 So we're going to continue our discussion of thermodynamics and look at adiabatic processes. So you've got a really important thing to deal with in thermodynamics is defining your system and your surroundings, all right? It sounds trivial, but it is not, okay? So we're going to do a couple of examples and you're going to see this. So when we're looking at our system, we want to know how much energy the system has. It's internal energy, U. So delta U is going to be the heat plus the work, Q plus W, okay? So let's talk about the sign convention. So in other classes, perhaps, you know, you've got off easy on sign errors just to point off something like that. But in thermodynamics, the sign is everything, okay? If you get the sign wrong in thermodynamics, it's the difference between making a refrigerator and a heater, okay? That's pretty wrong, right? So here we have a system and the surroundings. So let's say we have heat and work going from the system into the surroundings. Anybody know what kind of reaction this is called? All right, you got it, exothermic. Okay, so you've got heat going from the system into the surroundings and that's easy to see when your system is a purple box, all right? But let's talk about an example. Let's say combustion. You're at the beach with a bonfire. The log is on fire and you pick it up and you burn yourself, right? So there's an exothermic reaction going on. So you're saying heat's going out of the log. Well, if heat's leaving the log, why is it hot when I pick it up? See, now it's not quite so obvious, right? But if you think about it, the system is, say, the cellulose or whatever is burning in that log or combusting. So, and you are part of the surroundings. So the important thing with thermodynamics, it's not about you, right? It's you're part of the surroundings. So if you feel that log, you're feeling heat transferred from the burning log into your hand. So thus this still holds up. Even though it seems a little backwards, it's not. Okay, here we have one, wait, what? Okay, we just talked about that. Okay, now here's the system. We've got heat and work both going into the system. So if heat and work are both going into the system, what's it called? Good, endothermic. So, yeah, this is an endothermic reaction. Now let's think about this. So you've got an endothermic reaction going on in the hood of your lab. And you grab the container. Is it hot or cold? It's cold, good. Because with endothermic, it's pulling heat into it, right? So you say, well, if it's got all this heat in it, why is it cold? Well, it's pulling that heat energy from the surroundings. Your hand is part of the surroundings. That's why you feel cold. Okay. All right, so now heat's pretty clear, right? Which way heat's going? How do you know which way work is going? Well, here's the equation for work. So if it's got a negative in front of the P, what's the only way that this right side can be positive? If delta V is negative. So all you have to do is look at your change in volume and that'll tell you if there's work is going to be positive or negative. And when you calculate this work, you're always calculating the work performed on the system, not by the system. So if you think about like a compressed air cylinder, right, like a shock absorber on your car and you walk up and you push it down, you did work on it, right? And the volume inside the cylinder got smaller. So if you pay attention to what happens with volume if it's getting larger or smaller, that'll tell you your delta V. And hence you'll get, you'll know which way work is going. Okay, so PV work, as it's colloquially called, is just one type of work. We have a bunch of others, but we're only going to be dealing with PV work in this class. So you're not going to need to know how to do these others, but we just wanted you to be aware that that's not the only type of work you can describe. So that's work, how about Q. All right. So if we restrict work to PV work and we look at our equation here, we could substitute for work. We can put p delta V in there, right? And that little p there just means constant pressure. That's what that's there for. So this is a particularly interesting situation for chemists because most chemical reactions are performed under constant pressure, right? If you have reaction going on in a beaker that's open to the environment, your pressure is going to stay the same. It's just going to be 1 ATM, right? So hence, reactions carry out at constant pressure are a point of interest for chemists. They're not always carried out at constant pressure, but most of the time if you have a closed up system, that's a bad thing, right? Because you'll get glass in your eye when it explodes. So like it's not in your eye if you're wearing your glasses. But all right. So our delta U is our Q is minus our work. So here we're just explaining it, I'm expanding the deltas, right? So you've got delta U. Anytime you see a delta, you know that means final minus initial. So you've got U2 minus U1 and we expand volume in the same way. Then we do some simple algebra and we can rearrange it this way. So we've got the 2's together and the 1's together. The final and the initial are together here. And there's a difference. The reason we're doing that is U plus PV is also known as enthalpy. So if we were able to group those two together, you can say enthalpy final minus enthalpy initial is equal to our heat. So how can we define heat? Well it's our change in enthalpy assuming a constant pressure. And also assuming that the only work we're dealing with is PV work, which is the assumption that you can carry throughout this class. So now we've got a little example here. We've got two different minerals that conform from calcium carbonate, calcite and aragonite. Another thing calcium carbonate in the ocean, it's dissolved in the water and that's what mollusks pull out of the water to create their shells is calcium carbonate. Fun fact. Okay, so let's look at this problem. So the change in U when one mole of calcite is converted to aragonite is 210 joules per mole. Calculate the difference. Okay, so we want the difference between the enthalpy change and the change in the internal energy. So we want a difference, right away when you hear difference, you think subtraction, we're subtracting the two. And they give us the densities and the pressure at which we're working. So first we'll go to our enthalpy equation. And here's our enthalpy equation in terms of changing. So change in enthalpy, P doesn't get a delta, why? Constant pressure, right? So we define pressure as being constant so it doesn't get a delta. And we said we want the difference. So then we know we're just going to subtract the delta U from both sides and we have our difference is going to amount to P delta V or our PV work. Okay, so we've got, they gave us densities, but we're interested in the change in volume. So the way we get volume from density here, we know how many grams of stuff we have because we know how many moles and it's also, it's atomic weight or molecular weight. So if you divide that, the grams cancel, you're left with cubic centimeters. So we know our change in volume between these two forms of the calcite and the arachnite. And then we're going to rewrite it. We're going to rewrite it as Pascals times meters cubed. So we're going to convert our cubic centimeters to Pascals. We're going to convert our one bar to Pascals. And the reason for this is that Pascals times cubic meters gives you joules. And if you don't believe me, just Google Pascals times meters cubed. That's what I did. And yeah, it comes right up. It doesn't even give you a website. It just tells you. Google gets geekier by the day. I love it. So if we plug in everything we know, our change in temperatures and our, excuse me, our change in volumes and our pressure, we get negative 0.28 joules. So looking at this, this is only 0.1 percent of our total internal energy is represented by the PV work. So the reason we did this example is we want you to realize that if you're talking about delta U, change in internal energy, or the change in enthalpy, if you're dealing with solids, they're almost the same number. And it makes sense, right? Because solids, from one solid to another, the volume doesn't change much. The exception, of course, is with really high pressures, right? If you get that pressure really high, even though the delta V is small, we can get that number up overall. So that would be an exception. Okay. So now, here it is graphically. So on the bottom there, we've got internal energy. And up top, we've got the enthalpy. There we go. So what do you think that difference is between those two lines? The gap. What's the difference between these two numbers? That has that added to it, right? Yeah. So it's our PV. So this is just what it looks like graphically. So if you were to plot enthalpy as a function of temperature, these are the graphs you would get. And PV represents our difference. Okay. So we said it's the slope, right? So anytime you guys hear slope, you should always think derivative, right? At this point in your careers, you know slope derivative, automatic. So the important thing too is that it's a partial derivative, right? So anytime we take a partial derivative, we need to hold something constant and then find the slope in another direction, right? So you can hold one of two things constant. You can either hold volume constant and then you get a heat capacity of constant volume or you can hold pressure constant and then you get a heat capacity of constant pressure. So yeah, it's a partial derivative of internal energy with respect to temperature or enthalpy with respect to temperature. These are two forms of heat capacity. It's important you specify which one. Okay. So now we have an example. We've got this gadget here that you've got an electric heater hooked up to a block of copper and we're going to send some current through that and heat this copper up and we're going to measure the change in temperature. And from this, we should be able to figure out the heat capacity both of our specific system here, the block of copper and a molar heat capacity of copper in general. So by the way, Dr. Pener told me this is a kit that you guys can buy online. So if you want to do this at home and impress your friends at parties, you can show everyone the heat capacity of copper. Or not. So here's where the thermometer goes. There's the heater. It's 75 grams of copper, not rocket science. All right, so an important fact you need to know to do this problem is that 1 watt is equal to amps times the voltage. So you multiply the two together. Well, amps describes the flow of current, right? So that's charge per time. So how much charge is going by per time? So how fast the current is flowing through? This is the dangerous one, right? If you get electrocuted, high amps is trouble. Potential, eh, you know, it's the amps that kills you. So good thing to remember. Here we have our voltage here, and that's joules per coulomb, so energy per charge. Multiply them together, coulombs cancel, and we're left with joules per second. So now we know how much energy per second is being transferred from this electric heater to the copper. Yeah. And we also know that we ran the heater for 19 seconds, right? So you multiply by the time, you know, so it's like speed, right? If you're going 50 miles an hour for an hour, you go 50 miles. I think there's a YouTube video about that. I don't know if you've seen it. Same deal. If we got 50 joules per second, you go 19 seconds. You got 950 joules. You just multiply them. So we got 950 joules. Oh, okay, we should probably figure out how many moles of copper we're dealing with, right? Because we want to use both our heat capacity of our system and we want a molar heat capacity. So you just convert that to moles. I'm not going to walk you through that. Guys have done that a million times, I imagine. Okay, so heat capacity or constant pressure is a change in the energy over the change in time, right? So in this case, our change in energy is just heat, right? The volume's not changing, nothing like that, right? All we're worried about is the change in heat. And heat is the transfer of energy, right? So we transferred 950 joules and our temperature changed by 33 Kelvin, therefore our heat capacity is joules per Kelvin. So now down here, we want the molar heat capacity. So all you do is divide the heat capacity for your system by the number of moles that you're dealing with and you got it. All right? All right, nice easy one to start. Now what do we got? Now we want to figure out the change in enthalpy if you're heating 2 moles of oxygen from 24 degrees Celsius to 100 degrees Celsius at 1 ATM. Okay, so we're going to use heat capacity again. But this time, we're going to use it in kind of a little more of a slick way. So if I multiply both sides by DT, yeah, multiply both sides by DT, you can see how we get our DH by itself, we have our DT there, right? So now you can integrate both sides. So if you put an integral on a DT, that's just going to be final minus initial, right? Also known as delta H, right? Or DH, rather. So we integrate DH, we get delta H. Now over here, we have an integral from our initial temperature to our final temperature of our heat capacity and our constant pressure with DT. So if we can find a way to express our heat capacity with T temperature as our variable, we can solve that integral, right? As luck would have it, we can do this, right? I look kind of foolish right now. So there's our heat capacity. Now here's a formula where we express heat capacity in terms of our variable temperature. A, B, and C are just constants. So you guys know the ideal gas law. Well, van der Waals came up with a more precise ideal gas law that you can apply to actual gases. So each gas, whether it's oxygen, nitrogen, whatever, is going to have a different value for A, B, and C. So that's all this A, B, and C are. So that's our formula for heat capacity. So now we can throw that into our equation. There's the parameters we just talked about. That's what they are. So we put them in here. So the left side we already solved. The right side we have this integral now. So the integral of a constant is just going to be the constant times our delta T. Integral of T is just T squared over 2. And integral of T squared, 1 over T squared is minus 1 over T, right? Yeah, there's that formula. So we plug in everything we know. So here's our formula. We took those integrals. And then we just plug everything we know in here. Our temperatures, our constants, and voila. We know how much energy. Everybody got that? Okay. Okay, so now we're going to talk about adiabatic processes. Oh, okay. So for an adiabatic process. So first we're going to take our normal rule here. And then we're going to talk about a change. So if U is changing, so therefore Q is changing and W is changing. So we make it changes. But now we say for an adiabatic process that our Q is zero and our delta Q is also zero for these changes. But we can still make changes to the work. So if DQ is zero, well now we're saying the internal energy is equivalent, the change in internal energy is equivalent to the change in work for an adiabatic system, all right? So make sure these qualifiers are important. You don't want to just apply this blindly, right? So let's see here. That's my spot. Okay, this is just everything we've been saying. So now if we've got our DU as our heat capacity times DT, right? So same equation before, our heat capacity equation. We multiply both sides by DT again. But now we're going to take this fact that we know that if we're adiabatic, that DU and DW are interchangeable, right? So looking at this thing, we've got, that's the equation we knew. We're going to substitute DW in there. But DW, we said the only kind of work we're dealing with is PV work, right? So we put our PDV in there. But now if we know we're dealing with an ideal gas, for an ideal gas PV equals NRT every time, right? So instead of writing pressure there, well that's just the same as NRT over V, right? So I took the ideal gas law, divide both sides by V. So this thing right here is equivalent to pressure, NRT over V. And this is nice, right? Our variable is V. We've got a DV. Things are looking our way. So let's see, we substitute everything in. Our DU is equivalent to DW. Therefore we can bring this down here. So heat capacity DT is equivalent to minus NRT over V times DV. So now we can find a useful expression for this. So let's say we integrate both sides again. Well first of all let's divide both sides by T to get the T with the DT and the V with the DV, right? That's important. So we get all our variables where they need to be. We perform the integration. So the integral of 1 over X is ln of X, good. So let's see, yeah, that's all we did here. So we took the integral of both sides and so you've got the ln of our temperature here and of volume here. But it's going to be final minus initial, right? So the ln of T2 minus the ln of T1. But if you have two ln subtracted from each other, of course, we could put those together, right? And just put them in the same log but divide. So that's all we did here. All right, now to make this a little more usable even, let's divide both sides by CV. We'll get that over there. And then after we do that, we're going to put both sides as E to the X. And when you do this, you'll get rid of the log here, you get rid of the log here and then this will be the exponent. We don't have the algebra worked up here. It's pretty simple. Ask me afterwards if you want me to do it. But you're going to get this equation. So this is really handy because now we have an equation that if we know we're adiabatic and we can relate change in temperature to change in volume given that all we know is our R value which is a constant and our heat capacity with respect to a constant volume. So all we need to know are these things and we can relate these to each other. So now if you look at this graph here, what we're trying to show you, so an isotherm is a surface. Right? And what it means is that if you have this system and you're going to, so like a compressed gas cylinder and you're going to squeeze that volume together, the pressure is going to go up, right? So going along this path, this is if we hold temperature constant and you change this volume, that's what's going to happen to pressure. And this is at a different temperature. So now if we're talking about two different temperatures and we know it's adiabatic, then the change here is going to be larger for adiabatic than an isothermal process. So our change in pressure is going to be larger. So if you just, it makes sense if you look at this thing, right? If we're going along our isotherm and we want to see our change in pressure, well you just go here to there, right? But if you go along the adiabatic process, you've got to go there to there. So it's a longer path, just intrinsically because you're jumping from one surface to the other. It has to be. So let's talk about an isotherm a little bit. Here's our equation for adiabatic, for an adiabatic change. I already figured that out. Isotherm, if we look at this, that should be kind of obvious to you, right? Because down here we've got our delta V, but if that was just a regular V, we could bring that over there. You've got PV equals NRT. So again, isotherm just means you're sticking at the same temperature and you're following the ideal gas law. But your volume is changing. So that's all that is. Is it constant? Oh, for this equation? So yeah, so it's a derivative, but yeah. So each, a specific substance, if you look it up, you can get a value for it. Yeah, so now we're treating it as a constant. Very good point. Yeah, so, yeah, that was a really good point. So now we're just treating this thing as a constant, right? Okay, so now we've got one final problem here. So now we've got two moles of neon that expands adiabatically and we're actually going to figure it out for this one. Two moles of neon expands adiabatically and reversibly from 5 liters to 20 liters and we're staying at the temperature 3, oh no, no, we're not staying at the temperature. We're starting at the temperature of 373 Kelvin and we're going to end at some other temperature. So our temperature is changing. We're not going to use the isotherm equation, right? Isotherm, same temperature. This is not same temperature. So we're going to use our adiabatic equation. There it is. So what is our CV value? So in this case, we get to bring in statistical mechanics. So you thought you were done, you're not. We bring that back in. So we're talking about neon, which is an atomic gas, right? And as far as stat mechs concerned, we treat atoms as points, right? So any energy that these atoms have as they're bouncing around as a gas, you can't rotate a point, right? So it has no rotational energy and you can't vibrate a point either. So it has no vibrational. So all we're talking about is translational. So just to refresh your memory, if you look at the Hamiltonian for your gas, any value that's quadratic or squared is going to give us a value of r over 2, right? So we're going to get an r over 2 for our CV for any value, for anything that's quadratic. So 1 half mv squared or momentum squared over 2m, right? Both, they're squared, right? So those describe the kinetic energy of the atoms. So, and we can move in the x, y, and z direction. So we've got three total quadratic terms. So 3 times r over 2, that's going to be our heat capacity for this problem. So let's see. Well now we know everything. So we know our initial and final volume. We know our heat capacity. We throw that in there. And our change in temperature we have, or we solve for the change in temperature rather. And it's going to be 148 Kelvin. Is everybody good with that? Did I go a little fast? A little bit? Oh, I did. So that's all I got. If you have no questions, you're free to go. Have a good weekend. Would you have any questions? Okay. Have fun.