 There's a quiz Friday. That's quiz three. It's going to look just like quiz one and two, very similar. First couple problems will be pretty straightforward. Last two or three will be a little more difficult. All right, same rules. Open everything except computers including iPads and the like, okay? We'll have the scantrons for you. Now, G and Mark is going to give this lecture on Friday because I'm going to be gone. So you'll have the quiz as usual. You'll give the lecture as usual. I'm writing the lecture. It's going to be about examples pertaining to chapter 14 so it's important stuff. I think examples are very helpful in terms of understanding the material. Okay, so just to let you know, we'll post the lecture as usual and I'll be back on Monday. Okay, so on Monday, last Monday we were talking about this, this is one of the examples we looked at and we were calculating the partition function for OCLO. That's this molecule right here and we just said all we have to do to calculate the overall vibrational partition function is calculate the partition function for each one of these modes separately and multiply them together. The modes are independent of one another thermally in terms of whether they're occupied or not, right? Whether the continuum of states associated with each one of these modes is occupied. They act independently of one another so we can treat them independently. And so we can calculate this partition function using the equation we derived on Monday and here's what that looks like and we got these numbers 1.12, 1.01, 1.00 and these numbers make sense to us mainly because you know we know the thermal energy at room temperature is 200 wave numbers and so we don't expect these modes to be super occupied, right? Because 200 wave numbers is a lot less than 450 or 945 or 1100 and so these numbers sort of qualitatively make sense to us but we don't have a lot of intuition about this decimal place at this point. What should we expect these vibrational partition functions to be, right? It's easier to have intuition about translation and rotation in some ways because in both of those cases the temperature is always much higher than the characteristic temperature. So do these numbers make sense? It's a little hard to say. I mean qualitatively we expect them to be 1 and change but the change is the part that it's a little hard to know about. So let's calculate these characteristic temperatures at 450 wave numbers, 945, 1100, right? We expected excited vibrational to be only slightly accessible. Yes, okay. Well, so I did that. All right, I calculated the characteristic vibrational temperatures for each one of these modes and I got 647 Kelvin, 1300 Kelvin, 1500 Kelvin. Temperatures that are way higher than this temperature. In other words, we don't expect these modes to be, we don't expect the occupation of excited states to be very high because these temperatures where this occupation is expected to occur are way higher than the temperature we're paying attention to here. 298 Kelvin, we shouldn't get significant occupation of 450 until 600 Kelvin. We shouldn't get occupation of 945 until 1300 Kelvin. These are temperatures that are way higher than the one that we care about in this problem. Now, what partition function would we expect right at the characteristic vibrational temperature? Because we're going to be around the characteristic vibrational temperature a lot. So what I did here is I made a plot of the partition function on this axis as a function of the temperature on this axis. It's just a plot of this equation right here, right? And I did that for three different energies, three different age news, 10 wave numbers, that's this red data right here. 100 wave numbers, that's this yellow data and 1,000 wave numbers, okay? That's this green data right down here. 1,000 wave numbers, 100 wave numbers, and 10 wave numbers. I'm going to blow these up in just a second, okay? So let's calculate what the characteristic vibrational temperature is for a 10 wave number mode. I mean, of course, in real molecules, we hardly ever see a vibration that has this little energy, do we? This would be unusually low, all right? But let's do the calculation anyway. The vibrational temperature would be only 14 degrees Kelvin. And so if I look at this plot and I go to 14 degrees Kelvin and I follow it up to this red line here, all right? I find out that Q at 14 degrees Kelvin is 1.6. So if we're right at the vibrational temperature, right, if the temperature of the environment is the vibrational temperature, we're going to expect the partition function to be 1.6. In other words, we're occupying the ground state, of course. That's the 1, right? But we're not, that first excited state is not fully accessible by the system. Only 1.6 total states, including the ground state, are thermally accessible at this characteristic vibrational temperature. If we want that to be 2, indicating that that first excited state is fully accessible to the system, we have to go all the way up to 21 degrees Kelvin, all right? There's 2, if I follow that down, that turns out to be 21 degrees Kelvin, all right? So I have to go to a temperature that's about 30 percent higher than that characteristic vibrational temperature to get full occupation of that state. All right? Why am I showing this to you? Because we don't really have any intuition about what we expect this partition function to be at these characteristic temperatures, the characteristic rotation temperature, the characteristic vibration temperature. Those are the two that we care about the most. Usually in rotation, we're at temperatures that are much higher than the characteristic rotational temperature, but in vibration, not. We're often really close to it, but we don't really have any intuition about this. That's why I'm doing this. Now, what about, here's that, I blew this up now, all right, that's 500 degrees back here. It was 50, so now I made it 500. Now this is a 100-wave number mode, so now that yellow line that was down here that's blown up, boom, there it is. Okay, now the vibrational temperature for that mode is 144 Kelvin, all right? But notice, if I go to 144 and I go up to this plot, I get 1.6 again, all right? The partition function at the vibrational temperature stays the same, all right? Even though we change the energy of the mode by a factor of 10, and if I want to fully access the first excited state, I'm going to have to go up to 208, that's qualitatively exactly what we saw with the lower energy mode. I have to go to 30% higher temperature, roughly, in order to get full occupation of that excited state. And of course, if I do this with the 1,000-wave numbers, the same thing happens. 1.6, now the characteristic vibrational temperature is in order of magnitude higher, all right? But the partition function at that temperature is still 1.6, yes, so if I go 30% higher than 1400, I get 2,080, and that's what it would take to get to, okay? So that's the intuition we want to have, all right? If the temperature equals the characteristic vibrational temperature, we're going to expect the partition function to be 1.6, 1.5, okay? That's just the way the partition function works. It's counterintuitive, all right? Your intuition would tell you if you get to this characteristic vibrational temperature, then doesn't that mean that that first excited state should be fully accessible? Shouldn't the partition function be 2? Well, that's what my intuition tells me, all right? But what I'm telling you is that's not the way it actually works. You have to go to a temperature that's higher than this characteristic vibrational temperature to get to that partition function of 2, okay? All right, yes, that's what this shows. That's all this shows right here. Yes, yes, okay. Now, we're going to start talking more specifically about thermodynamics instead of statistical mechanics. Some of you may be happy about that. Some of you may be sad. I'm impartial. Both of these subjects are difficult for me, perfect to be perfectly honest. And I feel much the same way as Arnold Sommerfeld felt. The thermodynamics is a funny subject. The first time you go through it, you don't understand it at all. I think we can all relate to that. The second time you go through it, you think you understand it except for one or two small points. Not many of us have arrived at that point. And the third time you go through it, you know you don't understand it, but at that time you are used to it. It doesn't bother you anymore. This is what he said to a, he said to someone who asked him why he wasn't writing a textbook about thermodynamics. So this is a pretty famous guy. He made a lot of contributions to quantum mechanics, worked out x-ray, wave theory, was nominated for the Nobel Prize 81 times according to Wikipedia, but I don't think he ever won it. Nominated 81 times? I don't know how you can be nominated 81 times. Does that mean you were nominated like 10 times a year for eight years or Wikipedia could be wrong on that point? Okay, so assuming this is your first time through this material, the bar is set low, you don't understand it, that doesn't make you any different than this guy right here who's very famous. And I personally have had tremendous difficulty with this material through time. All right, maybe that's why I'm a good person to teach this to you because I appreciate how difficult thermodynamics is. It's not an easy subject. Okay, so you're ready for this? All right, so when we think about the universe, the total energy is assumed to be a fixed quantity. The total energy in the whole universe now, I'm not talking about planet Earth, is assumed to be a fixed quantity as a function of time. That's a hypothesis. This fixed amount of energy has been redistributed without being added to or subtracted from as far as we can tell for almost 14 billion years. Any physical quantity that cannot change with time is a conserved quantity. So energy would be an example of that. It's a conserved quantity in the universe. The first law of thermodynamics postulates that the energy is conserved in every process. Now there is no proof that this is true. It's like the Schrodinger equation. There's no proof for the Schrodinger equation. Schrodinger proposed the Schrodinger equation and we compare it with experimental observations that we make in quantum mechanics on how systems really work. We keep finding out that the Schrodinger equation predicts the right thing over and over again no matter how the system changes. And so we have more and more confidence that it's correct, but there's no proof that it's correct. Same thing is true in statistical mechanics. This equal weighting of states business, these microstates, each microstate is equally probable for a thermodynamic system. The only way to know that is to make many, many observations and to compare that hypothesis with what we see in the laboratory and over time we'd build up confidence in these hypotheses. That's what conservation of energy is in thermodynamics. It's a postulate. It's a hypothesis. So thermodynamics concentrates why it's called thermodynamics is it concentrates attention on the transfer of energy. It provides just two categories for this transfer of energy, work and heat. If we can understand flows of work and flows of heat in and out of the system that we're studying then we're going to understand the thermodynamics of that system. The thermodynamics of that system. Chapter 14 of the first law of thermodynamics, yes, yeah. Now, one of the things that makes thermodynamic confusing, thermodynamics confusing is the terminology. There's jargon and terminology and it's very confusing. One of the things that we talk about a lot in thermodynamics is the system represented by this purple rectangle. We distinguish between the system. This is the thing that we're going to be studying the thermodynamics of. It could be a microscope slide. It could be a beaker. It could be something else, like a cell, okay? And there's everything else around this system, right, which we're going to call the surroundings. So we, to make the theory simpler, we take the thing that we care about, we call it the system and everything else is the surroundings, okay? Now, there are three flavors of systems. There are open systems. A microscope slide is an open system, all right? There can be flows of matter in and out of the microscope slide, air can be dissolving in the liquid that's on the microscope slide, water can be evaporating from the microscope slide as we're looking at whatever's under there, okay? Matter can be going in and out, energy can be going in and out, we're shining a light on the microscope slide, we could be heating it up, all right? It's an open system, both matter and energy can flow in and out of the system if it's open. There's a closed system where matter can't move but heat and energy can't. Heat in the form of work, energy in the form of heat and work can flow in and out of the system. That's what I mean by this dash line here. All right? It's not going to allow matter to move in and out but heat and work can move in and out. And finally, there's a isolated system, all right? Now, this is a solid line, neither energy nor matter can flow in and out of this system, okay? And we'll talk later on about this isolated system. We're not going to talk a whole lot about open systems. There's a lot going on here and it basically ends up being a long Excel spreadsheet to understand flows of matter and heat and work in and out of such a system but we're going to pay a lot of attention to these systems right here, closed systems where we've made an effort to prevent matter from moving in and out, okay? Now, there's also different types of equilibrium just to make matters worse. There's mechanical equilibrium, in other words, the pressure of the system equals the pressure of the surroundings when we're in mechanical equilibrium. When the system is in mechanical equilibrium, the pressure is the same and inside the system and in the surroundings. In thermal equilibrium, the temperature is the same. In chemical equilibrium, the chemical potential, the total chemical potential is the same. We'll talk more about that later. All right, so we need to say something about the type of equilibrium that we're talking about often times. Sometimes it'll be obvious. If heat flows out of the system, so one of the most confusing things is the sign convention, all right, and we have to pay careful attention to it. All right, if heat flows out of the system, any process that transfers heat from the system to the surroundings is exothermic. If heat flows into the system, in other words, the total heat of the system gets higher as a function of that process, it's an endothermic process. Exothermic endothermic, all right, more thermodynamic jargon and the system has an internal energy which we're just going to call you, we've encountered you already, but now we're going to assign you to the internal energy of this system. Okay, so the change in you as a consequence of any process, work, heat, all right, the change in you is going to be just given by the final internal energy of the system, subtracted by the initial internal energy of the system. This is a property of any state function, all right. The internal energy is a state function, all right. This difference can always just be assessed by subtracting the final state from the initial state and this is true for any type of state function. We'll talk about others later on. Yes, it's the final U of the system after the process, yes, that's the initial and since a change in you can be affected by work or heat, it follows that the change in you is equal to Q plus W. Since that's the only way energy can enter or exit the system is in terms of heat or work, all right, those are our two options, all right. If Q is positive and W is positive then the internal energy of the system is going to get bigger. Now, we know intuitively what heat is, we've encountered it, all right, we have an intuitive feel for what it is, what's work? It's a little more abstract than heat, all right. It is, technically speaking, the force with which we act on the system multiplied by the displacement in the direction of this force and there's different kinds of forces that we can think about but let's just think now about a physical pressure, all right. So, technically it's the force which is a vector quantity now integrated over some path, all right, and so in general this work is going to be dependent on the path that we take. Now, that's a very abstract notion, we're going to make it concrete here in a minute, okay, force times displacement. Now, let's think about mechanical work, we'll recall mechanical work has to do with pressure, mechanical equilibrium has to do with pressure, pressure is the same inside and outside of the system. Now our system is going to be the gas that's inside this container here which has a volume V sub I that is controlled by the position of this piston along this axis here which is the x axis, this axis extends from x equals zero all the way up to x zero, that's the initial starting point of the piston, all right. The piston also has an area A that I'm not indicating in this diagram, okay, so there's a pressure inside the piston of 1 ATM, there's a temperature that we're going to maintain constant because we're going to maintain the temperature constant using this water bath and the initial volume is V sub I, okay. Now, we're in equilibrium here so the external pressure and the internal pressure, they have to be equal to one another, that's mechanical equilibrium. Now, here's a mass over here, I'm going to move this mass over to the piston, what's going to happen? Yes, isn't that beautiful? Now, we've got a final pressure, a final volume that's different from the initial pressure on the initial volume, let's work through this, see if we can figure this out, all right. The external pressure is the final pressure, that's the initial pressure that we started with plus the pressure that's being applied by this mass right here, that mass has mass M, that's the gravitational constant and that's the area of the piston, okay, and we started off with one ATM so that's MG over A plus one ATM. The external force is the force from the pressure that's outside of the system, all right, which is the same as the initial pressure acting on the area of the piston, that's the area that piston out here, that's its area, okay, and that's the mass times the gravitational constant so that's the total force, all right, that's the force from the external gas pressure which hasn't changed, same as it was initially and that's the force from the mass acting with the acceleration of gravity on the piston, boo, okay, so we said earlier work is just force times distance and so I can perform this integral right here, all right, I just have to substitute for the force, there's the force so I'm going to put that inside the integral, now it's not DR anymore, it's DX because X is the direction that the piston's moving in, all right, and we're going from X to X minus H, what am I talking about, we're going from X zero, all right, and when we put the mass on there the piston goes down to X zero minus H, those are the limits of the integration, okay, so when I evaluate that integral, whoops, I didn't say, why is that a minus sign, it's a minus sign because this force is directed in the negative X direction, it's directed down, all right, the positive X direction is up, all right, so the force is negative, yes? A is area, right, A is area, okay, so I can do this integral and the reason it's so easy to do is because the force is constant, it doesn't depend on X so I can pull it out of this integration and I can just plug these limits in, boom, boom, boom, and I get H, okay, so the work is just given by this really simple expression here, right, and it's multiplied, it's proportional to H, okay, with me so far, now, what happens to the volume? Well, the change in the volume is just the final volume minus the initial volume, right, here we are at the final volume, all right, and I can figure out what this volume is because it's this displacement multiplied by the area, right, area of the piston times the displacement that it's moving through and so that's the final volume right there, A times X zero minus H and that is the initial volume and so the difference between them is this minus A times H, that's the change in volume, okay, so now if I take that, delta V equals minus A H, I can make substitutions into this work expression up here, for example, A equals delta V over H, A equals delta V over H and so if I put delta V over H here, the H cancels like it minus P delta V for that first term, right, and when I make the substitution here, I'm going to get MG delta V over A, I'm just substituting from this expression right here, okay, so this is P sub I, by the way, I left off the subscript, okay, so I can factor out the delta V, all right, because there's delta V here, delta V here and so this is work is equal to minus P sub I plus MG over A, well, that's just the external pressure, the way we defined it earlier, that's just the total external pressure, right, that's the initial pressure and that's the pressure imparted by the mass that we stuck on the piston times delta V and so work equals minus P external times delta V, yes, we can use this equation to calculate the mechanical work, all right, this is a derivation of it, I don't think this derivation is in your book, it turns out to be a helpful way to think about this process as I'll show you in a second, okay, so we can use this equation, we can plug in chug, let's do that, calculate the work required to compress one mole of gas initially at 100 ATM pressure against the 200 ATM external pressure at constant temperature from 2 liters to 1 liter, assuming the gas is ideal, work, calculate the work, so we need work equals minus P external times delta V, I can plug in these change in volume from 2 liters to 1 liter, 1 liter is the final volume, 2 liters is the initial volume, we only need P external against the 200 ATM external pressure, my goodness, okay, and so when I do this calculation, I get 200 liter atmospheres. Now, these are energy units, but they're not energy units that we want to use too often, all right, we need to convert them into energy units that we care about, like joules, okay, so if I take liter atmospheres and I multiply by, here's just two, here's the gas constant, right, 8.31451, that's one gas constant and .08206, that's another gas constant, the Kelvin's per mole cancel, I get joules per liter atmosphere, well, that's pretty convenient, because we want to convert to joules, all right, so if I multiply liter atmospheres by this R over this R, I get rid of the, I convert to joules, I get 20.26 kilojoules, everyone see how I did that? So, that's the work, 20.26 kilojoules, we're done, it's plus, does that make sense? What does the plus mean? So, every single time you do one of these thermodynamic calculations, you've got to ask yourself if the sign makes sense, because the sign is going to mess you up, all right, the sign is very important, we've got to get it right, all right, it's a big delta, minus versus plus, all right, it tells us everything about what's going on, so we can't, it's more important, sine air, you know, in the past, not such a big deal, it's just a sine air, but here the sign is everything, all right, tells us whether energy is going in or energy is going out, okay, so what does the plus sign mean? It means that the volume got smaller, all right, we did work on the system, now that's not intuitive, all right, there's nothing intuitive about that, because work is not an intuitive concept, all right, but in PV work, if the volume gets smaller, you did work on the system, and work should be positive, all right, you're putting work into the system, is that intuitive? No, not to me it's not, let's look at this graphically, here's the initial state of this system, we've got a two liter container, we're at 100 ATM, now we're going to compress to this final state here, 200 ATM and one liter, that's what we're doing in this problem, right, where's the work on this diagram? Which area, it's the area of this rectangle, isn't it? No, all right, we imagine that this process occurs in two steps, first the pressure's increased to 200 ATM, then we compress the 200 ATM, that's the external pressure that we're acting against, this is the total work, all right, you might think it's just this, no, all right, does that mean that it makes no difference what pressure we started at, yes, all right, because no matter where we start, we're going to, in the first step of our imaginary process, we're going to increase the pressure to the external pressure that we're acting against and then we're going to compress, all right, doesn't matter if you start here, here, here, here or anywhere in between, that's odd, okay, that's the work, all right, it just happens that we started at 100 ATM. This means, in principle, we can arrive at the same final stage doing less work. If we divide this compression process into two steps, my goodness, it might take less work to get there, all right, work is not a conserved quantity, it's not like energy, all right, it's not a conserved quantity, use your imagination, you can do less of it. Many of you know that already. So, here's what we did before, we used one mass, we compressed the piston, that's tantamount to doing this, all right, one mass was used to accomplish this change in displacement of this piston from two liters to one liter, all right, so instead of doing that, let's take two masses, all right, we'll put one mass on at a time and after we put both masses on, we'll be at the same place, they'll all be the same final state of the system, but now we did it in two steps. We took the same mass, we sawed it in half and we used it in two steps. That's going to save us work, surprising, isn't it? Intuitively, that makes no sense, it seems like more work, all right, you had to saw the brick in half and transfer two parts of the brick to the piston, all right, here's what we did, we started here, we did this compression first and then we did this compression, I can calculate the work separately for each one of these two processes because I'm going to go to 101.5 liters and 150 ATM, boom, boom, okay, and there's the work for this guy, all right, the first guy, that was the work was that area, now the work is this area, all right, we can call these processes one and two, all right, when I add them together, I get 17.73 kilojoules, that's less than we calculated. We calculated 20.26, remember that, 17.73, that's less, we did less work and we got to the same final state, okay, the total work depends upon the path that we choose between two equilibrium states, this is a very important point, all right, the amount of work that you have to do to get to two, to traverse from an initial equilibrium state to some final equilibrium state, all right, that's not a conserved quantity, it depends on the path that you take, it's a path dependent quantity, that's a very important message of thermodynamics. So in principle, more steps means less work, all right, this is an example, so let me just make a few general comments. What we're talking about here is an example of expansion or compression against a constant pressure, yeah, this is mechanical work that we're talking about, mechanical equilibrium, there's other possibilities, there could be a reversible expansion, there could be an isothermal and reversible expansion, all right, in thermodynamics, a reversible process, here's more jargon for you, reversible, what's a reversible process? A reversible process is a process that is easily reversed, all right, what does that mean? It means that if I'm applying, did I say anything intelligent, it's a process of direction, I'm just saying it's a proud of me, so if I'm compressing a piston, right, and I apply a tiny extra increment of force, I can move that piston a tiny amount, okay, and if I take that force away, it'll move back by the same tiny amount, all right, and so that was an infinitesimal change in pressure that I applied to that piston, all right, and the process reversed, in other words, I compressed the gas a little bit, but then when I took the pressure away, the piston came right back, right, by the same amount, all right, only under these conditions can the system remain continuously in thermodynamic equilibrium, in other words, if I do, if I compress the piston at any macroscopically observable rate, even if you try to do it really slow, you're not doing it slow enough, you're introducing losses, right, dissipation, energy's getting lost, why? There's turbulence in the gas that you're exciting when you do this compression, for example, right, there's friction terms in the process, right, that suck energy out of it, all right, if you want the process to conserve energy, you've got to do it infinitely slowly, it turns out, all right, that's an odd notion, but that's the way it works. So, how do we do that here, all right, I cut the brick in half already, all right, and I know I can do less work with two halves of the brick, what I'm going to do is I'm going to take these two pieces of brick now and I'm going to put them in a mortar and pestle, how many people know what that is, all right, I'm going to grind them up and make a powder, all right, there it is, all right, powdered the brick and now I'm going to transfer one grain of this powdered brick onto the piston. It moved so that the volume is now smaller by DV, all right, I was careful not to lose any of the brick, mind you, all right, which is easy to do in this process, but I was careful, all right, and so now I'm taking the tweezers and I'm putting flake by flake all of the brick now onto this piston and sure enough, it ends up at the same final location, all right, there is no way to do this more slowly than this, all right, it takes months, all right, and now my diagram looks like this, all right, I put one tiny flake onto the piston and I moved right here and then two and moved right there and then three right there, all right, it took thousands to get all the way up to the final pressure, you with me? All right, there is no way, believe it or not, to do less work than this. Now, what can we understand about this process? Let's see if we can figure out mathematically what we just did because this is kind of important to understand this. Here's what we did, so as an example, if the external pressure is continuously adjusted, so it was infinitesimally higher than the initial pressure, that's what we did, we just made it infinitesimally higher by putting one grain at a time on the piston. Peace continuously adjusted, we can achieve this limitingly slow work, we're going to call that pressure that we're adjusting flake grain by grain, we're going to call that p-cis, all right, that's just to give it a name, all right, now we know work is minus p-cis dv, all right, work for each work associated with the addition of each grain, all right, because we're going to increase, we're going to decrease the volume rather by dv, all right, so if we integrate that, all right, this is the pressure that depends on the volume, that's what I mean by this notation, times dv, now what's p-cis? Well, if it's an ideal gas, p is just nRT over v, right, p's just nRT over v, nRT don't depend on v, all right, so they can go out front in this integral, so I've just got an integral from the initial volume to the final volume of dv over v, I can do that integral, that's pretty easy, all right, that's just log v final over v initial, all right, so that's the work, right there, how easy is that, all right, all I did is I write the work as a differential and I recognize that the volume change is I add each grain is going to be super tiny, all right, and then I do this integral, all right, I remember that I've got the ideal gas equation here to think about, so I've got RT over v, boom, that's what the work is equal to and so now I can figure out how much area is under this guy, that's the total work that I did in this case, right, and so when I do that calculation I got one mole times 8.31451 times 2,437 degrees Kelvin, where did that come from? Well, by golly that's the temperature, all right, I never told it to you but if I plug in all of these constants using the ideal gas equation, that's the temperature I get, so we haven't said anything about what the temperature was here but by golly it's pretty warm, all right, it's 2,437 degrees Kelvin, all right, and so if I plug that in and the volume went down by a factor of 14.04 kilojoules is the amount of energy involved, all right, 14.04 kilojoules, that's the area under this curve right here and by golly that's lower than either one of these, all right, we did it in one step, it was 20, we did it in two steps, it was 17 and now it's 14, all right, so this is reversible isothermal compression. Reversible means you crush the brick up into a million flakes and add one at a time, you with me? There's no way to do it in smaller steps than that and smaller steps wins in terms of minimizing work. Smaller steps wins, less work with more steps. Counterintuitive? Absolutely, all right, completely backwards, all right, from our normal way of thinking, all right, but now we got this equation for work, do they both agree in terms of this sign because we've got to pay attention to the sign while I already mentioned that if the volume gets smaller, work is positive? Well, you can see that that's true here because if V2 is smaller than V1 then this is going to be negative, that'll cancel that negative sign and I'm going to get a positive work, right? If V2 is smaller than V1 that's negative, that cancels that negative sign so I get positive work. If this V final is less than that V initial then that ratio is less than one and the log of less than one is less than zero so again we cancel that negative sign and work is positive, all right, so in both cases if V2 is smaller than V1 work is positive, try to keep that in the back of your mind because that's a check on your answer, all right? The last thing you need to check on that answer is is that sign right, did I make a sign error? So the sign is everything. Now that's mechanical work. There's other types of work that we're not going to talk about because we've only got 10 weeks to do everything in physical chemistry except quantum mechanics, okay? We're not going to talk about surface tensions. This is a very interesting subject. We're not going to talk about the extension of lines where there's an energy on the line. We're not going to talk about electrical work and 100 other types of work that we could talk about. We don't have time to. We're really just going to focus on this mechanical work right here. See if we can understand that pretty well. Okay, I think I'm going to let Gianmarc tell you about the rest of this. I think there's 130 slides in this lecture. I'm tired. Okay, so on Friday, quiz three, all right? Same drill, every other seat, and then a short lecture from Gianmarc.