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3.091 Introduction to Solid State Chemistry, Fall 2004 Transcript – Lecture 35

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3.091 Introduction to Solid State Chemistry, Fall 2004 Transcript – Lecture 35 MIT OpenCourseWare http://ocw.mit.edu 3.091 Introduction to Solid State Chemistry, Fall 2004 Please use the following citation format: Donald Sadoway, 3.091 Introduction to Solid State Chemistry, Fall 2004. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative Commons Attribution-Noncommercial-Share Alike. Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms MIT OpenCourseWare http://ocw.mit.edu 3.091 Introduction to Solid State Chemistry, Fall 2004 Transcript – Lecture 35 You think you're happy? I'm happy. I'm very happy. What we will do, today is wrap up phase diagrams, make some comments about the final exam, and then we're going to get some people that aren't associated with class are going to come in, they will pass out paperwork and do the course evaluations. And, we'll get you out of here at 11:55. Before I go any further, draw your attention to another IAP subject. If you're looking for something to do during January, and this will be offered by myself and my postdoc, Patrick Trapa. We are going to offer a lab subject without a lab. It's part of an experiment that we are conducting for the National Science Foundation to see if large classes such as these that don't have a lab experience can get much of the lab experience, that is to say, planning for discovery, preparation of the background information, and then mining the data from data sets that are in the public domain, and then analyzing the data and then presenting the data. All of that you get without turning the knobs. And, that's about three quarters of a lab experience. So, I think three quarters is better than none. That's what we want to try to do. It's a limited enrollment of 20 people, and it's going to run for two weeks during IAP in the mornings, and you can go to the IAP website. And, it's worth six credits as well. So, OK, enough said about that. My office hours, I'll have a little bit of extended office hours Friday 4:00-6:00, next Tuesday 4:00- 6:00. And, there will be a question and answer session on the evening of Tuesday, the 14th of December. Professor Ballinger is going to run that in 54-100. All of these, and I know the TAs, some of them are going to run review sessions. All of these are optional. They have strict instructions not to teach new material. If you don't want to go to those, you don't have to. If you want to, bring your questions, and I just want to make it totally value neutral. OK, last day we talked about phase diagrams. I want to continue the discussion today. We looked at type one, which involves very nearly similar materials. So, that gives a lenticular phase diagram shown here. And, then we looked at type two where we have a phase separation without change of phase, giving us a miscibility gap. Remember, whenever you're in a two-phase regime, two-phase regime, queen of diamonds, bingo, that means tie line, lever rule. So, phase separation will give us a single phase, whether it's a homogeneous solid solution breaking into two phases, or a homogeneous liquid solution breaking into two phases. Those two phases have different compositions, and the compositions of each of those two phases is given by the end members of the tie line. And the relative amounts of the two phases is given by the lever rule. Today I want to look at type three. And type three is characterized by sort of a mix of the two. And on the left, we have change of state, and on the right, we have no change of state. On the left, we have perfect solubility. On the right we have partial solubility. So, we're going to take the solubility characteristics of type two, and the change of state from type one, and put them together. And that's what we end up with as the overarching features here. And, this is what the phase diagram looks like, where we have something that's similar to, you see the bottom here? The bottom here looks like that syncline. And the top here looks like the beginning of the lens. So, we have liquid, and then beta here. Beta is a solid solution, alpha is a solid solution. So, let's designate those. Alpha is a solid solution of A and B, only it's A rich. It's very near pure A, but it's not exactly pure A. So, this is homogeneous solid solution just as you'd have in a lenticular diagram. And, beta is a solid solution of A and B, only it's B rich. And L, that's a liquid solution. That's a liquid solution of A and B. So, these are all single phase. These are all single phase. Let's put that up here. This is P equals one, single phase. This is P equals one, single phase. And, this is P equals one, single phase. Now, if I want to go from one single phase to another single phase, I have to go through a two-phase regime. So, this is slush, liquid plus solid alpha. This is slush, liquid plus solid beta, and now I have a solid solution on the right. I have a solid solution on the left, and I can't go from one single phase regime to another single phase regime. I had to go through a two-phase regime. So, this is alpha plus beta. And, now I've got two phases down here, one phase here. Look at this point here that I've indicated in purple. What is this? This is a special. This is alpha, beta, and liquid in equilibrium. And, this is called the eutectic. It's the equilibrium between solid alpha, solid beta, and liquid. And, this type of diagram is called a eutectic diagram. It's a eutectic diagram, and it has some remarkable properties that make it advantageous in processing. Look at here in the type one diagram. There's no point on the type one diagram where we see all liquid below the melting point of the lower of the two components. We essentially have a lens extending from the lower melting point to the higher melting point. If you look here, if I add A to B, the melting point drops. We go down this liquidus. That's not surprising. What's surprising in the case of the eutectic is when you add B to A, the melting point drops. So, we get freezing point depression from both ends with the result that, you see this whole liquid zone in here? This whole liquid zone has liquid lying at temperatures below the lower melting point. So, this gives us a much, much greater range in which to process. So, we have a depressed liquid region. So, let's call this the characteristic is freezing point depression of both components, of both A and B, which is something we didn't see in the case of the lenticular. And this eutectic point is the equilibrium of alpha, beta, and liquid. And, it's unique. It exists at only one value of temperature and composition. So, that's unique to every particular system. So, let's look at some examples. So, here's one apropos of the coming winter. This is water and ethylene glycol. And, we talked about water ethylene glycol in the coolant system with reference to controlling boil over. And, as the winter is coming, we want to avoid freezing of water in the channels of the engine block because the water is going to freeze from the outside in. That means the last water to freeze will already be plugged by ice at the ends. And, as you know, when water freezes, it expands. How do you know? You know for the negative slope on the PT diagram. And, that expansion is so energy-laden that it'll crack the iron of the engine block. So, you have to keep the water from freezing. So, what we can do is we can add a second component and give ourselves a eutectic type system. So, here's pure water on the left. Here's pure ethylene glycol on the right. It's a double alcohol. There is the chemical formula for it. And so, pure water, which we see the melting point of pure water is 0∞C. And, what this is showing is that this alpha region of ethylene glycol, and water is so narrow that for all intents and purposes this appears as a straight line. That's why you go automatically from seemingly pure water to this eutectic regime. So, if we add, typically, about 50 volume percent ethylene glycol, that's the recommended mixture, what do we see? We see we get freezing point depression down to about -35 or so Celsius which ought to take care of most driving needs. Now, if you want, you can get maximum freezing point protection by adding about two thirds. In other words, 68 volume percent glycol gives you a eutectic at -69∞C, or -92∞F. That ought to take care of everybody in this room.
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