 In the previous lecture, we were discussing about the thermodynamic basis of electrokinetics and we will continue with that. So just for a brief recapitulation, we will consider that what are the concepts that we had discussed. So we had discussed about the concept of chemical potential, chemical potential of a pure component and chemical potential of a component in a mixture. So in that way, we defined it through the partial molar Gibbs function expressed in terms of the fugacity for the pure component, fugacity of the pure component and for the mixture fugacity of the component in the mixture. Now to calculate the change in volume due to mixing, we considered the following expression. The expression that I was deriving in the board was for a special case when in the mixture there are 2 constituents A and B. Now whatever is presented in the slide is for a little bit more general case where there could be several numbers of components and we considered the ith component and write the chemical potential as mu i. So remember that we wrote that dG at TnB for the number of moles of the component B instead of nB we are writing nj for any j not equal to i. So it is like B not equal to A. So accordingly this expression we have written and the Maxwell's cross relation which is nothing but like the rule of partial derivatives that if dz is equal to mdx plus ndy then del m del y is equal to del n del x that is the Maxwell's cross relation. So that gives this expression. So eventually we get that d mu i is equal to vi bar dp which is the partial molar volume into dp at constant temperature and all ends fixed and by definition d mu i is equal to RTd ln f bar i that is the definition. So with this point let us go ahead. So we will write these expressions in the board and proceed further with that. So d mu a or at constant temperature is equal to or this mu a alternative notation is ga bar is equal to r bar Td ln is equal to va dp all the terms at constant temperature. Similarly you can write fugacity of the pure sorry the Gibbs free energy molar Gibbs free energy of the pure component a dga what is that this is RTd ln f a. What is this f a? This is fugacity of the pure component a. See fugacity is not a partial molar property. So that is why it is way of writing nomenclature of writing is different from the partial molar property. So this is equal to what? We calculated the change in volume due to mixing that is what is the change in volume due to mixing individual a is there and individual b is there you sum total the volume and after that you mix a and b you sum total the volume. The difference between these two is the change in volume due to mixing and that is what n a we will calculate individual terms va capital va bar – small va bar. So from the equations which are written at the top we can write va bar – small va bar dp at constant temperature remember all the things are at constant temperature. So you can emphasize it by writing a subscript t is equal to RTd ln f bar a by f a log of this – log of this. Now integrate it from a low pressure p star to the pressure p. Integrate it at constant temperature. Temperature is kept constant so this comes out of the t comes out of the integral from p star to p star means ideal gas state. So p star means a very low pressure. So then let us concentrate on the right hand side. So at p equal to p f a bar equal to f a bar and f a is equal to f a. So ln f a bar by f a – at p equal to p star what is fugacity of the component a in a mixture y a into p star mole fraction into the total pressure ln y a into p star divided by what is f a p star that is a pure component. For pure component fugacity at ideal gas state is same as pressure. So this is RT ln f bar a by y a f a. When will the change in volume due to mixing be 0? When this term is 0 and similarly capital V a bar – capital V b bar – small v b bar equal to 0. So for the change in volume due to mixing equal to 0 this should be 0 for delta v mix should equal to 0. That means for delta v mix equal to 0 you must have f bar a is equal to f a y a right and similarly f bar b is equal to f b y b. So you can relate the fugacity of the component in the mixture with the fugacity of the pure component purely through the mole fraction. It can be shown that for a ideal gas also the same thing is obeyed that means ideal gas is a special case of an ideal solution. Ideal solution is a more general case. So what is the definition of an ideal solution? The definition of an ideal solution is when the change in volume due to mixing equal to 0. So delta v mix equal to 0 this is the definition of ideal solution. This is not ideal gas. This is ideal solution. Ideal gas shows the same behavior. It can be shown that if you have two components a and b which are ideal gases and you mix them to form a mixture of ideal gases then the change in volume due to mixing is 0. That means ideal gas mixture is a special case of ideal solution. Ideal solution is a more general concept. Now we will find out what is the chemical potential of the mixture. So we will start with d g bar a is equal to r t d ln f bar a. We will start with this definition. Our objective is to find out an expression for the chemical potential in terms of the fugacity. So we will start with the corresponding differential form. Now we will integrate it p star to p but t is kept constant because these definitions are based on constant t. This is at constant t. I am not writing it all the time but it is constant t. So you integrate from p star to p. So if you do that you will get g a is equal to g a star plus ln f bar a divided by yes at p equal to p star what is f bar a? y a p star. So we will calculate. So what is our strategy? We have transformed the quantity g a bar from this state expressed in terms of an ideal gas state and this ideal gas state now we will write in terms of molar quantities instead of partial molar quantities. So now what is g? g is equal to h minus ts that is the definition of g no matter what whether it is mixture or single component or whatever g is h minus ts. So we have two components a and b. So g star is equal to h star minus ts star. Star means ideal gas. Now what is h star na cpa t plus nb cpb t? Now g star is equal to h star minus ts star therefore g star a bar a is equal to h star bar a minus ts star bar a. Just differentiate all the terms with respect to na by keeping tp and b fixed. So we will get this. Now what is h a star that is del h star del na at tp and b what is that na sorry del h star del na will be cpa into t and this is same as molar enthalpy of a. Now entropy a star minus s0 star this s0 star may be a hypothetical reference state is equal to na cpa ln t by t0. You are familiar with this expression you can use that basically where from this expression comes. You have tds is equal to dh minus vdp you can write molar quantities. So ds in for dh ideal gas dh is cp dt. So cp dt by t this cp these are molar cp. So cp should be there should be a bar at the top cp dt by t minus in place of v rt by p. So rdp by p this is for a pure component we are writing but you will understand this by the analogy. So if you integrate this you can write this as cp ln t2 by t1 minus r ln p2 by p1. Can you write it always when you cannot write this as cp ln t2 by t1 minus r ln p2 by p1 when you cannot write this when cp and when for example consider the first term if cp is a function of temperature then you cannot write it. Ideal gas means cp is a function of temperature and only temperature nothing else. So ideal gas does not mean cp is a constant but this expression you can write only when cp is a constant. So that is a special case of an ideal gas that is called as calorically perfect gas. So there is a difference between perfect gas and ideal gas. So this is called as calorically perfect gas for which cp is a constant. So you can take this out and here because of the mixture factor you are basically writing the partial pressure that is how why the total pressure times the mole fraction. So this is a star minus s0 star. So what is a star a? So del a star del na at tp n. So this is cp ln t by t0 minus r ln y a p star by p0. Similarly for the pure component what is small s star what is for the pure component s star na cp ln t by t0 minus r ln y a p star by p0. No question of mole fraction because partial pressure is not required. Always keep this in mind when you have a mixture in an ideal gas state always you refer to partial pressure and for an ideal gas that is total pressure times the mole fraction. So now if you subtract these two capital s star minus small s star is equal to sorry this na will not be there because this is per unit number of moles of a. So na is not there. This is molar property per unit number of moles. So s star capital s star minus small s star cp ln t by t0 term gets cancelled. This is also when you write this your reference state is the same. So s star minus s0 star actually. This is the reference state. For calculating entropy basically you require a reference state with respect to which you calculate the entropy because the second law best expressions talk about the change in entropy not the absolute value of entropy. For absolute value of entropy you have to refer to the third law of thermodynamics. So the change in entropy the reference states being the same. So this reference state you can choose in such a way that this is 0 at absolute 0. So that reference state is gone. So that reference state is gone. Otherwise you can consider some arbitrary reference state but when you subtract these two basically the reference state effects are gone. So you have s star minus s star cp ln t by t0 that term is gone minus r ln ya will be only remaining because then r ln p star by p0 that get cancelled from the terms. So to sum it up in this expression of g a star writing it with a different color so that it is understandable easily. So this is h a star minus t capital s a star is equal to small s a star minus r bar ln ya. So with this derivation see ideal gas states may be hypothetical. Always remember this this is a very important concept. When we are talking about a state star you can say I mean this thing should come to your mind automatically. We are talking about ions in a liquid. Now what about ideal gas? Will the ions become ideal gas? So the logic is that it refers to a hypothetical standard ideal gas state. It need not be that those ions which are there in the electrolyte for the electro kinetic phenomena they are in the ideal gas state. So this conceptual paradigm needs to be understood very carefully. Thermodynamics is a very beautiful subject. The thing is that more you think about the concept more and more actually doubts come into the come into the mind and we progress in thermodynamics as we resolve these doubts. So when you write g a in place of g a star so what is this? h a star minus t small s a star is small g a star plus r t ln y a plus r t ln y a plus r t ln y a plus r t ln f bar a by y a p star. From this expression we have written. So we can write g a is equal to g a star plus now r t ln y a gets cancelled. So you have r t ln f bar a by p star alternately you can write in place of g bar a you can write mu a chemical potential of a just a different way of writing the same thing. So I am writing alternative notations with the green color. This is mu a this is mu a star. We refer to a state which is called as standard state. The standard state will be designated by the superscript 0 which is a reference state. Some reference state we call as standard state. So 0 superscript means standard state. So mu a 0 is equal to mu a star. This is the standard state but of pure component a standard state not of mixture standard state of pure component a. So plus r t ln f a by p star. This is for the pure component. So you can get this by integrating d mu is equal to r t d ln f from the state star to the state p. You will get this pure component. So if you subtract these 2 then what you will get mu a is equal to mu a 0 plus r t ln f bar a by f a 0. This ratio f bar a by f a 0 is known as activity of the component a in the solution. This is called as activity. What is the value of activity if it is an ideal solution? Mole fraction because f bar a is y a f a. You can choose the standard state as same as this state. So we have already shown that for an ideal solution f bar a is y a f a. So the activity becomes the mole fraction at standard state, at standard state when you think of an ideal solution. So the summary of this finding is that for an ideal solution you can write mu a is equal to mu a 0 plus r t ln y a for an ideal solution at standard state. So we will summarize this discussion through the slides and then we will proceed further. So we had calculated, we will refer to the slides, we will refer to the slides, we will refer to the slides. So chemical potential of the component we calculated, we will come to this. We have calculated the chemical potential or the partial molar Gibbs function of the component i. So whatever I have written a in the board that is written as i in the slide for the ith component but in the slide it is generalized for a large number of components. So we can see that eventually we got that mu i equal to mu i star plus r t ln f i by p star and for component i in pure substance standard state mu i 0 is equal to mu i star plus r t ln f i 0 by p star. So if you subtract then you get mu i is equal to mu i 0. So what is written in the board as mu a equal to mu a 0 plus r t ln a i where a i is the activity. So you have for an ideal solution, so if we assume the ions to be ideal solution. So that I mean that is not a very impractical approximation if you consider the ions to be ideal solution then you have mu i is equal to mu i 0 plus r t ln y i. So you can rescale this y i is the number of moles of i by the total number of moles. Total number of moles being fixed you can suitably scare it by writing r t ln n i also without any problem. So it is just a matter of choosing the scale. Now what we have considered there is one mole right. So if you consider for each ion one mole of ion at standard state will contain how many number of ions? Avogadro number of ions. So each ion for each ion what you have to do you have to divide the expression by the Avogadro number. So whatever was r t ln y i that r divided by the Avogadro number will become the Boltzmann constant kb. So the Boltzmann constant is the universal constant divided by the Avogadro number. So it is basically to transform from one mole to a single entity. So let us see the expression. So eventually you can write the chemical potential of the ion for each ion mu i is equal to mu i 0 plus kb t ln n i right. So we have derived the expression. So you know what are the assumptions under which this is valid okay. Now before proceeding further I have already discussed that you know this is a system where you just do not have chemical species, chemical species have charges that is how they become ions right. So when you describe the total potential of the system you have chemical potential plus electrical potential because the entities the chemical entities have charges and in totality that is called as electrochemical potential. So if you go to the expression mu i bar we call it it is just a way of writing electrochemical potential. It is the chemical potential plus electrical potential. How do you write the electrical potential? See z i is the valency of the ith species and e is the charge of a proton. So the total charge of the ith species is z i into e okay and if psi be the potential within the electrical double layer we are discussing about the thermodynamics of the electrical double layer. If psi be the potential of the electrical double layer within the electrical double layer which is a function of y, psi is not a constant. Remember that there is a maximum value of psi at the wall then psi decreases as you go from the wall at the shear plane it becomes the zeta potential and outside the electrical double layer it becomes 0. So you have a distribution of psi as a function of y. So when you have that then basically the electrochemical potential the electrical potential part that is also a function of y. So what is the total potential mu i is equal mu i bar is equal to mu i 0 plus k b t l n n i plus z i e psi. So we will start with this expression and see what we get for equilibrium. Mu i bar is equal to mu i 0 plus k b t l n n i plus z i e psi. Now for the system to be in equilibrium what is the electrochemical potential? The gradient of electrochemical potential must be 0. So for equilibrium you must have d mu i bar d y equal to 0. What is the y direction? If this is the plate solid boundary at which the electrical double layer forms then this y direction is perpendicular to it or if you are forgetting about any direction that means you do not care about that whether the direction is y, x, z or whatever you want to generalize you can simply write d mu i bar equal to 0. So that means you can write k b t d n i by n i plus z i e d psi is equal to 0. Now if you integrate it if you integrate this expression what will you get? You can integrate it with an understanding that far away from the solid boundary located at hypothetical infinity. Infinity means what? This is just like boundary layer. Infinity means not literally at infinity. Infinity means at a distance from the plate where the surface charging effect of the plate is not failed that means the potential has dropped to 0. Where the potential has dropped to 0 the concentration is a bulk concentration that is n 0. And where the concentration is n i the potential is psi. So that means l n n i by n 0 is equal to minus z i e psi by k b t. So this means n i is equal to n 0 into e to the power minus z i e psi by k b t. So you have been able to successfully express the concentration as a function of potential the number density of ions of the is p c as a function of potential. This is known as Boltzmann distribution. So let us summarize this part through the slides. So you can see the final expression n i is equal to n i 0 which is n 0 in short notation what I have written in the board into e to the power minus z i e psi by k b t. So this is the ionic distribution under equilibrium. Now there is a different way of deriving this in fact there could be many different ways of deriving this. So we will derive this from the species conservation equation. If you recall that one of the in one of the early lectures we derived all the conservation equations. And there we derived the species conservation equation. We will refer to the species conservation equation but in this case you have to be careful because here there are species which have charges. So charged species we are talking about not in many chemical in many applications in chemical engineering we are talking about species which are not charged but here we are talking about species which are charged. So we will start with the expression look at the first equation general species conservation equation del n i del t plus del dot j equal to 0. The unsteady term and the flux the divergence of the flux. So the flux the special case is steady state when the unsteady term is 0 that is quite obvious. Now the flux will be typically related to what j i will be 1 is due to fluid flow this is equal to n i into u. Now you can write the flux directly from thermodynamics by noting that from a thermodynamics point of view the flux is proportional to the number of the number density because more the number density more will be the flux and the flux is proportional to the gradient of the chemical potential electro chemical potential rather not the chemical because here the chemical species have charges. So accordingly we have written j i is equal to n i u minus b i n i into grad of mu i m means basically molar quantity because like we are interested to write the expression for one mole of a system per unit mole. So basically whatever we define for per unit entity whatever we wrote for per unit entity we will multiply that by Avogadro number to get the expression for the total mole. Now what is this b i? b i is called as ionic mobility it is called as ionic mobility it is diffusivity by r t that is the definition of ionic mobility and d i is equal to k v t by f this is a very interesting relationship where you relate the fluid mechanics with the diffusion. So how is it possible or how do you get this expression I will quickly go to the board and there is a complicated way of deriving this but I will just give you a concept through a very simple case. So from the work energy principle so if let us assume that there is a sphere which is moving in a fluid. So there is a drag force on the sphere for some very special case why I am talking about drag force on a sphere because it very closely resembles the situation of an ion. The ion may be modeled as approximately a spherical entity moving in a liquid okay. So what is the drag force on the ion? The friction coefficient times the relative velocity. The friction coefficient can be derived by considering the Stokes law for example or some modification of the Stokes law. If it is if the Stokes law is valid then what is the expression for f? 6 pi we will write viscosity as eta because we have already used mu for chemical potential. So just to avoid duplicacy of symbols into the radius of the ion okay. So this f into v is the force. So this force see v scales with l by t right some characteristic length traversed by the ions over some characteristic time. This is the this is the drag force. What is the work done to overcome the drag force? This times this is force times the displacement right. This work where do you get the work? Ions of thermal energy by virtue of the thermal energy they can get the sufficient energy to overcome this drag. So what is the thermal energy that is k v into t? Boltzmann constant times temperature and what is this? L square by t this is nothing but the diffusion coefficient right. So f into diffusion coefficient is k v into t. So d is equal to k v t by f. This is the Stokes sign this is a very famous relationship called as Stokes Einstein relationship. Of course the actual derivation of the relationship is not this. I have just tried to give you a physical meaning of this through a scaling argument because the actual derivation is a little bit more involved and I do not want to spend more time on that okay. So now we have let me write the expression for j in the board. J i is equal to n i u minus B i n i into what is mu i m or mu i bar m. This is n a times mu i 0 plus k B t ln n i plus z i e psi right. This is what we have derived. Just it is a molar quantity so multiply by Avogadro number. So what is grad mu i m? What is B i? Okay. Then we are going to write this as mu i m will go step by step n i grad mu i m. So one n i gets cancelled n a k B t grad n i plus n a n i z i e grad psi. What is B i into n i grad mu i m? What is B i? B i is D i by R t. B i is nothing but let us write it somewhere D i by R t. So D i by R t n a into k B is R so R t grad n i plus D i by R t by R t. So this term becomes D i grad n i right. Diffusion coefficient times the concentration gradient. So this is just like the fixed law of diffusion right. This flux is just like the fixed law of diffusion because ions after all are chemical species. But this additional term is because of the electrical potential and that is because the ions have charges okay. So because of the electrical potential this additional term is there and physically this is because that if you apply an electric field or if there is an induced electric field the ions will have an additional transport beyond advection and diffusion because of the action of the electric field and this is called as electromigration okay. So in addition to advection and diffusion you also have electromigration of the ionic species. So you can write now this expression R by n a you can write as 1 by 2 by 2 by k B right. R by n a is k B so this becomes 1 by k B okay. So you have the electromigration given by D i n i Z i e grad psi by k B t okay. So now when you take the divergence of J you have also this the first term which is the advection term del dot n i u this will be minus right this is minus because it is minus B i n i grad mu i m. So this will become minus. So when you consider the species conservation equation del J i del t plus sorry del n i del t plus del dot J equal to 0. So you have del n i del t plus del dot n i u equal to the minus terms will become plus when they go to the right hand side. So how do they look? Let us go to the slides. So you have del n i del t plus del dot n i u this is the transient term this is the advection term del dot D i del n i plus del dot D i Z i e n i by k B t. Grad psi this is diffusion and this is electromigration okay. Now can we recover the Boltzmann distribution from this equation? So this equation by the way is the most general equation because here we have not made any assumption about whether irons are point charges or their ideal solutions we have not made any assumption right. We have started but the assumptions are implicit that the irons are ideal solutions why? It is a most general expression provided you do not write it in this form but you write it in this form this form that is J i is equal to n i e minus B i n i grad mu i m. Once you have specified the expression for mu i m like this then it is no more that general because it contains assumptions that you are considering the system as an ideal solution right. So that assumption is very much there so but it is more general than the Boltzmann distribution itself because the Boltzmann distribution considered a special case when the system is in equilibrium but here you are considering transport right you are considering basically a fluid flow field and you can also have unsteadiness. So all these additional effects are brought into picture in this equation so it carries the assumption of Boltzmann distribution no doubt but it includes the effect of advection and transiences. This differential equation is known as Knott's Planck equation. Now you have to keep in mind that you can derive more general expression by instead of see that is why the derivations are important. If you say that well I want to relax the constraint that the ions the solution is an ideal gas then simply what you do is you replace the mole fraction by activity then you can write a modified version of the Knott's Planck equation that no more considers an ideal solution okay. So it is possible to generalize sometimes we simplify for solving a problem but we must have a clear picture that we want to generalize what we should do. Now for the special case when there is no unsteadiness and there is no fluid flow this equation should boil down to the Boltzmann distribution because that is the same form of the electrochemical potential that has been taken so it should boil down to that. So let us see that how we can arrive at that conclusion. So you have del dot di if the unsteady and the fluid flow term is not there. For a one dimensional problem this will just become d dy of di d ni dy I am just writing a simple case of a one dimensional problem where everything is a function of y integrate it once if you integrate it once you will get di d ni dy plus di ni zi e by k b t d psi dy is equal to constant right. Now you know that wherever d ni dy is equal to 0 d psi dy must be 0 right that means c will be equal to 0 this is nothing but the differential form of the Boltzmann distribution right you just cross the various terms d ni by ni is equal to minus so this d i d i gets cancelled d ni by ni is equal to minus z i e by k b t d psi right that is the differential form of the Boltzmann distribution. So you can see that you can recover the Boltzmann distribution from this. Now many times we use the Boltzmann distribution even when there is fluid flow I will show you that when we are talking about a fluid flow in a channel which is taking place in the axial direction because that is the direction when we want where we want to have the flow we use still the Boltzmann distribution and there is an interesting reason behind this the interesting reason is that the orthogonality between the direction in which you are applying the Boltzmann distribution and the fluid flow direction. See you are having the fluid flow in this direction with the velocity u but the v velocity is approximately equal to 0. So it is an equilibrium picture that essentially governs the distribution in the transverse direction despite the flow taking place along the longitudinal direction. So despite having fluid flow the Boltzmann distribution can give a kind of reliable picture with certain approximations. So those approximations if those are not justified then we can no more use the Boltzmann distribution. We will discuss about more of this in the next lecture. Thank you very much.