Lecture 17

[MUSIC] [MUSIC] We will start lecture 17 and the topic we will cover in lecture 17 is solution thermodynamics we will continue and delji mix in condensation system means solid or liquid solution we will try to repeat infact we have already discussed delji mix in last class And we'll look at some more treatments on that and see how that varies with the composition, okay that's very important to see and after we look at that we'll slowly move on to the face diagram. It's related to mole fraction and temperature, but basically when we plot this, we're going to plot a particular temperature, pressure, and we're going to slowly keep the pressure constant, and we're going to plot different temperatures, and finally we're going to see a phase diagram plot. And we're slowly moving towards that, what we saw in the last class, we saw in the last class is a delgy mix and when A and B are the two components, if I take the alloy system, take the metallic alloy system, we can call it elements A and B that are being mixed. After mixing we get a solid solution or a liquid solution and there we also say that it is due to mixing and where we are doing the mixing and keeping the temperature and pressure constant and we get a formula that is Delta G Mix of that condensation phase Equal to RT X L N X A plus X B L N X B There X A I take X here X I That could be A Could be B X I being the mole fraction R being the gas constant RT being the temperature at which we bar is of our consideration of our consideration and it is also expressed as x a delta g bar delta g bar ma plus x b delta g bar m b where m is the mixture and calculating individually we get this formula then also We have seen that this is the change of free energy due to the formation of the mixture from the solution to the solution. This is the change of free energy. of one mole of for one mole of solution we expressed this as x a g bar a plus x b g bar b and we had our target which was the target from last class was that we have to multiply the two formulas and then we I understand where the need of those two formulas and those two formulas but one means important requirement is ok to draw face diagram those two formulas are one is G A times G plus X B Delta G Delta X A G times equal to G plus X A DG DXB we wrote these two formulas but we did not prove it we will try to prove it today what we did for that thought we will take the help of Givesom equation and a Givesom equation is taking an equation out of three equations fast one which will take help we know that Which is zero equal to x a d g a times plus x b d g times if we go to the old class we will see this formula but valid ok it got one ok another one is another formula what is g equal to two this is the formula that is written here G equals to x a g a bar plus x b g bar and we know that whenever we give g bar that's a partial quantity and that partial quantity is nothing chemical potential chemical potential is right in that solution of that opponent we can also write it as g bar equals We can also write two mu A or G bar B equal to mu B. We'll get to mu later, but for now, since we're taking g as a bar in this convention, we'll get into the chemical potential later. I can use either one. I will express it in different notations in different books, sometimes mu is given, sometimes G is given, but the matter is the same, the matter is the same. Okay, so we get this equation from these two equations, but we will get these two important equations. Let's see another formula that is we Assuming it is one mole case we know that first NA or NA amount of and NB amount of B element was taken and if I add these two then if I add these two then we get a number which is Avogadro's number which is the value of 6023 into 10 to the power my 23 is a number Hughes number so when we do this we assume it's n superscript subscript zero ok that can also be called n not fine ok this is the total of one mole of substance and from there but it will come x A and X B will come in. Okay, if that's taken, then we'll see that's molar and why we wrote these two later. Okay, from here we get one thing that's X A plus X B = 1. So what's going on with X A? is N A plus N this one which is X A plus NB Avogadro by Avogado number it should be one if I do this in this equation delta g equals to x a dg a bar plus g a bar d x a xb d g bar b plus g bar bishu b x b ok i know this formula i know allready if i add any two terms this term will be zero and if this term is zero then delta g would be equal to g a bar d x a plus g bar d x b now I'll see that since I 've written this over here, that means I can also write dx a plus dx b equals zero, so that's going to be dx a equal to minus dx b. It's dx a minus x a I've written minus dx a instead of dx b ok this time what I can do is I multiply xp by that ok first thing we'll do is delta g dx a equals to minus GB times this I'm going to multiply it by x b fine here I get an equation I'm going to number the equations this number is one this is two let's say three ok now if I do two plus three the equation is two plus three from there let's write it two is x a g a bar minus plus x b g bar and the three third equation is x b d g d x a equal to x b a bar minus x b bar If we add this then this term will be cut off so it stands for my g plus x p d g DX A is equal to X A plus X P G A times that means I can write G A when these two terms are added, it is 1. That means G A would be equal to G plus XP DG D X A. The equation I wrote this equation earlier I derived here how it was obtained OK this equation was obtained similarly I take the general equation times this is the star equation with this I take another this is not this let 's take another from this equation I'm the expression of GB GB bar Let's see how from this equation we can write Dg equals 2 I can write minus g A times x gives dxb because I saw this is plus gb times dxb because from this equation I substitute x for dx and I minus d I put X B, that means my delta is G DG D X B is equal to minus G A times plus GB times. Now with this, if I multiply X by X, fine, that means I've got the number four. Now I'm going to do two plus four. First, equation number two is G equal to X A G A bar plus X B GB bar and meanwhile X A DG D X B is equal to X minus X A G A bar plus X A GB bar. A DG D X P equal to two these two will be crossed X A plus X B GB times that means these two add up to one I know that means GB times equal to G plus X A delta G D X this is a formula found that means I It's a bar here, sorry. I missed the bar. Please put that bar. OK, that means I'm partial molar free energy of A and B. In that solution, I can express the molar free energy at that same composition and with that comes a multiplication factor with the slope at that. Point on that free energy composition diagram ok this is a very important thing that means I mean molar free energy we can find partial free energy molar free energy or chemical potential from these two equations if I take these two equations from these two equations we get from these two equations Let me take star as star one and take it as star two from star one and star two equation and its physical significance here is that at a particular composition slope of G is nothing but DG DXA or DG DXB depending on that composition. I'll take it with respect to or B with respect to and after multiplying this by a mole fraction with the sloper at the point where I've drawn the slope, I'll add the free energy value of the point at which I've drawn the slope, okay, then I'll get the partial But this is the slope, this slope helps me to graphically calculate the chemical potential, okay, I'll see, this time, let's see what it does graphically, but we're going to use these two equations to determine phase equilibria. It means that there is a very important relationship with the phase diagram. These two help us to understand phase equilibrium. This is our phase equilibrium. What is equilibrium? Two phases are actually two phases. Equilibrium is not equilibrium with respect to components. A particle component is in equilibrium in both phases. We'll say chemical equilibrium is set right because already thermal and mechanical equilibrium is set because we're at constant temperature pressure but we're doing the plot right now let's look at this graphically so the first thing we 're going to calculate is Let's look at the graphical measurement of mixing partial free energy and from there we'll calculate the partial free energy molar free energy for a particular component in solution. Let's look at that a little later in the stage. Equal to x a delta g a bar m plus x b delta g m bar b I already know this but one more thing to remember about one mole is that whenever I'm starting to write x a and x b at the level that means but I'm one mole Taking it as one mole, the solution that we are making is one mole because when we do one mole, we get xp. Yes, so this XB is coming up fine. If I do this, I can do the correlation of the previous relation. I can write Delta G A Wood B equal to plus XB Delta Delta Del G Mix DX. Another thing I can write is Delta G B times M equal to Delta G. MIX PLUS X A D DELTA G MIX BY D X B This is what I'm telling you, you try to do it yourself, it will move in the same way, but the way we did it, this is the way we did the calculation. There too but you can do the same thing you try ok fine now whenever I got it whenever I got it now let me try to represent it graphically ok now to represent it graphically what we need to do first let's see how our plot of this is coming out this I can also write x a r t x l n x a plus x b l n x b from this equation it will be easier for me to plot because I don't know if these two are correct but I know these two are measurable quantities whereas I x b Let's assume that means X A is equal to one here X B is equal to one that means pure and pure B. I'm doing that in one atmosphere and temperature is assumed to be a particle temperature. From this it will be zero value because I know whenever x a one or x p one the value of dell g mix but zero is standing ok fine that means if i do this i will get like this a symmetry plot i get symmetrical plot now i have Let's say at this point I have to determine delta G A bar M delta G B bar M to do that what I have to do is let's bring it down a little bit and read it a little bit then it will be convenient fine now if I draw a tangent on it I draw a tangent It's gone through, the tangent is OK, I've drawn the tangent to the curve, this time I draw a line parallel to the X axis to the X axis of the B at this point. Let's take this point and take this point Q corresponding to this point that over it I draw another line perpendicular to this I take this point P and this I take S OK this is N this is M OK this is my notations now let's look at this point That value is the value of DELGA MIX. This is the plot of DELGA MIX. The plot of DELGA MIX is the value of this point. The value prime is the value this is OK this point is fine and that value is standing nothing but from this point to this point that means I have to repeat G times delta G times M times its value is delta G mix prime because at this point that currency It stands for Plus X B X B up to this point the X B that will be X B means this composition is nothing but X B OK and just to pinpoint this composition let us put it S X B prime X B prime just to indicate This specific composition because this axis, I'm not holding x b is x b prime and that means this value is x b prime then delta g mix prime. This point at this point means what is the slope at point Q at point Q OK dx a dx at that means what I'm going to do now is where is this value coming from p q p q plus x biter value what is r q and this slope now let's look at the slope Nothing but the slope of this line is the slope of this line is the slope of this line how is RS by R Q equal to slope of this line that means I will write sorry RS by R cube OK RS divided by R Q OK that means RS R Q R Q is intersecting Wood B equal to this slope is delta delta g mix del x a that means it stands for p q plus rs is nothing but o r plus rs because these two values ​​are equal is equal to o s that means This value is nothing but delta g m delta g a bar m that means similarly i can calculate it another way this value is delta g b bar m what does that mean that tangent intercept tangent intercept at x equals to [music] one ok is nothing but delta g a bar m r tangent intercept at x b equal to one = delta g b bar m so graphically i but these two values ​​these two values ​​but can bar graphically that's what it says here fine Okay so that means we also saw that the partial molar quantity whether it's the mix value or the potential partial or the chemical potential value we can determine that. First what we determined is the mix partial quantity which is nothing but the tangent intercept at x is equal to one. The difference value when cutting it on the pure A or pure B side axis means the difference value from zero is the chemical potential for that particular component that I'm intercepting with the pure condition but we can graphically represent that value for now in the next class slowly Let's go in. How do we graphically find out delta G or GB times graphically this is these two quantities we will try to calculate and if we can calculate these two quantities but we will understand the phase equilibria then we will stop for now we will discuss it in the next class and the next. One more thing we will discuss in class is that we have repeated delza mix delji mix for now but enthalpy of mixing mean in case of solid solution or liquid solution and enthalpy of entropy of mixing solid solution or liquid solution will also try to repeat that and another I will repeat the thing that the mixing molar volume is ok, we will see what happens and our base is the same, the discussion base is the ideal solution, but we have not yet entered the real solution, the real solution will come later. H mix or del s mix bar I can understand ok stop for now we will discuss more in next class thanks [MUSIC] [MUSIC] [MUSIC]