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# PHYS 200: Fundamentals of Physics I

## Lecture 22

## - The Boltzmann Constant and First Law of Thermodynamics

### Overview

This lecture continues the topic of thermodynamics, exploring in greater detail what heat is, and how it is generated and measured. The Boltzmann Constant is introduced. The microscopic meaning of temperature is explained. The First Law of Thermodynamics is presented.

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html## Fundamentals of Physics I## PHYS 200 - Lecture 22 - The Boltzmann Constant and First Law of Thermodynamics## Chapter 1: Recap of Heat Theory [00:00:00]
And the way to find the Kelvin scale, you take the gas, any gas that you like, like hydrogen or helium, at low concentration, and put that inside a piston and cylinder. That will occupy some volume and there’s a certain pressure by putting weights on top, and you take the product of In other words, it appears that pressure times volume is some constant. I don’t know what to call it. Say The next thing I mentioned was, people used to think of the theory of heat as a new theory. You know, we got mechanics and all that stuff–levers and pulleys and all that. Then, you have this mysterious thing called heat, which has been around for many years but people started quantifying it by saying there’s a fluid called the caloric fluid and hot things have a lot of it, and cold things have less of it, and when you mix them the caloric somehow flows from the hot to the cold. Then we defined specific heat, law of conservation of this caloric fluid that allows you to do some problems in calorimetry. You mix so much of this with so much of that, where will they end up? That kind of problem. So, that promoted heat to a new and independent entity, different from all other things we have studied. But something suggests that it is not completely alien or a new concept, because there seems to be a conservation of law for this heat, because the heat lost by the cold water was the heat gained by the hot water. I’m sorry. Heat gained by the cold water was the heat lost by the hot water. So, you have a conservation law. Secondly, we know another way to produce heat. Instead of saying put it on the stove, put it on the stove, in which case, there is something mysterious flowing from the stove into the water that heats it up, I told you there’s a different thing you can do. Take two automobiles; slam them. This is not the most economical way to make your dinner but I’m just telling you as a matter of principle. Buy two Ferraris, slam them into each other and take this pot and put it on top and it’ll heat up because Ferraris will heat up. The question is what happened to the kinetic energy of the two cars? That is really gone. So, in the old days, we would say, well, we don’t apply the Law of Conservation of Energy because this was an inelastic relationship. That was our legal way out of the whole issue. But you realize now this caloric fluid can be produced from nowhere, because there was no caloric fluid before, but slamming the two cars produce this extra heat. So, that indicates that perhaps there’s a relation between mechanical energy and heat energy–that when mechanical energy disappears, heat energy appears. So, how do you do the conversion ratio? You know, how many calories can you get if you sacrifice one joule of mechanical energy? So, Joule did the experiment. Not with cars. I mean, he didn’t have cars at that time, so if he did he would’ve probably done it with cars. He had this gadget with him, which is a little shaft with some paddles and a pulley on the top, and you let the weight go down. And I told you guys the weight goes from here to here, the So, in the example of the colliding cars, take the ½ Okay. So today, I want to go a little deeper into the question of where is the energy actually stored in the car, and what is heat. We still don’t know in detail what heat is. We just said car heats up and the loss of joules divided by 4.2 is the gain in calories. Now, we can answer in detail exactly what is heat. That’s what we’re going to talk about today. When we say something is hotter, what do we mean on a microscopic level? In the old days when people didn’t know what anything was made of, they didn’t have this understanding. And the understanding that I’m going to give you today is based on a simple fact that everything is made up of atoms. That was not known, and that’s one of the greatest discoveries that, in the end, everything is made up of atoms, and atoms combine to form molecules and so on. So, how does that come into play? For that, I want you to take the simple example where the temperature enters. That is in the relation
So, if you put more gas into your piston you think it’ll produce more pressure. That’s actually correct. So, let’s go to that one particular sample in your laboratory that you did. So, put the mass that you had there. Then, you should put a constant still. I don’t know what you want me to call this constant, say, But here is what people found. If you do it that way, the constant So, you have to think about why is it that the mass directly is not involved. Mass has to be divided by a number, and the number is a nice, round number. 4 for this and 12 for that, and of course people figure out there’s a long story I cannot go into, but I think you all know the answer. But now we are allowed to fast forward to the correct answer, because I really don’t have the time to see how they worked it out, but from these integers and the way the gases reacted and formed complicated molecules, they figured out what’s really going on is that you’re dividing by a number that’s proportional with the mass of the underlying fundamental entity, which would be an atom. In some case a molecule, but I’m just going to call everything as atom. ## Chapter 2: The Boltzman Constant and Avogadro’s Number [00:11:54]So, if things, like, carbon, as atoms, weigh 12 times as much as things called hydrogen, then if you took some amount of carbon, you divide it by a number, like, 12 to count the number of carbon atoms. Okay, so hydrogen you want to count the number of hydrogen atoms. So then, what really you want here is not the mass, but the number of atoms of a given kind. We are certainly free to write either a mass or the number of atoms, because the two are proportional. But the beauty of writing it this way, you write it in this fashion, by this new constant So, you couldn’t have written it that way until you knew about atoms and molecules and people who are led to atoms and molecules by looking at the way gases interact, and it’s a beautiful piece of chemistry to figure out really that there are entities which come in discreet units. Not at all obvious in the old days, that mass comes in discreet units called atoms, but that’s what they deduced. So, this is called the Boltzmann Constant. The Boltzmann Constant has a value of 1.4 times 10 So, now what people like to do is they don’t like to write the number [N], because if you write the number, in a typical situation, what’s the number going to be? Take some random group gas. One gram, two grams, one kilogram, it doesn’t matter. The number you will put in there is some number like 10 For example, when you want to buy eggs, you measure in dozens. When you want to buy paper, you might want to measure it in thousands or five hundreds or whatever unit they sell them in. It’s a natural unit. When you want to find intergalactic distances, you may use a light year. You use units so that in that unit, the quantity of interest to us is some number that you can count in your hands. When you count people’s height, you use feet because it’s something between 1 and 8, let’s say. You don’t want to use angstroms and you don’t want to use millimeters. Likewise, when you want to simply count numbers, it turns out there’s a very natural number called Avogadro’s Number, and Avogadro’s Number is 6 times 10
If you write it this way, then you write this R will be PV, which is units of energy divided by T.In terms of calories, I’d remember this as a nice, round number. Two calories per degree centigrade. Degrees centigrade and Kelvin are the same. The origins are shifted, but when you go up by one degree in centigrade or Kelvin, you go the same amount in temperature. So, this [ ## Chapter 3: A Microscopic Definition of Temperature [00:18:50]Alright. Now, you start with this law and you ask the following question. On the left-hand side is the quantity So, for this purpose, we will take a cube of gas. Here it is. This is a cube of side So, who’s changing the momentum? Well, the wall is changing the momentum. It’s reversing it. For example, if you bounce head-on and go back, your momentum is reversed. That means you push the wall with some force and the wall pushes you back with the opposite force. It’s the force that you exert on the wall that I’m interested in. I want to find the force on the wall, say, this particular face. You can find the pressure on any face. It’s going to be the same answer. I’m going to take the shaded face to find the pressure on it. Now, if you want to ask, what is the force exerted by me on any body, I know the force has a rate of change of momentum, because that is Now, that’s a very complicated problem, so we’re going to simplify the problem. The simplification is going to be, we are going to assume that one-third of the molecules are moving from left to right. One-third are moving up and down and one-third are moving in and out of the blackboard. If at all you make an assumption that the molecules are simply moving in the three primary directions, of course you will have to give equal numbers in these directions. Nothing in the gas that favors horizontal or vertical. In reality, of course, you must admit the fact they move in all directions, but the simplified derivation happens to give all the right physics, so I’m going to use that. So, Next assumption. All the molecules have the same speed, which I’m going to call L. That is the force due to one molecule. That’s the average force. You realize it’s not a continuous force. The molecule will hit the wall, there’s a little force exchange between the two, then there’s nothing, then you wait until it comes back and hits the wall again. If that were the only thing going on, what you would find is the wall most of the time, has no pressure and suddenly it has a lot of pressure and then suddenly nothing. But fortunately, this is not the only molecule. There are roughly 10^{23} guys pounding themselves against the wall. So, at any given instant, even if it’s 10^{-5} seconds, there’d be a large number of molecules colliding. So, that’s why the force will appear to be steady rather than a sharp noise. It looked very steady because somebody or other will be pushing against the wall.This is the force due to one molecule. The force due to all of them would be L. N over 3 because of the N molecules, a third of them were moving in this direction. You realize the other two directions are parallel to the wall. They don’t apply force on the wall. To apply force on the wall, you’ve got to be moving perpendicular to the wall. For example, if the planes that walls are coming out of the blackboard, moving in and out of the blackboard doesn’t produce a force on this wall. That produces a force on the other two faces. So, as far as any one set of faces is concerned, in one plane, only the motion orthogonal to that is going to contribute. That’s why you have N over 3.We’re almost done. That’s the average force. If you want, I can denote average by some N over 3, mv over ^{2}L. Now, this is very nice because ^{3}L is just the volume of my box. So, I take the ^{3}L, which is equal to the volume of my box, and I send it to the other side and write it as ^{3}PV equals N over 3mv.^{2}This is what the microscopic theory tells you. Microscopic theory says, if your molecules all have a single speed, they’re moving randomly in space so that a third of them are moving back and forth against that wall and this wall, then this is the product kT. Now that guy deserves a box. Look what it’s telling you. It’s a really profound formula. It tells you for the first time a real microscopic meaning of temperature. What you and I call the temperature for gas is simply, up to these factors, 3/2 k, simply the kinetic energy of the molecules. That’s what temperature is. If you’ve got a gas and you put your hand into the furnace and it feels hot, the temperature you’re measuring is directly the kinetic energy of the molecules. That is a great insight into what temperature means.Remember, this is not true if T should be measured from absolute zero. It also tells you why absolute zero is absolute. As you cool your gas, the kinetic energy of molecules are decreasing and decreasing and decreasing, but you cannot go below not moving at all, right? That’s the lowest possible kinetic energy. That’s why it’s absolute zero. At that point, everybody stops moving. That’s why you have no pressure.Now, these results are modified by the laws of quantum mechanics, but we don’t have to worry about that now. In the classical physics, it’s actually correct to say that when the temperature goes to zero, all motion ceases. Now, this is the picture I want you to bear in mind when you say temperature. Absolute temperature is a measure of molecular agitation. More precisely, up to the constant
## Chapter 4: Molecular Mechanics of Phase Change and the Maxwell-Boltzmann Distribution [00:30:15]Now, if you have a solid–What’s the difference between a gas and a solid? In a gas, the atoms are moving anywhere they want in the box. In a solid, every atom has a place. If you take a two-dimensional solid, the atoms look like this. They form a lattice or an array. That’s because you will find out that, this is more advanced stuff, that every atom finds itself in a potential that looks like this. Imagine on the ground you make these hollows. Low points – low potential; high points – high potential. Obviously, if you put a bunch of objects here they will sit at the bottom of these little concave holes you’ve dug in the ground. At zero degrees absolute all atoms will sit at the bottom of their allotted positions; that’ll be a solid at zero temperature. So in a solid, everybody has a location. I’ve shown you a one-dimensional solid, but you can imagine a three-dimensional solid where in a lattice of three-dimensional points, there’s an assigned place for each atom and it sits there. If you heat up that solid now, what happens is these guys start vibrating. Now, here is where your knowledge of simple harmonic motions will come into play. When you take a system in equilibrium, it will execute simple harmonic motion if you give it a real kick. If you put it on top of a hotplate, the atoms in the hot plate will bump into these guys and start them moving. They will start vibrating. So, a hot solid is one in which the atoms are making more and more violent oscillations around their assigned positions. If you heat them more and more and more, eventually you start doing this. You go all the way from here to here; there is nothing to prevent it from rolling over to the next side. Once you jump the fence, you know, think of a bunch of houses, okay? Or a hole in the ground. You’re living in a hole in the ground, as you get agitated you’re able to do more and more oscillations so you can roll over to the next house. Once that happens all hell breaks loose because you don’t have any reason to stay where you are. You start going everywhere. What do you think that is?
Okay. So, this is the picture you should have of temperature. Temperature is agitated motion. Either motion in the vicinity of where you are told to sit. If you’re in a solid a motion all over the box with more and more kinetic energy. The next thing in this caricature is that it is certainly not true that a third of the molecules are moving back and forth. We know that’s a joke, right? Now, in this room there’s no reason on earth a third of the molecules are doing this than others are doing. That’s not approximation. They’re moving in random directions. So, if you really got the stomach for it, you should do a pressure calculation in which you assume the molecules of random velocities sprinkled in all directions, and after all the hard work, turns out you get exactly this answer. So, that’s one thing I didn’t want to do. But something I should point out to you is the following. So, suppose I give you a gas at 300 Kelvin. You go and you take this formula literally and you calculate from it a certain ½
-mv over ^{2}2kT. That’s the graph I’m trying to draw here. So, it looks like a Gaussian in the vicinity of this, but it’s kind of skewed. It’s forced to vanish at the origin. And it’s not peaked at v = 0. A real Gaussian peak at this point would be symmetric. It’s not symmetric; it vanishes here and it vanishes infinity. So, this is called a Maxwell-Boltzmann distribution. You don’t have to remember any names but that is the detailed property of what’s happening in a gas.So, a temperature does not pick a unique velocity, but it picks this graph. If you vary your temperature, look at what you have to do. If you change the number Now, this is another thing I want to tell you. If you took a box containing not atoms but just radiation, in other words, go inside a pizza oven. Take out all the air, but the oven is still hot, and the walls of the oven are radiating electromagnetic radiation. Electromagnetic radiation comes in different frequencies, and you can ask how much energy is contained in every possible frequency range. You know, each frequency is a color so you know that. So, how much energy is in the red and how much is in the blue? That graph also looks like this. That’s a more complicated law called the Planck distribution. That law also has a shape completely determined by temperature. Whereas for atoms, the shape is determined by temperature as well as the mass of the molecules. In the case of radiation, it’s determined fully by temperature and the velocity of light. You give me a temperature, and I will draw you another one of these roughly bell-shaped curves. As you heat up the furnace, the shape will change. So again, a temperature for radiation means a particular distribution of energies at each frequency. For a gas it means a distribution of velocities. Has anybody seen that in the news lately, you know, or heard about this?
And the way you determine that is you point your telescope in the sky. Of course, you’re going to get light from this star; you’re going to get light from that star. Ignore all the pointy things and look at the smooth background, and it should be the same in all directions. And plot that radiation, and now they use satellites to plot that, and you’ll get a perfect fit to this kind of furnace radiation, called Black Body Radiation. And you read the temperature by taking that graph and fitting it to a graph like this, but there’ll be temperature. In the case of light, this won’t be velocity squared, but it will be the frequency squared, but read off the temperature that’ll make this work and that’s what gives you 3.1 or something. Near 3 degrees Kelvin. In fact, the data point for that now if you got that in your lab then you will be definitely busted for fudging your data because it’s a perfect fit to Black Body Radiation. One of the most perfect fits to Black Body Radiation is the background radiation of the Big Bang. And it’s isotropic, meaning it’s the same in all directions, and this is one of the predictions of the Big Bang is that that’ll be the remnant of the Black Body Radiation. Again, it tells you there’s a sense in which, if you go to intergalactic space, that is your temperature. That’s the temperature you get for free. We’re all living in that heat bath at 3 degrees. You want more heat, you’ve got to light up your furnace but this is everywhere in the universe, that heat left over from creation. Okay. That’s a very, very interesting subject. You know, a lot of new physics is coming out by looking at just the Black Body Radiation because the radiation that’s coming to your eye left those stars long ago. So, what you see today is not what’s happening today. It’s what happened long ago when the radiation left that part of the universe. Therefore, we can actually tell something about the universe not only now, but at earlier periods. And that’s the way in which we can actually tell whether the universe is expanding or not expanding or is it accelerating in its expansion, or you can even say once it was decelerating and now it’s accelerating. All that information comes by being able to look at the radiation from the Big Bang. But for you guys, I think the most interesting thing is that when you are in thermal equilibrium, and you are living in a certain temperature, then the radiation in your world and the molecules and atoms in your world, will have a distribution of frequencies and velocities given by that universal graph. Now in our class, we will simplify life and replace this graph with a huge peak at a certain velocity by pretending everybody’s at that velocity. We will treat the whole gas as if it was represented by single average number. So, when someone says find the velocity of molecules, they’re talking about the average velocity. You know statistically that it’s the distribution of answers and an average answer. Because the average is what you and I have to know. Namely, ½ kT, on average.Okay. Now, I’m going to study in detail thermodynamics. So, the system I’m going to study is the only one we all study, which is an ideal gas sitting inside a piston. It’s got a temperature, it’s got a pressure, and it’s got a volume. And I’m going to plot here pressure and volume and I’m going to put a dot and that’s my gas. The state of my gas is summarized by where you put the dot. Every dot here is a possible state of equilibrium for the gas. Remember, the gas, if you look at it under the hood, is made up of 10
So, here is my gas. It’s sitting here. Now, what I do, I had a few weights on top of it. Three weights. I suddenly pull out one weight. Throw it out. What do you think will happen? Well, I think this gas will now shoot up, it’ll bob up and down a few times. Then after a few seconds, or a fraction of a second, it’ll settle down with a new location. By “settle down,” I mean after a while I will not see any macroscopic motion. Then the gas has a new pressure and a new volume. It’s gone from being there to being there. What about in between? What happened in between the starting and finishing points? You might say look, if it was here in the beginning it was there later, it must’ve followed some path. Not really. Not in this process, because if you do it very abruptly, suddenly throwing out one-third of the weights, there’s a period when the piston rushes up, when the gas is not in equilibrium. By that, I mean there is no single pressure you can associate with the gas. The bottom of the gas doesn’t even know the top is flying off. It’s at the old pressure. At the top of the gas there’s a low pressure. So, different parts of the gas at different pressure, we don’t call that equilibrium. So, the dot, representing this system, moves off the graph. It’s off. It’s off the radar, and only when it has finally settled down, the entire gas can make up its mind on what its pressure wants to be; you put it back here. ## Chapter 5: Quasi-static Processes [00:46:49]So, we have a little problem that we have these equilibrium states, but when you try to go from one to another you fly off the map. So, you want to find a device by which you can stay on the
In the old days, when I studied a single particular of the ## Chapter 6: Internal Energy and the First Law of Thermodynamics [00:50:19]Now, this is called a state. Two is a state and one is a state. Every dot here, that is a state. Now, in every state of the system, I’m going to define a new variable, which is called a quantity called So, now I’m ready to write down what’s called the First Law of Thermodynamics that talks about what happens if you make a move in the U. Let’s call it _{2} ΔU. We want to ask what causes the internal energy of the gas to change. So, you guys think about it now. Now that you know all about what’s happening in the cylinder, you can ask how I will change the energy? Well, if you wanted to change the energy of a system, there are two ways you can do it. One is you can do work on the gas. Another thing is you can put the gas on a hotplate. If you put it on hotplate, we know it’s going to get hotter. If it gets hotter, temperature goes up. If temperature goes up, the internal energy goes up. So, there are two ways to change the energy of a gas.The first one we call heat input. That just means put it on something hotter and let the thing heat it up. Temperature will go up. Notice that the internal energy of an ideal gas depends only on the temperature. That’s something very, very important. I mention it every time I teach the subject and some people forget and lose a lot of points needlessly. So, I’ll say it once more with feeling. The energy of an ideal gas depends only on the temperature. If the temperature is not changed; energy has not changed. So, try to remember that for what I do later. So, the change of the gas, this cylinder full that I put some weights on top and I’ve got gas inside, it can change either because I did, I put in some heat, or the gas did some work. By that, I mean if the gas expands by pushing out against the atmosphere, then it was doing the work and If I’ve got a piston here, it’s the force times the distance. But the force is the pressure times the area times the distance. Now, you guys should know enough geometry to know the area of the piston times the distance it moves is the change in the volume. So, we can write it as So, if you took the piston and you nailed the piston so it cannot move, and you put it on a hotplate, So, there are two ways to change the energy of these molecules. In the end, all you want is you want the molecules to move faster than before. One is to put them on a hotplate where there are fast-moving molecules. When they collide with the slow-moving molecules, typically the slow one’s a little more faster and the fast one’s a little more slower and therefore will be a transfer of kinetic energy. Or when you push the piston down, you can show when a molecule collides with a moving piston. It will actually gain energy. So, that’s how you do work. That’s the first law. So, let us now calculate the work done in a process where a gas goes from here to here on an isotherm. Isotherm is a graph of a given temperature. So, this is a graph
V to volume _{1}V, the work done, this is the work done by the gas. You can all see that gas is expanding and that’s equal to this shaded region._{2}By the way, I mention it now, I don’t want to distract you, but suppose later on I make it go backwards like this, part of the way. The work done on the going backwards part is this area, but with a minus sign. I hope you will understand, if you go to the right the area’s considered positive. If you go to the left, the area is considered negative. If you do the integral and put the right limit, you’ll get the right answer. But geometrically, the area under the graph in the V. That’ll automatically be the negative of the log of _{2}V over _{2}V. But geometrically, the area under the graph is the work, if you are going to the right. Yes?_{1}
Very good. So, this is now the work done by the gas. What is the heat input? The heat input is a change in internal energy minus the work done. Let me see. The law was V. What is _{1} ΔQ? How much heat has been put into this gas? How do I find that?
Maybe let’s ask the following question. His question is the following. You’re telling me you put heat into a gas, right? And you say temperature doesn’t go up. How can that possibly be? I always thought when I put heat into something, temperature goes up. That’s because you were thinking about a solid, where if you put in heat it’s got to go somewhere and, of course, temperature goes up. What do you think is happening to the gas here? Think of the piston and weight combination. When I want to go along this path from here to here, you can ask yourself where is the heat input and where is the change in energy, and why is there no change in temperature? If you take a piston like this, if you want to increase the volume, you can certainly take off a grain of sand, right? If you took the grain of sand and the piston will move up, it will do work and it actually will cool down, but that’s not what you’re doing. You are keeping it on a hotplate at a certain temperature so that if it tries to cool down, heat flows from below to above maintaining the temperature. So, what the gas is doing in this case is taking heat energy from below and going up and working against the atmosphere above. It takes in with one hand and gives out to the other, without changing its energy. So, when you study specific heat, which is my next topic, you’ve got to be a little more careful when you talk about specific heats of gases, and I will tell you why. There is no single thing called specific heat for a gas. There are many, many definitions depending on the circumstances. But I hope you understand in this case; you’ve got to visualize this. It’s not enough to draw diagrams and draw pictures. What did I do to the cylinder to maintain the temperature and yet let it expand? Expansion is going to demand work on part of the gas. That’s going to require a loss of energy unless you pump in energy from below. So, what I’ve done is that I take grain after grain, so that the pressure drops and the volume increases, but the slight expansion would have cooled it slightly but the reservoir from below brings it back to the temperature of the reservoir. So, you prop it up in temperature. So, we draw the picture by saying the gas went from here to here, and we usually draw a picture like this and say heat flowed into the system during that process. Alright, now I’ll come to this question that was raised about specific heat. Now, specific heat, you always say is Now, there are many, many ways in which you can pump in heat into a gas and heat it up and see how much heat it takes. But let’s agree that we will take one mole from now on and not one kilogram. Not one kilogram. We’ll find out if you do it that way, the answer doesn’t seem to depend on the gas. That’s the first thing. Take a mole of some gas and call the specific heat as the energy needed to raise the temperature of one mole by one degree. So, this should not be So, if I’m going to divide So, there’s one definition of specific heat called C is the one at constant volume. You don’t let the volume change. In other words, you take the piston and you clamp it. Now, you pump in heat from below by putting it on a hotplate. All the heat goes directly to internal energy. None of that is lost in terms of expansion. So, _{V} ΔV is zero. In that case, ΔQ over ΔT at constant volume, we denote that in this fashion, at constant volume, this term is gone, and it just becomes ΔU over ΔT. That’s very easily done. ΔU over ΔT is 3/2 R. So, the specific heat of a gas at constant volume is 3 over 2R. When I studied solids, I never bother about constant volume because a change in the volume of a solid is so negligible when it’s heated up, it’s not worth specifying that it was a constant volume process. But for a gas, it’s going to matter whether it was constant volume or not.Then, there’s a second specific heat people like to define. That’s done as follows. You take this piston. You have some gas at some pressure. You pump in some heat but you don’t clamp the piston. You let the piston expand any way it wants at the same pressure. For example, if it’s being pushed down by the atmosphere, you let the piston move up if it wants to, maintaining the same pressure. Well, if it moves up a little bit, then the correct equation is the heat that you put in is the change in internal energy plus So, the thing you have to remember, what I did in the end, is that a gas doesn’t have a single specific heat. If we just say, put in some heat and tell me how many calories I need to raise the temperature, that’s not enough. You have to tell me whether in the interim, the gas was fixed in its volume, or changed its volume, or obeyed some other condition. The two most popular conditions people consider are either the volume cannot change or the pressure cannot change. If the volume cannot change, then the change in the heat you put in goes directly to internal energy, from the First Law of Thermodynamics. That gives you a specific heat of 3 over 2 C because when you let the piston expand, then not all the heat is going to heat the gas. Some of it is dissipated on top by working against the atmosphere._{V}Then, notice that I’ve not told you what gas it is. That’s why the specific heat per mole is the right thing to think about because then the answer does not depend on what particular gas you took. Whether it’s hydrogen or helium, they all have the same specific heat per mole. They won’t have the same specific heat per gram, right? Because one gram of helium and one gram of hydrogen don’t have the same number of moles. So, you have to remember that we’re talking about moles. The final thing I have to caution you–very, very important. This is for a monoatomic gas. This is for a gas whose atom is the gas itself. It’s a point. Its only energy is kinetic energy. There are diatomic gases, by two of them [atoms] joined together, they can form a dumbbell or something; then the energy of the dumbbell has got two parts, as you learned long ago. It can rotate around some axis and it can also move in space. Then the internal energy has also got two parts. Energy due to motion of the center of mass and energy due to rotation. Some molecules also vibrate. So, there are lots of complicated things, but if you got only one guy, or one atom, whatever its mass is, it cannot rotate around itself and it cannot vibrate around itself, so those energies all disappear. So, we have taken the simplest one of a monoatomic gas, a gas whose fundamental entity is a single atom rather than a complicated molecule. And that’s all you’re responsible for. I’ll just say one thing. C, I want to mention it before you run off to do your homework. I don’t know if it comes up. It’s called _{V}γ, that’s five-third for a monoatomic gas. You can just take the ratio of the numbers. If in some problem you find γ is not five-thirds, do not panic. It just means it’s a gas which is not monoatomic. If it’s not monoatomic, these numbers don’t have exactly those values. We don’t have to go beyond that. You just have to know there’s a ratio γ, which is five-thirds in the simplest case, but in some problem, somewhere in your life, you can get a γ which is not five-thirds. Remember this problem set is due next Wednesday. But I would ask you to start attacking whatever you can, as and when I lecture.[end of transcript] Back to Top |
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