 In this sixth module on mass transfer across interfaces, we today talk about some introductory concepts. Then, we look at the resistances present in different steps and their magnitudes and finally, look at evaporation and how possibly it could be retarded. I suppose you have some idea from what I had mentioned about evaporation in some of the earlier lectures, but we look at that aspect a little more in depth today. So, before I talk about the resistances, a quick recap of what I covered in the last lecture would be in terms of how a solute molecule before it could arrive at a site of reaction would be influenced in its path and in its rate, would be dictated by how it comes to that spot in the first place. So, having just concluded the module on surface reactions, one could imagine that possibly for reactions to occur in the surfaces, the reactants may have to come from the bulk of gas phase, which is shown over here or from bulk of liquid or possibly if it is reaction of two species, then either reactant could come from each of these phases, but in coming to the interface where reaction would occur, it would have to necessarily be subjected to the resistance, one in the gas phase, gas phase is R G, the other in the liquid we represent by R L and R I may not matter if the reaction is in the surface, but if the reaction has to take place in the bulk of either gas or liquid, then even the interfacial resistance will come into picture. Generally, this is the resistance which is omitted in the introductory undergraduate discussions on mass transfer, but we will take that into account and I have briefly told you something about R I, we will look into it a little deeper today. So, a molecule when it passes across the gas liquid interface, it would have these three resistances added up R G in the gas phase, R I in the monomolecular layer or region which constitutes the interface and R L which is the resistance for mass transfer diffusion through the layer of liquid below the interface, that will be the total resistance. Although we are mentioning diffusion and mass transfer almost synonymously in this initial part, please remember a little fundamental feature of mass transfer, how a solute molecule can come to the site of reaction may be dependent on what kind of movement the solute is undergoing. The base level case is when the given phase is stagnant or stationary, stationary in the overall sense that there is no net movement visible for that phase. If that is the case, the only way molecules of a certain species could come to the reaction site would depend on diffusion. However, if there is a bulk movement possible on account of agitation or by intrinsic fluid movement which may be present either because of fluid flow under the influence of pressure difference which will be force convection or because of differences in densities on account of either differences in temperature or in concentrations which would then constitute the natural convection. So, convection which refers to bulk movement could be force convection or natural convection and natural convection can arise out of the expansion, cohesion, buoyancy or because of differences in densities created as a result of mass transfer itself. But whatever the mechanism the solute may be able to come to a particular point through a combination of some kind of convection if it is present and diffusion. But please remember that when we talk of mass transfer we must forever remember that diffusion is the signature it has to be there diffusion must be there in mass transfer its contribution may be minuscule, but it has to be always present. So, without diffusion being there we would not really call it mass transfer. And incidentally now that need not be over emphasized diffusion you all know is a molecular level transport it is arising out of may be concentration differences or more appropriately chemical potential differences, but it refers to the molecular level transport. And while individual molecules are free to move the way they want to there is a free will almost on account of thermal kinetic energy for each molecule. But when you look at the statistical picture a large ensemble of molecules you always find that there is a diffusion always taking place in the direction of decreasing chemical potential or more often less restrictively on account of differences in concentration. But this is only the molecular level transport and necessarily the slow step. Convection on the other hand brings about chunks of large elements of liquid or gas containing several tens of thousands of molecules. And therefore the transport rate contribution by convection is generally much too large compared to diffusion, but nevertheless diffusion is a must if the process is to be identified as a mass transfer one. So, with these three resistances resistance in the gas phase and resistance in the interface resistance in liquid we get the total resistance. Today we see how these resistances could be gotten feel of in terms of orders of magnitude rough numbers. So, that you have a perspective about the whole scenario of what are the contributing resistances. So, this picture is already covered. Usually the liquid phase resistance is the highest and that corresponds to the molecular diffusion of solute through a non turbulent liquid layer close to the surface. Non turbulent meaning you permit either the liquid to be stagnant or to follow a laminar orderly flow of liquid which does not correspond to any chaotic mixing as would be typical in turbulent flows characterized by a large Reynolds numbers. However, there can be exceptions we can sometimes make the liquid phase resistance. So, low that unexpectedly R i or R g the interfacial resistance or gas phase resistance could become rate controlling we will see how. So, to get an order of magnitude or of magnitude feel for these resistances we look at the basic differential equation corresponding to transfer across any plane surface and that is this d q by d t where q is the moles being transferred. So, d q by d t is the rate of transfer moles transferred per time that would be the area which is perpendicular to the direction of transport. A permeability cohesion or alternatively mass transfer cohesion and a certain concentration difference which acts as driving force. I had mentioned last time that if you take this as a rate and this as the driving force then the resistance will be 1 by a k for overall area a for unit area it will be k inverse. And that same generic form that rate is ratio of driving force to resistance will hold for any rate process it could be as different as let us say flow of electrons to a metal conductor. Incidentally delta c will be measured in moles per centimeter cube it is a bulk concentration, concentration per volume moles per volume. So, moles per centimeter cube gram moles per centimeter cube and that will mean the permeability cohesion or mass transfer cohesion is measured in centimeters per second. There is a variety of ways in which you can write the driving force will stick here to the simplest one moles per volume. Once you choose to express the driving force in terms of this basic concentration moles per volume the mass transfer cohesion of permeability cohesion comes out to be having the same dimensions as of velocity. Remember mass transfer cohesion has units of centimeters per second same units as of velocity. But only when the concentration driving force is used that is moles per volume that is the concentration and you use gram moles per centimeter cube as the units per concentration k inverse as I just remarked is the resistance and that will be always in seconds per centimeter provided delta c's gram mole per c c. We could rearrange that in that general manner and k could be identified as reciprocal of resistance and when we equate the rate of transfer what is given by here d cube by d t divided by a on the left hand side equal to k delta c. And if you write this in terms of the diffusional mechanism then k turns out to be equal to the ratio of diffusivity and delta x which is the thickness of the region which offers that resistance in that particular phase. So, k is reciprocal of r the rate constant or mass transfer cohesion or permeability cohesion is reciprocal of mass transfer resistance which is in turn given by d by delta x. This is with a very simple model for mass transfer which postulates that there is a steady state transport and we then have the film theoretic expression from mass transfer cohesion being proportional to diffusivity. Is this relation from experiment or? What is you may say in the to the extent that fixed law gives you the mass transfer flux only to that extent it could be regarded as experimental. But since I have already mentioned this earlier there is possible to derive fixed law itself from statistical mechanics we could regard this as completely theoretical term. It is a different matter that this theory may not be in general agreement with experiment across all ranges of parameters and you recall especially the people who have had second year undergraduate course with me that it is this particular dependence k proportional to d that is the basis of the variety of models that were that were required. In practice you do not find the proportionality of mass transfer cohesion to the diffusivity to the power 1 it could be anywhere between 0 to about 0.8, 0.9 that calls for different models some of those models will refer to here also alright. So, this is what is the genesis of that d cube by dt is written as a d by delta x time delta c or a times d into delta c by delta x that is your concentration gradient delta c by delta x. Now let us talk about R i the less talked of the resistances which are pertinent to any mass transfer across interfaces. If we have a clean liquid surface then we would expect the resistance attributable to that monomolecular region corresponding to the surface phase to be relatively low right. It would depends primarily on a certain probability a chance that a molecule may have enough energy or not to evaporate from the bulk of liquid escaping from the surface into the gas phase by overcoming that attraction of the neighbors. If it has that much energy it would make its way into the gas phase and that is a probability the chance in general. If we consider two way system at equilibrium the rate of evaporation should be also equal to the rate of condensation that we had seen in detail in kinetics of evaporation that we discussed in one of the early lectures. So, number of molecules which condense on to a unit area per second will be same as the rate of evaporation and that you know is given by a kinetic theory in terms of number of molecules per second per centimeter square from vapor as the equilibrium vapor pressure p divided by square root of 2 pi m k t where small m is the molecular mass and k is Boltzmann constant t is absolute temperature. And as you know this is a very high strike rate number of molecules impinging on the surface per area per time as per kinetic theory is very large actually. However, not all of these chance impinging molecules will be able to make their way into the liquid many of them will be reflected back. So, only a small fraction actually would manage to condense and that is where we had brought in this concept of a fraction alpha which would decide how many of those arriving molecules actually make it to become a part of liquid and this can be a pretty small number experimentally determined magnitude is 3.4 percent 0.034 for alpha. So, we modify our expression for rate of mass transfer as follows per area the mass of material condensing or evaporating w would then give you 1 by a d w by d t as rate of condensation that would be equal to the total number striking per area per time that is this p by root 2 pi m k t times alpha alpha is the fraction a small fraction of this and that number into mass of each molecule will give you the weight per area per time. So, we could express m in terms of Avogadro's number and rewrite this as follows first thing is this m if it is taken under net here root this root m cancels. So, we get p alpha root small m by 2 pi k t to the power 1 by 2 and if you multiply numerator and denominator by Avogadro's number small k times Avogadro's number is gas constant universal gas constant small m into Avogadro's number is a molecular mass. So, there we have the gas constant in irks per Kelvin and other units are CGS units. We could write it in terms of number of moles if required and that could be then from here we could write this as d cube by d t is equal to a alpha by d t is equal to a alpha by d t is equal p by 2 pi m r t under root all you need to do is divide through by capital M. So, we get d cube by d t as a alpha p by 2 pi m r t under root or a alpha and we strategically pull out r t from here. So, write this as p by r t and then this becomes a alpha square root of r t by 2 pi m all right. This is reason why we do this because you would see that this will bring in the concentration in the gas phase for the condensing material . So, d cube by d t is now a alpha square root of r t by 2 pi m into p by r t. Let us inch forward and see whether we can bring in the driving force. So, that we have a physical understanding of what this right hand side is for that we consider the condensation process at a molecular scale. What happens here? We have vapor condensing into the liquid. So, just by above the liquid surface the vapor concentration may be said to be C 0. What would be the vapor concentration in the liquid surface itself? Vapor concentration in the liquid surface itself in the liquid surface there is no vapor. So, concentration is 0 right. So, we could say concentration C 0 just above the liquid surface minus 0 is the concentration driving force and that you know is same as C 0 and C 0 is p by r t. p is the equilibrium vapor pressure of the solute or the condensing component divided by r t from the ideal gas law. p v equal to n r t gives you n by v equal to p by r t that is the gas phase concentration. So, we can write our equation 6 now in terms of this d cube by d t equal to a alpha by root of r t by 2 pi m into C 0 minus 0. We merely substituted for p by r t there a alpha the square root r t by 2 pi m and p by r t is replaced by C 0 minus 0. So, the resistance r i would be related to the permeability coefficient or mass transfer coefficient here based on the resistance in the interface k i is 1 by r i the rate constant is always inversely related to resistance and that is alpha root r t by 2 pi m we leave out the area we leave out the concentration driving force alright. So, now we have everything to estimate the magnitude of this resistance of the interface at given temperature. We could do it for water for instance and alpha will have to be taken from independent experiments as mentioned earlier and if you do this calculation then it works out to be two thousandth of a second per centimeter. So, let me just record this in this diagram so that you have an idea comparative idea of the resistances. So, we have an interface here let us say the gas phase here and liquid phase here. The interfacial resistance r sub i we are saying is 2 by thousandth of a second per centimeter, two thousandth of second per centimeter provided we are talking about clean water alpha equal to 0.034. With this value of alpha we get the free energy of activation of about 2300 calories per mole and this is the kind of free energy required to produce or find a hole in the liquid surface. In order for this condensing molecule to find its way into the interface you need to have this much energy expenditure 2300 calories per mole. What happens if we have contamination interface is not clean maybe let us say we have a monolayer covered surface of liquid. There can be a dramatic effect on the magnitude of alpha, alpha which is telling you how much of the arriving percentage of molecules is actually becoming part of the liquid surface that factor could reduce by a factor of 10,000. So, if there is a four orders of magnitude reduction in alpha we expect r i to go up correspondingly. So that means we have an idea about r i possibly increasing from this 2 by 1000 to about order of 10 seconds per centimeter. So, we have some idea about how small r i can be how large possibly it could be a few thousands of a second per centimeter to 10 seconds per centimeter that is a range of r i values. Next resistance in question is the gas phase resistance. So, we could talk about a situation where above water surface there is only air and we could think of measuring the net rate of evaporation. If we do that r g the gas phase resistance could work out to about 80 seconds per centimeter. So, let us say we talk of air this liquid is water and r g could be about 80 seconds per centimeter. Look at the highest magnitude of the interfacial resistance 10 second per centimeter gas phase resistance atmospheric air condition 80 seconds per centimeter. So, keep it in mind that in general under ordinary conditions the gas phase resistance itself can be quite large compared to interfacial resistance. One may question why is this gas phase resistance there? It is may be for evaporation of water into air or condensation of water from air into water. Water has to go through an overwhelmingly larger concentration of air components oxygen nitrogen whatever components of air those molecules. It is not that a water molecule starting from here will directly reach the interface. It will have these millions and billions of molecules in between of other kind with which it will necessarily collide. So, it will never be able to make it with straight rather it will go in a zigzag fashion traveling a great distance in total, but very little towards the interface. It is only a very small fraction of the total distance a molecule will actually travel which takes it towards its destination here right. That should remind you of the zigzag motion random motion of molecules in general and here it is a collision of water molecules against air component molecules. So, there is that idea that there is a collisional barrier before water can reach here it will have to overcome that resistance. So, if the collisional resistance is what is responsible for this high value of r g it should be conceivable that r g could be actually reduced somehow. If you understand that this is coming from the collisional barrier what is the most obvious way of reducing the gas phase resistance. Reducing the density of air reducing the density of air reducing the density of air reducing the pressure right. So, by simply starting with atmospheric pressure and evacuating the gas phase you will be able to bring down the collisional barrier and therefore, the resistance in the gas phase. So, we can reduce this quite a quite a lot we can actually bring it let us move to the slides. We can bring down the r g to only a few seconds per centimeter and you can do it by reducing pressure or possibly another way by improving the stirring of the gas phase in the vicinity of surface. If you have very very strong agitation of mixing of air then the stagnant layer which is offering that resistance that will be thin down in the film theory concept therefore, resistance can be decreased. You had a question ok. So, by reducing pressure or by increasing the stirring and therefore, mixing of the gas phase we could minimize the resistance and therefore, r g can be reduced to a few seconds per centimeter. So, that gives you an idea about the r g. We could say 80 to maybe a small number x seconds per centimeter remaining resistance is the liquid phase resistance. It might be safe to say that by and large that is the one which really matters in most cases and yet I start with this case that for a pure liquid r l is 0 for a pure liquid evaporating. If you look at the concentration of the liquid which is evaporating starting from bulk to the surface there is no change in concentration. If there is no drop in concentration there is no resistance then why do I say that in most cases r l is really the only resistance which matters. That is only a kind of a take home message a succinct message that in where most of the cases you will encounter in predominant number of those cases it is only r l which will matter. I am starting with the exception as usual evaporation of a pure liquid is an exception r l is 0. Most often you are interested in having certain species solute dissolving and diffusing into the liquid such as water. So, if you take this case like CO2 dissolving and diffusing into water the concentration gradient may extend over a very large spatial extent. The thickness delta x over which the concentration drop will occur is generally quite large. This is true for CO2 either diffusing from aqueous solution into air or the other way around. And if that resistance is to be characterized by the thickness of the liquid which is involved then obviously the liquid phase resistance would be quite high. In such a cases in such a case r l is as high as about 500 seconds per centimeter. So, let us put it down here the liquid phase resistance r l is of the order of say 0 to 500 seconds per centimeter. So, so far we have these ranges now r i is a few thousands to 10 second per centimeter r g about 80 seconds per centimeter could be reduced to just a few seconds per centimeter by reducing pressure by or by increasing mixing agitation. And r l will be depending on situation evaporation as 0 for transfer of a solute like CO2 could be as high as 0 500 seconds per centimeter. So, this gives you a comparative feel for the resistances involved while we talk about mass transfer across interfaces. R l depends on the way transfer is taking place. It will depend on the way transfer is taking place. It will depend on the thickness of the liquid which is involved. So, if it is unstirred liquid then you expect the resistance to be highest. If it is a stirred liquid then the stagnant layer close to the interface will be smaller here. It arises depending on the thickness of that stagnant layer and that thickness of stagnant layer will depend on the agitation in the bulk. If it is not present then for a steel liquid quiescent liquid you may say the entire liquid length theoretically from 0 to infinity everything comes and contributes to the resistance for mass transfer. If there is a stirring then it may be limited to a few hundred microns or lesser. Why is that R l is zero water? In case of the pure liquid is the same component which is getting lost. So, if it is let us say pure water then it is only water everywhere. Wherever the concentration is measured it is same. There is no drop in concentration and if there is no decrease in concentration effectively there is no resistance. Suppose a molecule is moving from right to interface then if the the agitation layer to interface moves there then the same molecule comes from back and replaces that. So, there is no drop in concentration at all. It is the same like molecules that you have. So, this delta x the thickness of the stagnant region, answered region in the vicinity of surface would depend on the mixing or hydrodynamics right and you would be able to see that through the dependence of delta x on the Reynolds number. Now, you just initiate the discussion of evaporation specifically something we have dealt with earlier. What if at the plain air water surface we have a monolayer a close packed monolayer of some long chain compound. We now expect a considerable increase in the interfacial resistance R i. Evaporation even with the monolayer present has bulk of liquid which is pure. So, R L is 0 here also necessarily with or without the monolayer present R L is 0. We expect therefore total resistance to be only coming from R G and R i. One would want to know what is the relative effect of monolayer on this total resistance. Obviously it will depend on this other component R G. How big R i is a fraction of R will depend on how big R G is and R G may be considerable if we have a stagnant gas film at atmospheric pressure just above the surface. Let us see and R G could be reduced by subjecting the system the gas phase to evacuation. It may be possible to reduce R G to become comparable to R i. In our diagram starting about starting at about 80 seconds per centimeter we may bring it down to 10 seconds per centimeter or lesser a few seconds per centimeter and therefore R G is reduced to almost similar level as R i which means some of the monolayers could actually in effect reduce the evaporation by a factor of 2 as demonstrated by radial in 1927 paper in general of physical chemistry. It is here we look at the difference between look at a slightly different way of measuring resistances as proposed by Langmuir. Langmuir suggested that the total resistance could be defined as reciprocal of measured rate. The measured rate is in grams per second per centimeter square and that reciprocal would have for clean surfaces with R L 0 R G plus R i. Of about 770 centimeter square second per gram. Experimentally if we have the water surface covered with a monolayer like of loric acid you find this resistance expressed as rate inverse of the rate of evaporation is about 1340 centimeter square per gram into seconds or the effect of monolayer is to increase R i by as much as 570 centimeter square second per gram. This is exactly the reason why people have thought of making use of monolayers for certain practical purpose. If we take the case of hexadecannol monolayer will offer very high resistance in these units 60,000 seconds centimeter square per gram or that will translate to about 1 second per centimeter. Experiments of these kind have been carried out by Seba's group Seba and Suteen what they did they put an entire Langmuir trough in a box which could be evacuated. So, you could actually carry out experiments for various different levels of gas phase resistance or they could be another method where you have a stream of dried air flow across the water surface and then you could measure the amount of moisture taken up and by that method again Seba with Brisco found the resistances as they vary with let us say chain length. At 20 dynes per centimeter they found that the resistances increase with the increase in the length of the hydrocarbon chain something expected the barrier is higher if the monolayer is sort of thicker because of higher chain length. For a C 22 alcohol this is dramatic the rate of evaporation reduced to just 19 percent of the rate for clean surface at that surface pressure 20 dynes per centimeter. What if we made this monolayer more compact compacted to double of that 40 dynes per centimeter and the evaporation rate would reduce to just 12 percent of that for clean surface or compress it further to about 48 dynes per centimeters and you could practically eliminate evaporation. It is a different matter how in practical system you could have that, but perhaps you do not need these dramatic figures we do not really need this C 22 alcohol. Take hexadecannol at 40 dynes per centimeter it can reduce the rate of evaporation to 60 percent of that for clean surface and the barrier for this interfacial transport R i in presence of long chain acids and alcohols is about 5 to 10 seconds per centimeter. This detectable even in steel layer when the gas resistance will be about 80 seconds per centimeter and yet with this hexadecannol we can get a very appreciable reduction in the rate of evaporation. If we use film of C 19 acid the enthalpy of activation is about 14600 calories per mole. You could use different chain lengths and find out what is the contribution per CH 2 group. It works out to be about 300 calories per mole of a contribution per CH 2 group to the enthalpy of activation for evaporation. A simple calculation here will indicate that of this total enthalpy of activation of 14600 calories per mole a very high value contribution of 9500 calories per mole will come from the carboxylic groups in the monolayer. You just have to subtract 300 times the number of CH 2 groups and you get from here about 9500 calories per mole from the CO H groups. If you think of long chain alcohols it will reveal even greater resistance from the OH groups. However, there are some pitfalls recalling your topics on spreading you remember that not all long chain molecules would spread quickly spontaneously on the liquid surface. We might have to edit it may spread but so slowly that is not practically useful. So, you might take recourse like you take that monolayer material dissolve it in a solvent which has a very high spreading coefficient and then let the solution spread over hoping that the solvent would be evaporated. We would have a spread film providing the resistance for evaporation retarding evaporation. Looks fine as an idea but if you use liquids like benzene as spreading aids somehow it remains in the surface and it causes this locations in an otherwise close packed monolayer making this barrier amenable to evaporation. And you could actually see that R I on account of these solvent molecules could be reduced by as much as 20 times. So, one has to be careful in choosing what material to use as a retarding agent and how we actually spread it. In any case one has to understand that evaporation is something one has to contend with. In hot and arid areas like many parts of a country we might have might be having annual lowering of the level of reservoir by as much as 10 feet. The losses might exceed the utilization. So, it makes sense to conserve water through some interfacial means like this. Even a monolayer of a polar oil can reduce the rate of evaporation and advantages will be will be small quantities required. And oxygen necessary for supporting the aquatic life would still be possible to diffuse into water. Coison surface of liquid has high liquid phase resistance and compared to the monolayer it may have a smaller effect. The argument which goes against thicker oil layers which could also be used for retarding evaporation. The chief argument against thick layer of oil is that it would retard oxygen diffusion to such low values that the aquatic flora and fauna might be endangered. Under natural conditions recalling our discussions on waves and ripples we know there is a lot of movement within the liquid which arises out of the transfer of momentum from the vent to water and that could contribute significantly to the mass transfer to reduction of resistance. But if you have a monolayer then it could retard the studying also the convection inside and besides offering resistance to diffusion of oxygen it would even decrease the uptake of oxygen. So, we need to be careful here and those considerations actually lead to a rough realization that we have RL almost doubled which means the oxygen deficit will be reduced also doubled. But the encouraging part is effect of placing a monolayer of hexadecannol on a reserve of water would reduce the dissolved oxygen contained from 90 percent saturation to about 80 percent saturation that does not have much effect on the living organisms in water. There by making this C16 alcohols hexadecannol as one of the most promising agents for retarding evaporation. Perhaps we can stop here we will begin with little more discussion on hexadecannol next time. Thank you.