I’ve already told you multipletimes that large-hearted, uppercase U is the internal energyof a arrangement. And it’s really everythingthrown in there. It’s the kinetic energyof the molecules. It has the potential energy ifthe molecules are throbbing. It has the chemical energyof the bonds. It has the potential energy ofelectrons that just wanted to get some locate. But, for our reason, andespecially if we’re kind of in an preparatory chemistry, physics, or thermodynamics course, let’s just assume thatwe’re talking about a organisation that’s an ideal gas.And even better, it’s a kindof a monoatomic paragon gas. So everything in on my systemare just individual atoms. So in that case, the only energy inthe system is all going to be the kinetic energy of eachof these specks. So what I want to do in thisvideo– it’s going to get a little bit mathy, but I thinkit’ll be quenching for those of you who stick with it– isto relate how much internal vitality there really is in asystem of a certain pressure, magnitude, or temperature. So we want to relate pressure, publication, or temperature to internal vitality. Notice all the videos we’ve doneup until now, I just said what’s the change ininternal energy. And we related that to the heatput into or made out of a method, or the workdone, or be done in order to, or done by the system. But now, let’s just say beforewe do any run or any heat, how do we know how much internalenergy we even is available in a plan? And to do this, let’s do alittle bit of a meditate experiment.There is a bit of asimplification I’ll oblige here. But I think you’ll find it OK, or reasonably filling. So let’s say– let me justdraw it– I have a cube. And something “ve been told” that Imight have already done this pseudo-proof in the physic playlist. Although, I don’t think I associated exactlyto internal vigour. So I’ll do that here. Let’s say my systemis this cube. And let’s say the dimensionsof the cube are x in every direction.So it’s x high, x wide, and x deep. So its magnitude is, of course, x to the third. And let’s say I haven particles in my organization, uppercase N. I could have written lowercasen moles, but let’s just keep it simple. I have N specks. So they’re all doingwhat they will. Now, this is where I’m goingto conclude the gross simplification. But I think it’s tolerable. So in a regular plan, everyparticle, and we’ve done this before, is just bouncing offin every which way, every possible random direction. And that’s what, when theyricochet off of each of the sides, that’s what causesthe pressure.And they’re always bumping intoeach other, et cetera, et cetera, in all randomdirections. Now, for the sake of simplicityof our mathematics, and really to be able to do itin a tolerable amount of period, I’m going to makean assumption. I’m going to make an assumptionthat 1/3 of the corpuscles are going– well, 1/3 of the molecules are going parallel to each of the axes. So 1/3 of the molecules aregoing in this direction, I predict we have been able to say, left to right. 1/3 of the corpuscles aregoing up and down.And then 1/3 of the particlesare going forward and back. Now, we know that this isn’twhat’s going in reality, but it becomes our matha lot simpler. And if you actually were to dothe statistical mechanics behind all of the particlesgoing in every which way, you would actually end up gettingthe same result. Now, with that said, I’msaying it’s a gross oversimplification. There is some infinitesimallysmall chance that we actually do fall onto a system wherethis is already the bag. And we’ll talk a little bitlater about entropy and why it’s such a small probability. But this could actuallybe our plan. And this system wouldgenerate pressure.And it becomes our matha lot simpler. So with that said, let’sstudy such systems. So let’s take a sideways examine. Let’s take a sidewaysview right here. And let’s just studyone particle. Maybe I should havedone it in lettuce. But let’s say I haveone particle. It has some mass, m, andsome velocity, v. And this is one of the capitalN corpuscles in my plan. But what I’m curious is howmuch pressure does this particle exert on thiswall right here? We know what the area ofthis wall is, right? The domain of this wallis x goes x. So it’s x squared region. How much coerce is being exertedby this particle? Well, let’s think aboutit this course. It’s going forward, or leftto right just like this.And the force will be exertedwhen it changes its momentum. I’ll time a little of reviewof kinetics right here. We know that force is equal tomass era acceleration. We know acceleration can bewritten as, which is equal to mass seasons, change in velocityover change in time. And, of course, we know thatthis could be rewritten as this is equal to– mass is aconstant and shouldn’t change for the physic we dealwith– so it’s delta. We could frame that insideof the conversion. So it’s delta mv overchange in time. And this is just changein momentum, right? So this is equal to change inmomentum over change in time. So that’s another wayto write force. So what’s the changein force going to be for this particle? Well, it’s going to bumpinto this wall.In this direction, right now, it has some force. Its force is equal to mv. And it’s going to bump into thiswall, and then going to ricochet straight back. And what’s its momentumgoing to be? Well, it’s going tohave the same mass and the same velocity. We’ll usurp it’s a completelyelastic collision. Nothing is lost to heator whatever else. But the velocity is inthe other direction. So the new momentum is goingto be minus mv, because the velocity has switcheddirections. Now, if I come in with amomentum of mv, and I rebound off with a momentum ofminus mv, what’s my change in momentum? My change in momentum, off ofthat ricochet, is equal to– well, it’s the differencebetween these two, which is just 2mv. Now, that doesn’t giveme the force.I need to know the change inmomentum per unit of day. So how often does this happen? How routinely? Well, it’s going to happenevery time we come here. We’re going to hitting this wall. Then the speck is going tohave to travel now, bounce off of that wall, andthen come back here and touched it again. So that’s how frequentlyit’s going to happen.So how long of an interlude dowe have to wait between the collisions? Well, the corpuscle has totravel x going back. It’s going to crash. It’s going to have to travelx to the left. This length is x. Let me do that in adifferent colour. This interval right here is x. It’s going to have totravel x to go back. Then it’s going to haveto travel x back. So it’s going to have totravel 2x distance. And how long will it take itto travel 2x length? Well, the time, delta T, isequal to, we know this. Distance is equal torate eras season. Or if we do distance divided byrate, we’ll get the amount of time we made. This is just our basicmotion formula. Our delta T, the distancewe have to travel is back and forth. So it’s 2 x’s, dividedby– what’s our rate? Well, our proportion isour velocity. Divided by v. There “theres going”. So this is our deltaT right here.So our change in momentum pertime is similar to 2 eras our occurrence impetu. Because we ricocheted back withthe same magnitude, but negative force. So that’s our changein momentum. And then our change in timeis this value over here. It’s the total distance wehave to travel between conflicts of this wall, dividedby our velocity. So it is, 2x divided by v, whichis equal to 2mv ages the reciprocal of this– sothis is just fraction math– v over 2x. And what is this equal to? The 2′ s cancel out. So that is equal to mvsquared, over x. Interesting. We’re getting someplaceinteresting once. And if it doesn’t seem toointeresting, just hang on with me for a few seconds. Now, this is the force beingapplied by one particle, is this– force from one particleon this wall. Now, what was the field? We care about the pressure. We wrote it up now. The pressing is equal tothe force per locality. So this is the forceof that particle. So that’s mv squaredover x, divided by the area of the wall. Well, what’s the areaof the wall? The orbit of the wall here, each sideis x. And so if we choose the wallthere, it’s x periods x.It’s x squared. So divided by the area ofthe wall, is x squared. And what does this equal? This is equal to mv squaredover x cubed. You can just say, this is times1 over x squared, when this all becomes x cubed. This is just fraction math. So now we have an interestingthing. The persuade due to this oneparticle– let’s just call this from this one particle–is equal to m v squared over x cubed. Now, what’s x cubed? That’s the work ofour receptacle. Over the work. I’ll do that in abig V, right? So let’s see if we can relatethis to something else that’s interesting.So that means that the pressurebeing exerted by this one particle– well, actuallylet me just take another step. So this is one particleon this wall, right? This is from one particleon this wall. Now, of all the particles– wehave N particles in our cube– what fraction of themare going to be bouncing off of this wall? That are going to be doingthe exact same thing as this particle? Well, I just said. 1/3 are going to be goingin this direction. 1/3 are going to begoing up and down. And 1/3 are going to gobe going in and out. So if I have N total corpuscles, N over 3 are going to be doing exactly what thisparticle is going to be doing. This is the pressurefrom one molecule. If I missed the pressure fromall of the specks on that wall– so the total pressureon that wall is going to be from N over 3 ofthe particles.The other specks aren’tbouncing off that wall. So we don’t have toworry about them. So if we want the full amounts of the pressureon that wall– I’ll precisely write, pressuresub on the wall. Total pressure on the wall isgoing to be the pressure from one speck, mv squared, overour volume, times the total number of particleshitting the wall. The total number of particlesis N divided by 3, because merely 3 will be goingin that direction.So, the full amounts of the persuade on thatwall is equal to mv squared, over our magnitude of ourcontainer, times the full amounts of the molecules divided among 3. Let’s see if we can manipulatethis thing a little. So if we multiply both sidesby– let’s see what we can do. If we multiply both sides by 3v, we get pv terms 3 is equal to mv squared, eras N, where Nis the number of particles. Let’s divide both sides by N.So we get 3pv over– actually, no, let me leave the N there. Let’s partition both sidesof this equation by 2. So we get, what do we get? We get 3/2 pv is equal to–now this is interesting. It’s equal to N, the number ofparticles we have, terms mv squared over 2. Remember, I only dividedthis equation right here by 2 to get this. And I did this for a veryparticular ground. What is mv squared over 2? mv squared over 2 is the kineticenergy of that little corpuscle we started off with. That’s the formula forkinetic energy. Kinetic energy is equalto mv squared over 2. So this is the kinetic energyof one particle. Now, we’re multiplying thattimes the number of members of corpuscles we have, ages N. So N hours the kinetic energy ofone particle is going to be the kinetic energy ofall the particles. And, of course, we also madeanother assumption. I should state that I assumedthat all the corpuscles are moving with the same velocityand have the same mass.In a real situation, theparticles might have very different velocities. But this was one of oursimplifying presumptions. So, we just assumed theyall have that. So, if I multiply N occasions that–this statement right here– is the kinetic energyof the system. Now, we’re almost there. In fact, we are there. We only established that thekinetic energy of the system is equal to 3/2 durations thepressure, occasions the loudnes of information systems. Now, what is the kineticenergy of the system? It’s the internal exertion. Because we said all the energyin the system, because it’s a simple ideology monoatomic gas, all of the vigor in the system is in kinetic energy. So we could say the internalenergy of the system is equal to– that’s just the totalkinetic energy of the system– it’s equal to 3/2 epoches ourtotal stres, ages our total volume.Now you might say, hey, Sal, you only figured out the pressure on this area. What about the pressure on thatside, and that line-up, and that line-up, or on everyside of the cube? Well, the pressure onevery side of the cube is the same value. So all we have to do is findin terms of the pressure on one side, and that’s essentiallythe pressure of the system. So what else can wedo with that? Well, we know that pv is equalto nRT, our ideology gas formula. pv is equal to nRT, where thisis the number of moles of gas. And this is the idealgas constant. This is our temperaturein kelvin. So if we determine that permutation, we’ll say that internal energy can also bewritten as 3/2 terms the number of moles we have, timesthe ideal gas constant, epoch our temperature.Now, I did a lot of work, andit’s a little bit mathy. But these results are, one, interesting. Because now you have adirect relationship. If you know the pressure and thevolume, you know what the actual internal vigor, orthe total kinetic energy, of the system is. Or, if you know what thetemperature and the number of molecules you have are, you alsoknow what the internal force of information systems is. And there’s a couple of keytakeaways I want you to have. If the temperature does notchange in our principle statu here– if delta T is equal to0– if this doesn’t convert, the count specks aren’tgoing to change. Then our internal vigor doesnot change as well. So if we say that there issome change in internal vigor, and I’ll use this infuture proofs, we could say that that’s equal to 3/2 timesnR times– well, the only thing that can change , not thenumber molecules or the ideal gas constant– timesthe change in T.Or, it could also be written as3/ 2 days the change in pv. We don’t know if eitherof these are constant. So we have to say the changein the commodity. Anyway, this was alittle bit mathy. And I apologize for it. But hopefully, it gives you alittle bit more sense that this really is just the sumof all the kinetic energy. We relevant it to some of thesemacro territory variables, like influence, loudnes, and time.And now, since I’ve done thevideo on it, we can actually use this result infuture proofs. Or at least you won’t complaintoo much if I do. Anyway, see you inthe next video ..
