 So, we come to the lecture number 4 of this module 1 still in the introduction, but now focusing on the gas liquid interfaces. Today we discuss properties over curved surfaces. I am certain some of the results that we will be deriving here you would have probably come across at least tangentially sometime in your studies so far. To begin with let us say we going to talk about vapor pressures. So, I need you to imagine that a drop of a liquid a small very small drop of liquid like water is created from a large body of water with planar surface. Now supposing that this little drop has radius small a, we ask ourselves the question as to what would be the relation between the chemical potential of the drop vapor system as against the planar water with associated equilibrium vapor that system. So, what would be the kind of relation between these two chemical potentials? Presuming that drop is in equilibrium with vapor and on the other hand we have planar water surface in equilibrium with the associated vapor. How would the two chemical potentials compare? Now before we going to theoretical deduction of an interrelationship between these two chemical potentials, I would like you to think about it first. So, if we have to basically just compare the magnitude of chemical potential in the drop vapor system against the planar water vapor system, what would be your reflection? We are not really looking at this stage at the exact mathematical relationship, but maybe in terms of inequality inequality, how would mu a compare against mu? If it is not obvious to you, you can think of the way the drop might have been created. After all we picked up a small quantity of water from under a planar surface and then brought it into the being that is a small drop. So, we expect we would have done certain energy expenditure in going from under planar surface to a drop form. That is exactly what I am asking you to imagine here. And as we go along through this lecture, I would want you to keep your paper and pencil ready so that you can start working out things along with me here. So, here we have that small drop of radius a. If it is in equilibrium with the vapor associated surrounding it, let the chemical potential be equal to mu a and the corresponding chemical potential for a planar water surface in contact with equilibrium vapor is mu. So, we now looking at some sort of relation between mu a and mu. We expect the chemical potential for the drop vapor system to be greater than for the planar water vapor system. And that would be indicative of how much work has been done in creating the that creating that drop from a certain number of molecules which were originally under a planar surface. The difference between these two chemical potentials will be expectedly dependent on these two quantities the radius of the drop that we created and the surface tension. And the precise relation is given by Kelvin equation. As we will see later as we go along, the relationship could be obtained from following considerations. Let us say we take a small number of molecules d n from water at constant temperature, constant volume and constant surface area from within the interior of a large mass of liquid beneath the flat surface and add these to the interior of the chosen drop of radius small a. Now, the work done could be written in terms of chemical potential as d f is equal to mu a minus mu into d n right. So, this could be also seen to be equal as the work that must be done in having these d n molecules added to the drop which would increase the surface area of the drop. So, the work done according to the increased surface area will be gamma times d a. So, equating these two we would say d f is equal to mu a minus mu d n equal to gamma d a and presuming the drop is spherical we will have this equal to gamma times differential of 4 pi a square or that could be completed as 8 pi a gamma d a. Now, we need to eliminate d n and d a from this equation. The way to do that would be to consider the increase in volume when we add d n molecules of water to the drop if by small v we denote molecular volume this should be v d n and this should be equal to the increase in the volume of the drop d v where v is equal to 4 third pi a cube. So, here is a relation which will permit us eliminating from equation 15 d n and d a. So, you can work along with me here this is v d n equal to d v or when you substitute for v as 4 third pi a cube and complete the differentiation we will get d n is equal to 4 pi a square d a divided by small v. Now, substituting this in equation 15 for d n and simplifying we get mu a minus mu is equal to 2 gamma v by a. This bears us right bears out us right we had anticipated mu a to be greater than mu. So, mu a minus mu should be positive surface tension ordinarily for stable liquid is a positive quantity molecular volume p and radius a all are positive quantities. So, that is a general form of Kelvin equation mu a minus mu equal to 2 gamma v by a. Since we want to learn something more about vapor pressures we could for ideal vapors write the following pair of equations mu a can be equated to mu 0 plus k T ln p a likewise mu is equal to mu 0 plus k T ln p where p a and p are the vapor pressures that one would be able to measure for the drop of radius a if it is an equilibrium with vapor around it. Similarly, mu is the chemical potential for the planar water surface in contact with the vapor which would have vapor pressure small p. Now, if we substitute for mu a and mu in the general form of Kelvin equation we will get mu 0 eliminated mu a minus mu will be equal to k T ln p a by p and when you transpose that factor k T to the other side you would obtain equation 20 ln p a by p equal to 2 gamma v by a k T. We repeat p a and p respectively stand for equilibrium vapor pressures for the curved surface against that for a flat surface. So, we understand here everything on the right hand side being positive the implication is the vapor pressure p a is greater than vapor pressure p. So, when we go from a flat surface to a curved surface we have an increase in the vapor pressure. Many a time in this series of lectures we will be talking about water it is partly by way of convention adopted in literature on interfacial science and engineering and partly because this is one liquid we have in such large abundance on a planet and we ought to know more and more about water and perhaps air. So, I will I will expect this class to try and learn more about water and air I will be guiding some of those ideas for your considerations. So, little bit of a digression historically composition of water which is two parts hydrogen to one part oxygen was discovered by Henry Cavendish in London in 1781 and his findings were reported in terms of phlogiston which was later proved to be hydrogen and deflogisticated air which was proven to be oxygen and interesting fact about Cavendish was that he died in his laboratory about 30 minutes walk from present site of London south bank university. This is the picture of that old gentleman. We return to the vapor pressures over curved surfaces and apply to get some numerical estimates. If we use this relation between P and P by plugging in for water different droplet sizes we come to these observations. For a drop size of 10th of a micron the increase in vapor pressure is about 1 percent. At 100th of a micron we note a 10 percent increase in the vapor pressure at a small size of 50 angstroms the increase in vapor pressure is 100 percent. At 10 angstroms the increase is 200 percent and if we go down to a very small cluster of 6 angstroms the increase in vapor pressure is about 450 percent. All calculations have certain assumptions and this simple calculation here presumes the surface tension to be constant. As we will see later this can be justified to give you a flavor of simple theories and their abilities to predict quantities of interest to a sufficient accuracy will be time again will be checking the results of the theory time and again against external data. So, Kelvin equation 2 has been tested experimentally by measuring growth of aerosol droplets to equilibrium. They have been 2 independent experimental approaches here both relating to growth of aerosols. We will take these in turn in next couple of slides. The first approach was to make a mono disperse aerosol of di-octyl phthalate. This aerosol was allowed to reach an equilibrium over a mixture of di-octyl phthalate and toluene having a flat surface. You can begin thinking about what kind of measurements would have been made. It is to be expected that the vapors of toluene would condense on the aerosol droplets until the vapor pressure of toluene over these DOP droplets or particulates becomes equal to that over flat surface of the mixture at which point the growth of the aerosol particles would cease. That would be the equilibrium situation. Toluene will keep leaving the surface of the planar system and the toluene molecules would get condensed on the aerosol droplets. There by changing the composition of the aerosol at equilibrium there should be no growth of aerosol particles, but taken with this part of knowledge that we had reduced that the vapor pressure over curved surface is greater than over a planar surface. What would we expect to measure here? Obviously, one could measure the size of aerosol droplets when equilibrium is reached that is one measurement we obviously can make. What else could be measured? We would know what is the initial composition of the mixture at the planar surface toluene and diocotylthalate. If we take a sample of the aerosol one would again be able to measure the composition of toluene concentration of toluene in the aerosol. It is here that we expect to see a difference in composition that is exactly what was measured the vapor pressure over the curved surface being greater than that over the flat surface rather small mole fraction of toluene is found in aerosol droplets at equilibrium when compared to that in the bulk under the planar surface. Now using these data comparing the actual growth of aerosol droplets with that which would give the droplets the same composition as in the bulk mixture under the planar surface we would be able to check the excess vapor pressure over the surface of various curvatures. And by doing this it could be checked how good Kelvin's equation would be in predicting the effect of curvature on the vapor pressures. It turns out the experimental results bear Kelvin equation to be accurate to within plus or minus 2 percent that is a very good accuracy. In another independent approach the Kelvin equation was verified using aqueous sodium chloride droplets in equilibrium with a plain sheet of water. Once again those measurements reveal the theory to be in agreement with experiment to within a few percent. So, that confirms the basic soundness of Kelvin equation. It is at this stage that I would like you to think about what else could be affected by curvature. After all the title for this lecture has been chosen as properties over curved surfaces. So, vapor pressure has been the first of the property properties. So, we could think of other properties, but before we come to that maybe I can give you certain amount of homework to do. You can use the same approach for 2 other geometries. First is a cylinder of radius small a and the other one is a trough of liquid of radius a. For a cylinder of liquid the vapor pressure is enhanced or increased as per equation 21. Lawn P a by P is equal to gamma v by a k t whereas, for a trough of liquid of radius a we have Lawn P by P a equal to gamma v by a k t. I would like you to try and prove equations 21 and 22 or rather derive these equations 21 and 22 from a similar approach as we just seen. It is interesting how other properties 2 are influenced by curvature. Once we are focusing on gas liquid systems the next entity we come across is excess pressure inside bubbles. The work necessary to expand spherical gas bubble surface is gamma d a or 8 pi a gamma d a on similar lines as we did for drops. And if an external pressure is applied to the inside of the bubble maybe with the help of a fine capillary the work done in increasing the size of the bubble will be P e times d v or that is equal to substituting v equal to 4 third pi a cube P e d v is equal to 4 pi a square P e d a and that is equated to the work of enlarging the spherical gas bubbles 8 pi a gamma d a. So, cancelling of 4 pi a d a we get excess pressure as 2 gamma over a. I am sure as children you would have played with soap bubbles. So, next think about the situation when we blow a soap bubble in air unlike the bubble created in the interior of water. Now, we will have 2 surfaces which need to be extended and by following similar arguments we can show that the excess pressure for a soap bubble would be twice of what is shown in equation 24 P e is now equal to 4 gamma by a. Like in case of vapor pressures you should be able to obtain a field for magnitudes of excess pressures by plugging in numbers like surface tension and different redive for bubbles, but for the time being we could see an example you could play with this equation and accustomed yourself to the magnitudes. But to give an example here right away if a soap bubble of 1 micron radius is created from a solution aqueous solution typical surface tension value could be taken as 25 dynes per centimeter. And it is interesting to know that note that excess pressure in the bubble as per this 4 gamma by a will work out to be about 1 atmosphere. A small soap bubble could have excess pressure of the order of atmosphere inside it. It is important to get feel for these numbers because these facts could be useful in practice. We will look at more implications in times to come. For the time being one of the important consequences of excess pressure would be seen in breakdown of foams. We realize that smaller bubbles will have enhanced excess pressures inside. Those of you who have had some exposure to mass transfer would be able to see immediately that air will tend to diffuse through liquid lamelly from small bubbles in foam into larger bubbles because the pressure in larger bubbles, excess pressure in larger bubbles will be smaller. How dramatic can this be? Here is an example. In foams made of rubber latex or tipol solution one can observe that the total number of bubbles would decrease to just about 10 percent of the original in as little as 15 minutes. And that without any single film ever rupturing. This mode of foam breakdown which is occurring through combination of excess pressure and therefore, increased permeability which in turn is a result of increased solubility and therefore, increased driving forces for diffusion from smaller bubbles to larger ones is called disproportionation in the literature on foams. It is so simple as that smaller bubbles have higher excess pressure. So, the air inside will have higher solubility therefore, higher permeability and therefore, greater permeation or diffusion rates and therefore, smaller bubbles will keep on losing their gas contained to the larger bubbles until they vanish all together. So, that should not be surprising in about 15 minutes only 10 percent of the bubbles remain without actually having any so film rupturing. So, that is this proportionation for you as is known in the modern parlance on literature in foams. Yet another property you could think of as being influenced by curvature. We sort of refer to this the solubility solubility from small droplets or particles. For the sake of derivation we analyze a case of a very fine emulsion. We take an extremely fine emulsion of oil in water will generically represent oil in water type of emulsions as O slash W type. We need to think about the solubility of oil in water. We expect this solubility to be slightly enhanced and if it is a fine emulsion that we are talking of we likely to have very large interfacial area and associated with this large interfacial area. We will have a very large interfacial energy. This in turn would tend to raise the chemical potential of the oil effect of large interfacial energy would be to raise the chemical potential of oil appreciably. We could conduct a similar thought experiment as we had in the case of vapor pressure concentration. We let small mu to be chemical potential of oil in contact with water in a system where we have a flat interface and by mu sub e we denote chemical potential of oil dispersed in water in the form of droplets in that emulsion form droplets are of radius a. Next we imagine that a certain small number of molecules d n of oil are picked up from the interior of the aqueous solution in contact with a planar oil water interface and place them on a small droplet of oil of radius a which is in contact with aqueous solution of oil in the emulsion. So, we are charting out the same kind of thought pattern. So, when we move this d n molecules from under a planar surface and put it into a small droplet the work done could be estimated from the change of chemical potential for this collection of d n oil molecules. Once these are placed on the tiny oil droplet they would cause expansion of the drop surface. Now, against the interfacial tension in the vapor pressure case we are talking of surface tension. Here we are placing these d n oil molecules inside a small oil droplet of radius a in an emulsion. So, the work of expansion will be against the interfacial tension. Equating these two we can write mu a minus mu d n is equal to 8 pi a d a times gamma i. Within the bracket we have differential of area d a. Gamma i is interfacial tension or interfacial energy. I will repeat what I said about surface tension. Surface tension we said will have units of dimes per centimeter, but then dimensionality same as oaks per centimeter square. So, we see the likeness of surface tension and a sort of energy. So, when we proved the Helmholtz free energy is equivalent to surface tension we had similar understanding. Now, we are talking of interfacial tension once again in dimes per centimeter against interfacial energy which would be in oaks per centimeter square. We also make an auxiliary balance corresponding to volume change of droplet. If we place d n molecules each of volume small v then v d n is equal to d of capital B which is 4 pi a square d a. Substituting for d n in the previous equation 26 we find after cancelling 4 pi a d a mu a minus mu is equal to 2 gamma i v by a. This result we see is similar to what we had in case of the vapor pressure concentration. Now, presuming the solution of oil in water behaves like an ideal solution we can substitute for mu a minus mu the right hand side of equation 29 k t ln c a by c. And therefore, we can write by combining equations 28 and 29 ln c a by c equal to 2 gamma i v by a k t. Once again we see the solubility c a will be higher for the smaller particles or smaller droplets. What we are stating for an oil in water emulsion we can also extend to particles in water or any solvent. So, the solubility for smaller particles c a will be greater than the solubility that you will be able to measure over flat surfaces. I may ask you to think about this are we likely to have any experimental consequence observable for this result. Is it simply a theoretical deduction or will it have any real consequence? The power of theory and the ability of mathematics to analyze what is common among diverse phenomena is sometimes amazing. So, it is not just the vapor pressures are enhanced solubility is also affected significantly by the size scheme or curvature. Now, in context of solid liquid systems we can put this in perspective solubility of oil for a small droplet is higher than for a planar oil surface in contact with water exactly similarly as the situation is with vapor pressure over curved surface. In context of solid liquid systems this has an interesting effect. When you look through literature on crystallization you would see a similar effect for the higher solubility for smaller crystals. Now, this extent is not difficult to perform one could do this if the crystals are formed very rapidly from a super saturated solution. So, rapidly that upon filtration you would not be able to retain any crystals on the filter paper. That quick crystallization would produce very small crystals and if you observe over period of time what happens to the population of these small crystals what you see is interesting. You find upon standing there would be only a small number of large crystals left. Clearly large crystals seem to grow at the expense of the smaller crystals. That is not now difficult to imagine understand. The smaller crystals dissolve due to their higher solubility exactly according to this equation. And if the smaller crystals have higher solubility they lose the solute molecules to the mother liquor. Through the mother liquor they make their way to the interface of larger crystals which have higher which have a smaller solubility. Larger crystals will have smaller solubility. So, the solute molecules make their way from small crystals through the mother liquor to the large crystals making the larger crystals grow further. Given sufficient time only a few large crystals would remain. This well known phenomenon is known in literature as host wall ripening. So, if we take stock of the three situations that we have discussed so far. We see higher vapor pressures for smaller droplets something very similar to higher solubility for smaller particles or smaller droplets in a solid liquid or liquid liquid systems. And a higher higher excess pressure for smaller bubbles in gas liquid systems that probably will lead you to anticipate that in a fog the much finer water droplets would tend to make way for larger droplets. In case of foam smaller bubbles would disappear in disproportionation and in case of crystallization the smaller crystals would vanish as a result of host wall ripening. Behind all this phenomena there is one common theme that is the enhanced property for smaller size scale or greater curvature. We continue to alternate with the theme of water. This is picture of water as normally visualized. I made a remark last time that water is highly associated there is strong hydrogen bonding among water molecules. So, glass full of water could be imagined to be a single very large molecule that would be a model for it. The structure of water again is given here. We see the strong hydrogen bonding over successive layers of water. We made a mention of certain implicit assumption in our derivation for effect of curvature on vapor pressure or effect of curvature on solubility. And that implicit assumption was that the surface tension in the liquid vapor or liquid gas system was constant or interfacial tension in the liquid-liquid emulsified systems was taken as constant. One could question every assumption and it is not surprising that that has been done. So, let us deal with this assumption here. How good would we do if we treat surface tension to be constant? There are certain calculations made in literature which I would like to have a recapitulation here. For liquid surfaces with the radius of curvature A large compared to thickness of surface region, the surface tension is close to the value which would be measured or expected for planar interface. Question then comes up if for water present in the form of such a small drop as to have the radius A comparable to the thickness of the surface layer, will the surface tension still be same or appreciably different from that for a planar surface. And that is the sense of small size scales. We have much of what constitutes a small particle or droplet very close to the surface. There is not much of bulk there. So, will surface tensions still be same or will it differ appreciably? There are number of attempts to answer this question. First we have Tolman's work in 1949 who used quasi thermo dynamical method independently Kirkwood and Buff used a statistical mechanical treatment and in 1951 we had Benson and Shuttleworth using a molecular interaction method. And all these methods predict that the surface tension will be less than that for planar surface when the radius of the droplet is comparable to a few molecular radii. So, very rigorously speaking treating the surface tension constant would not be correct. But if you take a pragmatic view even when the radius of the droplet is just about a few molecular radii large, the surface tension decrease is relatively small. One could make judgments about these independent methods. To be critical we could say among these three approaches the molecular interaction method is the most reliable one. And that is because there is a very small number of molecules which would be required to make such a tiny droplet. For example, for water if the droplet radius is 4.6 angstroms it would require only 13 molecules and it is based on the scale molecular model which counts the number of bonds broken. The prediction is that the fall in surface tension is just about 15 percent or less. So, for the above water droplet the surface tension predicted would be about 61 dynes per centimeter against the 72 dynes per centimeter is the common value that we have always come across per water for planar surfaces. It therefore seems safe to conclude here that for very small droplets such as embryos and nuclei which are responsible for phase transformation the surface tension indeed decreases slightly, but the decrease is not very large. And in everyday life of a scientist and engineer on usual scales of measurement as are used in the measurement of surface tension of liquids such as by a droplet method which we will be addressing later. The effect of curvature of surface tension effect of curvature of droplet on surface tension can always be ignored. So, the take home message here is ordinarily we can regard surface tension to be independent of curvature, but when you go down to very small size scales like for nuclear embryos which are basic to phase transformations these effects could be somewhat considerable. It is here that I would like you to be aware of another property. In the course of these lectures we will be alternating between some practical considerations and some fundamental or abstract concepts. We will also have to have certain elementary models to estimate properties which are pertinent to an accuracy which is acceptable on an order of magnitude basis. So, after having talked about surface tension abate, vapor pressures, excess pressures, solubilities, Helmholtz and Gibbs free energy, we next think about total surface energy. The total surface energy is greater than the Helmholtz free energy for the surface per area. And the relation is what is summed up in equation 31. F s is equal to U s minus product of T and S s, where U s is the total surface energy and the other thermodynamic quantities refer to surface excesses. To find the entropy S s, we make use of this relation S s is equal to minus dou F s by dou T at constant N and V. We had earlier proved in the last lecture that F s is equal to gamma 0. So, S s is now equal to minus dou gamma 0 by dou T at constant N and V number of molecules is equal to minus N volume. Or we can rearrange our earlier equation to get U s equal to gamma 0 plus T times minus dou gamma 0 by dou T at constant N and V. So, at 25 degrees centigrade for water, U s will work out to 118 ergs per centimeter square taking the surface tension as 72 dynes per centimeter and the temperature coefficient of surface tension as minus 0.154 and for mercury, U s works out to be 541 ergs per centimeter square. We will stop here for today and we will begin next time with a recapitulation of the total surface energy. So, we conclude this lecture here.