22. The Boltzmann Constant and First Law of Thermodynamics
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Fundamentals of Physics I22 / 24
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DayPlanetScale (map)HeatGasWater vaporAtmospheric pressureRailroad carCartridge (firearms)TemperatureMechanicData conversionModel buildingCollisionTagesschachtAbsolute zeroVideoGemstoneConcentratorYearSunrisePerturbation theoryCrystal habitContinuous trackFinger protocolWeightCogenerationGameCaliperAlcohol proofNanotechnologyPresspassungCaliberKardierenFerryField-effect transistorFood storageFoot (unit)Kitchen stoveKelvinMaterialAutomobilePlatingAtomismDrehmasseMelting pointArbeitszylinderMolding (process)StonewareLeadMassPotentiometerHydrogen atomKickstandFACTS (newspaper)RRS DiscoveryParticleEnergy levelControl rodMeasurementHeißwasserspeicherTiefdruckgebietJoule heatingPistonPaddle steamerVolumetric flow rateSpecific weightWärmekapazitätLecture/Conference
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Antenna diversityMassMeasurementCartridge (firearms)Alcohol proofHydrogen atomReciprocal latticeGasBoltzmann constantMultiplizitätKelvinEffects unitAtomismAtmospheric pressureKilogramHood (vehicle)Noise figureFoot (unit)Joule heatingElectronDayRestkernTemperatureUniverseMaskCapital shipFACTS (newspaper)Mast (sailing)TelephoneCardboard (paper product)LightMetreHybrid rocketPaperCosmic distance ladderCommodore MAX MachineDrehmasseBahnelementGentlemanLecture/Conference
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CollisionTemperatureAtomismAtmospheric pressureGasForceBallpoint penMicroscopeHeat exchangerNanotechnologySwitchThrust reversalBlast furnaceSchwimmbadreaktorBasis (linear algebra)Direct currentTurningOrder and disorder (physics)Cosmic distance ladderTypesettingFACTS (newspaper)Gas turbineCartridge (firearms)Water vaporScoutingSnowPlatingDrehmasseLacePlane (tool)TARGET2AmplitudeGround stationPhase (matter)MeasurementSway barAtmosphere of EarthPaddle steamerField-effect transistorShock waveNoise (electronics)Lecture/Conference
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TemperatureAbsolute zeroGasKelvinMassWhiteDirect currentHot workingMelting pointFoot (unit)FahrgeschwindigkeitAmplitudeBird vocalizationElectric power distributionGentlemanWater vaporCogenerationAtmospheric pressureStream bedGround (electricity)Alcohol proofAtomismHarmonicThermodynamic equilibriumFACTS (newspaper)Negativer WiderstandLocherHydrogen atomPlatingMeasurementSquid (weapon)Median filterRemotely operated underwater vehicleProzessleittechnikController (control theory)Ship breakingCommitteeSolidLiquidOrder and disorder (physics)Group delay and phase delayTiefdruckgebietBending (metalworking)Curved mirrorMembrane potentialStomachHollow EarthSharpeningTrainMeltingArray data structureQuantumCarburetorShort circuitRoots-type superchargerRoll formingLecture/Conference
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Hulk (ship)Atmospheric pressureAudio frequencyGasAngeregter ZustandSupernova remnantThermodynamic equilibriumKelvinPresspassungElectric power distributionBlast furnaceFahrgeschwindigkeitUniverseRadiationSizingBig BangAtomismHot workingCosmic microwave background radiationTemperatureFACTS (newspaper)LappingYearNewton's laws of motionDirect currentSeeschiffIdeal gasAtmosphere of EarthHood (vehicle)Pitch (music)ColorfulnessPlant (control theory)TiefdruckgebietHauptsatz der Thermodynamik 2SatelliteSpare partMassCartridge (firearms)Alcohol proofForceCogenerationDoorbellLightRemotely operated underwater vehicleBird vocalizationWeightOrbital periodHeatStarRadarBending (metalworking)MinutePistonCardboard (paper product)HochfrequenzübertragungShip classStagecoachRoll formingAmplitudeCurrent densityLecture/Conference
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GaleTrajectoryHeatParticleTemperatureIdeal gasFullingArbeitszylinderGasCogenerationThermodynamic equilibriumPistonSpare partAtmospheric pressureCosmic distance ladderForceHot workingProzessleittechnikAngeregter ZustandHauptsatz der Thermodynamik 2Plane (tool)SolidFrictionPerturbation theoryVideoNorthrop Grumman B-2 SpiritPlatingLiquidCylinder blockLiberty shipRainStellar atmosphereWeightWill-o'-the-wispBauxitbergbauParking meterDVD playerDayYearField-effect transistorLecture/Conference
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PistonGasWeightHeatTemperatureTransfer functionHot workingCombined cycleInitiator <Steuerungstechnik>Spare partCartridge (firearms)PlatingSolidFACTS (newspaper)WärmekapazitätTrajectoryArbeitszylinderSpecific weightContour lineLumberNegationProzessleittechnikDirect currentReciprocal latticeNanotechnologyLimiterQuality (business)Vertical integrationPerturbation theoryNightTiefdruckgebietIntensity (physics)SpaceportSeparation processPlane (tool)Stellar atmosphereSeeschiffGround stationSeat beltQuadraphonic soundLecture/Conference
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GasTemperatureHot workingSpecific weightAtmospheric pressureHeatStellar atmospherePistonPlatingIRASCartridge (firearms)Key (engineering)SolidBallpoint penRoman calendarHauptsatz der Thermodynamik 2EveningExpansionsturbineKette <Zugmittel>Cocktail party effectSpare partMassChemical substanceGamma rayEnergy levelAmateur radio repeaterField-effect transistorLocherhome officeHydrogen atomProzessleittechnikKilogramNoise (electronics)Answering machineBelt (mechanical)Alcohol proofWärmekapazitätIdeal gasCaliperLecture/Conference
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Computer animation
Transcript: English(auto-generated)
00:01
So, I have to leave you in the middle of something pretty exciting, so I'll come back and take it from there. So, what is it you have to remember from last time? You know, what are the main ideas I covered? One is, we took the notion of temperature for which we have an intuitive feeling and turned it into something
00:21
more quantitative. So, you cannot only say this is hotter than that, that's hotter than this. You can say by how much, by how many degrees? And in the end, we agreed to use the absolute Kelvin scale for temperature. And the way to find the Kelvin scale, you take the gas, any gas that you like,
00:42
like hydrogen or helium, or low concentration, and put that inside a piston and cylinder where it occupies some volume, and there's a certain pressure by putting weights on top, and you take the product of P times V, and the claim is for whatever
01:02
gas you take, it'll be a straight line. Remember now, this is in Kelvin. Your centigrade scale is somewhere over here. I've shifted the origin to the Kelvin scale. So, somewhere here will be the boiling point of water, somewhere with the freezing point of water, that may be the boiling point of water.
01:22
And if you took a different amount of a different gas, you'll get some other line, but there will always be straight lines if the concentration is sufficiently low. In other words, it appears that pressure times volume is some constant, I don't know what to call it, say C, times this temperature.
01:47
And you can use that to measure temperature, because if you know two points on a straight line, then you know that you can find the slope, and then you can calibrate the thermometer, Then for any other value of P times V that you get, you can come down and read your temperature.
02:03
That's the preferred scale. And we prefer the scale because it doesn't seem to depend on the gas that you use. I can use one, and you can use another one. People in another planet who've never heard of water, they can use a different gas. But all gases seem to have the property that pressure times volume is linearly proportional to
02:22
this new temperature scale, measured with this new temperature, with the new origin at absolute zero. There is really nothing to the left of this T equal to zero. The next thing I mentioned was, people used to think of the theory of heat as a new theory. You had mechanics and all
02:41
that stuff, livers and pulleys and all that. Then you had this mysterious thing called heat, which had been around for many years, but people started quantifying it by saying there's a fluid called a caloric fluid, and hot things have a lot of fluid, 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,
03:04
law of conservation of this caloric fluid that allows you to do some problems in calorie-metry. You mix so much of this with so much of that, where will they end up, that kind of problem. So, that promoted the heat to a new and independent entity, different from all other things we have studied.
03:22
So, this is a new theory, but something suggests that it is not completely alien or new concept because there seems to be a conservation 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.
03:42
Secondly, we know another way to produce heat. Instead of saying, 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,
04:02
and of course 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, 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.
04:22
So, in the old days we would say, well, we don't apply the law of conservation of energy because this is an inelastic collision. 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 produces this extra heat. So, that indicates that
04:44
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? How many calories can you get if you sacrifice one joule of mechanical energy? So, joule did the
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experiment, not with cars. I mean, he didn't have cars at that time, so if he did, he would have probably done it with cars. He had this gadget with him, which was a little shaft with some paddles and a pulley on the top, and you let the weight go down, and I told you guys if the weight goes from here to here, the mgh loss will not be the
05:24
gain in ½ mv². Something will be missing. Keep track of the missing amount, so many joules, but meanwhile you find this water has become hot. You find then how many calories should have gone in, because we know the specific heat of water, we know the rise in temperature, we know how many calories were produced. And then you compare the two
05:42
calories. So, when you find that 4.2 joules is equal to 1 calorie. So, that is the conversion ratio of calories to joules. One joule, 4.2 joules of mechanical energy. So, in the example of the colliding cars, take the ½ mv² for each
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car, turn it into joules, slam them together. If they come to rest, you've lost all of that. And then, you take that and you write it as divided by 4.2, and that's how much calories you have produced. If the car was made of just one material, it had a specific heat, then it will go up by a certain temperature you can
06:21
actually predict. Okay, so today I want to go a little deeper into the question of where is the energy actually stored in the car? 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.
06:41
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 at 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
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based on the simple fact that everything is made up of atoms. That was not known, and that's one of the discoveries. 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.
07:22
That is in the relation PV equal to some constant times temperature. Do you know what I'm talking about? Take some gas, make sure it's sufficiently tight, you would put it into this piston, measure the weights on top of it, divide by the area to get the pressure, that's the pressure, that's the volume.
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The volume is the region here. Multiply the product. Then, if you heat up the gas by putting it on some hot plate, you will find the product PV increases. And as the temperature increases, PV is proportional to T. We want to ask, what is this proportionality
08:00
constant? Suppose you were doing this in the old days. This is what people did. What do we think should be on the right-hand side? What is going to control this particular constant for your given experiment? Do you know what it might be proportional to? Yes?
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Student 2, Inaudible Professor Ramamurti Shankar Amount of gas. That's true, because the amount of gas is zero, we think there's no pressure. When you say amount of gas, that's a very safe sentence, because amount measured by what means, by what metric? Right, suppose you were not
08:42
aware of particles, then what would you mean by amount of gas? The mass. Now, if you guys ever said moles, I was going to shoot you down. You're not supposed to know those things. We are trying to deduce that. So, put yourself back in whatever stone ages we were in. We don't know anything else. Mass would be a reasonable
09:01
argument, right? What's the argument? We know that if you had some amount of gas producing a pressure, and you put twice as much stuff, you think it'll produce twice as much pressure. The same reason why you think the expansion of a rod is proportionally changing temperature times the starting length. So, this mass is what's doing it, so it's proportional to the mass.
09:21
It's a very reasonable guess. 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, but I'll call it C prime. This constant contains
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everything, but I pulled out the mass, and the remaining constant I want to call C prime. This is actually correct. You can take a certain gas and you can find out what C prime is. But here is what people found. If you do it that way,
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the constant C prime depends on the gas you are considering. If you consider hydrogen gas, let's call that C prime for hydrogen. Somebody else puts in helium gas. Then, you find that C prime for helium is one-fourth C
10:21
prime for hydrogen. If you do carbon, it's another number. C prime for carbon is C prime for hydrogen divided by 12. So, each gas has a different constant.
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So, we conclude that yes, it's the mass that decides it, but the mass has to be divided by different numbers for different gases to find the real effect it has in terms of pressure. In other words, one gram of hydrogen and one
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gram of helium do not have the same pressure. In fact, one gram of helium has to be divided by 4 to find its effect on pressure. So, you have to think about why is it that the mass directly is not involved. So, you have to figure out why the mass has to be divided by a number, and the number is a nice, round number.
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Four for this and 12 for that, and of course, people figured out there's a long story I cannot go into, but I think you all know the answer. But now, you're 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 gases reacted and formed complicated molecules,
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they figured out what's really going on is that you're going to find the mass that's proportional to the mass of the underlying fundamental entity, which would be an atom, in some cases molecule, but I'm just going to call everything as atom. So, if things like carbon, as atoms, weigh 12 times as much as things called hydrogen,
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then if you took some amount of carbon, you divide it by a number like 12 to count the number of carbon atoms. In the case of hydrogen, you want to count the number of hydrogen atoms. So that what really you want here is not the mass, but the number of atoms of a given kind.
12:21
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, if you write it in this fashion, where there's a new constant, k, k is independent of the gas. So, you want to write it in a
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manner in which it doesn't depend on the gas. You can write it in terms of mass, if you did, for each mass you've got to divide by a certain number. Then, once you divide it by the number, you can put a single constant in front. Or, if you want a universal constant, what you should really be counting is the number of atoms or molecules. So, you couldn't have
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written it that way until you knew about atoms and molecules. And people were led to atoms and molecules by looking at the way gases interact, and it's a beautiful piece of chemistry to figure out, and it's not really that there are entities which come in discrete units. Not at all obvious in the old days that mass comes in discrete units called atoms,
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but that's what they deduced. So, this is called the Boltzmann constant. The Boltzmann constant has a value of 1.4 times 10 to the minus 23, let's see, joules, kelvin, that's it,
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or joules, kelvin, or degrees centigrade. So, this is a universal constant.
14:07
So now, what people like to do is they don't like to write the number, because if you write the number in a typical situation, what's the number going to be on it? Take some random gas, one gram, two grams,
14:20
one kilogram, it doesn't matter. The number you will put in there is some number like 10 to the 23 or 10 to the 25. That's a huge number. So, whenever a huge number is involved, what you try to do is to measure the huge number as a simple multiple of another huge number, which will be our units for measuring large numbers. For example,
14:41
when you want to buy eggs, you measure it in dozens. When you want to buy paper, you may want to measure it in thousands or 500s or whatever units 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.
15:03
When you count people's height, you use feet because it's something between one and eight, 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.
15:23
Avogadro's number is 6 times 10 to the 23. It has no units, it's simply a number and that's called a mole. So, a mole is like a dozen. You wanted to buy 6 times 10 to the 23 eggs, you would say, get me one mole of eggs.
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A mole is just a number, it's a huge number. You can ask yourself, what's so great about this number? Why would someone think of this particular number? Why not some other number? Why not 10 to the 24? Do you know what's special about this number? Yes?
16:05
Yes, if you like, a mole is such that one mole of hydrogen weighs one gram. And hydrogen is the simplest element. The nucleus is just a proton and the electron mass is negligible. So, this, if you like, is the reciprocal of the mass
16:23
of hydrogen. In other words, one over Avogadro's number is the mass of hydrogen in grams, of a hydrogen atom in grams. So, you basically say,
16:41
I want to count this large number, so let me take one gram, which is my normal unit, if you're thinking in grams, then I ask, how many hydrogen atoms does one gram of hydrogen contain? That's the number, that's the mole. So, if you decide to measure the number of atoms you have in a given problem, in terms of this number, you write it as some other small
17:01
number called moles times the number in a mole, and you are free to write it this way. If you write it this way, then you write it as nRT, R is the universal gas constant. It's n times the Boltzmann
17:22
constant, n zero times the Boltzmann constant. So, that happens to be 8.3 joules per degree centigrade or per kelvin, right? The units for R will be PV, which is units of energy, divided by T.
17:43
In terms of calories, I remember this 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 with the same amount in temperature.
18:03
So, this is what they found out first, because they didn't know anything about atoms and so on. But later on, when you go look under the hood of what the gas is made of, if you write it in terms of the number of actual atoms, you should use a little k, or you can write it in terms
18:21
of number of moles, in which case you use big R. And the relation between the two is simply this. If you're thinking of a gas and how many moles of gas do I have, for example, one gram of hydrogen would be one mole, then you will use R. If you've gone right down to
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fundamentals and say, how many atoms do I have, and you put that here, you will multiply it by this very tiny number. Now, you start with this law and you ask the following question. On the left-hand side is the quantity P times V.
19:00
On the right-hand side, I have nRT, but let me write it now as nkT. You guys should be able to go back and forth between writing it in terms of number of moles or the number of atoms. You like this because all numbers here will be small, of order one. R is a number like, what, eight in some units,
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n will be one or two moles. Here, this n will be a huge number like 10 to the 23, k will be a tiny number like 10 to the minus 23. Think in terms of atoms, that's what you do. Big numbers, small numbers, small constants. If you think of moles, small numbers, I mean moderate numbers and moderate value of constant. We want to ask ourselves, is there a microscopic basis for
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this equation? In other words, once we believe in atoms, do we understand why there is a pressure at all in a gas? That's what we're going to think about now. So, for this purpose, we will take a cube of gas. Here it is.
20:05
This is a cube of side L by L by L. Inside this is gas and it's got some pressure and I want to
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know what's the value of the pressure. So, you've got to ask yourself, why is that pressure? Remember I told you what pressure means. If you take this face of the cube, for example, it's got to be nailed down to the other faces, otherwise it'll just come flying out because the gas is pushing you out. The pressure is the force on this face divided by area. So, somebody inside is trying
20:43
to get out. Those guys are the molecules or the atoms and what they're doing is constantly bouncing off the wall and every time this one bounces on a wall, its momentum changes from that to the other one. So, who's changing the momentum?
21:00
Well, the wall is changing the momentum. It's reversing. 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
21:21
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 is rate of change of momentum because that is d by dt of mv and m is a constant and that's just
21:43
dv dt, which is mv, which is ma. I'm just using the old f equals ma, but I'm writing it as rate of change of momentum. Now, I have n molecules or n atoms randomly moving inside the box, each in its own direction, suffering collisions with the box, bouncing off like a
22:03
billiard ball would at the end of the pool table, going to another wall and doing it. That's a very complicated problem. So, we're going to simplify the problem. The simplification is going to be, we're 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
22:22
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 number in each direction. There's nothing in the gas that favors horizontal and vertical. In reality, of course, you must admit the fact they move in all directions, but the simplified derivation
22:41
happens to give all the right physics. I'm going to use that. So, n over 3 molecules are going back and forth between this wall and this wall. I'm showing you the side view. The wall itself looks like this. The molecules go back and forth. Next assumption,
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all the molecules have the same speed, which I'm going to call v. That also is a gross and crude description of the problem, but I'm going to do that anyway and see what happens. So, now you ask yourself the following question. Take one particular molecule. When it hits the wall and it
23:22
bounces back, its momentum changes from mv to minus mv. Therefore, the change in momentum is 2mv. How often does that change take place? You guys should think about
23:42
that first. How often will that correlation take place? Once you hit the wall here, you've got to go to the other wall and come back. So, you've got to go a distance 2L and you're going at a speed v. The time it takes you is 2L over v. So, Δp over Δt is 2L divided by v. That gives me mv
24:05
squared over L. That is their force due to one molecule. That's the average force. You realize it's not a continuous force. The molecule hits the wall, there's a little force exchange between the two, then there's nothing,
24:21
then you wait till 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 to the 23 guys pounding themselves against the wall. So, at any given instant, even if it's 10 to the minus
24:42
5 seconds, there'll be a large number of molecules colliding. So, that's why the force will appear to be steady rather than a short noise. It'll look very steady because somebody or other will be pushing against the wall. So, if you ask now, how many molecules? This is the force due to one molecule. The force due to all of them
25:02
will be N over 3 times mv squared over L. N over 3, because of the N molecules, a third of them are moving in this direction. You realize the other two directions are parallel to the wall.
25:21
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 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
25:43
orthogonal to that is going to contribute. That's why you have N over 3. Well, we're almost done. That's the average force. If you want, I can denote average by some F bar. Then, what about the average pressure? The average pressure is the average force divided by the area of that face,
26:01
F over L squared. That gives me N over 3 mv squared over L cubed. Now, this is very nice because L cubed is just the volume of my box. So, I take the L cubed, which is equal to the volume of
26:24
my box, and I send it to the other side and write it as PV equals N over 3 mv squared.
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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 PV. Experimentally, you find PV is equal to NkT. So, you compare the two
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expressions and out comes one of the most beautiful results, which is that mv squared over 2 is 3 over 2kT. Now, that guy deserves a box because look what it's telling
27:22
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 of a gas is simply up to these factors, 3 over 2k, simply the kinetic energy of the molecules.
27:43
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
28:02
T is measured in centigrade. If T were measured in centigrade, at freezing point of water, mv squared would vanish. But that's not what's implied. 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 is decreasing and
28:21
decreasing and decreasing, but you cannot go below not moving at all. 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.
28:40
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. If you divide by two, more precisely up to the constant k, 3 over 2k,
29:02
the kinetic energy of a molecule is the absolute temperature. That's for a gas. If you took a solid and you say, what happens when I heat the solid? Well, you've got a question. Yes? You divide by two because
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½ mv squared is a familiar quantity, namely kinetic energy. That's why you divide by two. Another thing to notice is that every gas, whatever is made of, has a kinetic energy at a given temperature as a given kinetic energy, because the kinetic energy per molecule on the left-hand side is dependent on absolute
29:42
temperature and nothing else. So, at certain degrees, like 300 Kelvin, hydrogen kinetic energy will be the same, carbon kinetic energy will also be the same. The kinetic energy will be the same, not the velocity. So, the carbon atom is heavier, it'll be moving slower at that temperature in
30:01
the same way as the kinetic energy. So, all molecules, all gases at a given temperature, all atoms, let me say, at a given temperature in gaseous form will have the same kinetic energy. 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
30:22
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
30:41
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.
31:02
At zero degrees absolute, all atoms have set 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,
31:21
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 motion comes into play. When you take a system in equilibrium, it'll execute simple harmonic motion if you
31:40
give it a real kick. If you put it on top of a hot plate, 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. To heat them more and more
32:00
and more, eventually, you start doing this. You go all the way from here to here, there is nothing to prevent you from rolling over to the next side. Once you've jumped the fence, think of a bunch of houses, okay, all whole 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
32:20
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. So, what do you think that is? Pardon me? That's melting. That's the definition of melting. Melting is when you can leap over this potential barrier, potential energy barrier, and go to the next side. The next side is just like
32:42
this side. So, if you can jump that fence, you can jump this one, you go everywhere and you melt. That's the process of melting. And once you have a liquid, atoms don't have a different location. Now, between a liquid and a solid, there is a clear difference, but a liquid and a vapor is still more subtle. So, I don't want to go into that, but at some point, liquids, if you look at a liquid
33:02
locally, it will look very much like a solid, in the sense that inter-atomic spacing is very tightly constrained in liquid. Whereas, in a solid, if I know I am here, I know if I go a hundred times the basic lattice spacing, there will be another person sitting there. That's called long-range order. In a liquid, I cannot say that. In a liquid,
33:20
I can say if I am here, locally, the environment around me is known, but if you go a few hundred miles, I cannot tell you at a precise location if some other atom will be there or not. So, we say liquid has short range, position a lot, but not long-range order. And a gas has no order at all. If I tell you there's a gas molecule here, I cannot tell you where anybody else is,
33:41
because nobody has any assigned location. 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, or motion all over the box with more and more kinetic energy. The next thing in this caricature is that,
34:04
certainly not true that a third of the molecules are moving back and forth. We know that's a joke, right? Right now in this room, there's no reason on earth a third of the molecules are doing this and others are doing that. That's an 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 have random
34:21
velocities sprinkled in all directions. And after all the hard work, it turns out you'll 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 mv². If you know the mass of the
34:42
atom, say it's hydrogen, we know the mass of hydrogen. Then you find the velocity and you say, okay, this man tells me that anytime I catch a hydrogen with an atom, it'll have this velocity at 300 Kelvin. They have random directions, but he tells me that's the velocity squared. Take the square root of that, that's the velocity. It seems to us to have a unique velocity to each
35:02
temperature. Well, that's not correct. Not only are the molecules moving in random directions, they're also moving with essentially all possible velocities. In fact, there are many, many possible velocities. And this velocity I'm getting in this formula is some kind of average velocity,
35:22
or the most popular one, or the most common one. So, if you really go to a gas and you have the ability to look into it and see for each velocity, what's the probability that I get that velocity? The picture I've given you is the probability is zero, except at this one magical velocity controlled by the
35:42
temperature. But the real graph looks like this. It has a certain peak. It likes to have a certain value. So, if you know enough about statistics, you know there's a most probable value, there's a median, there's a mean value. There are different definitions. They will all vary by factors of order one,
36:02
but the average kinetic energy will obey this condition. Yes, it's not really a Gaussian because if I draw the nature of this curve, it looks like v square e to the minus mv square over two kt. That's the graph I'm trying to
36:22
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 equal to zero. A real Gaussian peaked at this point would be symmetric. It's not symmetric. It vanishes here and vanishes at infinity. So, this is called the
36:43
Maxwell-Boltzmann distribution. You don't have to remember any names, but that is the detailed property of what's called the Maxwell-Boltzmann distribution. 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 t
37:03
here, if you double the value of t, that means if you double the value of v square here and there, the graph will look the same. So, at every temperature, there is a certain shape. If you go to your temperature, it will look more or less the same. So, it's a certain velocity, if you go to a higher temperature.
37:23
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
37:42
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, how much is in the blue? That graph also looks like this.
38:01
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 the light. You give me a temperature and I
38:21
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.
38:44
Now, that is, has anybody seen that in the news lately or heard about this? Pardon me? About this particular graph for radiation, the probability at each frequency of finding radiation
39:01
of that frequency in a furnace at some temperature T. Yes? No, but in current news, in the last few years, what people did was the following. It's one of the predictions of the Big Bang theory, that the universe was formed
39:25
some 14 and a half billion years ago, and at the early stages, the temperature of the universe was some incredibly high degrees. Then, as it expanded, the universe cooled, and today, at the current size, it has got a certain average temperature, which is a remnant of the
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Big Bang. And that temperature means there is, we are sitting in a furnace of the Big Bang, but the furnace has cooled a lot over the billions of years. The temperature of the universe is around 3 degrees Kelvin. And the way you determine that is you point your telescope in the sky.
40:02
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 nowadays, they use satellites to plot that. And you get a perfect fit to this kind of furnace radiation
40:21
called black body radiation. And you read the temperature by taking that graph and fitting it to a graph like this, but that'll be temperature. In the case of light, this won't be velocity squared, but it'll 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.
40:40
In fact, the data point for that, if you got that in your lab, 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 from the Big Bang. And it's isotropic, meaning it's same in all directions. And this is one of the
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predictions of the Big Bang, is that that'll be the remnant of the black body radiation. Again, it tells you, that's a sense in which the – 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
41:21
the universe, that heat left over from creation. Okay, that's a very, very interesting subject now. A lot of new physics is coming out by looking at just that black body radiation, because the radiation that's coming to your eye left those stars long ago.
41:42
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
42:02
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
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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
42:40
pretending everybody is at that velocity. We will treat the whole gas as if it's represented by a single average number. So, when someone says, okay, let's find the velocity of molecules, they're talking about the average velocity. You know statistically there is a distribution of answers and an average answer, because the average is what you and I have to know. Namely, ½ mv²
43:02
is 3 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,
43:26
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
43:41
dot. Every dot here is a possible state of equilibrium for the gas. Remember, the gas, if you look at it under the If the gas is good, it's made up of 10 to the 23 molecules. The real, real state of the gas is obtained by saying, giving me 10 to the 23 locations and 10 to the 23 velocities.
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Accordingly, Newton, that's the maximum information you can give me about the gas right now, because with that and Newton's laws, I can predict the future. But when you study thermodynamics, you don't really want to look into the details. You want to look at gross macroscopic properties, and there are two that you need. You need pressure and volume.
44:21
Now, you might say, what about temperature? Why don't they have a third axis for temperature? Why is that also not a property? Yes? Yeah, because PV equal to nkT. I don't have to give you
44:41
T if I know P and V. It's not an independent thing you can pick. You can pick P and V independently, you cannot pick T. Let me tell you, by the way, PV equals nkT is not a universal law. It's a law that you apply to dilute gases. But we are going to just study only dilute ideal gas. Ideal gas is one in which
45:01
the atoms and molecules are so far apart that they don't feel any forces between each other unless they collide. 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,
45:21
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 to the new location. By settle down, I mean after a while I will not see any macroscopic motion. Then, the gas has a new
45:42
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
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beginning, it was there later, it must have followed some path. Not really. Not in this process, because if you do it very abruptly, by suddenly throwing out one-third of the weights, that's the period when the piston rushes up, when the gas is not in equilibrium. By that, I mean there is no single pressure you can
46:21
associate with the gas. The bottom of the gas doesn't even know the top is flying off. It's at the old pressure. And the top of the gas is at low pressure. So, different parts of the gas are different pressure. We don't call that equilibrium. So, the dot representing the system moves off the graph. It's off. It's off the radar. And only when it has
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finally settled down, the entire gas can make up its mind on what its pressure wants to be. So, we'll put it back here. 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 PV diagram as you change the state of the gas.
47:02
That brings us to the notion of what's called a quasi-static process. So, a quasi-static process is trying to have it both ways in which you want to change the state of the gas and you don't want it to leave the PV diagram. You want it to be always at equilibrium. So, what you really want to do
47:21
is not put in three big fat blocks like this, but instead take a gas where you have many, many grains of sand. They can produce the pressure. Now, remove one grain of sand. It moves a tiny bit and very quickly settles down. It is, again, true, during the tiny bit of settling down,
47:40
you didn't know what it was doing, but you certainly nailed it at the second location. Remove one grain at a time, and you get a picture like this. Then, you get a picture like this, and you can see where this is going. You can make the grain smaller and smaller and smaller, and at a mathematical sense, you can then form a continuous line. That is to say, you perform a process that leaves the system arbitrarily close to equilibrium,
48:02
meaning give it enough time to readjust to the new pressure, settle down to the new volume, take another grain and another grain, and in the spirit of calculus, you can make these changes to the new pressure. You can make the changes to the new pressure vanishing, so that you can really then say, it did this. Yes? Pardon me? Yes, such a process is also
48:24
called, you can call it quasi-static, but one of the features of that, it is reversible. You've got to be a little careful when you say reversible. What we mean by reversible is, if I took off a grain of sand and it came from here to the next dot, and I put the grain back, it'll climb back to where it
48:42
was. So, you can go back and forth on this. But now, that's an idealized process, because if you had a friction, if you had any friction between the piston and the walls, then if you took out a grain and it went up, you've put the grain back, it may not come back to quite where it is. Because some of the
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frictional losses, you will never get back. You cannot put the Humpty Dumpty back. So, most of the time, processes are not reversible, even if you do them slowly, if there is friction. So, it's assumed it's a completely frictionless system. Because if there is friction, there is some heat that goes
49:21
out somewhere and some energy is lost somewhere and we cannot bring it back. If you took a frictionless piston and on top of it, moved it very, very slowly, you can follow this graph. So, that's the kind of thermodynamic process we're talking about. In the old days, when I studied a single particle in the xy plane, I just said the guy goes from here to here to there.
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So, that's very easy to study and there is no restriction on how quickly or how fast it moved. Particles have trajectories no matter how quickly they move. For a thermodynamic system, you cannot move them too fast because they are extended. And you're giving a huge gas a single number called pressure. So, you cannot change one part of the gas without waiting for all of them to communicate
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and readjust and achieve a global value for the new pressure. Then, you can move gradually. That's why it takes time to figure out how to drag along 10 to the 23 particles as if they had a single number or two numbers characterizing them. So, we'll be studying processes like this. Now, this is called a state. Two is a state and one is a
50:22
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, I'm going to call it u, maybe I'll start fresh, a quantity called u, which stands for the internal
50:44
energy of the gas. The internal energy is simply the kinetic energy of the gas molecules. For solids and liquids, there's a more complicated formula. For the gas, internal energy is just the kinetic energy.
51:02
And what is that? It is three-halves kT per molecule times n. I'm sorry, three-halves kT. Or you can write it as three-halves
51:22
nRT, but nRT is PV. You can also write it as three-halves PV. So, internal energy is just three-halves PV. That means at a given point in the PV diagram, you have a certain internal energy. If you are there, that's your internal energy. Take three-halves of PV and
51:43
that's the energy and that's literally the kinetic energy of all the molecules in your box. 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 PV plane from one place to another place. If you go from one place to
52:03
another place, your internal energy will change from u1 to u2. Let's call it Δu. We want to ask, what causes the internal energy of the gas to change? So, you guys think about it energy of the gas now. Now that you know all about what's happening in the cylinder, you can ask,
52:22
how will I change the energy? Well, if you want 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 hot plate, because if you put it on a hot plate, we know it's going to get hotter. If it gets hotter, temperature goes up. If temperature goes up, the internal energy goes up.
52:42
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. The temperature will go up. Notice that the internal energy of an ideal gas depends only on the temperature. That's something very,
53:03
very important. I mention it every time I teach a subject and some people forget and lose a lot of points needlessly. I'll say it once more with this feeling. The energy of an ideal gas depends only on the temperature. The temperature has not changed, energy has not changed.
53:20
So, try to remember that for what I do later. So, the change of the gas, this cylinder full of, but I put some weights on top and I 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,
53:43
if the gas expands by pushing out against the atmosphere, then it is doing the work and ΔW is the work done by the gas. That's why it comes to the minus sign because it's the work done by the gas. If you do work, you lose energy.
54:00
So, what's the formula for work done? Let's calculate that. If I 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.
54:21
It's the change in the volume. So, we can write it as P times dV. That leads to this great law. Let me write it in a new blackboard because we're going to be playing around with that law. This is law number one. The change in the internal energy of a system is equal to
54:42
ΔQ minus P ΔV. What does it express? It expresses the law of conservation of energy. It says if the energy goes up, either because you pushed the
55:02
piston or the piston pushed you, then you decide what the overall sign is, or you put it on a hot plate. We are not equating putting it on a hot plate is also equivalent to giving it energy because we identify heat as simply energy. So, if you took the piston and you nailed the piston so it
55:21
cannot move, and you put it on a hot plate, PdV part will vanish because there is no ΔV. That's the way of heating it is called ΔQ. Another thing you can do is you can thermally isolate your piston so no heat can flow in and out of it, and you can either have the volume increase or decrease. If the gas expanded,
55:41
ΔV is positive and the Pd, minus PdV is negative, and the ΔU will be negative, the gas will lose energy. That's because the molecules are beating up on the piston and moving the piston. Remember, applying a force doesn't cost you anything, but if the point of application moves, you do work. And who's going to pay for it? The gas. It'll pay for it through
56:02
its loss of internal energy. Conversely, if you push down on the gas, ΔV will be negative and this will become positive, and the energy of the gas will go up. 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
56:21
hot plate where there are fast-moving molecules. When they collide with the slow-moving molecules, typically the slow ones will move faster and the fast ones will move slower, and therefore there will be a transfer of kinetic energy. Or when you push the piston down, you can show when a molecule collates with a moving piston, it will actually gain energy. So, that's how you do work. This is the first law.
56:48
So, let us now calculate the work done in a process where a gas goes from here to here on an isotherm.
57:02
Isotherm is a graph of a given temperature. So, this is a graph, P times V equal to constant because PV equal to nRT. If T is constant, PV is a constant, it's a rectangular hyperbola. The product of the x and y coordinates is constant. So, when the x coordinate
57:21
vanishes, the y will go to infinity, the y coordinate vanishes, x will go to infinity. So, you want to take your gas for a ride from here to here throughout this at a certain temperature T. What work is done by you? That's a very nice interpretation. The work done by you is the
57:42
integral of PdV. But what is the integral of PdV? That's P and that's dV. PdV is that shaded region. In other words, if you just write PdV, it makes absolutely no sense. If you go to a mathematician and say, please do the integral
58:00
for me, can the mathematician do this? What's coming in the way of the mathematician actually doing the integral? What do you have to know to really do an integral? You have to know the function. If you just say P, you'll say maybe P is a constant, in which case I'll pull it out of the integral. But for this problem,
58:22
because PV is nRT and T is a constant, P is nRT divided by V, and that's the function of V that you would need to do the integral. And if you did that, you will find it's nRT. All of them are constants. They come out of the integral, dV over V, and they integrate from the
58:42
initial volume to the final volume. You guys know this is a logarithm, and the log of upper minus log of lower is the log of the ratio. And this gives me nRT log V2 over V1. So, we have done our first work calculation.
59:03
When the gas goes on an isothermal trajectory from start to finish, from volume V1 to volume V2, 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. By the way, I mentioned it
59:28
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'll understand.
59:41
If you go to the right, the area is 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 PV diagram is the work done if you're moving in the direction of increasing volume, if you decrease the volume. For example, if you
01:00:00
It just went back from here. The area looks the same, but the work done is considered negative. You don't have to think very hard. If you do the calculation going backwards, you'll get log of v1 over v2. That'll automatically be the negative of log of v2 over v1. But geometrically, the area under the graph is the work if you're going to
01:00:21
the right. Yes? Oh, this one? This will be a graph, PV equal to essentially a constant. So, you take your gas, you see how many moles there are, you know R, you know the temperature,
01:00:41
you promise not to change the temperature, so you'll move on a trajectory so that the product PV never changes. And in any xy plane, if you draw a graph where the product xy doesn't change, it'll have this shape called the rectangular hyperbola. It just means whenever one increases, the other should decrease, keeping the product constant.
01:01:02
That's why P is proportional to the reciprocal of V and you should be able to do the integral. Very good. So, this is now the work done by the gas. What is the heat input? The heat input is the change in internal energy minus the work done. Let me see.
01:01:22
The law was ΔU is equal to ΔQ minus ΔW. Now, let's go back to this law. In this problem, ΔW is what I just calculated, nRT, whatever log V2 over V1. What is ΔQ?
01:01:44
How much heat has been put into this gas? How do I find that? Pardon me? For the heat input you mean? Yeah, you can use mCΔT, but you don't have to do any more work.
01:02:00
By that I mean you don't have to do any more celebration. What can you do with this equation to avoid doing further calculations? Do you know anything else? Yes? Yes, this is what I told you. It's a fact that people do not constantly remember, but you must.
01:02:21
This gas did not change its temperature. Go back to equation number, whatever I wrote down, U equals 3 halves nRT or something. T doesn't change, U doesn't change. That means the initial internal energy and final internal energy are the same, because the initial temperature and final temperature are the same. So, this guy has to be 0. That means ΔQ is the same as
01:02:42
ΔW in this particular case. So, yes, when you say mCΔT, you've got to be careful what formula you want to use.
01:03:03
I'll tell you why you cannot simply use mCΔT. If you've got a solid and you use mCΔT, that is correct, because when you heat a solid, the heat you put in goes into heating up the temperature of the solid. In the case of a gas, you can say, maybe let's ask the following question. This question is the following. You're telling me you put heat into a gas,
01:03:22
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?
01:03:40
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
01:04:00
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 actually it will cool down. But that's not what you're doing. You are keeping it on a hot
01:04:21
plate 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 there and working against the atmosphere above. It takes in with one hand and gives out the other without changing its energy. So, when you study specific
01:04:47
heat for a gas, and this is the next topic, you've got to be a little more careful when you talk about specific heats of gases, and I'll 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
01:05:01
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 loss of energy unless you pump in energy from below. So, what I've done is that I
01:05:20
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 the process.
01:05:42
All right, now I'll come to this question that was raised about specific heat. Now, specific heat, we always say is ΔQ over ΔT, over ΔT divided by the mass of the substance.
01:06:03
Now, it turns out for a gas, you've already seen that what you want to count is not the actual mass, but the moles. Because we have seen at the level of the ideal gas law, the energy is controlled by not simply the mass, but by the mole, because every molecule gets a certain amount of energy,
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namely 3 halves kT, and you just want to count the number of molecules or the number of moles. Now, there are many, many ways in which you can pump and 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,
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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 m, this should be the number of
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moles. So, if you take one mole, you can say, okay, one mole of gas, I was told, has energy U equals 3 halves RT, because it was 3 halves nRT, but n is one mole. Now, you want to put in some
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heat, and you really want the ΔQ over ΔT. So, I remind you the ΔQ is ΔU plus BΔV. The heat input into a gas is the change in energy plus BΔV. That's just from the first law.
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So, if I'm going to divide ΔQ by ΔT, that's the problem here. Did you allow the volume to change or did you not allow the volume to change? That's going to decide what the specific heat is. In other words, when a solid is heated, it expands such a tiny amount,
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we don't worry about the work done by the expanding solid against the atmosphere. But for a gas, when you heat it, the volume changes so much that the work it does against the external world is non-negligible. Therefore, the specific heat is dependent on what you allow the volume term to do.
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So, there's one definition of specific heat called CV. CV 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 hot plate. All the heat goes directly to internal energy.
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None of that is lost in terms of expansion. So, 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 over 2R.
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So, the specific heat of a gas at constant volume is 3 over 2R. When I studied solids, I never bothered about constant volume because the 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
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process. But for a gas, it's going to matter whether it's 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.
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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 P times ΔV,
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where now P is some constant pressure, say the atmospheric pressure, ΔV is the change in volume. So, ΔQ needed now will be more because you're pumping in heat from below and you're losing energy above because you're letting the gas expand,
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because you're letting the pressure be controlled from the outside at some fixed value. So now, ΔQ, this ΔU will be 3 over 2R ΔT. Now, what's the change in P times ΔV? Here is where you should know your calculus. The P times change in V is the same as change in
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PV if P is a constant. Right? Remember long back when I did rate of change of momentum, it's d by dt of mv, it's m times dv dt. So, if m doesn't change, you can take it inside the change. But now we use PV equals RT. I'm talking about one mole, PV equals RT. That's a change in the
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quantity RT, so R is a constant, that's R times ΔT, so I put in here R times ΔT. This is the ΔQ at constant pressure. So, the specific heat at this point is ΔQ over ΔT, keeping the pressure constant.
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You divide everything by ΔT, you get 3 over 2R plus another R, which is 5 over 2R. So, the thing you have to remember, what I did in the end, is that a gas doesn't have a single specific heat.
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If you 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.
01:12:02
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 2R. If the pressure cannot change, you get 5 over 2R. You can see Cp is bigger than Cv, because when you let the piston expand,
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then not all the heat is going to heat the gas. Some of it is dissipated on top by working against the atmosphere. 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,
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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
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we're talking about moles. 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. So, if you have two molecules that are diatomic
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gases, but two of them join 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.
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Some molecules also vibrate. There are lots of complicated things, but if you've 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
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than a complicated molecule. That's all you are responsible for. I'll just say one thing. Cp over Cv, I want to mention it before you run off to do your homework. I don't know if it comes out. I'm going to write it down. It's called γ. That's five-thirds for a monoatomic gas. You can just take the ratio of the numbers. If in some problem you find γ is not five-thirds,
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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. We just have to know this ratio γ is five-thirds in the simplest case, but in some problem somewhere in your life, you get a γ which is not five-thirds.