on water, the surface of which is covered by hydrophobic hair. The development of artificial surfaces with engi- neered wetting properties is an extremely active field of research. If that topic is of interest to you, you will find some suggested readings hereunder.

Exercises Figure II–16: The legs of water strid- 1. Capillary action is one of the two mechanisms responsible for ers are covered with hydrophobic hair, which confers them super- the ascent of sap in plants from the roots to the leafs, in addition hydrophobicity. Picture from from to osmotic pressure. Try to guess the order of magnitude of the Wikipedia. diameter of xylem vessels, and then compare its with actual data.

2. A bucket of plastic is ﬁlled with water, and you punch with a a) b) needle a hole about half a millimetre wide twenty centimetres below the surface. Does water ﬂow through that hole? You may assume the contact angle of water with the plastic making up the bucket is about 140 °.

Figure II–17: (a) The so-called capillary 3. The drawing in Fig. II–17a is a historical proposal for a perpetual cup perpetual motion setup, and (b) motion machine. I have proposed as simpliﬁed version of it in a simpler version of it, whereby a Fig. II–17b: it basically consists in a Jurin-like experiment with the capillary tube shorter than the capillary height is dipped in water. height of the tube shorter than the height predicted by Eq. II–10. Debunk this.

4. Think about Jurin’s law in the case where the spreading parameter S is positive. Does Eq. II–14 hold?

5. Next time you use a straw for drinking, pay attention to the small drops of water that may remain stuck in the straw, while the larger ones ﬂow under the effect of gravity. The reason for this is contact angle hysteresis. Write down a balance of force to determine the maximum length of a column of water that may be blocked in a straw, in terms of the advancing and receding contact angles θa and θr. Do some experiments at home with water and a straw to try learning something quantitative about θa and θr.

6. Think about the double-sided floater shown in Fig. II–18. Doesn’t Figure II–18: A double-sided floater with each side made up of a specific something puzzle you about the horizontal equilibrium of such material, having different contact angles an object? The force to the right has to be equal to the force to with water. the left, otherwise it could serve as a perpetual motion machine. Enumerate all the forces involved. III–0 c.j. gommes, chim0698

7. In Wilhemy’s plate experiment for measuring γ (see Fig.I– 4), the plate is made of roughened platinum. Explain why this is important.

Further reading

• Here is a website hosted at MIT, on the general subject of wetting phenomena, with many videos;

• David Quéré gave two lectures about the effect of surface texture on wetting, which you can find on Youtube the first here and the second here. This is a fascinating and very timely field of research;

• Here is a paper published in Science magazine, about Surface Tension Transport of Prey by Feeding Shorebirds: The Capillary Ratchet (Science, 320 (2008), 931). It is about how a seabird exploits contact angle hysteresis;

• Here is another paper about a desert beetle having hydrophobic and hydrophilic patches on its back, which it uses to collect water from fog-laden wind (Nature 414 (2001) 33). Solved problems involving wetting

And so castles made of sand, melts into the sea eventually.

Jimi Hendrix

The present chapter is about some aspects of wetting phenomena that would not justify a theoretical analysis in themselves, but which are interesting to be aware of. They are treated here in the form of problems. Most of them touch on the mechanical aspects of wetting. An illustrative example is the cohesion of wet sand, which justifies the epigraph of the present chapter and is treated in Problem 3. An- other one is capillary stresses in partially saturated porous solids. Figure III–1: Jimi Hendrix (1942-1970) That situation is presented in Problem 4 in the medical context of infant pulmonary distress but, the physical phenomenon is quite general. On an almost philosophical note: you may have noticed from your student’s life so far, that in many instances the creative part of problem-solving is taken out of your hands. The following example is borrowed from Eric Mazur5: You drive to the supermarket and you 5 I am pretty sure of it but I cannot find find no free parking spot; would you better wait until a car leaves and frees a the exact reference back. Eric Mazur is a physics professor at Harvard, who is spot, or try to park your car elsewhere? This is real-life decision that you famous notably for the development of have to take, ideally based on some engineering-like rational analysis. Peer Instruction. However, if that question was posed in a teaching context it would probably be stated as follows. Given that there are 120 cars on the parking lot, and that every customer stays about 20 min in the store, how long will you have to wait on average for a free spot? In that form, the answer is straightforward: you can expect to wait about 10 s. You should definitely wait! Note the leap between the actual problem and the way in which the question is formulated for students. Of course the question is easier, and I understand why you would rather have that type of question to solve at the exam. However, all the creative and interesting part has been taken out of it. Don’t fool yourself: what will be socially expected of you as a scientist is to answer the first type of question, not the second! III–2 c.j. gommes, chim0698

In each problem hereafter, I state clearly the question that you should try to answer. Finding the answer to that question is what is really expected of you. You should try to answer it with no additional hint. If only to understand what is really expected of you, and to measure its difficulty. It is generally acknowledged that However, because the purpose of the present notes is teaching, it personal problem solving is central to the development of mathematical realise I have to guide you somehow in that process. This is the rea- skills. See the role of mathematical son for which you will also find a series of subquestions that should puzzles. For some reason people seem to think that this does not apply help you get the final answer, if you cannot find one by yourself. That to other sciences... I really do not list of subquestions is by no means the only way to get the answer. It understand why. is just one possible way, that is either the only I found or the easiest for me to explain. There is room is science and engineering for personality. If you find your own idiosyncratic approach to solving scientific problems, wouldn’t that make it more interesting? solved problems involving wetting III–3

A. Wetting in zero gravity You can ﬁnd on Youtube some movies It is often reported in popular culture that a glass would not be able involving liquids in zero gravity. I to hold any liquid, should gravity be absent. A famous illustration found this one particularly interesting. is in Fig. III–2. This is, however, not accurate because there is an adhesion energy between water and glass (γ γ ) that is expected SL ≤ L to keep the liquid inside the glass. Of course, the shape of the free surface has no reason to be ﬂat.

What is the equilibrium shape of a liquid in a glass in absence of gravity, in relation to the surface energy of all relevant interfaces?

The unimaginative steps that you may follow if you feel com- Figure III–2: This is roughly what this pletely lost are the following exercise is about

1. Prove that the shape of the free surface of water is a spherical cap.

2. Assume a cylindrical glass with radius r. Given that the free surface is a spherical cap and that the total volume of liquid V is a h r given, express the total surface energy of the system in terms of a single well-chosen geometrical parameter. And ﬁnd the equation R that the parameter has to satisfy at equilibrium.

3. Express the single geometrical parameter in terms of the contact angle θ and express everything in a simple way in terms of the Figure III–3: Deﬁnition of a spherical cap. The area is A = π r2 + h2 and its Young-Dupré relation. π 2 2 volume is V = 6 h 3r + h . III–4 c.j. gommes, chim0698

Solution

1. Of course, the sphere is the geometrical shape that minimises the area for a given volume. Note, however, that this is not directly relevant here because there are other contributions to the energy than the water/air interface. The simplest way to prove that the surface has to be spherical is via Laplace’s law: in absence of gravity the pressure has to be is uniform in the drop, so that the curvature at each point of the surface has to be the same. The only possibility is a spherical surface.

2. The overall configuration is shown in Fig. III–4, where some notations are also introduced. The free surface is a spherical cap, which is uniquely specified through its radius R for a given radius r and height h2. Equivalently, the radius R can be replaced by the height h1. For a given volume of water V, the height h1 and h2 are not independent because they are related through π V = πr2h h 3r2 + h2 (III–1) 2 − 6 1 1 the second term in this equation is the classical expression for the volume of a spherical cap. In absence of gravity the only contributions to the energy are those from the various surfaces involved. Those are (1) the free surface of water having surface tension γ and area A,(2) the surface of the wet glass with surface energy γ and area A , and (3) the SL w Figure III–4: Shape of the free surface surface of the dry glass with surface energy γS and area Ad. The and geometrical meaning of symbols. water spontaneously adopts the configuration that minimises the following energy

U = γA + γSL Aw + γS Ad (III–2)

The areas of the wet and dry parts of the glass are simply expressed as

A = 2πrh and A = 2π(H h ) (III–3) w 2 d − 2 with the notations of Fig. III–4. The area of the cap is calculated by the classical formula

2 2 A = π r + h1 (III–4)

Using Eq. III–1 to express h2 in terms of h1, the expression of the energy takes the form 2 π U = π r2 + h2 γ + V + h (3r2 + h2) (γ γ ) + 2πrHγ 1 r 6 1 1 SL − S S (III–5) solved problems involving wetting III–5

The equilibrium condition dU/dh1 = 0 leads to

h 2 γ h 1 2 1 + 1 = 0 (III–6) r − γ γ r S − SL from which h1 can be calculated.

3. Instead of calculating h1, it is more intuitive to calculate the contact angle between the water and the glass that corresponds to Eq. III–6. Simple trigonometric calculations lead to the relation

2h1/r cos(θ) = 2 (III–7) 1 + (h1/r) This can be put as

h 2 1 h 1 2 1 + 1 = 0 (III–8) r − cos(θ) r

Comparing Eqs. III–6 and III–8 shows that the value of h1 that minimises the energy is exactly the one satisfying the Young- Dupré equation. The practical problems of astronauts is more often to get water out of the glass, The answer is therefore simply that the conﬁguration is such that than to put it back in as in Fig. III–2. Look here to see the contraption they the free surface of the liquid is a spherical cap, and the contact angle use for drinking. satisﬁes the Young-Dupré equation.

Thinking further

• In absence of gravity, the pressure in the liquid is uniform. What is the value of the pressure?

• Equation III–6 has two solutions. What do they correspond to? III–6 c.j. gommes, chim0698

B. The sticking together of two wetted plates Look here for a Youtube movie of a You have probably already played with microscope slides, which capillary bridge between two plates. Be stick strongly together when they are wet. If you did so, you proba- careful, however, that their explanation they give for the origin of the force is bly also noticed that the further apart the two plates are, the easier it not accurate. is to draw them separate them further. The attraction force between the two plate is therefore inversely related to the separating distance. The typical amplitude of roughness6 of glass is lower than 1 µm, 6 See e.g. here for the roughness of which means that this is about how close you can bring two glass some common materials. plates together. Imagine two microscopy slides with area 2 cm2 and water in between.

r What is the value of the force that pushes them together assum- h ing γ = 75 mN/m and θ = 60 °? R

The unimaginative steps that you may follow if you feel completely lost are the following Figure III–5: Two ﬂat plates, with a thin liquid layer in between. In this view, 1. Looking at Fig. III–5, what is the pressure in the liquid relative to the liquid layer is assumed to have the outer pressure P , as a function of h? Simplify that expression the shape of a disk with radius R and 0 thickness h. for the case R h; 2. You are pulling the two plates apart with a force F. In order to determine the value of F that can overcome the capillary forces, imagine cutting the liquid layer horizontally and account for all forces exerted on, say the upper half of the system. Simplify again that expression for R h. solved problems involving wetting III–7

Solution

1. The two radii of curvature of the interface are r and R. The pressure in the ﬁlm is therefore 1 1 P = P + γ (III–9) 0 R − r

where we have assigned a different sign to each curvature, as we should. From elementary trigonometry, the radius of curvature r is related to h by h r = (III–10) 2 cos(θ) The ﬁnal expression is therefore

2 cos(θ) P = P γ (III–11) c 0 − h where we have taken account of R h. The subscript c stands for capillary pressure.

2. If we imagine cutting the system horizontally in two identical parts through the centre of the layer, the forces pushing the system up are 2 Fup = F + PcπR (III–12) The forces pulling down are

2 Fdown = P0πR + 2πRγ (III–13)

where the second term is the direct contribution of the surface. 7 7 If this is not clear to you, go and Equating both terms leads to read again our analysis of the forces in Wilhemy’s plate experiment on page I–2. 2 cos(θ) F = πR2γ + 2πRγ (III–14) h Note that the second term is approximately R/h times smaller than the ﬁrst one, so that it is of the same order of magnitude as the term that we neglected in Laplace’s pressure. To be consistent, the second contribution has to be neglected. The rest is just a numerical application. The capillary pressure is Pc = 75 kPa. Multiplying this by the area leads to a force of 15 N. That is about the weight of 1.5 kg. I am not surprised by that ﬁgure. Are you?

Thinking further

1. If the spacing h is increased indeﬁnitely, while preserving the volume of the liquid, how does the force F depend on h? III–8 c.j. gommes, chim0698

2. The general expression of the force that you found in point 2 changes sign for a particular value of the contact angle θ. Explain why.

3. The calculation of F that we discussed was based on Laplace’s pressure. Redo the same calculation based on energies. It may help you to go back to chapter II and read again the two derivations of Jurin’s law. solved problems involving wetting III–9

C. The capillary force between two spheres A Youtube movie showing a capillary It is a fact known to all kids that wet sand is better for making cas- bridge between two glass beads is tles than dry sand. The reason is that water binds the sand grains visible here. together. At the level of the sand grains, the situation is similar to that shown in Fig. III–7. For the purpose of the present problem you may assume that the contact angle between the water and the sand is θ = 0°.

What is the force that pulls the two sand grains together in the limit of small amounts of water? Figure III–6: Wet sand is stickier than dry sand because of Laplace’s pressure in the liquid bridge that forms at the We shouldn’t give you any hint for the present problem, except contact between two grains. that it looks very much like the previous one, doesn’t it?

r Figure III–7: Two touching spheres R with a liquid meniscus at their contact point. x III–10 c.j. gommes, chim0698

Solution

The quantity x in Fig. III–7 is a useful measure for the amount of liquid in the bridge. The limit of very small amounts of water is equivalent to the limit x 0. We shall therefore ﬁrst express all → relevant quantities in terms of x. The smallest radius of curvature in the system in r, which we may express as a function of x using the Pythagorean theorem as

(x + r)2 + R2 = (R + r)2 x2 + 2xr = 2Rr (III–15)

Because the point of contact between the two spheres is locally ﬂat, r decreases more rapidly than x for x 0. In the case where r x, → the relation between r and x simpliﬁes to

1 2R = (III–16) r x2

The pressure in the liquid is then

1 1 γ 2γR P = P + γ P = P (III–17) c 0 x − r ' 0 − r 0 − x2

The equilibrium is analysed in the usual way: cut mentally the system by a vertical plane passing between the two spheres and express that each sphere is mechanically equilibrated. This leads to

2γR P πx2 + 2πxγ = P πx2 + F (III–18) 0 0 − x2

The ﬁrst terms on the left- and right-hand sides (proportional to πx2) are the integrals over the corresponding surface of the sphere of the pressure force. This is not as simple as it may seem: it is the result of an integral over a spherical surface of a projected force, which happens to be proportional to the projected area of the surface. Be sure you understand it. The ﬁnal result is then simply

F = 2πxγ + 2πRγ 2πRγ (III–19) ' When the quantity of water in the meniscus becomes vanishingly small (i.e. for x 0), the force converges to a constant value 2πRγ, → which only depends on the radius of the particle. The reason for this is that the capillary pressure happens to be inversely proportional to the surface over which it is exerted, so that the force is constant. This is is a geometrical accident. solved problems involving wetting III–11

Thinking further

1. In practice, a surface is never so smooth enough for the asymptotic limit of the previous paragraph to hold. At a very small scale, the surface of contact looks more like many small spheres touching each other than two large spheres. How does that affect the force?

2. Generalise the analysis for arbitrary values of θ.

3. Try to apply an analysis based on energies to calculate the force between the two spheres.

4. Do the same analysis for the capillary force between two cylinders. This might be relevant to the sticking of your hair together when you get out of a shower. III–12 c.j. gommes, chim0698

D. Capillary stresses and surfactant deﬁciency disorder

The final structure in the respiratory tracts of mammals is the pulmonary alveolus III–8, which is a locally spherical cavity, with approximate diameter d = 100 µm. The inner surface of alveoli are coated with a liquid having a finite surface tension, which would tend to make the structure of the lung collapse. To prevent this from happening, the alveolar cells produce a lipoprotein complex that acts as a surfactant, and reduces the surface tension of the liquid. In premature newborns, the alveolar cells are unable to produce Figure III–8: Alveoli are the final and smallest structures in the lung, at which that protein. This is the origin of so-called Surfactant Deficiency scale the exchange between between the Disorder, which may be fatal. air and the blood takes place (Image from Wikipedia).

What is the order of magnitude of the (negative) pressure that has to be exerted on a lung to prevent its alveoli from collapsing?

For a fuller account of SDD, including The unimaginative steps that you may follow if you feel com- some historical aspects, see e.g. S. Bubbles, Babies and Biology: pletely lost are the following Wrobel’s The Story of Surfactant published in the FASEB Journal, 18 (Oct. 2004). 1. The free surface of the liquid in the lungs has a given total area Av per unit volume. Imagine cutting off a cubic piece of lung with side l, and deforming it by a small quantity dl in all directions. How does the surface energy change? What is the energy cost of that change?

2. Once you have solved point 1, all you have to do is propose a crude structural model of a lung, to estimate the area AV. For example, you may assume spherical alveoli with diameter d ﬁlling a fraction e of the total volume of the lung. The value e = 0.5 is probably close to reality. solved problems involving wetting III–13

Solution

1. For small linear deformations dl/l 1, the relative change in the area of individual alveoli satisﬁes da/a = 2dl/l. This implies in turn that the change in the total area of the all alveoli in the volume is also dA/A = 2dl/l. As a consequence, the energy cost of dilating a lung is dl γdA = γA2 (III–20) l For that deformation to be possible, the change in surface energy has to be equal to the mechanical work of the forces F exerted on the faces of the cube, namely

dW = 3Fdl (III–21)

This leads to the following simple relation

F 2 A = γ (III–22) l2 3 l3

3 where A/l is noting but what we have called AV. Note also that F/l2 is the tensile stress, equivalent to a negative pressure.

2. All is left to do is to estimate the surface area AV of a lung. If the number of alveoli per unit volume of the lung, nV, were known, you would calculate the area as

2 AV = nV4πR (III–23)

The number of alveoli can be determined from the porosity e via

4 e = n πR3 (III–24) V 3 This results in an area per unit volume

3e A = (III–25) V R Note that you may have used another 4 With the values R = 50 µm and e = 0.5, one ﬁnds AV 3 10 model for the lung, for example a cubic 1 ' packing of alveoli. In that case you m− . If the liquid was pure water, with surface tension γ = 75 would have a surface 4πR2 in a volume 3 mN/m, the capillary stress would be of (2R) . That would be AV = π/(2R). Compare that with Eq. III–25 with F e = 0.5. = 1500 Pa (III–26) l3

That pressure is equivalent to about 1.5 kg exerted on a surface of 10 10 cm2. Imagine a newborn’s lung having to exert that at each × breath! III–14 c.j. gommes, chim0698

Thinking further

1. The purpose of the alveoli is to enable mass transfer between the air and blood. What is the total area of the alveoli in an adult’s lung?

2. Estimate the order of magnitude of the capillary stress for a nanoporous material with, say A 300 m2/cm3. V ' 3. In the analysis here above, we have neglected the mechanical stiffness of the material. How can that be taken into account? solved problems involving wetting III–15

E. The mechanical equilibrium of a meniscus

The present problem is motivated by the double-sided ﬂoater case presented in Fig. II–18, the mechanical equilibrium of which may seem paradoxical. We shall consider the simpler case of just a wall wetted by a liquid, as in Fig. III–9, and ask the question

What is the total force exerted by the liquid on the wall? And R(z) h how does it depend on the various surface energies involved? z There is a straightforward way to estimate that force, which does require any calculation. Try to find it. If you do find it, use the result to calculate the height h of the triple line above the surface of the liquid. Figure III–9: Shape of the meniscus On the other hand, if you are running out of imagination, you may close to a flat vertical wall. also go through the following steps

1. As a starting point in any analysis, it is often advisable to do a dimensional analysis. Use Buckingham’s Π theorem to ﬁnd the general expression for the height h of the meniscus as a function of the contact angle θ, the surface tension γ, the density ρ, and the acceleration of gravity g.

2. Expressing that the pressure in the liquid is hydrostatic, calculate the local radius of curvature of the meniscus R(z) as a function of the height above the free surface.

3. The surface can be parameterized as z(α) where α is the local slope of the surface, as shown in Fig. III–9. The angle α increases from 0 far from the wall to π/2 θ on the wall. Assume for now − that θ π/2. Use the answer to point 2 to ﬁnd the differential ≤ equation that governs dz/dα and solve it. What is the height h that the liquid climbs on the wall? Compare your answer to that of point 1.

4. What is the total force that the liquid exerts on the wall? III–16 c.j. gommes, chim0698

Solution

The straightforward way consists in noting that the force exerted on the wall by the liquid is equal to the force exerted on the liquid by the wall. Because any vertical slab of liquid has to be equilibrated in its own right. Consider then a thick slab of liquid extending from the wall to far from it, where the free surface is practically horizontal. The force pulling that slab to the right is simply γ. Therefore, the horizontal force exerted by the liquid on the wall is also γ. You can use that result to ﬁnd the height h by expressing explicitly the equilibrium of horizontal forces on the liquid slab as

Z h γ sin(θ) = γ + ρgz dz (III–27) 0 −

In the left-hand side are the force pulling towards the wall, and in In this analysis, we have assumed that the right-hand side the forces pulling/pushing away from the wall. the ambient pressure is P0 = 0, but this does not matter. If you keep P0 in This leads immediately to the calculation, you will ﬁnd that its effect cancels out in the end. With some q experience, you should progressively h = l 2(1 sin(θ)) (III–28) c − feel comfortable with this. p where lc = γ/(ρg) is the capillary length.

As for the unimaginative way, here it is.

1. There are four dimensional variables h, ρ, g, γ, and a dimensionless variable θ. The dimensional variables are based in three dimensions: length, time, and mass. You can therefore only form one dimensionless variable (4 3 = 1). Let Π be that variable. The − 0 general relation that you look for can only be of the type

Π0 = f (θ) (III–29)

where f (.) is can a priori be any function.

As for Π0, any suitable combination would do. Because h is what you are interested in, and you know that the capillary length p lc = γ/(ρg) has the dimension of a length, it is natural to chose Π0 = h/lc. The mathematical expression for the height can therefore only be of the form

h = lc f (θ) (III–30)

where f (θ) is a yet unknown function. It is important that you understand that, whatever the ﬁnal answer you ﬁnd for this problem, it has to be compatible with Eq. III–30. solved problems involving wetting III–17

2. The pressure in the liquid obeys

P(z) = P ρgz (III–31) 0 − and the pressure in the gas phase is simply P0 (it is much lighter than the liquid so the vertical pressure gradient in it can be neglected). The pressure discontinuity at the interface is accounted for by Laplace’s law, which leads to γ R(z) = (III–32) ρgz Just a check: R ∞ for z 0, as it should. → → 3. Draw an inﬁnitesimal part of the surface with height dz and hypothenuse dl (along the surface). The length of the hypothenuse is related to the local radius of curvature by Rdα = dl, and we also have the relation dz = dl sin(α). The relation between dz and dα is therefore dz = R sin(α)dα (III–33) Combining this with Eq. III–32 leads to the following differential equation dz γ sin(α) = (III–34) dα ρgz the solution of which is q z(α) = l 2(1 cos(α)) (III–35) c − where we chosen the integration constant in such a way that z(α = 0) = 0. The ﬁnal answer is q h = l 2(1 sin(θ)) (III–36) c − which is indeed compatible with Eq. III–30.

4. There are two component to the horizontal force exerted on the wall by the liquid: the direct term, equal to γ sin(θ) per unit length of the triple line, and the pressure term. Because the ambient pressure is irrelevant, we may set it equal to 0, in which case the latter contribution is equal to Z h 1 (0 ρgz)dz = + ρgh2 (III–37) − 0 − 2 Pay attention to the signs: this is a negative pressure on the right- side on the wall, so the net effect is a positive force in the right direction. The total force is ﬁnally 1 F = ρgh2 + γ sin(θ) (III–38) 2 Using Eq. III–36, one ﬁnds F = γ independently of the contact angle θ. The fact that the total force is independent of the contact angle solves the apparent paradox raised by Fig. II–18. III–18 c.j. gommes, chim0698

Further thinking

1. Look back to our analysis of Wilhelmy’s plate (Fig.I– 4) with the present problem in mind. IV–0 c.j. gommes, chim0698

As a general conclusion for these few problems I would like to insist again that the skill that you should develop is to ﬁnd you own way to answer the questions in boldface characters in each problem. Answering the subquestions is not what is expected of you, this is just useful for learning. Reﬂecting on the general way in which every problem here was solved, it seems that Pólya’s general steps of problem-solving (see Fig. III–10) can be followed. The steps are 1. Understand the problem. That seems obvious, but this means here understanding the physics of the problem. In the case of problem D on capillary stresses, that step was realising that any macroscopic change of volume of the lung is accompanied by a change in the surface area of liquid free surface, and that this change is surface area required some energy;

2. Devise a plan. In that case, the plan is (i) calculate the change in surface area resulting from a macroscopic change of volume, and (ii) equate the resulting change of energy to the mechanical work of the macrosopic forces; Note that step 3 (carry out the plan) is the only one that you do if you 3. Carry out the plan. That step is self-explanatory. This is the only try to apply formulae in a rote way. one where some mathematics is involved; Most of the time, that step is the least interesting one, but for some reason it 4. Look back. Have you answered the question? Does the result seems to be the one on students spend make sense? In the case of the lung, estimating the total area of the more time. lung, and relating the pressure to a force is a way to look back. The solution of each problem is ended with a section I have entitled "Further thinking". Some of the questions raised there are amenable to a Pólya-like analysis. See if you can answer those questions.

Further reading

• Many interesting applications of ﬂuids involve some type of capillary instability, which is the general term for Plateau-Rayleigh-like type of phenomenon, by which a system evolves under the inﬂu- ence of capillary forces. There is collection of solved problems on capillary instabilities in the book Hydrodynamique physique: Prob- lèmes résolus avec rappels de cours by Marc Fermigier, Dunod (1999).

• The hungarian mathematician George Polya published in 1945 a book entitled How To Solve It which sold over one million copies, in more than 15 languages. The topic of the book is to propose general principles for problem solving. If you get a chance, it is Figure III–10: George Pólya (1887-1985). Picture from Wikipedia. interesting to read. For a glimpse of what is in the book, you can look here. Intermolecular forces

The itsy bitsy spider climbed up the waterspout. Down came the rain and washed the spider out. Out came the sun and dried up all the rain And the itsy bitsy spider climbed up the spout again.

Folk song, ca 1910

Intermolecular forces are forces exerted between molecules, as opposed to intramolecular forces which exert within individual molecules. All the forces and energies that we have mentioned when discussing surface energies and surface tension in terms of missing bonds are intermolecular forces. There are entire books written about Figure IV–1: Johannes Diderik van intermolecular forces. The present chapter is just a very basic intro- der Waals (1837-1923) . Picture from duction that should provide you with a minimal working knowledge. Wikipedia. The intermolecular forces are often referred to as van der Waals forces, after the Dutch chemist who hypothesised them to explain the properties of non-ideal gases though his celebrated equation of state a P + 2 (Vm b) = kBT (IV–1) Vm − where the constants a and b account for the interaction between molecules and their ﬁnite volume, respectively. To make a long story short, when vapours are compressed close to their dew point, the pressure is lower than predicted by the ideal gas law, which hints at attractive forces between molecules. Although intermolecular forces were discovered in the context of non-ideal gases, their effects extend in many different areas. The epigraph of the present chapter is a reminder that van der Waals forces are strong enough to enable insects to walk on vertical surfaces, apparently defying the laws of gravity. In the case of geckos, the contact of the animal’s feet with the surface is maximised thanks to micrometer-sized setae. The intermolecular forces are electromagnetic in nature. If molecules carry an electric charge, they are expected to attract or repel each Figure IV–2: The feet of the gecko are other in agreement with Coulomb’s law. Less intuitively, even elec- covered with innumerable setae that maximise the contact area with the trically neutral molecules inﬂuence each other; unlike Coulombic underlying surface. This the animal to climb on vertical walls, despite his relatively large size. Picture from Wikipedia. IV–2 c.j. gommes, chim0698

interactions, however, the forces between neutral molecules are always attractive, as we now discuss. Consider the case of two polar molecules, carrying an electrical dipole µ1 and µ2. If the molecules are free to rotate, they will prefer- entially orient themselves in such a way that their parts with opposite charges will face each other. This will result in an average attractive energy given by 2 2 µ1µ2 CK wK = 2 6 = 6 (IV–2) − 3(4πe0) kBTd − d

where e0 is the vacuum electrical permittivity, kB is Boltzmann’s Figure IV–3: Willem Hendrik Keesom constant, T is the temperature, and d is the distance between the (1876 -1956). Picture from Wikipedia. two molecules. The interaction energy depends on the temperature Keesom had been a student of van der because the lower the temperature is, the more likely the two dipoles Waals and of Kamerlingh-Onnes. Does the name Kamerlingh-Onnes ring a bell will be aligned. The interaction described by Eq. IV–2 is called the to you? Keesom interaction, after the Dutch physicist who ﬁrst calculated them in 1921. Another case is the situation where a molecule carrying dipole µ interacts with a non-polar, yet polarisable molecule with polarisability α. If the molecule is free to rotate, the interaction energy is

2 µ α CD wD = 2 6 = 6 (IV–3) − (4πe0) d − d which bears the name of Peter Debye. Note that here too, the depen- 6 dence in the distance is proportional to d− . Yet another situation is that of two neutral molecules bearing no electrical dipole whatever. A handwaving explanation for the attraction between two such molecules is the following. Electrons in each molecule move around very rapidly with a frequency compara- 8 ble to the ionisation frequency ν. As a consequence, the molecules Figure IV–4: Peter Debye (1884 -1966). are non-polar only on average: at every moment they carry a small Picture from Wikipedia. 8 dipole that fluctuates very rapidly. If the fluctuations in each molecules Ionisation frequencies are of the order of 1015 1016 Hz. The ionisation − were independent of one another, this would not result in any force process by which an electron is stripped between them. The force would be now attractive now repulsive and out of a molecule a photon, i.e. by an oscillating electric field, can be thought the two would cancel out exactly. However, when the two molecules of as a resonance phenomenon. The are brought close together, the fluctuations of the dipoles in each one frequency ν of the field has to match of them become anti-correlated: if one points, say up the other will the natural frequency of the electron. be more likely to point down. This results in a net attraction between the two molecules. Imagine compressing gaseous nitrogen The quantitative analysis of the attraction between two non-polar at a constant temperature. When you reach a given volume the gas will molecules requires quantum-mechanical calculations. The calcula- start to condense, which means that tions were done for the first time by Fritz London in 1921, and the molecules have come sufficiently close to each other to attract each other. result can be written as Have you ever thought about why two 3 α1α2 hν1ν2 CL nitrogen molecules attract each other at wL = 2 6 = 6 (IV–4) all? − 2 (4πe0) d (ν1 + ν2) − d intermolecular forces IV–3

where α1 and α2 are the polarisabilities of the molecules, and ν1 and ν2 are their ionisation frequencies. Once again, the distance depen- 6 dency is d− . This contribution to van der Waals forces is called the London force, and sometimes the dispersive force. The three contributions - Keesom, Debye, and London - are ad- ditive. Imagine approaching two polar molecules A and B. The molecules will naturally undergo the Keesom attraction. However, the molecules having a permanent dipole µ does not prevent them from being polarisable too. This means that if the molecules were put in an electrical ﬁeld E, their dipoles would have a contribution There is a factor 2 in Eq. IV–5 in front proportional to E in addition to their permanent dipoles µ. The of the Debye contribution. This is not a molecules will therefore undergo the Debye and London attraction typo. Think about it. too. The resulting energy is therefore C + 2 C + C C w = K × D L = tot (IV–5) − d6 − d6

where we have introduced the obvious notation Ctot. The values of the various contributions to the van der Waals forces are given in tab. IV–1 for a few common molecules. In the particular case of non-polar molecules such as He or N2 the Keesom and Debye contributions are absent, and one is left only with the London force. In the table, the value of the constant Ctot is compared to an experimental value derived from the van der Waals coefﬁcients a and b via

9ab C = (IV–6) exp 4π which results from assuming that the molecules behave like hard spheres at short distances. Globally the agreement between the elec- Figure IV–5: Fritz London (1900 - 1954). tromagnetic calculation of the actual intermolecular forces is fair. It Picture from Wikipedia. particularly interesting to compare the various contributions to the intermolecular forces in the case of polar molecules, such as HCl. The largest contribution by far is generally the dispersive London force! A notable exception is water, for which the Keesom contribution is dominant.

µ α/(4πe0) hν CK CD CL Ctot Cexp Table IV–1: The values of the Keesom, (D) (10 30 m3) (eV) Debye, and London contributions to − van der Waals forces between a few 79 6 He 0 0.2 24.6 0 0 1.2 1.2 0.86 usual molecules, in units of 10− Jm . Adapted from Butt, Graf and Kappa, CH4 0 2.59 12.5 0 0 101.1 101.1 103.3 Physics and Chemistry of Interfaces, HCl 1.04 2.7 12.8 9.5 5.8 111.7 127.0 156.8 Wiley: 2005. CH3OH 1.69 3.2 10.9 66.2 18.3 133.5 217.9 651.0 H2O 1.85 1.46 12.6 95.8 10.0 32.3 138.2 176.2 N2 0 1.74 15.6 0 0 56.7 56.7 55.3 Equation IV–5 describes the attraction between two individual molecules. To see how this converts to attraction between two macro- IV–4 c.j. gommes, chim0698

scopic bodies, say A and B, one has to sum over all pairs of molecules with one in A and the other in B. Mathematically, this means that we have to calculate a six-dimensional integral. Let us write the Eq. IV–5 6 as w = CAB/d where CAB is speciﬁc to the types of molecules in A and B. The macroscopic interaction would be Z Z CAB ρAdVA ρBdVB WAB = (IV–7) 2 2 2 3 − VA VB [(x x ) + (y y ) + (z z ) ] A − B A − B A − B where ρA and ρB are the densities of media A and B expressed as a number of molecules per unit volume, in such a way that ρAdVA and ρBdVB are the number of molecules in the elementary volumes.

a) b) Figure IV–6: a) Van der Waals interac- D tions between two semi-inﬁnite bodies y x d r A and B separated by a distance D, z z and b) interaction between a single D z x y molecules A at a distance D from a A B B semi-inﬁnite body B.

We shall calculate explicitly the integral in the simple case of two semi-infinite media separated by a spacing D, as sketched in Fig. IV–6. Let us calculate that integral in two steps, first by calculating the interaction between a single A-type molecule and a semi- infinite medium of type B, and then by integrating that energy over all molecules in medium A. The first step is equivalent to calculating a three-dimensional integral on volume B, which is best calculated in cylindrical coordinates as

Z ∞ Z ∞ CABρB WA B = dz 2πrdr (IV–8) | − 0 0 [(z + D)2 + r2]3 where D is the distance between the A-like molecule and the semi-

infinite medium B. Note that we have used the notation WA B be- | cause this is not the same as WAB. This leads to the following expres- To get easily from Eq. IV–8 to Eq. IV–9 sion make a change of variable η = r2, and π CABρB integrate first with respect to η then to WA B = 3 (IV–9) | − 6 D z. 6 Note that the very strong d− dependence becomes here a weaker 3 d− dependence. To obtain the van der Waals energy between two semi-infinite media, Eq. IV–9 has to be integrated over all the molecules in the semi- infinite volume A. Of course, in the strict case of two (semi-)infinite media the interaction energy is infinite, unless it is expressed per unit area of the two bodies. This leads to the following expression

Z ∞ π ρAdz WAB = ρBCAB (IV–10) − 6 0 [D + z]3 intermolecular forces IV–5

Two ﬂat surfaces Two spheres Two parallel cylinders Two crossed cylinders

R1 D R 1 R1 R2 R2 D R2

q A A R1R2 A R1R2 A W = 2 √R1R2 − 12πD − 6D R1+R2 − 12√2D3/2 R1+R2 − 6D

Figure IV–8: Van der Waals interaction energies between bodies of various and therefore to shapes. In all cases A is the Hamaker A constant, and D is the spacing. W = AB (IV–11) AB − 12πD2 where AAB is the so-called Hamaker constant, deﬁned as

2 AAB = π CABρaρB (IV–12)

The Hamaker constant characterises the two interacting media. Note that D does not have the same meaning in Eq. IV–9 and in Eq. IV– The expression of WAB in Eq. IV–7 is quite general; you can apply 10. In the former equation, it is the it in principle to calculate the van der Waals interactions between distance between a molecule of A and bodies of any size and shape. Depending on the shapes of the two the surface of B; in the latter it is the distance between the surfaces of A and bodies, the calculation can be cumbersome, but the calculations are B. conceptually as simple as those leading to Eq. IV–11. A few expressions of the energies are gathered in Fig. IV–8 for interactions between spheres and cylinders. Note that these are expression for the energy. The corresponding forces are calculated as

dW F = (IV–13) dD We shall come back to these expressions later. Before we proceed to see what Eq. IV–11 can be good for, it might be useful to elaborate a bit on Hamaker’s constant. You may have noticed that the relevant constant in Eq. IV–11 characterises the inter- Figure IV–7: Hugo Christian Hamaker (1905-1993). Picture from Langmuir 7 action of two different media A and B, while we have only discussed (1991) 209. In that same paper the au- the forces between identical molecules so far in Eqs. IV–2, IV–3 and thors mention the following anecdote. Hamaker left the field of colloids in IV–4. The fact that the dispersion forces is the largest of the three the forties to focus on the development contributions for non-polar (or weakly polar) molecules simplifies the of statistical methods for the Phillips analysis. From Eq. IV–4, the following approximation is seen to hold company, and he stopped following the developments in that field. In 1965 p one of his sons who was studying soil A α α ρ ρ A A (IV–14) AB ' A B A B ' A B science asked him: "Dad, there is something called a Hamaker constant. Is it In other words, all you need to know to estimate the interaction named after some relative of ours?" between two bodies via van der Waals forces is the Hamaker constant of individual bodies, and to compose then according to Eq. IV–14. Eq. IV–14 is sometimes referred to as the Berthelot composition rule. IV–6 c.j. gommes, chim0698

20 2 A (10− J) γ (mJ m− ) from Eq. IV–19 Liquid He 0.057 PTFE (teﬂon) 3.8 18.5 Pentane 3.8-4 Table IV–2: Hamaker constants of a Ethanol 4.2 20.5 few materials (both liquid and solid), sorted in increasing order, and inferred Acetone 4.1 20 value of the surface energy. Compare Hexane 4.32 those values to those from Tab.I– 1 on pageI– 3. Octane 4.5-5.02 Cyclohexane 4.8 - 5.2 Decane 5.45 Water 3.7 - 5.5 18-27 Toluene 5.4 Formamide 6.1 30 Fused Quartz 6.3 Silica (SiO2) 6.5 24-29 Diiodomethane 7.2 - 7.8 Polyvinyl chloride (PVC) 7.5 7.8 38.0 Quartz 7.93 Polyethylene (PE) 10 Mica 10-11 Aluminium 15 ZnS 15-17 73 - 83 Iron oxide (Fe3O4) 21 Silicon 25.5 Zirconia (ZrO2) 27 Diamond 28.4 Copper 28.4 Germanium 30 Silver (Ag) 40 Rutile (TiO2) 43 Gold 45 Graphite 47 intermolecular forces IV–7

19 The order of magnitude of Hamaker constant in air is 10− J. It is interesting to use that value to see how large van der Waals forces can be. For that purpose, consider the force between two solids with 2 It is a better strategy for the gecko to a contact area of 1 cm . The force per unit area is calculated from the minimise the distance between his feet derivative of Eq.IV–11, which leads to and the surface than to maximise the contact area, because the dependence is 3 A D− . F = (IV–15) 6πD3 The typical distance D between solids in contact is D 0.2 nm. ' With these values, the van der Waals force across a surface of 1cm2 is found to be of the order of 7 104 N. This corresponds to a weight of 7 tons! This means that the contact area between a gecko’s feet (see Fig. IV–2) and the surface on which it is walking does not have to be large to stand the gecko’s weight. Let us make another estimation of van der Waals forces, this time from the nanometer world. In many practical situations, one has to deal with metal nanoparticles dispersed on a support (see e.g. Fig. 0–5 on page 0-4). A legitimate question is how strong is the force that keeps the particle on the surface. This can be calculated from the energy given in Fig. IV–8 for two interacting spheres, in the limit where one sphere becomes inﬁnitely larger. In that case the energy becomes AR W = (IV–16) − 6D and the force is AR F = (IV–17) 6D2 In the case of metal on silica, the Hamaker constant can be calculated Figure IV–9: A metal nanoparticle, say, more accurately than in the case of the gecko. Using the geometrical 4 nm across sits on a silica surface: how mean as in Eq. IV–14 and the values from Tab. IV–2, one ﬁnds strong is the force keeping the particle on the surface ? 20 A √6.5 35 = 15 10− J (IV–18) metal/SiO2 ' × With a radius R = 2 nm, and a distance D = 0.2 nm, the force is found to be F = 1.25 nN. 9 The only way to make sense of a force of the order of 10− N is to compare it to another force. For example, one may estimate the weight of the 4 nm nanoparticle. Assuming a density of ρ = 20 g/cm3, the weight of the nanoparticle is found to be of the order of 20 11 10− N. In other words, the van der Waals force is about 10 times stronger than that. Another way to get a feeling for this would be to consider that you would need to centrifuge with an acceleration of a hundred billion g’s if you wanted to detach the nanoparticle from the support. Let us push this type calculations to still smaller sizes. If you think Figure IV–10: Marcelin Berthelot (1827- 1907). Picture from Wikipedia. of it, the van der Waals forces are responsible for what we loosely IV–8 c.j. gommes, chim0698

referred to as the missing bonds when discussing the physical origin of surface tension. There should therefore be a relation between the surface tension/energy γA of a given substance and its Hamaker constant AA. This is illustrated in Fig. IV–11. Consider for now the case where A and B are identical media, say, the same liquid. In B B that case, the energy it takes to cut the liquid into two halves is 2γA

per unit area, because two new surfaces are created. On the other A hand, this energy can also be calculated as the van der Waals energy A between the two halves when they were in tight contact. Assuming an initial spacing D0 of molecular dimensions, and using Eq. IV–11 Figure IV–11: The energy cost of mov- the relation we ﬁnd is therefore ing a liquid B away from a solid A, γ + γ γ , is the van der Waals in- A B − AB AA teraction between the two macroscopic 2γA = 2 (IV–19) liquid bodies. 12πD0

The value of D0 that is universally accepted for most liquids and solids alike for this type of calculation is D 1.65 Å. Some values 0 ' of Hamaker ’s constants are gathered in Tab. IV–2, together with the values of γ inferred from Eq. IV–19. Comparing those values with Tab.I– 1, one sees that the calculations are reasonably accurate in the case of weakly-polar substances. This is, however, not the case for polar molecules and for metals, for which dispersive forces That a single value can be used for D0 for a broad variety of substances is a responsible for only a fraction of the cohesion. For example, it the bit of a mystery. You can read some case of water disperse forces account for only about 25 % of all the explanations in the excellent book of J. Israelachvili on Intermolecular Forces, intermolecular forces. in section 13.13. The reasoning leading to Eq. IV–19 based on Fig. IV–11 can be repeated in the case of two distinct media. In that case, equating the change in surface energy to the energy that is needed to overcome the van der Waals attraction between the two media leads to

AAB γA + γB γAB = 2 (IV–20) − 12πD0 Expressing that A √A A leads to the following relation AB ' A B

γAB γA + γB 2√γAγB (IV–21) ' − You can ﬁnd in the literature expres- which can be used to estimate interface energies, when only the sions similar to Eq. IV–21 with a semi- empirical factor in front of √γAγB to surface energies are known. Of course you do not expect this relation account for the polarity of the sub- to be very accurate in the case of polar substances, but if you don’t stance. That is of course numerically more accurate... but also less general. know better, this relation can be a good starting point. So far, we have considered only the interaction between two bodies in vacuum. In that case, the van der Waals forces are always attractive. This is not necessarily the case if the two interacting bodies are immersed in a liquid. To make things more practical, consider two bodies A and B and a liquid L. Imagine the situation where the Hamaker constant between A and B would be weaker than between A and L. In that case, you may expect that the liquid will try intermolecular forces IV–9

to squeeze between A and B, thereby leading to an effective repulsion. This situation is described using the concept of disjoining pressure, introduced by Boris Derjaguin. In the case of two flat plates A and B facing each other at distance D, the disjoining pressure Π(D) is the pressure that has to be exerted on the plates to prevent a specific liquid to squeeze in the interstice and increase D. To determine the energy related to the insertion of a liquid layer of thickness D between two media A and B, consider the various steps sketched in Fig. IV–14, from 1 to 4. We refer to that energy as Wwetting. If Wwetting is positive, the process will not spontaneously take place. On the contrary, if it is negative a positive pressure will Figure IV–12: Boris Vladimirovich need to be exerted on A and B to prevent the liquid from pushing Derjaguin (1902-1994). Picture from them apart. Colloid Journal 64 (2002) 648. Before we proceed with that particular calculation, it is necessary to slightly generalise Eq. IV–11, in order to calculate the interaction energy between a semi-infinite medium B and a slab of A of finite thickness extending, say, from distances z0 to z1 from the surface of B. Starting from Eq. IV–9 the calculation is straightforward:

Z z 1 π CABρB W = ρAdz (IV–22) − z 6 z3 z1 0 B z which leads to 0 ! A 1 1 W = AB (IV–23) 12π 2 2 − z0 − z1 A Note that this energy reduces to Eq. IV–11 for z ∞, as it should. 1 → We can now turn to Fig. IV–14. The change in energy from 1 to 2 Figure IV–13: The interaction energy between a slab of B extending from z0 corresponds to increasing the distance between A and B from D0 to to z1 above a semi-inﬁnite medium A is D, i.e. ! calculated as Eq. IV–23. AAB 1 1 W12 = 2 2 (IV–24) 12π D0 − D according to Eq. IV–11. Note that this energy is positive. The energy change from 2 to 3 can be calculated using Eq. IV–23 with z0 = D0 and z1 = D. The result is ! AL 1 1 W23 = 2 2 2 × 12π D0 − D ! ! ALA 1 1 ALB 1 1 2 2 2 2 (IV–25) − 12π D0 − D − 12π D0 − D

The positive contribution on the first line is the energy it takes to remove the slab of liquid aways from the two semi-infinite L-like media, one on the top and the other on the bottom. The two negative contributions result from the slab being put in contact with the semi- infinite A and B. The last contribution is the putting back together IV–10 c.j. gommes, chim0698

A A A A L L L L 1) 2) 3) 4) B B B B

of the two semi-inﬁnite L-like media, from 3 to 4. This results in a Figure IV–14: The insertion of a liquid change in energy layer L between two solids A and B has an energy associated with it, which can ! be calculated via the following steps. AL 1 1 W34 = 2 2 (IV–26) − 12π D0 − D which is negative as it should. Putting all contributions together results in the following expression for Wwetting = W12 + W23 + W34: ! AALB 1 1 Wwetting = 2 2 (IV–27) 12π D0 − D where the effective Hamaker constant is given by

A = A + A A A (IV–28) ALB AB L − AL − BL The subscript highlights that it corresponds to the interaction of media A and B, with medium L in between. Whether the liquid layer L will tend to thicken and move the two media A and B apart depends on the sign of the effective Hamaker constant. If AALB is negative, the film tends to increase in thickness. The disjoining pressure is defined as the pressure that has to be exerted on the film to prevent it from thickening further. It is therefore positive when AALB is negative. It is calculated as the derivative of the wetting energy

dW A Π(D) = = ALB (IV–29) − dD − 6πD3 If A 0 then the disjoining pressure is negative, which means ALB ≤ that a tension (negative pressure) has to be exerted to prevent the ﬁlm for becoming thinner. It is interesting to relate this analysis to the macroscopic approach of Chapter II via the deﬁnition of a spreading parameter S as in Eq. II–2. Clearly the two approaches should lead to the same result in the macroscopic limit, i.e. for large values of D. In other words, one should have

AALB lim Wwetting = = γAL + γBL γAB (IV–30) D ∞ 12πD2 − → 0 intermolecular forces IV–11

I will let you ﬁnd out on your own9 that this is indeed the case. 9 Hint: use Eqs. IV–19 and IV–20. Therefore, the present analysis enables us to ﬁnd the correction to the macroscopic wetting energy γ + γ γ in the case where AL BL − AB the layer is not macroscopically thick. The result can be written as

" 2# D0 W(D) = (γAL + γBL γAB) 1 (IV–31) − − D ) 1 AB γ The value obtained with the generally accepted values D0 = 1.65 is 0.8 − plotted in Fig. IV–15. The ﬁgure shows that the macroscopic values

BL 0.6 of the energies are reached for thicknesses of about 2 nm. For objects γ

+ 0.4 smaller than about that size, you expect signiﬁcant deviations from AL

γ 0.2

the predictions of the macroscopic laws of wetting. ( 0 W/ 0 5 10 15 20 Let us now discuss the sign of A ALB and of the disjoining pres- D (A˚) sure in a few characteristic situations. Using the geometric-mean rule Figure IV–15: Normalized wetting Eq. IV–14 for composing Hamaker constants, enables one to rewrite energy, as a function of the thickness D Ae f f in the following way as a function of the Hamaker constants of of the liquid layer. This was plotted by A, B and L: assuming D0 = 1.65 Å. p p p p A = A A A A (IV–32) ALB A − L B − L Based on that general equation, a few interesting situations can be discussed. In the case where media A and B are identical, the effective Hamaker constant is p p 2 A = A A (IV–33) ALA A − L which can only be positive. Two identical media can only attract each other.

When one of the two media, say, B is vacuum, the situation is The rule of thumb for spreading is simply that of a liquid spreading over a surface. The question is that a liquid spreads on any solid about whether the liquid will spontaneously thicken or not. In that more polarisable than itself. This is the reason why oil spreads on metals, and case, the effective Hamaker constant is obtained by setting AB = 0, water does not spread on most plastics. i.e. If you look at the Hamaker’s constant p p p of liquid helium in Tab. IV–2, you will A = A A A (IV–34) ALO − L A − L also understand why liquid helium is a universally wetting ﬂuid. According to this analysis, the liquid spreads on the solid A if the Hamaker constants satisfy A A . If you recall that the ori- L ≤ A gin of A is the London dispersive forces described by Eq. IV–4, this criterion is equivalent to stating that the liquid is electrically less polarisable than the solid. The criterion for wettability based on In the general situation, the effective Hamaker constant is positive Hamaker’s constant looks very much like Zisman’s criterion based on a except when AL has a value intermediate between AA and AB. In critical surface tension γC discussed in that case, it is energetically favourable for the liquid to enter in the in- Chapter II. terstice between A and B, thereby exerting a pressure that pushes A IV–12 c.j. gommes, chim0698

Figure IV–16: Direct van der Waals force measurements using an Atomic Force Microscope (AFM), published in Measured long-range repulsive Casimir-Lifshitz forces, Nature, 457 (2009) 170. The voltage is a measure of the force. On top: attractive force between a gold particle and a gold substrate in bromobenzene, and bottom: repulsive force between the same particle and silica in the same solvent.

and B apart. This is illustrate on Fig. IV–16. Using Atomic Force Mi- croscopy (AFM) the authors of that study measured directly the force between a gold nanoparticle and a support, in bromobenzene. When the support is also made of gold, the force is attractive as expected based on Eq. IV–33. By contrast, when silica is used as a support a repulsive force is observed. Silica and gold were not chosen randomly for that experience. For a repulsive force to be observed, the Hamaker constant of bromobenzene has to be intermediate between silica and gold. From Tab. IV–2, the Hamaker constants of silica and gold are very different (A 6.5 10 20 J and A 45 10 20 J), SiO2 ' − Au ' − which is a favourable condition to observe repulsion. Figure IV–17: Hendrick Casimir (1909- The calculation of the effective Hamaker constant via Eqs. IV–28 2000). Picture from Wikipedia. or is interesting conceptually because it enables one to understand physically the sign of AALB, but is not necessarily practical because it requires the three Hamaker constants AA, AB and AL to be known. Another way has been developed by Lifshitz based on electromagnetic calculations. The following formula is a good approximation for AALB 3hν (n2 n2 )(n2 n2 ) A = + e A − B L − B ALB 2 2 2 2 2 2 1/2 2 2 1/2 8√2 (nA + nB)(nL + nB) (nA + nB) + (nL + nB) 3k T e e e e + B A − B L − B (IV–35) 4 eA + eB eL + eB where kB is Boltzmann’s constant, h is Planck’s constant, νe is the Figure IV–18: Evgeny Mikhailovich Lifshitz (1915-1985). Picture from main electronic absorption frequency, and ei and ni are the static Wikipedia. dielectric constant and refractive index of phase i. Remember that the dielectric constant e and the refractive index n both characterise the electrical polarisability of a medium. There- intermolecular forces IV–13

fore, from a physical perspective, Eqs. IV–35 and Eq. IV–4 are not so different after all. A practically important situation where repulsive van der Waals forces play a role is the freezing a water. It is well known that growing ice crystals are able to to push away solid material initially in contact with water. This is notably striking in frost heave, by which entire layers of soil are deformed by growing ice crystals (see Fig. IV– 19). If you think of it, this is possible only if a repulsive force exists between the growing ice and the suspended material. In absence of Figure IV–19: Stone rings is an example any repulsive force, the growing ice crystal would simply engulf the of structures that can form further to solid and grow around it rather than push it. frost heaving. Picture from Wikipedia. To analyse phenomena like frost heave, the Hamaker constant between ice and a few oxides across water are reported in Tab. IV–3. These values have been calculated using the general relation given Eq. IV–35, which is the reason for which the dielectric constants and refractive indices are reported in the table. For the calculations, the value ν 3 1015 Hz was used for the ionisation frequency. The e ' values in the table show that ice attracts ice across water, as testiﬁed Table IV–3: Van der Waals interaction by a positive value of the Hamaker constant A. This was expected energies between ice and various from our previous analysis based on Eq. IV–33 because two identical oxides. Taken from G.W. Scherer, Cement and Concrete Research, 30 media can only attract each other, and it is also confermed by the (2000), 673.

Lifshitz expression. 21 Substance e n A (10− J) More interestingly, the values in the table also show that ice repels Water 80 1.33 - all the considered oxides (quartz , mica, and alumina) as testiﬁed by Ice 110 1.309 1 the negative values of A. The van der Waals repulsion between ice Quartz 3.8 1.448 -1.20 Mica 7.0 1.60 -1.211 and these oxides across water is the nanometer-scale phenomenon Alumina 11.6 1.75 -2.94 that is responsible for such spectacular effects as frost heave (see e.g. Fig. IV–19). An interesting technical application of repulsive van der Waals forces between a solidifying solvent and suspended particles is the so-called freeze-casting methods for shaping ceramic and metallic materials. This is illustrated in Fig. IV–20. With this type of process, one can chose appropriately the solvent to have a selective repulsion of some suspended particles (not necessarily all of them), to design porous materials with speciﬁc morphologies. IV–14 c.j. gommes, chim0698

Figure IV–20: Examples of microstruc- tures formed by freeze-casting, i.e. by the repulsion of suspended material by a solidifying solvent. The pores in those ﬁgures were initially occupied by the solvent crystals, which were subsequently freeze-dried (sublimated). Picture taken from S. Deville, Freeze- Casting of Porous Ceramics: A Review of Current Achievements and Issues, Advanced Engineering Materials, 10 (2008) 155.

Further reading

• The comprehensive reference for many topics covered in this chapter is Jacob N. Israelachvili Intermolecular and Surface Forces, 2011, of which you can ﬁnd an online version here. To have full access, however, you will have to connect from your university account.

• The mechanism by which the gecko climbs on vertical surfaces is fraught with beautiful physics, some related to wetting phenomena and van der Waals forces, some to the mechanics of the setae which are responsible for the close contact between the animal’s feet and the surface. One of the first papers on the subject is by K. Autumn et al. Evidence for van der Waals adhesion in gecko setae Proc Nat Acad. Sci. USA 99 (2002) 12252. A more comprehensive You will find many things to read on study was published by the same authors under the title Properties, the Internet about geckos. I love the title of this blog post How Geckos Stick Principles, and Parameters of the Gecko Adhesive System in Biological on der Waals! Adhesives (Smith and Callow Ed.), 2006, Springer, of which you can find a preprint here. • Repulsive van der Waals forces play an important role in the dam- aging of porous rocks by the crystallisation of salts in their pores. You can have a look at this lecture by George Scherer on that topic, which he gave on the occasion of his receiving the Onsager medal in 2011.

Exercises

1. Use dimensional analysis to ﬁnd the D-dependence of the van der Waals interaction energy between two semi-inﬁnite bodies.

2. Compare the van der Waals forces between two spheres with the capillary force calculated on Page III–9.

3. Deﬁne a dimensionless number to characterise the relative importance of van der Waals forces compared to gravity. V–0 c.j. gommes, chim0698

4. According to Eq. IV–33, for very large values of the Hamaker constant AL, the effective attraction of the A-like media is larger than in vacuum. How is this possible physically?

5. Carbon nanotubes synthesised by the chemical vapour decomposition of gaseous hydrocarbons often come in the form of bundles comprising several thousand nanotubes arranged in a hexagonal packing. Assuming that the tubes are about 10 nm in diameter and a few microns in length, calculate the energy needed to disperse the nanotubes (i.e. to destroy the bundle). Gibbs-Thomson Effects

Deﬁnition - If to any homogeneous mass we suppose an inﬁnitesimal quantity of any substance to be added, the mass remaining homogeneous and its entropy and volume remaining unchanged, the increase of the energy of the mass divided by the quantity of the substance added is the potential for that substance in the mass considered.

J. W. Gibbs On the Equilibrium of Heterogeneous Substances, Transactions of the Connecticut Academy of Arts and Sciences 111 (1878) 108 .

This chapter is an incomplete draft, which I included here for your convenience. Please rely on your own notes to study this chapter. Figure V–1: Josiah Willard Gibbs (1839- 1903) . Picture from Wikipedia.

In technical terms, the subject of the present chapter can be put as that of the thermodynamical consequences of dispersion. In plain english, this is equivalent to determining how the size of an object influences its physicochemical properties such as its vapour pressure, its solubility, its melting point, etc. It so happens that the size-dependence is identical for phenomena apparently as different as vaporisation, dissolution, and melting, among others. That particular dependence, independently of the specific phenomenon it is applied to, is known as the Gibbs-Thomson relation. A spectacular example of size-dependent property is provided by the melting of nanometre-sized crystallites, which occurs at a temperature significantly lower than the melting temperature of the bulk material. This effect is illustrated in Fig. V–2 in the case of gold nanoparticles, which was already touched on in the epigraph of the introduction section. For very large particles the melting temperature approaches that of the bulk (Tb 1330 °C), and it decreases dramati- m ' In this chapter, the word bulk is used as an opposite to dispersed, and it is almost synonymous to macroscopic. What this means will soon become clear to you. See in particular how it is used in Fig. V–2.

V–2 c.j. gommes, chim0698

cally for very small particles. For particle of 4 nm in diameter (similar to that we considered in Fig. IV–9) the melting temperature is as low a 1000 K.

Figure V–2: The melting temperature of gold nanoparticles measured as a function of their diameter. The ﬁgure is taken from Phys. Rev. A 13 (1976) 2287.

The most obvious characteristic of nano-materials is their large surface-to-volume ratios, which scale like 1/R (see the Introductory Chapter). The contribution interfaces to the thermodynamic properties, compared to the volume, is therefore also expected to scale in that way. Can this be the origin of the trends in Fig. V–2? It could be, if only because the divergent behaviour at small sizes is remi- niscent of 1/R. In order to analyse this in detail, i.e. to eventually have a quantitative relation describing Fig. V–2, we will apply Polya’s method, namely: (i) Understand the physics; (ii) Devise a plan; (iii) Carry out the plan; (iv) Look back.

To try solving a new problem, it is always convenient to start solving a related but simpler problem. What other situation do you know, in which the melting temperature changes? The ﬁrst example to pop in your mind might be the effect of salt on ice, but this is a complicated situation involving two substances. In Fig. V–2 we just have pure gold. With a pure substance, the simplest example that I have of a changing melting temperature is the sheer effect of pressure. In general, the melting temperature of solids increases when the pressure is increased. The case of gold is shown in the phase diagram of Fig. V–3. You most probably studied this in an introductory thermodynamics course: the mathematical relation relevant here is called the Clausius-Clapeyron equation, but do you remember what it meant physically?

Here is a simple way to look at it. Melting consists in the pass- Figure V–3: Phase diagram of gold, from D.A. Young, Phase diagrams of ing of the substance from a crystalline state with low energy uC and the elements, technical report from the low entropy sC to a liquid state with higher energy uL and larger US Department of Energy (1975). gibbs-thomson effects V–3

entropy s . The energy u u because the atoms in the crystal sit L C ≤ L in fixed positions that minimise the energy, while in the liquid they constantly move around each other. On the other hand, because of that motion there are more degrees of freedom in the liquid state to store thermal energy than in the crystal state, i.e. s s . As a first L ≥ C approximation, we may assume that the energies and entropies uC/L and sC/L are independent of the temperature. With that particular model in mind, the transition from C L is possible only if the → temperature T is sufficiently large, so that the heat taken up from the environment δQ = T(s s ) is enough to account for the change in L − C energy u u . In other words, the melting temperature is given by L − C u u T0 = L − C (V–1) m s s L − C To be fully accurate, we would have to account for the fact that in general the volume is larger in the liquid state vL than in the crystalline state vC. In this case, there is an addition energy cost to melting, that is the mechanical work exerted by the melting substance on its environment, namely δW = P(vL vs). In that case, the heat taken − The lower-case letters in the notations from the environment is needed for uL uC as well as for δW. This − uC/L, sC/L and vC/L refer to molar eventually leads to quantities. Implicitly, we are making here energy balances on one mole of u u v v T = L − C + P L − C (V–2) substance. m s s s s L − C L − C This simple physical picture is all is needed to understand the data in Fig. V–3. In other word, the melting temperature increases with the pressure because the larger the pressure, the more energy is required by the melting-induced swelling of the particle. With that in mind, the data in Fig. V–2 suggests that the energy cost for melting a nanoparticle is lower than u u . This would explain why the melting temperature L − C is lower, because less heat has to be taken from the environment. Moreover, the smaller the particles are the lower that energy is. This clearly hints at a surface energy contributions. Before we proceed to explain the thermodynamic origin of Fig. V–2, it is useful to put this analysis in a more abstract form. In terms of chemical potential, the condition for the equilibrium of the crystal and liquid state is

µC = µL (V–3) It might not strike you, but Eq V–3 is an equation describing the shape of the transition region between the crystal and the liquid in

the phase diagram, because the chemical potentials µC/P are func- tions of P and T. Eq. V–3 is not true for all values of P and T. It is only true at equilibrium, i.e. when T = Tm(P) where Tm(P) is the Eq. V–3 is something you knew: at V–4 c.j. gommes, chim0698 an equilibrium phase transition, the chemical potentials are the same in the two phases. However, do you remember where that comes from? melting temperature at pressure P. In order to actually use that equation to practically determine how Tm depends on P, however, one has to know exactly how µC/P depends on P and T.

The equation that tells you how the chemical potential depends on P and T is a cornerstone of thermodynamics. It is called the Gibbs-Duhem equation. For a pure substance, it states

dµ = vdP sdT (V–4) − where v and s are the molar volume and entropy, respectively. Imag- ine we knew the chemical potential, say, of the liquid state L at a given pressure P0 and temperature T0. Using a Taylor development, one could estimate the value of µL for a value of P and T not very far from P0 and T0 using the following Taylor development ∂µL ∂µL µL µL (P0, T0 ) + [P P0 ] + [T T0 ] ' ∂P T − ∂T P − 2 2 1 ∂ µL 2 1 ∂ µL 2 + 2 [P P0 ] + 2 [T T0 ] 2 ∂P T − 2 ∂T P − ∂2 µ + L [T T ][P P ] + (V–5) ∂T∂P − 0 − 0 ··· Using that general development and limiting yourself to the ﬁrst- order accuracy, the relation can be written as follows in the particular case of the crystal and liquids C and L

µ (P, T) µ (P , T ) + v [P P ] s [T T ] (V–6) C/L ' C/L 0 0 C/L − 0 − C/L − 0 where we have used the Gibbs-Duhem relation Eq. V–4 to express (∂µ /∂P) = v and (∂µ /∂T) = s . Putting that C/L T C/L C/L P − C/L particular dependence in Eq. V–3 leads to

µ (P , T ) + v [P P ] s [T T ] C 0 0 C − 0 − C − 0 = µ (P , T ) + v [P P ] s [T T ] (V–7) L 0 0 L − 0 − L − 0

Of course, you can chose in here any value of P0 and T0 that would make your life easier. A convenient choice here is to take P0 = 0 and 0 for T0 the melting temperature at that particular pressure Tm. With that particular choice, one has µC (P0, T0 ) = µL (P0, T0 ) so that Eq. V–7 becomes signiﬁcantly simpler. One ﬁnds v v T = T0 + P L − C (V–8) m m s s L − C which is identical to Eq. V–2, as it should.

We are now facing a typical situation: how do we generalise this type analysis to determine the melting point of nanoparticles?

Figure V–4: Pierre Duhem (1861-1916). Picture from Wikipedia. gibbs-thomson effects V–5

Does Eq. V–3 still apply? There is no way you can answer that question unless you know exactly where that equation comes from. Think about this before proceeding with the studying/reading of this chapter: where does it come from? As a matter of fact, the equality of the chemical potentials of two phases at equilibrium is not a ﬁrst principle of thermodynamics. The general law of thermodynamics is that the free energy of a system is minimal, subject to whatever external constraint is imposed on it. The particular form of the free energy that is relevant depends on the variables that are used to characterise the system. For example, we are currently interested in analysing the melting/freezing of a substance at a given temperature T and pressure P. With these variables, its is the Gibbs free energy G that is relevant. The general differential form of Gibbs equation in the case of a system comprising a crystalline phase C and a liquid phase L can be written as follows

Figure V–5: William Thomson, ﬁrst dG = V dP S dT + µ dN + γ dA C − C C C C C Baron Kelvin (1824-1907). Picture from + V dP S dT + µ dN + γ dA (V–9) Wikipedia. L − L L L L L

where P is the pressure, T is the temperature, and VC/L, SC/L, µC/L and NC/L are the volume, entropy, chemical potential and number of molecules in phases C and L, respectively. Compared to the Gibbs equation that you encountered in introductory thermodynamics, there are here two new contributions in the form of γdA accounting for the changes in surface energies. Note that the other terms in the Gibbs equation, such as VdP and SdT are proportional to volumes, so that the relative contribution of γdA compared to the latter scale like area-over-volume ratios. In other words, the γdA contributions are important only when the systems are small; when they reach nanometer sizes they are crucial. Figure V–6: Joseph John (JJ) Thomson The particular process that we are interested in now is the melt- (1856-1940). Picture from Wikipedia. ing/freezing of nanoparticles, which takes place at constant temperature and pressure. The only variable left are NC and NL and AC and AL. Because any nanoparticle in the system is either frozen or molten, the total number of atoms is a constant: dN = dN . C − L Moreover, if we assume that the particles are spherical, their area and the number of atoms they contain are not independent quantities. They are related by

dA 2v dA = vdN = dN (V–10) dV R where v is the molecular volume, and the second equality results from assuming that the particles are spherical. With that in mind, the equilibrium state of the system comprise molten and frozen nanopar- V–6 c.j. gommes, chim0698

ticles, expressed as dG = 0, becomes simply

2 2 µ (P, T ) + γ v = µ (P, T ) + γ v (V–11) L m L R L C m C R C This equation generalises Eq. V–3 to dispersed system, i.e. in which surface energies are not negligibly small. In order to convert Eq. V–11 into a more usable equation relating the melting temperature to the particle size, all we need to do is express the chemical potentials explicitly in terms of the temperature. This can be done using Eq. V–6 with suitable values of P0 and T0. A convenient choice is to take P0 as the ambient pressure and b T0 = Tm as the melting temperature of the bulk material at that par- b b ticular pressure, i.e. such that µC (P0, Tm ) = µL (P0, Tm ). With that particular choice, Eq. V–11 takes the following simple form

T 2 γ v γ v m = C C L L b 1 − (V–12) The choice we made of the reference T − R ∆hm m temperature and pressure T0 and P0 has no consequence on Eq. V–12. b where we have used the notation ∆hm = Tm (sL sC ) for the en- It simply enabled us to express the − thermodynamic quantities in terms thalpy of melting of the substance. Eq. V–12 is the Gibbs-Thomson b of the bulk melting temperature Tm. equation. It relates the melting temperature Tm of particles to their As an exercice, repeat the analysis size. Note that the dependence is in 1/R, and that Tm converges to without making any explicit choice P T the bulk melting temperature Tb for R ∞. for 0 and 0. You will then obtain an m → expression similar to Eq. V–12 with a With respect to the Polya’s steps, we now reached the point where cryptic combination of µC/L(T0, P0). If we are supposed to look back, and see if we achieved what we were you express that for large particles, the melting temperature should coincide aiming at. The purpose of our analysis, which eventually lead to Eq. b with Tm, it will all simplify into Eq. V–12, was to describe the data in Fig. V–2. Does it make sense? In V–12. many respects it does. However, we still have to be sure we understand the sign of T Tb . As a ﬁrst approximation, consider the m − m simple case where v v . In that case, the numerator in the Gibbs- C ' L Thomson equation is proportional to γ γ , which determines C − L whether the small particles melt at a higher or lower temperature. To understand the sign of γ γ , I encourage you to go back to C − L Chapter 1 and be sure you are comfortable with Stefan’s interpre- tation of surface energy (see Eq.I– 12) before you continue reading. For the sake of a back-of-the-envelope calculation, consider again a 6-coordinated cubic arrangement of atoms, both in the crystalline and in the liquid states. Because we assumed v v , we may also C ' L assume that the spacing between atoms d is the same in both phases. The order of magnitude of d is the Angstroem. In the crystalline (the liquid) phases, each atom has a binding energy eC (eL) with each of its six neighbours, except at the surface where one bond is missing. The surface energies of the crystalline and liquid phases are therefore something like e γ C/L (V–13) C/L ' d2 gibbs-thomson effects V–7

where d2 is the area occupied by an atom at the surface. On the other hand, you know that it takes some energy to melt a solid. In the context of the simple cubic model, the enthalpy of melting ∆hm is related to the binding energies via

∆h = 6 (e e ) (V–14) m C − L

The direct consequence of this analysis is that γC can only be larger than γL. Accordingly, Eq. V–12 does indeed predict that small particles melt at a lower temperature. We can push the back-of-the-envelope calculation further and estimate how small the particle ought to be for their melting point to be signiﬁcantly impacted. For that purpose, we may use Eqs. V–13 and V–14 to rewrite the Gibbs-Thomson relation in the form

Tm 1 d b 1 (V–15) Tm ' − 3 R

where we have used v = v d3. Of course this equation is a C L ' crude approximation, if only because of the ridiculous cubic model. However, its physical insight is interesting: the melting point depression is signiﬁcant when the size of the atoms is not negligibly small compared to that of the particles. There is an alternative way to understand the thermodynamics of Gibbs-Thomson phenomena based on Laplace’s pressure inside the nanoparticles. As we saw in Chapter 1, Laplace’s pressure applies to liquids as well as to solids (see e.g. Fig.I– 12). For a nanoparticle with diameter 4 nm and a surface energy of 1 J/m2 the Laplace pressure is as large as 104 atmospheres. Instead of taking explicitly the interface energy into account in the Gibbs relation (Eq. V–9), you may alter- natively take the Laplace pressure into account when calculating the chemical potentials via Eq. V–6. This results in additional terms of the type 2γ v C/L (V–16) C/L R in the chemical potentials of the crystalline and liquid phases. If you do that, of course you should not put the γdA terms in the Gibbs equation, so that the equilibrium condition is Eq. V–3. Basically, in order to account for surface energies you have two options:

• Either, you write Gibbs equation explicitly with the γdA terms as in Eq. V–9;

• Or you use the usual thermodynamic relations, e.g. Eq. V–3, but you account for the change in chemical potentials resulting from Laplace’s pressure. V–8 c.j. gommes, chim0698

Sometimes the second option is more intuitive, but it is also less general. For example, Exercice XX at the end of the chapter is devoted to the calculation of melting point depression in lamellar crystals, which have no curvature. In that case the Laplace pressure is irrelevant but the experimental melting point is nevertheless found to depend on the thickness of the lamellae. The rest of the chapter is a generalisation of Gibbs-Thomson equations to

• other phase equilibria than melting-freezing: solubility and liquid- vapor (Kelvin equation);

• phases conﬁned inside porous materials (role of contact angle, capillary condensation and evaporation);

I had no time to type this down; you will have to rely on your own notes to study that. Next years students will be luckier than you, perhaps. VI–0 c.j. gommes, chim0698

Further reading

Exercices

1. Use the chemical potential formalism to explain the phase diagram in Fig. V–3. In other words, start from Eq. V–9 with no γdA term and derive Eq. V–2.

2. Similarly, use the same type of basic energy balance approach as in Eq. V–2 to derive the Gibbs-Thomson relation Eq. V–12.

3. In semi-crystalline polymers, crystalline regions have the shape of lamellae with a thickness of a few nanometers surrounded by amorphous (molten) regions. In that case, the observation is that the melting temperature depends on the thickness of the lamellae. Derive the expression for the melting point depression, assuming that the crystals grow or melt laterally, while keeping a constant thickness.

4. Use the Gibbs-Duhem equation to ﬁnd a Taylor development of µ Figure V–7: Example of lamellar struc- similar to Eq. V–6 but valid to the second order in P and T. tures in semi-crystalline polymers: in that case the melting temperature depends on the thickness of the lamellae. Picture from Macromolecules 42 (2009) 2135. Thermal ﬂuctuations

If, in some cataclysm, all of scientiﬁc knowledge were to be destroyed, and only one sentence passed on to the next generation of creatures, what statement would contain the most information in the fewest words? I believe it is the atomic hypothesis (or the atomic fact, or whatever you wish to call it) that all things are made of atoms: little particles that move around in perpetual motion, attracting each other when they are a little distance apart, but repelling on being squeezed into one another. In that one sentence, you will see, there is an enormous amount of information about the world, if just a little imagination and thinking are applied.

Richard Feynman, Lectures on Physics

Figure VI–1: Ludwig Boltzmann (1844- 1906). Picture from Wikipedia. This chapter is an incomplete draft, which I included here for your convenience. Please rely on your own notes to study this chapter.

Very few scientific items, let alone equations, have the privilege of becoming cultural symbols. Einstein’s E = mc2 is definitely one such item. The second equation on that list is probably Boltzmann’s definition of entropy S = k ln[W], which can be read on his tombstone in Vienna. To the present day, Boltzmann’s definition of entropy is the best link we have to relate the microscopic and the macroscopic worlds, i.e. the motion of atoms and molecules to the property of matter at our scale. Richard Feynman necessarily had that equation somewhere in his mind when he wrote that sentence in the epigraph. The goal of the present chapter is to provide you with a crash course on a few topics that are usually part of a broader course in statistical thermodynamics. In other words, this chapter is about the type of imagination a thinking that Feynman refers to. VI–2 c.j. gommes, chim0698

Microstates and the ergodic hypothesis

The distinction between microstate and macrostate is central to statistical mechanics. A macrostate is typically what is studied in thermodynamics. In this context, the state of a system is defined by its macroscopic properties such as volume its V, its total number of molecules N, the pressure, the chemical potential, etc. By contrast, a microstate is a complete microscopic description of the system, at the level of the molecules or atoms, or of whatever building block is necessary for the type of system considered. For example, describing the microstate of a monoatomic gas consists in specifying the 3 coordinates and the 3 velocities of each atom. Depending on the degree of accuracy of the description, the state of the electrons in each atom might also have to be specified, etc. When molecules collide with each other, or simply interact through van der Waals-like forces, they constantly exchange energy. During any evolution of a system, its total energy is a constant. What fluctuates, however, is the way in which the energy is split among all the degrees of freedom of the system. When two molecules collide, the translation kinetic energy of the two molecules may decrease, if one molecules starts spinning thereby increasing its rotational energy. Similarly, some of the energy may be transferred to the electrons of a molecule, which may later result in a photon emission, etc. Through any of these processes, the energy is constantly transformed from one form to another. Keeping track of the microstate of a system comprising of the order of 1023 molecules, i.e. checking whether any particular molecule is spinning rather than another one is vibrating, is a daunting task beyond any possible analysis. A breakthrough we owe to the genius of Boltzmann consists in assuming that what we call thermal equilibrium is a situation where the energy of a system is shared evenly over all its possible degrees of freedom. Put differently: at thermodynamic equilibrium, all the microstates with the same energy E are equiprobable. This is sometimes referred to as the ergodic hypothesis. You can find on the Internet many Java A central characteristic of a macrostate is the number of mi- applets dealing with simple molecular dynamic simulations. See e.g. this one, crostates compatible with it. Given the ergodic hypothesis, a sys- and use toi check that each degree of tem is more likely to be found in macrostate that corresponds to freedom. One Helium atom (atomic weight 4) ∼ numerous microstates than the other way around. This is exactly the and one Neon atom (atomic weight meaning of Boltzmann’s equation; on Bolztmann’s grave, it is writ- 20) are trapped inside a small box ∼ and they collide repeatedly with each ten as S = kB ln(W), where the W stands for Wahrscheinlichkeit, i.e. other. Assuming ergodicity, can you probability. determine the average kinetic energy of A more common notation for the number of microstates having each atom? energy E is Ω(E). Boltzmann equation for the entropy is

S = kB ln [Ω] (VI–1) thermal fluctuations VI–3

which can also be written as S = k ln [p] (VI–2) − B where p = 1/Ω is the probability of any microstate. V,N Es Sub-system

Boltzmann’s Distribution heat

Imagine a small system with a given volume and a given number of Heat Bath T, Eb molecules, which is in contact with a much larger system with which it can exchange energy under the form of heat. For example: a tiny Figure VI–2: A subsystem in con- dust particle floating in a large room filled with air. The particle is tact with a heat bath, with which it constantly hit by air molecules, which makes its every atom jiggle exchanges energy. around some equilibrium position. Each time an air molecule hits the dust particle, it may loose some of its energy in favor of the particle. The opposite may also happen, an atom of the particle may hit an air molecules in its vicinity, thereby loosing some of its energy. The dust particle is therefore constantly fluctuating between different microstates. We want to characterize these fluctuations. In the following, we refer to the dust particle as the subsystem s and the air filling the room as the heat bath b. The system comprising the subsystem and the heat bath is the composite system. Although the energies Es and Eb do fluctuate in a complex way, the energy of the composite system E = Es + Eb is a constant. Whenever the subsystem has energy E , the heat bath has energy E E . i − i The clever trick for analyzing the fluctuations of s consists in noting that the ergodic hypothesis does apply to the composite system. Accordingly, the probability that the subsystem be in state s is

1 Ωb(E Es) ps = × − (VI–3) ∑ Ω (E E ) i b − i where the denominator accounts for the fact that the probabilities should add up to 1. One interesting observation that can be done at this stage is that ps does only depend on the energy Es. This is naturally a consequence of the ergodicity of the composite system. To express the dependence of ps on Es in a more practical form, one can write

Sb(E Es) ln[ps] = ln[Ωb(E Es)] C = − C (VI–4) − − kB − where the constant C is simply the logarithm of normalizing constant, and the second equality results from Boltzmann’s equation. The thermodynamic relation 1 = T Remember that the subsystem is much smaller than the heat bath, ∂S can be considered as the ∂E V,N so that E E. This enables to write a Taylor development, limited deﬁnition of the temperature. s to the ﬁrst term as

Sb(E) 1 dSb ln[ps] Es C (VI–5) ' kBT − kB dE − VI–4 c.j. gommes, chim0698

Remembering the thermodynamic deﬁnition of the temperature, the result takes the ﬁnal form

1 Es ps = exp[ ] (VI–6) Z − kBT where we have incorporated all the constants in the single constant Z. In many statistical mechanics texts you The seemingly irrelevant normalization constant Z happens to be a will ﬁnd the notation β = 1/(kBT). central quantity in statistical physics. It is called the partition function

E Z = ∑ exp[ i ] (VI–7) i − kBT Knowing the dependence of the partition function on the macroscopic variables V, N, T, is equivalent to knowing all the thermodynamical properties of the system, and even slightly more.

Relation to thermodynamics

The average energy E can be calculated as ∑ E p , which can be h i i i i expressed in terms on the derivatives of Z as

1 Ei ∂ ln[Z] E = ∑ Ei exp[ ] = (VI–8) h i Z i − kBT − ∂β The average entropy is calculated in the same way, starting from Eq. ??, i.e. as S = kB ∑ pi ln[pi] (VI–9) h i − i Note that this deﬁnition coincides with the microcanonical entropy (Eq. ??) in the particular case where pi = 1/Ω for all states. Using Boltzmann’s distribution (Eq. VI–6) for pi, the deﬁnition of the average entropy can be put under the form

E S = h i + k ln[Z] (VI–10) h i T B This expression enables us to identify Helmholtz’s free energy F = E T S with h i − h i F = k T ln[Z] (VI–11) − B Therefore, knowing the partition function is equivalent to knowing a thermodynamic potential of the system. All the relevant thermodynamic properties can then be calculated in the usual way, as

∂F ∂F ∂F S = P = µ = (VI–12) − ∂T V,N − ∂V T,N ∂N T,V The derivatives of the free energy result from the differentiation of F = E TS, − when using the general relation dE = TdS pdV + µdN. − thermal fluctuations VI–5

The statistical mechanical approach enables to go further than the classical thermodynamics. For instances, the energy ﬂuctuations of the system can be estimated as

D E D E2 σ2 = E2 E (VI–13) E − Quantifying the amplitude of the thermal ﬂuctuations as

2 D 2E 1 2 Ei 1 ∂ Z E = ∑ Ei exp[ ] = 2 (VI–14) Z i − kBT Z ∂β

and using the fact that ∂ ln[Z]/∂β = E , one reaches the ﬁnal − h i result p 2 σE = kBT CV (VI–15)

where CV = ∂E/∂T is the heat capacity of the subsystem. Can you imaging an experimental An interesting observation in Eq. VI–15 is that ∆E increases only setup to put statistical ﬂuctuations in evidence? as N1/2

A statistical derivation of the ideal gas law

We consider here the behavior of a large collection of N molecules in a box of volume V. We know that this system should obey the ideal gas law P = NkBT/V. The purpose of this section is to illustrate the method of statistical mechanics on a well known physical system. Gaussian integrals are integrals of the The states of the system are deﬁned here over a continuous space, type Z +∞ r π both for the 3N positions and the 3N momentum coordinates. The exp[ ax2]dx = ∞ − a partition function is calculated by replacing the sum by an integral. − 1/2 Could you guess its a− dependence Z Z " N 2 !# from dimensional considerations? 1 1 3N 3N 1 pn Z = 3N d x d p exp ∑ k k (VI–16) N! h¯ − kBT n=1 2m

The ﬁrst factor accounts for the fact that interchanging any two particles does not change the state. The volume integral is simply equal to VN. The momentum integral can be calculated as

3N Z 1 p2 dp exp (VI–17) − kBT 2m which can be calculated by Gaussian integration. The partition function of the ideal gas is therefore

N 1 V 3N/2 Z = (2πmkBT) (VI–18) h¯ 3N N! from which the free energy is found to be

3 F = k T N ln[V] ln[N!] + N ln[2πmk T] 3N ln[h¯ ] (VI–19) − B − 2 B − VI–6 c.j. gommes, chim0698

To estimate the chemical potential, All the properties of the ideal gas can then be obtained by simple we have used Stirling’s approxima- differentiation, following Eq. VI–12. In particular, one ﬁnds immedi- tion ln[N!] N ln[N] N, which ' − ately holds for large values of N. Nk T P = B (VI–20) V as one should. The chemical potential can be put under the form

µ = k T ln[P/ ] (VI–21) B P0 where (T) is a function that depends only on the temperature P0

1 3/2 5/2 0 = (2πm) (kBT) (VI–22) P h¯ 3 VII–0 c.j. gommes, chim0698

Further reading

1. There are many good textbooks on statistical mechanics. An excellent starting point is the relevant chapters in the Feynman lectures on Physics;

2. If you like online courses, a few good lectures by Leonard Susskind are available on Youtube. These lectures are more general, and also more theoretically-oriented, than what I saw in this chapter.

Exercices

1. About Macrostrate and Microstates: estimate the memory space required to store on a computer the microstate of 1 mg of oxygen (in double precision variables). Compare that ﬁgure to the total computer memory space available on Earth.

2. Determine the vertical density profile of an isothermal ideal gas in a gravitational field. Do you find some similarity with Boltzmann’s distribution ? Metastability and nucleation

A lot of people like snow. I ﬁnd it to be an unnecessary freezing of water.

Carl Reiner

This chapter is an incomplete draft, which I included here for your convenience. Please rely on your own notes to study this chapter.

When studying macroscopic equilibria the speciﬁcs of how a phase is created are generally irrelevant, as far as its equilibrium properties are concerned. This is because the state of a system is uniquely deﬁned by its thermodynamic variables. This is not necessarily the case for nanometer-sized systems, for which the size and shape of the phases matter, in addition to the macroscopic variables. This is incidentally the reason why nanometer-sized systems often exhibit hysteresis: several states are possible for a given set of thermodynamic variables. We shall come back to this later. For the moment, consider a supersaturated system free to evolve isothermally. It can a priori be any type of system: a gas compressed isothermally above its saturation pressure, a vapour cooled below its dew point, a melt quenched into the unstable region of it phase diagram, a salt solution exceeding its solubility product, whatever. What all these systems have in common is that they are unstable: a phase transition is possible according to macroscopic thermodynamics. The reason why it does not necessarily take place is that the formation of germs of the new phase would necessarily come with an interface having a thermodynamic cost (γ per unit area of the interface). Because the new germs have to be small initially, the relative importance of surface tension is initially very large. Although a phase transition may be possible macroscopically, the thermodynamic cost of creating the interface may be prohibitive, so that the transition VII–2 c.j. gommes, chim0698

does not necessarily take place. An example of such a metastable system is freezing rain: it consists of undercooled rain drops that remain liquid while in the air, and that freeze upon touching a preexisting icy surface.

Supersaturation

Let us see how this can be analysed thermodynamically. We need a thermodynamic measure of how far we are from equilibrium. The natural measure would be ∆µ, the difference in chemical potential between the metastable and the equilibrium phases. Note that, by deﬁnition ∆µ 0. It is customary to deﬁne the supersaturation β as ≥ ∆µ ln(β) = (VII–1) kBT We now consider a few particular cases, and explicit the values of ∆ and β in these cases.

Vapour compressed above its saturating pressure The chemical potential of a vapour at a given temperature is

P µ = µ0(T) + kBT ln (VII–2) P0(T) where P0(T) is the saturating pressure and µ0(T) is the chemical potential of the liquid 10. In this case ∆µ = µ µ (T), and the super- 10 Note that this relation satisﬁes µ = µ − 0 0 saturation is simply for P = P0, as it should. P β = (VII–3) P0

Undercooled liquid The chemical potential of the liquid and solid phases are equal at the meting temperature Tm. To calculate their values at other temperature, we can use the general relation (∂µ/∂T)P = s and write − µ = µ (T ) s (T )(T T ) l l m − l m − m µ = µ (T ) s (T )(T T ) (VII–4) s s m − s m − m The subscripts stand for solid and liquid. These expressions are expected to hold for temperature not too far from Tm. Expressing that µl(Tm) = µs(Tm), we ﬁnd

µ µ = (T T )(s s ) (VII–5) l − s − m s − l Taking account of the general relation

∆hm sl sl = (VII–6) − Tm metastability and nucleation VII–3

where ∆hm is the enthalpy of melting, one has the ﬁnal relation

∆h T ln(β) = m 1 (VII–7) kBT − Tm

Supersaturated salt solution Consider a salt solution with the following simple equilibrium

+ A + B− AB(s) (VII–8) ↔ where AB is a solid. The difference in chemical potential is

∆µ = µ µ µ (VII–9) AB − A − B where µA and µB depend on the concentrations CA and CB through

0 µA = µA + kBT ln(CA) 0 µB = µB + kBT ln(CB) (VII–10) and µAB is a constant. The equilibrium corresponds to the situation where ∆µ = 0, which can be put in the familiar form

CACB = KS (VII–11) where the solubility product KS is given by

k T ln(K ) = µ µ0 µ0 (VII–12) B S AB − A − B The supersaturation can therefore be written in this case C C β = A B (VII–13) KS which can be easily generalised to more complex equilibria than Eq. VII–8.

Nucleation barrier and critical radius

We consider now a supersaturated system. It can be any type of system, but it might be convenient at this stage to think of it as compressed vapour. Because ∆µ 0 the vapour should spontaneously condense ≥ into liquid, according to macroscopic thermodynamics. If one looks carefully into this, however, one realises that in order to create a liquid region one also has to create a liquid/vapour interface with a given thermodynamic cost γ per unit area. In other words, the extra free energy associated with the creating of a liquid droplet in a vapour at pressure P and temperature P0 is

∆G = ∆µN + γA (VII–14) − VII–4 c.j. gommes, chim0698

where N is the number of molecules in the droplet, which is also the number of molecules taken out of the vapour, and A is the surface area of the droplet. Particularising this to a spherical drop, with the vapour chemical potential given by Eq. VII–2, one has

4πR3 ∆G = ∆µ + γ4πR2 (VII–15) − 3vm

where R is the radius of the droplet, and vm is the molecular volume. The first term in Eq. VII–17 is the one we had in mind when thinking of macroscopic thermodynamics: according to this term the free energy is decreased when R is made as large as possible, i.e. when as much vapour as possible is condensed (as expected for P P ). ≥ 0 However, the second term balances this effect by assigning a positive thermodynamic cost to the creating of a droplet. What you should expect to happen therefore depends on more than merely the com- parison of P and P0. When you have a situation such as described by Eq. VII–17, it is always useful to put it first in a dimensionless form, in order to understand the role of each of the dimensional parameters vm, γ, and kBT (which we take as an order of magnitude for ∆µ = KBT ln(β)). For that purpose, on may first notice that the first term is of the order 3 2 kBTR /vc; it is balanced by the second term, which is of order γR . This suggests using v γ λ = m (VII–16) kBT as a unit for R and, say 4πγλ2 as the unit of ∆G. Using these units, the extra free energy associated with the creation of a droplet of radius R can be written in the simpler way

∆G R 2 1 R 3 = ln (β) (VII–17) 10 γ4πλ2 λ − 3 λ The situations sketched in Fig. VII–1. For β < 1 the volume contribu- ) tion is positive, so that the minimum of ∆G is reached for R = 0, i.e. 2 5 no condensation is expected whatever. By contrast, in the case of su- πγλ (4

persaturated vapour (β > 1), the state with R = 0 is only metastable. G/

∆ 0 A state with a lower fee energy can be reached for large values of R, but this requires passing a ﬁnite free-energy barrier.

The critical radius of the droplet that has to be formed for conden- −5 d G dR = 0 1 2 3 4 5 sation to proceed is obtained by expression that ∆ / 0, which R/λ leads to 2λ R = (VII–18) Figure VII–1: Free energy correspond- ∗ ln(β) ing to the creation of a spherical object with radius R for various supersatu- and the associated free energy barrier is rations: β = 1/2 (red), β = 1 (green), β = 2 (blue), and β = 4 (black). The 4 4πγλ2 1 ∆G = = γ4πR2 (VII–19) length unit is λ = γvc/(kBT). ∗ 3 [ln(β)]2 3 ∗ metastability and nucleation VII–5

It is easy to remember that the free energy barrier is simply one third of the critical nucleus surface energy. To make these results more "real", let us consider the practical case of undercooled water. In this case, the critical radius can be written as v γ 1 R = 2 m (VII–20) ∗ ∆h 1 T/T m − m In the case of water ice with v 1.09 10 6 m3/g, ∆h = 333 J/g, m ' − m and γ 30 10 3 J/m2, one has ' − v γ R (0) = 2 m 2Å (VII–21) ∗ ∆hm ' This is the critical radius extrapolated to T 0 K. Remember, → however, that the assumptions underlying Eq. (VII–7) hold only for T T . Anticipating on the next section, the relevant quantity for ' m nucleation is ∆G /(kBT). In the case of undercooled water, it can be ∗ written as Reaching 50 ◦C of supercooling is not rare. A famous case is sodium acetate ∆G 1 γ4πR (0)2 1 used in hand warmers. The melting ∗ ∗ temperature of sodium acetate is 58 ◦C, = 2 (VII–22) k T 3 k Tm T/T (1 T/T ) B B m − m but it can remain in a metastable liquid state down to about 0◦C. with γ4πR (0)2 ∗ 4 (VII–23) kBTm ' The critical nucleus size and the corresponding free energy barrier are plotted in Fig. VII–2. A typical value for the supercooling of water is 220 K. This is e.g. the temperature of freezing rain. This ' corresponds to T/T 0.8; The critical nucleus is about 2 nm across. m '

2 4 10 10 Figure VII–2: Size of the critical ice radius and the corresponding free energy barrier in the case of supercooled water 103 1 (Tm = 273 K). 10 ) T ) B

k 2 ( nm 10 / ( ∗ ∗ R G 0 10 ∆ 101

10−1 100 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 T/Tm T/Tm

Let us consider now the case of a water vapour at T = 373 K compressed beyond P0 = 1 atm. The molecular volume of liquid water is v 2.99 10 29 m3, the surface tension is γ 75 10 3 N/m, m ' − ' − which leads to λ 4.3 Å. The free energy barrier is therefore ' ∆G 1 4π(2λ)2 1 ∗ = 2 (VII–24) kBT 3 kBT [ln(P/P0)] VII–6 c.j. gommes, chim0698

with 4π(2λ)2 139 (VII–25) kBT ' The critical nucleus size and the corresponding free energy barrier are plotted in Fig. VII–3.

2 4 10 10 Figure VII–3: Size of the critical ice radius and the corresponding free energy barrier in the case of water 103 1 vapour at 373 K compressed above 10 ) T ) P = 1 atm. B 0

k 2 ( nm 10 / ( ∗ ∗ R G 0 10 ∆ 101

10−1 100 0 0.5 1 1.5 2 0 0.5 1 1.5 2 ln(P/P0) ln(P/P0)

The free energy barrier is directly related to the nucleation rate. Imagine the case of nucleation from a supersaturated solution of a simple molecule, say sugar. Sugar molecules in the solution are constantly assembling into fleeting clusters. If these clusters are smaller than the critical size R , they can only dissolve back into the solution. ∗ If one cluster forms larger than the critical size, then it is unstable towards growth and it eventually leads to the crystallisation of a macroscopic quantity of sugar. The probability of seeing a fluctuation is directly related to the corresponding change in the free energy ∆G. In fact, it is proportional to11 11 This is Einstein’s thermodynamic ∆G/k T approach to fluctuations. We shall come Prob( fluctuation) e− B (VII–26) ' back to it in details in a further chapter. The sign means that many factor have not been written. In partic- ' ular, it is clear that the probability has to depend on the total volume that is being considered, as well as of the observation time. Clearly, the probability of seeing a fluctuation of a given amplitude has to be two times larger is the volume of the solution is two times larger. Also, if you wait two times longer, you expect to see two times as many critical nuclei forming. What we are interested in is therefore the nucleation rate, i.e. the number of critical nuclei forming per unit volume of the solution and per unit time. The missing factor in Eq. 3 1 VII–26 has the dimensions of m− s− . 3 1 Physically, one expects the nucleation rate (in m− s− ) to depend on the mobility of the molecules in the solution. If the molecules were very slow, the statistics of the fluctuations would still be described by eq. VII–26 but the time it would take to see the clusters appear and disappear would be very long. A convenient quantity to use is therefore the viscosity of the solution η, the unit of which metastability and nucleation VII–7

3 is Pa.s = J m− s. The other dimensional quantities that may enter the expression are the molecular volume vm and the thermal energy kBT. The only combination of these quantities with the correct di- 2 mension is kBT/(vmη). One therefore reaches the conclusion that the homogeneous nucleation rate has to be of the form

2 kBT 16π γλ I 2 exp 2 (VII–27) ' vmη − 3 kBT[ln(β)] where the missing factor is now dimensionless and of the order of one.

Heterogeneous nucleation

The situation we have implicitly analysed in the the previous section is that of homogeneous nucleation. In case there is a solid surface in the system, the metastability can be signiﬁcantly reduced as we now show. Imagine we are condensing, say water vapour. If the solid surface is hydrophilic, clearly the thermodynamic cost of the free surface is reduced if the vapour condenses on the surface, rather than as a free droplet. In this case the shape of the nucleus is like in Fig. VII–4, and the term γA in Eq. VII–14 would have to be replaced by h FA = γAL + (γSL γS)ASL = γ (AL ASL cos(θ)) (VII–28) − − r where AL is the surface area of the liquid, ASL is the surface area of the wetted solid and the γ’s are the corresponding surface energies. R For the RHS, we have used the Young-Dupré relation. Let us do some geometrical calculation to express the volume and the relevant surface areas in terms of the radius of curvature R of the droplet12. The volume of the droplet is Figure VII–4: Drop on a surface, with π contact angle θ. Cf. the ﬁrst exercise we V = R3(1 cos(θ))2(2 + cos(θ)) (VII–29) 3 − did on surface tension. 12 The calculations are straightforward. and the areas are They are based on the formulae V = πh(3r2 + h2)/6 and A = π(r2 + h2) for A = 2πR2(1 cos(θ)) the volume and area of a spherical cap. L − A = πR2(1 cos(θ)2) (VII–30) SL − Using these expressions, the free energy of formation of a nucleus on a surface can be put as

4πR3 ∆G = ∆µ + γ4πR2 f (θ) (VII–31) − 3vm with 1 3 1 f (θ) = cos(θ) + cos3(θ) (VII–32) 2 − 4 4 VIII–0 c.j. gommes, chim0698

Eq. VII–31 is identical to Eq. VII–17 except for the factor f (θ). This means that Eq. VII–18 remains valid in the case of heterogeneous nucleation, but the radius has to be understood as a radius of curvature

2λ R = (VII–33) ∗ ln(β) On the other hand, the free energy barrier has to be multiplied by a factor f (θ), the dependence of which on θ is plotted in Fig. VII–5. As 1 you may have expected, f (θ) is close to 1 for θ close to 180 , because ◦ 0.8 in this case one has a spherical nucleus sitting on a surface, with a 0.6

vanishingly small contact area. By contrast, f (θ) is close to 0 for θ ) θ (

close to 0. In this case case, the change of surface energy γ γ is f SL − S 0.4 so negative that it almost balances the γL contribution. Remember that the very reason why there is a nucleation barrier is 0.2

the ﬁnite thermodynamic cost of the interface between the medium 0 0 45 90 135 180 and the new phase being created. In the case of heterogeneous nucle- θ(◦) ation on a surface with θ 0, the cost of the liquid/vapor interface ' 5 f (θ) γ is almost exactly compensated by the reduction of the solid sur- Figure VII– : Function deﬁned in L Eq. VII–32. face energy, when passing from γS to γSL. This is the reason why the nucleation barrier vanishes for θ 0. Condensation occurs then → exactly at saturation. It is interesting to analyse also the nucleation rate in the case of heterogeneous nucleation. In this case, the quantity you are interested in is no longer the number of nucleation events per unit time and volume, but per unit time and area of the solid surface. So the 1 2 unit of I has to be s− m− . The same dimensional analysis as in the case of homogeneous nucleation leads now to

k T 16π γλ2 I B exp f (θ) 34 5/3 2 (VII– ) ' vm η − 3 kBT[ln(β)]

which differs from Eq. VII–27 though the function f (θ) and through the exponent of vm in the pre-exponential factor. Another thing that I encourage you to think about is the case of Fig. VII–6 which is more realistic that a flat surface. For a given contact angle θ and a given volume of the droplet, the ratio of the liquid/solid to the free liquid surface area is larger than on a flat surface. You expect this to decrease still the nucleation barrier. Figure VII–6: Surfaces are seldom perfectly flat. Remember that the length scales we are talking about are just a few nanometers. On that scale, any surface has defects, which may be modelled as ditches: you may therefore expect this configuration to be more realistic than Fig. VII–4. Adsorption on surfaces

There are no beautiful surfaces without a terrible depth.

Friedrich Nietzsche

This chapter is an incomplete draft, which I included here for your convenience. Please rely on your own notes to study this chapter.

An adsorption isotherm describes the mass-transfer equilibrium Figure VIII–1: Irving Langmuir (1881- (a given temperature) between a reservoir and a surface on which 1957) is the ﬁrst industrial researcher to receive a Nobel Prize (in 1932). molecules can adsorb. The general relation takes therefore the form

n = f (µ) (VIII–1)

where n is the amount of molecules adsorbed (e.g., in moles of molecules per gram of adsorbent), and µ is the chemical potential of the adsorbate molecules in the reservoir. Just a summary of standard thermody- If the reservoir is gaseous, it can often be considered as made up namics of open system. Gibbs relation of ideal gases with chemical potential dG = SdT + Vdp + µdN (VIII–2) − µ = µ0(T) + kBT ln(P/P0) (VIII–6) which notably enable one to write ∂G P P µ = (VIII–3) where is the partial vapour pressure, 0 is the saturating pressure ∂N T,p µ (T) and 0 is the chemical potential at saturation. Because, at a given One of Maxwell’s relations is temperature the chemical potential only depends on P, the adsorp- ∂µ ∂V = (VIII–4) tion isotherms are often expressed as n = f (P). In case the reservoir ∂p ∂N is a dilute solution, one can approximate its properties by those of an In the case of an ideal gas, this leads to ideal solution. In particular, the chemical potential is ∂µ k T = B (VIII–5) ∂p p µ = µ0(T) + kBT ln(C/C0) (VIII–7) which leads directly to Eq. VIII–6.

where C is the solution and C0 is the saturating solution. In this case, the adsorption isotherm is expressed as n = f (P). In both cases, the VIII–2 c.j. gommes, chim0698

chemical potential is of the form

µ = B(T) + kBT ln(x) (VIII–8) where B is a temperature-dependent constant, and x is an intensive variable that controls the composition. Note that Eqs. VIII–6 and VIII–7 are the chemical potentials per gas phase molecule. The same expressions are valid for the chemical potentials per mole if Boltzmann’s constant kB is replaced by the ideal gas constant R. This is merely a question of units. adsorbed phase

Langmuir’s adsorption Figure VIII–2: Adsorption is a phase equilibrium between a gaseous phase Langmuir’s model of adsorption consists in assuming that the sur- (or a solution) and an adsorbed phase. face of the adsorbent is equivalent to a checkerboard, on each square of which a single molecule can come and stick with a speciﬁc adsorption energy ua. Because the molecules are attached to the surface, the corresponding energy ua is negative. A possible origin of that energy is dispersive forces, but the nature of the forces is irrelevant here. Let us assume that the adsorbent has a total of A adsorption sites on which the molecules can adsorb. If we want to describe quantitatively adsorption in this model system, we have to answer the following question. When the adsorbent is put in contact with a reservoir with chemical potential µ and temperature T, how many sites become occupied by a molecule? The answer to that question is a famous result in surface science and bears the name of Irving Langmuir. If you Google the words "Langmuir’s isotherm" you will ﬁnd many different mathematical derivations of it. This is the characteristic of important results: there are many different ways in which they can be understood. The derivation that we present here is not the most common, but it is probably the most general. Its structure is the following

1. We ﬁrst use the fundamental principles of statistical mechanics to

calculate the chemical potential µads of molecules adsorbed on the surface, as a function of the temperature T and of the total number of molecules adsorbed N. This will lead to a general expression of the form

µads = Function(N, T) (VIII–9)

The fact that µads depends on N is very natural: the more molecules you have in a system, the larger their propensity to leave that system, and vice versa.

2. We will then express that the surface with N adsorbed molecules is in mass-transfer equilibrium with a reservoir with chemical adsorption on surfaces VIII–3

potential µres. The equilibrium condition is expressed as µres = a) b) c) µads, which will eventually enable us to ﬁnd an expression for N vs. µres, at a given T.

To calculate the chemical potential µads one has to calculate first the Helmholtz free energy Figure VIII–3: On a given surface with A adsorption sites, there are a) Ω1 = A F (N, T) = U (N, T) TS (N, T) (VIII–10) ads ads − ads different configurations with N = 1 molecules, b) Ω = A (A 1)/2 2 × − where U is the energy and S is the entropy, and using the general configurations with 2 molecules, and c) in general Ω = (A) for N adsorbed thermodynamic relation N molecules. ∂F µ = (VIII–11) ∂N T to calculate µads. The energy is easy to calculate: it is simply given by

Uads = Nua (VIII–12)

The entropy is barely more difficult to calculate. All you have to remember is the microscopic definition of the entropy in terms of the The difference between the Helmnholtz number of configurations Ω compatible with whatever macroscopic free energy F and Gibb’s free energy G is important only for systems that variables define your system. This is the celebrated Bolzmann’s equa- can undergo large changes of volume. tion In the present context, the difference is irrelevant. S = kB ln(Ω) (VIII–13) In our case, Ω is the number of ways in which N molecules (macroscopic variable) can be arranged on a surface with A adsorption sites. As we show in Fig. VIII–3, the total number of different configurations with N adsorbed molecules is A A! Ω = = (VIII–14) N N!(A N)! − In principle, the problem is now solved. In practice, however, the derivative of F is going to be difficult to calculate if we keep the fac- torials. Luckily enough, any macroscopic number of molecules, such as N is extremely large, so that the factorial can be approximated with a very good accuracy using Stirling’s formula

ln(N!) N ln(N) N (VIII–15) ' − Using this approximation, the number of conﬁgurations can be written as N N N N ln(Ω) = A ln + 1 ln 1 (VIII–16) − A A − A − A which is plotted in Fig. VIII–4. Note that N/A is nothing but the fraction of the sites that are occupied by a molecule, which we shall VIII–4 c.j. gommes, chim0698

refer to as θ. The quantity ln(Ω) is sometimes referred to as the con- ﬁgurational entropy. The latter quantity is found here to be maximal 0.8 for θ = 1/2, and to vanish for θ = 1 and θ = 0. These two values 0.6

simply mean that there is a single way of having all sites occupied or /A 0.4 empty, respectively. ln(Ω) The chemical potential of the adsorbed phase is then obtained by 0.2 taking the derivative of F with respect to N, which leads to 0 0 0.2 0.4 0.6 0.8 1 N/A N µads = ua + kBT ln (VIII–17) A N Figure VIII–4: Number of configura- − tions Ω of N molecules adsorbed on a It is plotted in Fig. VIII–5. The chemical potential has two contribu- surface with A sites, as calculated from Eq. VIII–16 . tions: the first (ua) is energetic and the second is entropic. Because of the latter, the chemical potential diverges for N A and for N 0. → → This means that molecules would tend to desorb when N A and ' tend to adsorb when N 0, even if it is energetically unfavorable. It ' is interesting to note that the entropic term is proportional to kBT, the thermal energy. At the absolute zero, the chemical potential would An order of magnitude that is useful to know is the following: at room simply be ua because a molecule could only go to a state with lower temperature, the thermal energy is 1/40 energy. At finite temperature, thermal fluctuations enable a molecule eV. Think of it. How does that compare to move to a state of higher energy by taking the missing energy in a with the typical energy involved in a chemical reaction? reservoir under the form of heat. All this is in Eq. VIII–17.

5 Figure VIII–5: Chemical potential µads

) as function of surface coverage N/A, as T calculated from Eq. VIII–17 . B k ( / ) b

u 0 − ad s µ ( −5 0 0.2 0.4 0.6 0.8 1 N/A

If the surface deﬁned by Eq. VIII–17 is put in contact with a reservoir deﬁned by Eq. VIII–8, molecules get transferred from the surface to the reservoir and vice versa until the equilibrium is reached, which corresponds to µads = µres. This can be written as N Kx = (VIII–18) A 1 + Kx where K is a constant, given by

B u K = exp − a (VIII–19) kBT adsorption on surfaces VIII–5

and x is a composition variable (e.g. concentration or partial vapour pressure). Equation VIII–18 is the celebrated Langmuir’s adsorption isotherm.

1 Figure VIII–6: Langmuir’s adsorption isotherm for K = 1 (blue), 5 (green) and 0.8 20(red).

0.6

N/A 0.4

0.2

0 0 1 2 3 4 5 x

In the limit of small concentrations or vapour pressure, i.e. if Kx 1, it reduces to a simple proportionality law N Kx (VIII–20) A ' which is sometimes referred to as Henry’s law. The two parameters of the Langmuir’s isotherm are A and K. The total number of adsorption sites A is proportional to the speciﬁc surface area. When the concentration or the vapour pressure is very large, all adsorption sites tend to be occupied. In this case N A. → The other parameter is K, which depends on the energetics of the adsorption (relative to the thermal energy kBT.) The isotherms are plotted in Fig. VIII–6. Note the effect that a change in K has on the overall shape of the adsorption isotherm . Think about how a change in temper- To obtain a more precise, even if qualitative, meaning of the con- ature affects K and the isotherm. Does that make sense to you, based on your stant K, one needs a better understanding of the various terms in intuitive understanding of thermal the reservoir chemical potential, Eq. VIII–8. Consider for instance ﬂuctuations? a solution. If the solution is dilute, the interactions between solute molecules can be neglected. In this case, we could use a cartoon model of a solution similar to the Langmuir model of a surface (see Fig. VIII–7). In this case the number of adsorption sites A is replaced by the number of pseudo-sites in the solution V, the adsorption energy is replaced y the solvation energy us, and N is replaced by x V, where x is the molar fraction of the solutes13. Remembering 13 We make here an implicit assumption × that the solution is diluted, i.e. that x 1 , Eq. VIII–17 becomes about the sizes of the solute and solvent molecules. Can you state it explicitly?

µsol = us + kBT ln(x) (VIII–21) which is exactly of the type of Eq. VIII–8 , with B = us. In this case VIII–6 c.j. gommes, chim0698

the constant K takes a very explicit meaning

u u K = exp s − a (VIII–22) kBT

K is therefore just a Boltzmann factor of the usual type exp( ∆E/(k T)) − B . In order not to let you think that Lagmuir-like adsorption is so simple as what we have presented here, I would like to end the dis- Figure VIII–7: A cartoon model of cussion with Fig. VIII–8. In order to adsorb a molecule from a solu- a solution: the solute molecules are tion, one has to remove solvent molecules adsorbed on the surface positioned in a 3D cube. This leads to to free some space for the solute, one also has to remove salvation the same mathematics as Fig. VIII–3. molecules from around the solute, etc. The expression of K that we have obtained remains essentially valid, but the change of energy ∆E is naturally not simply u u . Can you think of a more accurate s − a expression, and enumerate all the energies that should enter it?

a) b) c) Figure500 VIII–8: The situation is actually a bit more involved. For example, to adsorb a solute molecule on a surface (a), one400 has ﬁrst to remove the solvent

molecules) adsorbed on the surface, as well as those solvating the solute /g (b),3 and300 only then can adsorption take place. cm ( 200

Multilayer adsorption ad s V 100 When nitrogen adsorption is measured on carbon nanotubes the isotherms look like in Fig. VIII–9, which is clearly different from the 0 Langmuir-type isotherm that we last considered. Isotherms of that 0 0.5 1 type are quite common, and they are referred to as "type II". P/P0 Unlike for "type I" (Langmuir-like) isotherms, the quantity adsorbed does not saturate when the pressure is increased. The most Figure VIII–9: Example of nitrogen adsorption at 77 K measured on a obvious explanation for this type of behaviour is multilayer adsorp- carbon nanotubes sample. Note the tion. different shape with respect to the type I isotherms described by Langmuir’s To describe quantitatively this type of adsorption, the assump- equation. tions of the Langmuir model have to be relaxed. The BET adsorption model is identical to Langmuir in all respects, except that we now allow several molecules to adsorb on top of each other in each adsorption site. adsorption on surfaces VIII–7

Figure VIII–10: The three authors of the BET equation on the occasion of a later reunion (picture taken from I. Hargittai, The Martians of Science, Oxford University Press). From left to right: Paul Emmett (1900-1985) was an American Chemical Engineer, Stephen Brunauer (1903-1986) is a Hungarian physicist who emigrated to the USA, Edward Teller (1908-2003) is also Hungarian. Besides the BET theory, Teller was also the director of the American fusion (H) bomb program, as well as of the so-called Starwars program. Teller regretted not to receive a Nobel prize for the BET; he was, however, the ﬁrst recipient of the Ig-Nobel prize for peace in 1991 for his contribution to the hydrogen bomb.

In principle, the equilibrium properties of the model in Fig. VIII– 11 can be calculated in the same statistical way that we used for Langmuir’s isotherm. In practice, however, the evaluation of Ω is extremely complicated when multi-layer adsorption is permitted. Just to convince you that this is a difficult problem, we shall calculate the first few values, up to N = 4. To make things even simpler, we shall suppose that the adsorption energy is the same whether a Figure VIII–11: In the BET model several molecules may adsorb on top molecule adsorbs on a surface or on another molecule. Therefore, we of each other on each adsorption site. do not have to keep track of the energy of each configuration because This is the only difference with the they all have the same energy. Even so, this is a difficult problem. Langmuir model.

1. For N = 1, one has Ω = A.

2. For N = 2, one has Ω = A(A 1)/2 + A. The first term is the − number of configurations with a single molecule per site (as in Langmuir), and the second term is the number of configurations with 2 molecules on top of each other. So far so good.

3. For N = 3, one has Ω = (A) + A + A(A 1). The first term is 3 − the same as in Langmuir, the second is the configurations with 3 molecules piled up in a single site, and the last term is for the configurations having one site with 1 molecule and another with 2 molecules. We call this the [1, 2] configurations. Note that [1, 2] is distinguishable from [2, 1], so that term doesn’t come with the VIII–8 c.j. gommes, chim0698

factor 1/2.

4. Things start to get tricker for N = 4. In that case you expect

1 A 1 A Ω = A + A(A 1) + A(A 1) + A − + (VIII–23) − 2 − 2 4

The first term is for 4 molecules in a site, the second is for [3, 1] configurations, the third for [2, 2], the fourth for [2, 1, 1] and the last one for the Langmuir-like configurations.

5. Etc.

As you can see, things become rapidly intractable. Remember that we need a general relation, then to estimate its asymptotic approximation for large values of A and N, and ﬁnally to calculate its derivative! I give up. There is fortunately another possible approach. The validity of the calculations is limited to gas phase adsorption. But this is anyway the conditions in which multi-layer adsorption is generally observed. Here is how it works.

Langmuir again, but differently

We start by considering again the Langmuir adsorption scenario, to build some conﬁdence in the approach. The approach here is dynamical. Each adsorbed molecule desorbs at a given rate that we can calculate. At the same time, the surface is constantly hit by molecules from the gas phase, and we can also calculate the adsorption rate. We shall then calculate the equilibrium as the condition under which the adsorption rate exactly balances the desorption rate. The surface has A sites, of which a fraction θ0 is free and a fraction θ = 1 θ is occupied by one molecule. At any ﬁnite temperature, 1 − 0 each molecule adsorbed on the surface vibrates as a result of all the collisions with other molecules as well as of the vibration of the surface. Occasionally one collision may bring an energy larger than the adsorption energy ua, and the molecule desorbs (or evaporates). The rate is expected to be proportional to the number of adsorbed molecules N = θ1 A and we may write the proportionality constant as 1/τa where τa is a characteristic desorption time.

N˙ e = Aθ1/τa (VIII–24)

We shall come back later to the temperature dependence of τa. During the same time, molecules come from the gas and hit the v dt surface, and may therefore adsorb. The number of collisions per Figure VIII–12: Simpliﬁed calculation unit time and unit area can be estimated in the following way. The of the collision frequency. Because the molecules have an average thermal velocity v, the number of collision in a time laps dt is roughly equal to the number of gas molecules in a layer of thickness vdt over the surface. adsorption on surfaces VIII–9

thermal energy being kBT, the average velocity of a molecule in the gas is p v k T/m (VIII–25) ' B where m is the molecular mass14. The number of collisions of the 14 This is obtained by equating the molecules with any surface in a time laps dt is roughly equal to the thermal energy kBTwith the kinetic energy of a molecule1/2mv2. The number of molecules in a layer of thickness vdt that covers the sur- symbol means that a factor of the ' face. Therefore order of 1 has been neglected. In this case the factor is √2 P p #collisions/time/area kBT/m (VIII–26) ' kBT

15 15 where the ﬁrst factor is simply the concentration of the gas Of From the perfect gas law: PV = nkBT course there are many numerical factors missing in this estimation16, and therefore n/V = P/(kBT). 16 We could make the calculation more but the important result is that the rate is proportional to the pres- accurate by noting that only half the sure. In the following, we shall simply write molecules in the vdt-thick layer are going towards the surface. Etc. #collisions/time/area = κP (VIII–27) where κ is a constant that depends only on the gas. The rate of adsorption of molecules can now be written as

N˙ a = eaθ0 AaκP (VIII–28) where a is the area occupied by a single adsorption site. In this equation, starting from the right: κP is the collision rate per unit area, Aa is the total surface area, θ0 is the fraction of the area that is available for adsorption, and ea is a factor lower than one that we have introduced to account for the fact that not every collision results in adsorption. You may think of ea as the probability that a molecule hitting the surfacer does adsorb. Some molecules may simply bounce on the surface; they do so with probability 1 e . − a Equating the adsorption and desorption rates N˙ a = N˙ d one ﬁnds

θ1 = τaeaaκPθ0 (VIII–29) which eventually results in

KaP θ1 = (VIII–30) 1 + KaP with Ka = τaeaaκ (VIII–31) Equation VIII–30 is exactly Langmuir’s adsorption isotherm.

Multilayer adsorption

The analysis leading to Eq. VIII–30 assumed that only one molecule at most could exist at a given site, so the system was completely char- acterised by the fraction of occupied sites θ1. If multi-layer adsorption VIII–10 c.j. gommes, chim0698

takes place, we have to calculate the fraction of sites occupied by one molecule θ1, by two molecules θ2, three molecules θ3, etc. Once all theses quantities will be known to us, we will be able to calculate the average number of molecules per site as

n = θ + 2θ + 3θ + . . . (VIII–32) h i 1 2 3 from which the total number of molecules adsorbed will be calculated as N = A n (VIII–33) h i Our aim now is therefore to calculate n . Before we start the analy- h i sis, note ﬁnally that the θ’s satisfy

θ0 + θ1 + θ2 + θ3 + ... = 1 (VIII–34)

because each site is either bare or occupied by some molecules. To calculate the θ’s, we will proceed as in Sec. , but we now have to write for each layer a balance equation similar to Eq. VIII–29 . Because the parameters e and τ may vary from one layer to another, th we shall use the notation Ki = τieiaκ for the characteristics of the i layer, starting the numbering at 0. The equilibrium of each layer is written as

K0Pθ0 = θ1 K1Pθ1 = θ2 ...

Ki 1Pθi 1 = θi − − ... (VIII–35)

where K0 is the quantity we called Ka in Sec. . The second equation, for example, means that the number of molecules condensing on the sites that host already one molecule is equal to the number of molecules evaporating from the sites that host two molecules. The other lines have similar meanings. A standard BET assumption is that only the ﬁrst layer (in direct contact with the solid) is different from the others. The properties of the latter layers are assumed to be indistinguishable from those of

the bulk liquid. In the following, we use the index b (e.g. Kb) for the values of the bulk, and index a for the values of the first layer (as in Ka) for the. With that assumption in mind, it proves convenient to use the notations Ka β = KbP and C = (VIII–36) Kb i With these notations the solution to Eq. VIII–35 is simply θi = Cβ θ0. Using that result, the average number of molecules per site is17 17 This part here is almost recreational mathematics! The first step itself is ∞ i already nice: To estimate ∑i=0 β for any β 1, you can write it as ≤ (1 + β + β2 + ...)(1 β) − (1 β) − Of course you haven’t changed anything by multiplying and dividing by (1 β). If you develop the numerator, − however, you will find that it is equal to 1. This leads to ∞ 1 βi = ∑ 1 β i=0 − which ends the first step of the derivation. Here comes the second. Taking the derivative of both sides of this last equation with respect to β, and multiplying them by β leads to

∞ β iβi = ∑ (1 β)2 i=1 − This is the result you need to derive Eq. VIII–37. adsorption on surfaces VIII–11

∞ Cβ n = iθ = θ (VIII–37) h i ∑ i (1 β)2 0 i=1 −

This is not the solution yet because θ0 is still unknown. The value found by expressing that ∑i θi = 1, which leads to (1 C)(1 β) + C 1 = − − θ (VIII–38) 1 β 0 − The ﬁnal result therefore takes the form Cβ n = (VIII–39) h i (1 β)(1 + (C 1)β) − − which is the BET equation, for the derivation of which Edward Teller considered he deserved a Nobel Prize. To put the result under a more familiar form, we have to understand the physical meaning of β. Remember that

β = τbebaκP (VIII–40)

Because β is dimensionless and proportional to the pressure, all the factors in front of P can be written as 1/Pb where Pb is some yet to be speciﬁed pressure that characterises the bulk adsorbate at the considered temperature. To gain some insight into this, consider for instance what happens for P = Pb, which leads to β = 1. This can be written as 1 = ebaκPb (VIII–41) τb

This equation means that, at the pressure Pb, the number of molecules that evaporate from the surface of the bulk adsorbate (left-hand side) is equal to the number of molecules that condense on it from the gas phase (right-hand side). This is exactly the deﬁnition of the saturation. In other words Pb = P0! The ﬁnal form of the BET equation is therefore the following

N CP/P = 0 (VIII–42) A (1 P/P )(1 + (C 1)P/P ) − 0 − 0 which is the one that we shall use in Sec. to determine speciﬁc surface areas from experimental adsorption data. We postpone to Sec. the thorough discussion of the BET constant C. Examples of BET isotherms are plotted in Fig. VIII–13 for three different values of C. At low relative pressure, the same Henry-like linear adsorption as in the Langmuir isotherm is observed. When the pressure approaches the saturation, the number of molecules adsorbed per site diverges. This corresponds to the condensation of a thick liquid layer on the surface. The effect of the constant C is signiﬁcant only when the thickness is relatively small, say n 5. h i ≤ VIII–12 c.j. gommes, chim0698

5 Figure VIII–13: The BET adsorption 20 isotherm, corresponding to Eq. VIII–42, for C = 1 (blue), 10(green) and 100 4 (red). 10

3 0 N/A 0 0.5 1 = i

n 2 h

0 0 0.2 0.4 0.6 0.8 1 P/P0

Note in passing that no condensation at all would be observed for P P in absence of a solid surface. What Fig. VIII–13 is showing is ≤ 0 an example of how a nanostructure18 can shift a phase equilibrium. 18 Remember that a large surface Imaging you had a porous solid with pores that would be just large area can only be achieved with nanostructured materials. enough for about 10 molecules to ﬁt in it, 5 on each side. Looking at Fig. VIII–13 you can already guess that these pores will be ﬁlled with condensate at P 0.8P . We will analyse this is a more quantitative ' 0 way when discussing the Gibbs-Thomson effect.

Discussion

The BET constant

The constant C in the BET equation is referred to as the BET constant. Remember that it is deﬁned as

K e τ C = = (VIII–43) Kb eb τb

where the e’s are the fraction of collisions from the gas phase that result in condensation (on the surface or on the bulk liquid surface), and the τ’s are characteristic evaporation times. Evaporation is a thermally activated process. A molecule close to the surface of a bulk liquid (but inside the liquid) has a binding energy u 0 that b ≤ keeps it in the liquid. Because of the thermal motion, the molecule constantly receives energy from its neighbours or by being hit by an incoming gas molecule, etc. As a consequence, the energy of the molecule is ﬂuctuating randomly in time, and the probability of ﬁnding it in a state of energy ∆E above its ground state is given by adsorption on surfaces VIII–13

Boltzmann’s factor ∆E Proba(E) exp − (VIII–44) ' kBT where the missing factor19 is a normalising constant. If that extra- 19 That is the meaning of the symbol . ' energy happens to be larger than the binding energy of the molecule, i.e. if ∆E + u 0 then it can evaporate. One therefore expects the b ≥ following temperature-dependency for the evaporation rate 1/τb 1 ub = νb exp (VIII–45) τb kBT where νb is a typical vibration frequency. Be sure you understand this equation, in particular the sign in the exponential: ub is negative so that if the binding energy is very large (i.e. very negative). A similar expression can of course be given for adsorbed molecules. This leads to e ν u u C = a b exp b − a (VIII–46) ebνa kBT The most important factor in there is the exponential. The prefactor involves νb/νa the ratio of the natural vibration frequencies of the molecules in the bulk and on the solid surface, as well as ea/eb the ratio of the "probabilities of capture" when a gas molecule hits the solid surface or the surface of the bulk liquid. One can reasonably expect that ratios are close to unity. Moreover, we do not expect them to be temperature-dependent. The important physical quantity that controls the value of the exponential factor is (u u )/(k T), having b − a B a clear physical meaning. The constant C therefore characterizes the binding of a molecule to the surface. It has exactly the same meaning as the constant K in Langmuir’s equation. The relation can be made even more explicit by considering the general expression that we have found earlier: K = exp((B − ua)/kBT) with B = µ0 the chemical potential of the liquid, in the case of vapour adsorption. Since the chemical potential is Gibb’s energy per molecule, one can write µ = u + P v Ts (VIII–47) 0 b 0 b − b where ub, vb and sb are the molar energy, volume, and entropy in the bulk state. Taking this into account, the expression of K can be written as P v Ts u u K = exp 0 b − b exp b − a (VIII–48) kBT kBT where the similarity with Eq. VIII–46 is now obvious. The relation between the two approaches (statistical and dynamical) could be pushed even further. However, what we have done here should suf- ﬁce to convince you that the two approaches are consistent with each other, and that they actually give the same result. VIII–14 c.j. gommes, chim0698

Measuring speciﬁc surface areas

The BET model is extremely famous, mostly because it provides an very practical way of analysing adsorption data to determine speciﬁc surface areas. Many people actually refer to nitrogen adsorption measurement as a "BET measurement": "let’s do a BET!" I wouldn’t be surprised if half of you would end up one day calling a nitrogen adsorption measurement a "BET". That’s ﬁne in the long run, but in the coming year or so I will not let you do that. BET is a model, not an experimental method. Anyway, the idea is that you expect molecules to occupy a given area on a surface, so that if you know the number of adsorption sites, you can determine the total area of the surface. So determining the surface area is tantamount to determining the number of adsorption sites, which we have referred to so far as A. What you measure, on the other side, in N as a function of P/P0. Here is the usual way to extract A from such a datatset. From Eq. VIII–42, you can write

P/P 1 C 1 P 0 = + − (VIII–49) (1 P/P )N AC AC P − 0 0 The left-hand side is something you can calculate in a straightforward way from the measured data, e.g. with a spreadsheet; we refer to this quantity as the BET function. Equation VIII–49 tells you that if you plot the BET function against the reduced pressure, you should obtain a straight line. Calling the intercept α and the slope β, the BET constant C and the number of adsorption sites A are then calculated as β 1 C = 1 + and A = (VIII–50) α αC As simple as that. The BET function corresponding to the data in Fig. VIII–9 is shown in Fig. VIII–14. The values of α and β obtained from a least- square fit from P/P = 0.05 to P/P = 0.35 are C 79 and 0 0 ' A 53cm3/g. The latter value corresponds to a volume in STP ' conditions20 which is just a weird unit for a number of moles. The 20 Standard temperature and pressure 3 3 conditions, T = 300 K, and P = 1 atm. value 53 STP cm /g converts to 2.37 10− mol/g. This is the number One mole of gas occupies 22.4 liters in of molecules needed to cover the entire surface of the nanotubes with STP conditions. a layer that is exactly one molecule-thick. Looking on the isotherm itself (Fig.VIII–9), one sees that this value is the position of the plateau in the isotherm. In principle, the value could be read directly on the isotherm, but the BET fit is a more accurate way of doing this. If we accept the usual figure that one nitrogen molecule occupies about 16.2 Å2 on a surface. The value of A coverts to 230 m2/g. This is the specific surface area of the nanotubes in Fig. VIII–9. adsorption on surfaces VIII–15

0.03 Figure VIII–14: BET plot corresponding to the nitrogen adsorption data of Fig. )

3 VIII–9. The red line is the least-square ﬁt of the data for 0.05 P/P 0.35 . ≤ 0 ≤ g /cm ( 0.02

0.01 BETfunction

0 0 0.5 1 P/P0

Exercices

1. The modelling of adsorption presented here is extremely simpli- ﬁed. Enumerate some ways in which adsorption could in principle differ from what is presented here, and things that we might have neglected.

2. We have often mentioned energies in this chapter: adsorption energies, salvation energies, etc. What is the nature of the physical forces responsible for those energies?

3. In principle, a nitrogen molecule can adsorb in two different ways on a surface: perpendicular to it, or flat on it. How is the chemical potential of an adsorbed molecule modified if it can adsorb in two different configurations? Assume that the two configurations have the same energy. How is the adsorption isotherm modified?

4. If a solution is not very diluted, deviations from ideality have to be considered. How is the adsorption isotherm modiﬁed if a Margules equation is used?

5. Same question for a Van der Waals gas.

6. One classically advocates that the BET model should only be used from P/P0 = 0.05 to P/P0 = 0.35. Figure VIII–15 shows how the BET model actually compares with the data. To your opinion, why is there any deviation between the data and the model? The reasons need not be the same at low and high pressure. IX–0 c.j. gommes, chim0698

500 Figure VIII–15: Real volumes calculated from the BET ﬁt in Fig. VIII–14. 60 Can you think of reasons (possibly many) for the deviation between the 400 data and the BET model at low and at 40 large pressures ? The reasons need not ) be the same in both cases.

/g 20 3 300

cm 0

( 0 0.05 0.1 200 ad s V 100

0 0 0.2 0.4 0.6 0.8 1 P/P0

7. When analysing the nitrogen adsorption data in Fig. VIII–9 we found a speciﬁc surface area of 230 m2/g. Assuming that nanotubes are closed, so that adsorption only takes place on their outer surface, make a rough estimate of their diameters. (Hint: the density of the graphitic carbon that makes up the nanotubes walls is ρ 2 g.cm 3). ' − 8. Can you guess the order of magnitude of the area occupied by water molecule adsorbed on a surface.

9. Typical values for the BET constant are C 20 for adsorption on ' polymeric materials and around C 200 for silica. Can you make ' sense of this? What does this mean?

Langmuir (statistical mechanical approach) BET Broekhof-de Boer Metastability.

Further reading

Exercices