Lecture 18

[MUSIC] [MUSIC] We will start lecture 18 the topic of lecture 18 will be solution thermodynamics we will continue what we will do today is delta v mix means molar volume mix means after mixing two components that we are solid solution or liquid I'm going to try to find out the molar volume of the solution, the molar volume of mixing, what is the enthalpy of mixing, and what is the entropy of mixing. The component partial mole is the mixing value of the bar. It's graphically correct and what we did is that we first saw that the delta G mix means the molar quantity. That value is the molar quantity. One mole value means the solution or the liquid solution or the solid solution. We saw that. x a delta g a bar m plus x b delta g b bar m and from there we get delta g a bar m equal to delta g mix plus x b d delta g mix d x a delta g b bar m equal to delta g mix Plus X AB looked at these two formulas from there we also saw graphically how that is we plotted a free energy plot it's zero value it's zero value and there when I at this point if I can multiply these two quantity values ​​that's graphically seen how to do Can I draw a tangent here after drawing the tangent x a = 1 xb = 1 when looking at the pure substance pure component there is the point of interaction of the tangent intersection that point we can see the value of that point is nothing but delta g a bar m r this value is delta g BAR M This is what we saw in class OK. Our final goal is Delta G AR Delta GB How is it expressed G Plus X B Delta G BY Delta G Delta X This is G Plus X A DG D X B But we want to repeat these two quantities and before that let us see how we can express these three quantities in terms of mole fraction or temperature and it is important to remember that I have assumed constant pressure and temperature is the temperature we have seen at this point. At this point, when I'm measuring the mixing partial molar free energy quantity of that particular element, we're plotting the graph of del g mix as a function of x b. What I thought was that we will now calculate del V mix del h mix and del s mix and after calculating that but we will understand how we can graphically do these two times ok so let's do our next stage which is delvi mix We're going to calculate the DelV mix to calculate the DelV mix and here V is the molar volume molar volume and here if I give I a then the molar volume of the rectangle component OK what do we have to do to do this we have that equation again that we did n't see. Let's go back to the equation that DG is equal to VDP minus SDT and the other thing we'll use for this is we'll take the partial molar quantity of the fine off component and that's the value that's right in the mixture. Since we have assumed binary components A and B and since we have assumed the alloy system to be a metallic alloy system, we can capture these two as elements. del p a constant temperature and composition fine we can do that and this is the equation if i add the composition there then i get delta g delta ni summation t p other composition dna this we knew this time here i fixed the composition ok so For what will it stand is Vi will stand and not just VI now it will stand for me partial molar volume ok because it's taking it in solution now if I take it in pure system G delta PT composition if I take it pure then I'm going to take it and pure because standard Value VI Not these two things I can write O If I can write these two then I can I bar from these two delta GI bar minus GI not del PT and composition fixed equal to VI bar PI not We know that GI bar minus g i not equal to nothing but r t l n x i we 've seen this before so that means here i can put it equal to delta g r t l n x i del pt composition equal to r this is how i write delta V bar i in solution as we denote the free energy and we know that it is constant composition also constant at the same time I know that x means mole fraction is independent of p if independent of p means delta vi bar m is Equal to zero Now if we find out the delta V mix in a similar way as we saw the delta G mix equal to x a delta g bar m plus x b delta g bar b m a similar pattern I can write molar volume change molar volume change is my After making the solution, I can write it like this. I can write it fine and I know that individually it's also zero and it's also zero and this two is non-zero quantity means these two but non-zero quantity because what I'm getting is that means after mixing now if I mix Suppose there must be a finite composition. If I is zero, then there will be no element. Whenever there is a finite composition, that means X A and X B cannot be zero. Well, that means if these two are zero, then what does it stand for? Can do x a del rtl n x a dpt composition plus xp dpt composition I have already checked it is zero that means my quantity will also become zero this quantity will also become zero finally my stand is delta v mix equal to Zero means I have no volume change after mixing it means no volume change after mixing is ok I also understand that the rule of mixture that follows is the rule of mixture I take a particle volume particle element A and B are both molar volumes VM and VA and VB and the volume will change in the ratio of my mole fraction so if I plot the molar volume here is let's say A pure B pure and let's let this axis be Delta V mix sorry this is my final volume. Which would be the V of the solution I can take the V solution molar volume as if I had given the G solution the molar free energy change for the solution V solution if I do if it's at V if it's V then I can connect it straight line OK and V t will be x a v a plus x b v ok this means the rule of mixture will follow it is straight line linear and x b is increasing in this direction this plot will mean i see an important relationship this relationship will be seen now let's see if this relationship is maintained Well, where we will use it is the most important thing that we will probably apply it when we take the quasi chemical approach. Del V Mixta which is equal to zero fine ok this time it became Del V Mixta I saw no volume change after solution form and volume change volume which is the total volume which will stand in the solution that is rule of mixture will follow now our del to find Let's look at Dale H. Mix OK now what I can do for this is Gives Helmold formula we can equation formula we can use Gives Helmerz equation we'll use we've already seen that Del g bar i by t by dt t dt of g i bar by t pressure constant and composition constant that I can write minus H bar by t squared that means I can write another one when it is pure and standard state then I can also do this from these two I will get fine I can write it now see this value Let me know that G I bar minus G I is not equal to R T L N X I . If I replace this here, I'm going to be T R T L N X I by T DT DT OF DT Composition equal to minus I can write this upper part. This part I can write delta h times i am by t square t and see that x i independent of temperature r to constant is always constant as seen in previous case when we passed the value of del v mix still r was constant x i If it's independent then this value will be zero equal to minus squared and any time if I take a binary binary system A and B system then I can write del H mix which is molar enthalpy change due to mixing as molar volume change due to mixing if I write this So I can write x a delta h a bar m plus x b delta h bar bm I can replace this with this formula x a del g by t I can replace that with rl n x a dtp pressure constant and composition constant x p d dt r l n x b pressure composition constant it is zero it is also zero that means delta h mix equal to zero I see that in condensate system when we are making solution we are making a mixture but if the system is ideal then del h mix It's going to zero we already saw in ideal system delvy mix also zero means volume change also mixing volume change but zero is ok that means two equations are obtained two one relation is obtained I see in case of ideal solution means i no we can say that when mixing gas Or mixing in condensation system, my volume change or enthalpy change is not happening. Mixing time is fine. It stands and that means it has a physical significance. Let's see what the physical significance is. But the same idea will be that these two elements A and B will not have any preference between them, there will be no preference, no preference attraction or repulsion. no compound no compound formation no face separation no face separation ok just a completely random solution will be formed at every place if i look at the composition we will find the composition is equal and the elements will be mixed equally why this compound will not face separation why To understand that it won't, but we need to look at the regular solution model for that we need to understand these two we need the regular solution model we'll do that later and then we'll see that the delta H mix could be negative it could be positive it could be zero okay And in all three cases I will create different alloy systems ok we will see that later for now we stay with the ideal approach we stay with the ideal solution model ok now what we need to do last is to find delta s mix ok what do we do to get delta s mix As we saw earlier, we know that G delta G is equal to V DP minus S DT plus summation delta G I N I T constant P constant NJ other components composition is off fixed only N I T is changing D N I this formula we know now if I do partial differentiation delta g delta t p and if we keep the composition fixed then I get minus s so now I look at delta g mix what is delta h mix minus t del s mix it's all ready we looked at ideal solution zero ok for ideal solution Ok so that means my delta S mix equals to delta minus delta G mix by T so ok it can be done this way another way it can be done is if I do this then what I get is minus RT X L N X A plus Xplnx p by t but the t is getting cut off so i would be minus r xl nx a plus xbl n xp r using this formula but it can be done ok that is delta delta gx delta p delta t Rather delta t constant pressure and composition, it will be delta s mix with a negative sign. The mixing entropy change is what we can do with this formula. Since we're taking a binary system, we can do this del s mix in another way. Okay, let's see how we can do that. Now to do that, the first time we look at a little distribution. Granted, we'll try a little more later to see how the distribution is coming along, but for now, we know one thing about the distribution. That is, we'll discuss this equation of Kln Omega later. For now, let's see how the Dell S mix is ​​coming along. Here's who is the Borzman constant. Omega is the number of distinct microstates. Okay, microstate means there can be many distributions in that system. Okay, if I Let's assume two red balls, two black balls, two red balls and a black ball. Let 's say there are four black balls on one side and four red balls on one side. When these two are mixed, what combinations can happen? We will try to understand this in more detail later four red balls and four black balls now each ball but identical red balls each identical black balls each identical now whenever we look at the system like this it means this can be a system where all red balls on one side are all black balls on one side It can also be like this: one black ball goes here and the rest are red balls and here one red ball goes and the rest of the black balls are fine so this is a system this is a system this time this black ball can take any site and every site is identical and the black balls but It can take here it can take it here it can take it here it can take it every site can take it so I'm going to get a new distribution OK I'm taking that distribution as a microstate that means this is a microstate this is a microstate there can be many variations of this microstate like here it can be the black ball on the left A separate distribution like this goes here and the red balls stay here and so my black balls go here the red ball goes here and the black balls stay here ok like this I also got a microstate these are all one by one microstates how many microstates can there be like this and there After removing the common ones I will get several distinct microstates ok if I know the number of those microstates that is my omega ok how can we repeat that omega can repeat the omega if I assume A Number of Atoms and Nb Number of Atoms A and B are in case of adding these two what I get is NV Number of Atoms or N Zero N Zero is Avogadro Number Avogadro Number When will this happen when will we mix and how will we mix What we are trying to see in this situation of pressure and temperature constant means we are just trying to think about the distribution of atoms and B atoms and how many microstates there can be. No. But I can have microstates and they will be independent. OK, but I'll come back later. I'll come to the next lecture because we have less time now. So I'm showing it like this now. In class or later let's see it this time whenever we do this we can write s equal to 2 now we are talking about s what is s if i take s as final s of final means after solution that means i can say it except final i can write s of solution is equal to n zero avogadro number factorial divided by n a factorial and nb factorial i did it but i have to bar del s mix what do i do for that i started where did i start being s a means pure a atom when there is When SB pure B atom is present it is before mixing now after mixing is S solution and if we subtract the two we get del s mix ok s let's look at del s mix sorry here will be Ln I missed an LN Because S is equal to KLN omega fine is KLN Avogadro's number factorial divided by N A factorial NB factorial minus KLNNA factorial Omega here Let's see what the factorial omega is here. And when we get How Many Ways Can We Distribute That Means How Many Independent Ways Can We Distribute We'll See That The Distribution Stands Out The Same Way We Distribute Because Everything Looks Common So It's Not Identical If I Take These Four With 100 If this atom goes here, if it goes here, if it goes round once, but I'm not going to get a new distribution, I'm going to get the same distribution, I'm assuming this is the same. Then we can distribute only one way, that means omega one will be omega equal to one omega a rather minus kln omega b equal to one because omega is the number of distinct microstates that means these two terms will become zero these two terms will become zero So that means this part is this and this part is this and this is right so that means I'm standing del s mix is ​​also kl n avogadro it's fine this time we'll do a little starling approximation as starling approximation I can write l n not factorial equal to n not L N N Not Minus N Not L N Factorial N A Equal To N L N N A Minus N L N Nb Factorial Equal To Nb Ln Nb Minus Nb Now I'm going to put these here these three things here now I can write del s mix Equal to K LN factorial minus NN factorial LNN A factorial minus lon NB factorial k I'll replace this this here N not LNN not mean Avogadro's number minus N not minus NL N lon N A plus N A minus NB L N N B Plus N B We know that if we add these two terms add that leads to n knot that means these two terms will cross we are left with k n knot L N N knot minus N N L N N A minus N B L N NB This stands now I'll break it down so I can write minus one term out I can write N A minus L N N NOT minus N B L N N N A plus N L N N A plus N B L N N B minus K N L N L N N N not plus NB lon NB divided by N not I take the N side to one side and the NB side to the other side this is nothing but x this is nothing but xb so that's minus k n l n x a plus n b l n x p Now here's a I multiply that by N not so it becomes NN not L N X A plus N N N not L N X B. I can write this as X A is X B so that's minus K N Not X L N X A plus X B L N xb ok that means and this value is this value is going to be and bicz is equal to avogadro number avogadro is equal to universal gas constant equal to boardman constant into avogadro number ok that means i can replace it so that mean is equal to minus rx L N X A plus X B L N X B OK and this is always positive because these two term fractions vary between zero and one which means it will be positive which means this is the entropy of mixing but always positive. We see in solid solution or liquid solution condensation system is positive so that means I always mixing always but random mixing in ideal case I but always favorable because it is positive we know entropy if entropy change is positive we know system is happening and that means mixing is spontaneous so that means What I stand for is delta v mix equal to zero delta h mix equal to zero and delta s mix equal to minus r x l n x a plus x b l n x b ok our very important derivation of these three will be useful for all three Need ideal system face diagram and when I take regular solution regular solution when I use it will be seen later Dell S mix will be same Dell H mix will be not equal to zero and Dell V mix will be zero ok this idea is regular Let's do this for the solution, okay, we'll stop here for now, we'll extend this solution thermodynamics, but we're slowly getting into the phase diagram, and we've already got an idea of ​​how to calculate the partial molar free energy, but it's We have done the partial molar free energy foreign element foreign element when mixing is the mixing part but we have been able to do it from the intercept ok now we will do the partial molar quantity in the next class which is called the chemical potential thank you [music] ] [music] [music]