 All right in this lecture 4 we will talk about measurement of spreading cohesion through the sessile drop method which is used for negative spreading cohesion and then we will think about the measurement of positive spreading cohesion. One will have to think of a clever way of doing a measurement in case the spreading cohesion is positive and the background for that is already known to you, but before we do that we will have a quick recap of the weighting balance method of finding the work of addition and spreading cohesion. This method proposed by Guastalla is known as weighting balance method using willyme plate where we obtain the work of addition or the work of weighting without measuring the contact angle directly and as expected this would correspond to the work of addition related to a monolayer of water or surfactant on the solid in adsorb form. A thin slide of perimeter 1 centimeter is used it is immersed in the liquid and then raised to the surface till the external vertical pull f becomes constant. After buoyancy correction that force should be equal to gamma L a cos theta. Work done is now of withdrawing the slide and that is further equal to work of devating the slide. If the slide is hydrophobic then cos theta is negative and the work is gained when the slide is withdrawn. If the plate is immersed then work will have to be done. I tried to explain this through a mnemonic analog of a plastic ball being dipped in water if it were to contain a hole. In the process of immersing this ball in water you will have to overcome the buoyancy force if the ball contains a tiny hole through which water can enter the ball then it will help in immersing this solid object inside water. In the real practice that corresponds to filling of crevices in the surface by water and also involves the work of leaving a thin film of water on top of solid. As we had seen earlier in systems of this kind hysteresis will be observed because the advancing and receding contact angles would not be same and we tried to assign significance to the area of hysteresis loop that would correspond to the work lost in removing adsorbed gas from the surface cracks. Sometimes that involves also the work of leaving this thin film in the cracks or scratches after withdrawal of the slide. So, to get a feel for numbers if we take a paraffin wax slide of perimeter 1 centimeter F would be obtained as about minus 24 dynes and that means gamma LA cos theta when equated to minus 24 dynes per centimeter gives a contact angle of 110 degrees nearly equal to the measured value. Should you have a surfactant like meristille tri-methyl ammonium bromide dissolved in water then this wax solid surface can be made weightable. This will immediately give you F of 15 dynes per centimeter 15 dynes this is for unit perimeter. So, 1 centimeter is already there incorporated and if we equate that to the gamma LA cos theta we will find that here the contact angle necessarily will correspond to the weightable situation. The positive F shows that after the adsorption of the surfactant monolayer the wax slide will be immerse spontaneously unless the restraining external pull F can prevent that. Waiting will be complete if F tends to gamma LA that is theta will be then less than or equal to 0 and this is attained at higher concentrations of MTMAB greater than very small value of 1 milli normal and together with the concentration of a solid surface like a glass we would see that the glass will get de-weighted. So, what can weight paraffin surface will be de-weighting glass and it could leave the surface dry. So, by measuring gamma LA as well as F 1 can obtain the magnitude of contact angle. So, that is the first thing that is the result of the weighting balance method we can measure the contact angle from the measured vertical pull F and the measured surface tension and accidentally one would do this experiment for two kinds of solid surfaces one for which the liquid shows a weighting behavior in which case theta will be 0. In other case theta is non-zero by measuring the vertical pull in the two cases F 1 and F 2 and using equality of F 1 2 gamma L cos theta 1 and F 2 equal to gamma LA cos theta 2 where cos theta 2 will become 1 because theta 2 is 0 we should be able to get gamma LA and therefore, cos theta and therefore, theta. Once we know the contact angle then by using this pair of equations S equal to gamma LA cos theta minus 1 and W S L equal to gamma LA cos theta plus 1 for theta greater than 0 we can obtain the spreading cohesion and the work of addition. Remember this is the spreading cohesion which is negative. So, with that quick recap we move on to the main topic of discussion here one. This one is first to address the Caesile drop method. Once again the Caesile drop method yields you a negative spreading cohesion. So, the situation is one of equilibrium. This method involves measurement of the height of a large Caesile drop lying in equilibrium on a smooth solid surface. Before we go any further I would like you to look at some lighter side of this word Caesile. The dictionary meaning of Caesile drop the adjective Caesile has come from the life sciences. We will take two versions the botanical version is that is something attached by the base or without any distinct projecting support. So, in the context of leaves it would be a leaf directly issuing from the stem alright. So, that is the first interpretation. The zoological meaning of this is it is something permanently attached and not freely moving ok. We may not go into the origin of this word, but rather look at the next slide. Incidentally this picture is taken from a promotional still from 1937 feature film of Laurel and Hardy way out waste. What is the connection of the two gentlemen here to the Caesile drop? What I wanted to indicate was here you do not see any neck. In a sense this head could be described as Caesile head it is directly supported by the base. Coming back to our measurement in our experiment this is an exaggerated figure of a liquid drop placed on a horizontal smooth solid surface. So, we have a drop with radius r and height h. This density this liquid L has density rho L and we presume the volume of the drop capital V to be constant. Now we could imagine for a large drop of this sort a differential infinite symbol change in the equilibrium position of the drop. For instance one could imagine that this drop is let us say spread slightly onto the solid surface. It is an equilibrium drop, but we allow it to spread infinite similarly in which case the radius r will increase by an amount delta r. Because the volume of the drop is constant the height of the drop h will diminish by a magnitude delta h alright. It is here that we propose to make the energy balance. You may want to reflect on this figure a little more. What all things can come into consideration? What all factors merit accounting here? First let us say we look at what happens to the potential energy of this drop when it is spread slightly so that the center of gravity moves. Original location of the center of gravity should be at h by 2 from the base, but since the height has dropped by delta h the center of gravity would have moved a small distance downwards. So, there will be loss of potential energy that is the first consideration. The second consideration is we will have some additional area on the solid which was originally in contact with air here. I am not indicated, but the rest of the space here is air. That additional area which was in contact with air would now be in contact with this given liquid. So, there should be a change in free energy from the initial surface energy of the solid to the final interfacial energy between the solid and liquid. Is that all or would there be anything else? That is the question I can pose before you. Solid liquid interfacial area we said what was already there that has become slightly larger. So, we are taking care of that, but do you suspect there is something else operative here? Remember in all analysis it is important to know what is your relevance list, what all factors relevant in your problem must not be missed out rate of increase of r. We are talking of let us say just the differential increase and it is an it is a differential change in an equilibrium system. So, by very tenets of thermodynamics we know that time is not to be worried about here, but there is something else perhaps what is it? Total surface area of the drop changes. Total surface area of the drop changes. So, what would that bring in? How would one account for surface energy of the liquid if it changes? Let me give you a hint besides energies may be it is worthwhile to always keep in mind as far as we deal with liquids its surface tension and already large drop when it is flattened a bit little more we would have to do certain work and if it is an equilibrium system which is not receiving external work then that work will come at the cost of the energy of the system itself. So, that should account for it with that question prompting you to think perhaps it will be easier for you to grasp the next equation. We are looking at the total change in energy in this equilibrium system that must be equal to 0. So, right hand side is clear. The first factor that I talked about is the loss in potential energy. So, if there is a differential change in the height h by dh it will mean the loss in potential energy corresponding to lowering of the center of gravity from h by 2 to h minus delta h by 2 right. So, total height is decreased by delta h. So, center of gravity will have shifted by this amount h minus delta h the total height of I mean the height of the CG will be h minus delta h by 2 initially it is h by 2 finally, it is h minus delta h by 2. So, the displacement of CG is by delta h by 2. So, loss in potential energy can be given as minus delta h by 2 and the product of this rho L V times G M G delta h by 2 V is constant. So, rho L V is the mass of the drop times G is the weight of the drop M G h M G delta h by 2 is the loss in potential energy clear. When we have allowed the drop to spread a bit the increase in the area will be 2 pi r is the perimeter of the drop in the beginning times delta r is here I think I must clarify a bit more here. Please remember that this drop is relatively large and in our simpler analysis we are taking an approximation that this can be represented as a cylinder. So, the curvature at the edge of the drop is neglected in a more precise analysis we can allow for the actual shape, but supposing it is a drop which you can regard as a cylinder then the perimeter will be 2 pi r and when the radius increases by delta r then 2 pi r delta r will be the additional area. It is that additional area that we have to worry about in estimating the change in energy. Since we are looking at the change in energy free energy we can look at it this way this is a loss term. So, we have put a negative sign loss in potential energy. So, that should guide us writing the other terms of this equation this is the area over which spreading has occurred 2 pi r d r f s a s minus f s l s is the change in free energy f s a s is the free energy of the surface in contact with air that is initially. Finally, it will be f s l s. So, f s a s minus f s l s is the decrease in the free energy of the solid when it has been invaded by the liquid ok. So, these terms I think were obvious to you from the beginning at least the factors responsible for these 3 terms. Potential energy loss change in the free energy when the solid was initially in contact with air later in contact with liquid right. The other thing is when this drop has expanded a bit you can look at work done against the surface tension. The surface tension force was operative over a perimeter 2 pi r. So, when the perimeter has increased we would have equivalent energy loss where this force is 2 pi r gamma l a and the action of this force is through distance delta r or d r. So, because it is an expansion that has to come at the cost of energy this will correspond to the loss of energy in having this infinite simile expansion of the drop against the restraining action of the surface tension ok. So, once that is noted then I think this equation is self-explanatory. Once we have this in this slide I think I may add this additional part the first term is the product of additional area which is this term and the energy change in spreading over the additional area which comes principally from the solid being now exposed to liquid from initial exposure to air and the loss of energy due to the expansion against action of surface tension and the second one is already explained to you. Now comes the approximation that this very flat large drop can be taken as a cylinder. So, if it is a cylinder then the volume of the drop V which is constant is going to be pi r square pi r square into H where H is the height of the drop. So, area times height that is the volume for you. Since V is constant we could differentiate this equation and write it as 2 pi r d r times H plus pi r square d H that is equal to 0. First differentiate this so you get 2 pi r d r times H and next plus pi r square times d H V is constant so this should be equal to 0. Now here we can replace for pi r square V by H pi r square is V by H. So, if you substitute for pi r square V by H then we get 2 pi r d r is equal to minus V by H by this H. So, minus V by H square d H right and now substituting this 2 pi r d r replaced by minus V by H square d H in equation 13 minus V by H square d H minus V by H square d H we have V d H cancelling and this we note f s s minus f s l s minus gamma l a is the spreading question right. So, substituting for 2 pi r d r as minus V by H square d H and cancelling V d H we obtain s equal to minus half rho l g H square clear. We have V d H cancelling from here only H square is left in the denominator. So, that goes on this side and therefore, s is equal to minus rho l g H square look at the right hand side density acceleration due to gravity and the height of the drop all are positive quantities. So, this is a negative s as has to be there for the equilibrium non-spreading drop to begin with. Question now is what are the precautions one has to take while making use of this analysis or the method of sysile drop for estimating this negative spreading question. We only need this height of the drop density of the liquid can be easily measured. To measure the height of the drop we have some options we will come to those options, but even before that I would like to tell you something more about how large this drop can be or should be. Obviously, it cannot be such a small drop that we cannot take the cylindrical approximation for its shape ok. So, sysile drop typically is a large drop yeah. So, according to that equation how will we accounting for the change in the solid like if we change the solid will be estimate the same. Interesting question the spreading question the spreading question is if you recall a difference of the work of addition of the liquid to the solid and the work of cohesion of the liquid right. So, whatever is the interaction between the liquid and solid and whatever is the intramolecular interaction for the molecules of liquid would all be reflected in this single entity H. So, if we were to take a different solid you would expect a change here. Correspondingly the spreading question the difference of work of addition of work of and work of cohesion will be different. So, experimental is very simple you do not really actually measure the contact angle you do not need to measure the contact angle you straight away measure only the height of the drop. Provided the approximations made are complied with bio precautions this would give a very quick reliable way of estimating negative spreading question. So, what does one experimental do one places a drop of liquid on a clean horizontal solid surface and just waits for equilibrium to be established. Then the height of the drop is measured in two ways you could use a spirometer or a long focus cathetometer. So, very quick introduction to spirometer and long focus cathetometer can be provided by the next few slides. This is just a schematic of a spirometer you see here there are three legs third one is not visible that is behind here there is a central one here. And we have this capable of rotating in the process the vertical displacement can be measured two turns of this disc will correspond to about tenth of a centimeter on this centimeter scale. So, this is going to move up and down because it is supported with a screw here. So, when you rotate this it will correspond to upward movement and that is measured related to this. So, in typical measurement of the radius of curvature what one does this is a rough photograph of the apparatus here these three legs visible this is the central one that is the disc and this is the vertical centimeter scale. This schematic diagram will tell you that here we are looking at the two legs over here this is the central one and this circle is supposed to be what is circumscribing the three legs this one this one then one behind radius of that is to be read as R this is written E here it should be R. And the difference between the heights of the central leg and these three other legs which are in the same plane that will be H radius of curvature then can be found out as R equal to R square by 2 H plus H by 2. Once you know this the height of the drop can be found out or simpler still will be a method that most of you might have come across in the experiments that you might have done in the mass transfer laboratory while determining the gas phase diffusivity of volatile liquids. What you do is you contain the liquid in a thin capillary the liquid should be volatile and it is laid vertically and a stream of air which does not contain any vapour of this liquid flows past the top of the tube. In the process the liquid evaporates at the interface between air and the meniscus of the liquid and the evaporated liquid then makes its way through the diffusional path between the interface and the top of the tube. If one measures the drop in level of the liquid with time one would be able to use a steady state diffusion analysis to calculate the diffusivity. And to measure the change in the height one uses the long focus cathetometer or a microscope cathetometer which is shown here. So, it allows you to focus on the meniscus at a distance and any change in the level of the meniscus can be measured accurately with this cathetometer. So, you could use the same thing for measuring the height of the drop. Once the height of drop is then known then minus rho L gh square gives you that spreading cohesion. So, this is the only thing which is necessary to measure density of liquid of course, is of course, straight forward. Now how do we comply to the approximation that this drop can be taken as a cylinder? We have to ensure that the radius is large compared to the height of the drop or should be much greater than h. For most liquids if you take a drop volume of about 1 ml this should be a safe choice. And for measurements to be reproducible of course, with the same requirements that the solid should be smooth and clean. So, what kind of values do we get and what is the accuracy of this method? For paraffin wax water system from the height measurement we obtain the spreading cohesion as minus 100 dyns per centimeter by this method. If we were to use the wetting balance method we would get values from minus 99 to minus 100 dyns per centimeter. And if we take the help of a goniometer to measure the contact angle and thereby measure the spreading cohesion from that gamma L e cos theta minus 1 kind of equation we get minus 99 dyns per centimeter. So, this should be pretty much convincing that societal drop actually offers a useful direct method of measuring the negative spreading cohesions. That kind of concludes all the major discussion about negative spreading cohesions that I intended to do in this lecture. That brings us then to the bigger question how do we how do we make measurement of positive spreading cohesion? How do we get the positive s values directly? In the backdrop of what we have seen so far I would like you to think now. How do we measure a positive s for a given spreading liquid on solid? If we take that solid and place the liquid just spreads all over forming a very flat film leads to no measurement. Even societal drop method is not adequate now that was only for negative values. I will like you to connect this problem with at least two earlier notions. We did have two or three occasions in which we needed to talk about a liquid spreading on solid. Because in this system theta will be equal to 0 we had difficulties. I am just trying to connect this question with as many things as possible that you already know so that you get a good hangout what this idea is all about. May be the first to come to your mind could be when we talked about an adsorb monolayer on surface of pure water. We know that if there is a surfactant monolayer on top of water the surface tension drops and it drops by a magnitude which corresponds to the surface pressure or film pressure of the monolayer. So, if the pure water has surface tension gamma 0 and the surface pressure of the monolayer is pi then the actual surface tension that you get to measure is gamma equal to gamma 0 minus pi. It is a diminished value below gamma 0 by magnitude pi right. So, this is the first point of contact second point of contact was in relation to the spreading. We talked about different liquid drops being placed on water and that was the discussion when we are trying to get started with exposure to contact angles. You take a heavy paraffinic oil drop it rests on water giving you equilibrium shape. If you take a lower boiling paraffin oil we might be able to just manage spreading. If you take a small chain associated with alcoholic group we will have a very large spreading question. So, the liquid can spread quickly on the solid surface and when you have that strong positive spreading tendency even if there is there is contamination on the surface this spreading liquid highly spreading liquid will just push out the impurities and occupy the position on the liquid surface. So, this is second point that I want you to recollect. If these things are clear then many things become integral in your mind. Third thing what you should you should be thinking of will be about the piston oils and indicator oils. Once again that was the concept to characterize the contamination of on water bodies. So, they were two different approaches. In the indicator oil you could think of spreading liquids which which have different spreading pressures. So, that you can you can have a thick enough film to yield interference pattern which can be related to thickness. In case of piston oil it should have sufficient surface pressure to be able to overcome the surface pressure of the contamination. So, while taking different piston oils with different spreading pressures you could be able to bracket the existing pressure right. Yet another point which will be a relevant to this discussion conceptually is when we talked about determining the work of addition of a given liquid with respect to a given solid when this liquid spreads spreads on the solid. So, we talked of let us say a liquid L which is capable of weighting a solid S. There is no hope of measuring the contact angle here. So, we would not be able to use the Laplace Young equation. But then a little reflection revealed that we could always think of a two liquid system where the second liquid M is chosen such that it does not weight the solid. And now looking at the measurable contact angle for the given solid between these two liquids we could establish an energy balance which allowed us to extract WSL from all other measurements right. We also said that the liquid L which weights a solid will be capable of pushing the liquid M from the surface although it can take time. All these have one generic principle that is operational here that is given a strongly spreading tendency of a liquid we might be able to offset it by the action of opposing surface pressure of another entity. It could be a layer of another liquid or it could be a surface layer. Here in particular we will look into the action of a monolayer. So, now I think we are in a position to go further and understand this method quickly. The spreading cohesion S is positive and we try to restrain it with the help of surface pressure of a monolayer at air water surface. So, what we do is maybe I can show you the diagram first and then come back to this slide. In this schematic diagram we see water, air, solid and this liquid L ok. It has not been indicated by any line here, but there is a thin layer of water on the solid even to begin with before we place the drop over here. Once the drop is placed the drop will squeeze out this thin film of water push it out because it has affinity for the solid and in the process that pushed out water will get accumulated around its periphery. What we do is to offset the spreading tendency of this drop over the solid we do not start with water, but an aqueous solution of a surfactant whose molecules are indicated here as these chains coupled to heads right. So, there is a surface pressure associated with this monolayer of surfactant on water. So, when this drop tries to spread it will push out this surfactant solution and it will pile up around this drop. In order to spread further it will push this layer of surfactant solution further it may happen to an extent, but after certain extent of spreading it would reach an equilibrium situation provided the surface pressure here is sufficient to counteract the spreading pressure of this drop. So, we have few notations to contain here with as usual air, water and solid are indicated by AWS liquid L by this liquid is L, the surface tension acts along this direction gamma L A, the surface tension of the surfactant solution is gamma which acts as a tangent to this surface and then the contact angle here that is theta 1 other notations are FSLS for the interfacial energy between solid and liquid and between water and solid FSWS. So, what we done is we started with the solid surface kept just below the surface of this aqueous solution of surfactant in a Langmuir trough and then the oil drop could be prevented from spreading with the help of a monolayer which could be steric acid or satil alcohol. Now, this is the situation that we need to analyze, there could be some criticism for this method, one of the strong ones could be is there no possibility of contamination after all you are using a surfactant or a surface activity like steric acid or satil alcohol, would it not contaminate the solid and would it not change its interfacial energy with the liquid? It turns out the experiments reveal that other monolayers than steric acid also give you the same values, accidental values. So, that effectively counters the criticism that they may be contamination of the solid surface altering its interfacial energy. How do we analyze this? First, the basic definition of spreading cohesion for the given liquid on the solid S is FSAS minus gamma LA plus FSLS. In your mind you can replace FSAS by gamma SA FSLS by gamma SL and it is the definition of spreading cohesion of one liquid on another. For solids we replace the surface tension interfacial tension by the surface and interfacial free energy per area. This we know is positive, we need to measure it. So, we look at our equilibrium situation when our drop is in contact with this monolayer covered water all around it. At this point how would we set up the force balance? You know that there will be surface tension between solid and liquid operative. There will be surface energy between solid and liquid operative, surface tension of liquid, surface interfacial energy between the solid and water, surface tension of this aqueous solution, contact angle and one more entity which you might not foresee is the hydrostatic head of this water layer which will be trying to restrain the drop here. So, if we take all these factors together, keep this in mind with this diagram in mind we should be able to understand this equation. G is the hydrostatic spreading force of water layer that is helping the attempt of the liquid to keep to itself in vicinity of the solid FSLS plus gamma LA. Some of you might think why is there no contact angle associated with gamma LA. Remember that this was a kind of highly spreading liquid on solid. So, even when you have this restraining monolayer the drop will still exhibit considerable spreading tendency. It will just be overcome by the surface pressure of the monolayer. So, that angle we take as quite small close to 0. So, we have gamma LA. This is what is trying to help the drop keep to itself the left hand side. The right hand side on the other hand is gamma cos theta 1 in the opposite direction here component of gamma operating in this direction plus FSWS right. There should be 2 things in the gamma water layer and gamma water solid. Why is not why is there. No, no hold on. I while I try to help you to make the force balance and the vectorial concepts visualize the equations never let go the fact that solid is ultimately different from liquid ok. So, we have 2 liquids here L which has surface tension of gamma LA. Another liquid which is the surfactant solution in water which has the surface tension gamma and then there are 2 interfaces with the solid of L with S and of W with S. Those should be expressed in terms of the surface energies. So, free and this will be interfacial energy FSLS and FSWS. What you are trying to think of is why is there no interfacial tension between solid and water that is represented here by the energy right. Instead of dynes per centimeter we are talking of urges per centimeter square. So, we have this action trying to restrain the drop and this is what is trying to pull it apart. So, the factors which are trying to pull it apart are the interfacial energy between solid and water and the surface tension of the surfactant solution with this component. And then these 2 plus the hydrostatic pressure of this water layer over here. Now, how do we go further? Maybe we have to think of yet another situation. This oil L has a positive spreading tendency for this solid over this solid which means we could think of a pure water drop resting on the solid that would give you equilibrium shape ok. So, let see what is the consequence of that. If we place a pure water drop on the solid surface same solid surface surface tension of water is gamma 0 acting along the tangent. F s a s is the free energy of the solid in contact with air. F s w s is the free energy of the solid in contact with water and now that the oil spreads on the solid water would not spread. So, equilibrium contact angle is theta 2 this should be measurable right. So, we repeat gamma 0 is now for pure water theta 2 is the contact angle for water drop placed on the same solid surface in absence of any monolayer. In principle then we should be able to make a force balance here. So, that is the next thing. This equation 17 is the force balance for system of pure water drop resting at equilibrium on the same solid. Here the hydrostatic pressure is trying to help the spreading. F s a s is the free is trying to pull the drop outward. The hydrostatic pressure of this drop is trying to help it. Gamma 0 and F s w s act in the opposite direction counteracting. So, we have gamma 0 cos theta 2 plus F s w s ok. Now, this is all we require. We have the definition for spreading cohesion. We have the balance of the surface energy and surface tension acting. For the drop trying to spread on a monolayer covered water on the solid and the last one is just the drop of water pure water on solid. All we need to do is combine these three equations. For F s a s we substitute from equation 17 from pure water drop. For F s l s we substitute from water drop or like oil drop, l drop resting on monolayer covered water surface trying to push it. So, once you substitute for F s a s and F s l s from equation 17 and 16 in this spreading cohesion, the result you get is this. Gamma 0 cos theta 2 plus F s w s minus g this is what F s a s is that is from this equation gamma 0 cos theta 2 plus F s w s minus g. So, we substitute for F s a s here and over here we have gamma l a which we leave as such for F s l s we substitute from here as gamma cos theta 1 plus F s w s minus g minus gamma l a that is over here gamma cos theta 1 plus F s w s minus g minus gamma l a and then we see that gamma l a will cancel off and F s w s minus g will also cancel off leaving only the terms containing gamma 0 cos theta 2 and gamma cos theta 1. So, that is our result s is now gamma 0 cos theta 2 minus gamma cos theta 1. In principle this is the equation which one can use for measuring the spreading cohesion because gamma 0 is known surface tension for the monolayer covered surfactant solution is also known and the two contact angles. Maybe we will just take a logical break here and resume from here. In principle this is to be simplified for different materials and we have to check how good the predictions of this equation is prediction of this equation is in the face of external values.