Proof: U = (3/2)PV or U = (3/2)nRT

I've already told you multiple times that big uppercase u is the internal energy of a system as really everything thrown in there it's the kinetic energy of the molecules it has the potential energy if the molecules are vibrating it has the chemical energy of the bonds it has the potential energy of electrons that want to get someplace but for our for our sake especially if we're kind of an introductory chemistry physics or thermodynamics course let's just assume that that we're talking about a system that's an ideal gas and even better it's a kind of a monoatomic mono atomic ideal gas so everything in on my system are just individual atoms so in that case the only energy in the system is all going to be the kinetic energy of each of these particles so what I want to do in this video it's gonna get a little bit of Matthieu but I think it'll be satisfying for those of you who who stick with it is to relate how much internal energy there really is in a system of a certain pressure volume or temperature so we want to relate pressure volume or temperature to internal energy notice all the videos we've done up until now I just said what's the change in internal energy and we related that to the heat put into or taken out of the system or the work done or done to or done by the system but now let's just say before we do any work or any heat how do how do we know how much internal energy we even have in a system and to do this let's do a little bit of a thought experiment and there there is a bit I know there is a bit of a simplification I'll make here but I think you'll you'll find it okay or reasonably satisfying so let's say let me just draw it I have a cube and something tells me that I might have already done this pseudo proof in the physics playlist although I don't think I related it exactly to internal energy so I'll do that here so let's say my system is this cube and let's say the dimensions of the cube are X in every direction so it's X hi X wide and X deep so it's volume is of course X to the Q X to the third and let's say I have n particles in my system and capital N and particles I could have writen lowercase and moles but let's just keep it straightforward have n particles and particles so they're all doing what they will now this is where I'm gonna make the gross simplification but I think it's reasonable so in a normal system every particle and we've done this before just bouncing off in every which way every possible random direction and that's why you know when they ricochet off of each of the sides that's what causes the pressure and they're always bumping into each other etc etc and all random directions now for the sake of simplicity of our mathematics and just to be able to do it in a in a reasonable amount of time I'm gonna make an assumption I'm gonna make assumption that one third of the particles are going well one third of the particles are going parallel to each of the axes so one third of the particles are going in this direction are going I guess we could write say left to right one third of the particles are going up and down so are going up up and down one third and then one third of the particles are going forward and back forward and back now we know that this isn't what's going on in reality but it makes our math a lot simpler and if you actually were to do the the statistical mechanics behind all the particles going in every which way you would actually end up getting the same result now what that said I'm saying it's a gross oversimplification there is some infinitesimally small chance that we actually do fall on to a system where this is already the case and we'll talk a little bit later about entropy and why it's such a small probability but this could actually be our system and this system would generate pressure and it makes our math a lot simpler so with that said let's study this system so let's take a sideways view let's take a sideways view right here and I have and let's just study one particle maybe I should have done it in green but let's say I have one particle it has some mass m and some velocity V some particle right there and what I'm curious and this is one of the capital n particles in my system what I'm curious is how much pressure does this particle exert this wall right here on this wall right here we know what the what the area of this wall is right the area of this wall is x times x so it's x squared area how much force is being exerted by this particle well let's think about it this way it's going forward and or left-to-right just like this and the force will be exerted when it changes its momentum I'll do a little bit review of kinetics right here we know that force is equal to mass times acceleration when we know acceleration can be written as which is equal to mass times I'll just write change in velocity over change in time and of course we know that this could be rewritten as this is equal to mass as a constant shouldn't change and for the physics we deal with so it's Delta we could put that inside of the change so it's Delta m/v over change in time and this is just change in momentum right so this is equal to change in momentum over change in time so that's another way to write force so what's the change in momentum going to be for this particle let's go to bump into this wall we're in this direction right now it has some momentum its momentum is equal to mV then it's gonna bump into this wall and they're gonna ricochet straight back and what's it what's its momentum gonna be we're gonna have the same mass and the same velocity we'll assume it's a completely elastic collision nothing is lost a heat or whatever else but the velocity is in the other direction so the new momentum is going to be minus MV because the velocity has switched directions now if I come in with them with a momentum of MV and I ricochet off with a momentum of minus MV what's my change momentum my change in momentum off of that ricochet is equal to well it's the difference between these two which is just 2 m/v now that doesn't give me the force I need to know the change in momentum per unit of time so a change in momentum per unit of time per unit of time so how often does this happen how frequently well I have it's gonna happen every time you we come here we're gonna hit this wall then the particles to travel here bounce off of that wall and then come back here and and and hit it again so that's how frequently it's going to happen so how long of an interval do we have to wait between between the collisions well the particle has to travel X going bat is going to collide that's going to have to travel X to the left this distance is X let me do that in a different color this distance right here is X it's going to have to travel X to go back then it's gonna have to travel X back so it's going to have to travel to X distance and how long will it take it to travel to X distance well the time delta T is equal to we know this distance is equal to rate times time or if we do distance divided by rate distance divided by rate we'll get the amount of time we took right this is just our basic motion formula so the our delta T the distance we have to travel is back and forth so it's two x's divided by what's our rate well our rate is our velocity divided by V divided by V there you go so this is our delta T right here so our change in momentum per time change in momentum per time is equal to two times our kind of incident momentum because we were getting ricocheted back with the same magnitude but negative momentum so that's our changed momentum and then our change in time is this value over here it's the total distance we have to travel between collisions of this wall divided by our velocity so it is 2x 2x divided by V which is equal to 2m V times the reciprocal of this so times this is just fraction math V over 2x and what is this equal to the twos cancel out so that is equal to M V squared over X interesting or already we're getting someplace interesting already and if it doesn't seem too interesting just hang on with me for a second now this is the force being applied by one particle is this force force from one particle on this wall force from one from from particle now what was the area we care about the pressure right the pressure the pressure we wrote it up here right we wrote the pressure is equal to the force per area right so the pressure is equal to the force per area so this is the force of that particle so that's M V squared over X divided by the area of the wall well that's the area of the wall the area of the wall here it's each side is X and so if we draw the wall there it's x times X it's x squared so divided by the area of the wall is x squared now what is this equal this is equal to M V squared over X cubed right I just divide you can just say this is x one over x squared when this all becomes x cubed this is just a fraction math so now we have an interesting thing the pressure due to this one particle pressure from one particle pressure from let's just call this from the particle from this one particle is equal to MV squared over X cube now what's X cubed that's the volume of our container over the volume I'll do that in a big in a big in a big V right so let's see if we can relate this to something else that's interesting so that means that the pressure being exerted by this one particle well that actually let me just take another step so this is one particle on this wall right this is from one particle on this wall now of all the particles we have n particles in our in our cube what fraction of them are going to be bouncing off of this wall that are going to be doing the exact same thing as this particle well I just said one third are going to be going in this direction one third are going to be going up and down and one third are going to be going in and out so if I have n total particles and over three you're going to be doing exactly what this what this particle is going to be doing so if I want it this is a pressure from one particle if I wanted the pressure from all of the particles on that wall so the total pressure on that wall is going to be from n over three of the particles the other particles aren't bouncing off that wall so we don't have to worry about them so if we want the total pressure on that wall pressure I'll just write pressure sub on the wall total pressure on the wall it's going to be the pressure from one one particle MV squared over our volume times the total number of particles hitting the wall so times the total number of particles is n divided by three because only three are gonna be going in that direction so the total pressure on that wall is equal to MV squared or our volume of our container times the total particles divided by three let's say you couldn't manipulate this thing a little bit so if we multiply both sides by let's see what we can do if we multiply both sides by three if three V we get P V times three right is equal to M V squared times n where n is the number of particles let's divide both sides by n so we get three P V over over actually no let me leave the end there let's divide both sides of this equation let's divide both sides of this equation by two so we get what do we get we get three halfs PV is equal to is equal to now this is interesting it's equal to n the number of particles we have times MV squared over 2 remember I just divided this equation right here by 2 to get this and I did this for a very particular reason what is MV squared over 2 MV squared over 2 is the kinetic energy of that little particle we started off with that's the formula for kinetic energy kinetic energy is kinetic energy is equal to MV squared over 2 so this is the kinetic energy of one particle right this is kinetic energy of one particle now we're multiplying that times the total number of particles we have times n so n times the kinetic energy of one particle is gonna be the kinetic energy of all the particles and of course we also made another assumption I should state that I assume that all the particles are moving with the same velocity and have the same mass in a real situation the particles might have very different velocities but this was one of our simplifying assumptions so we just assume they're all have that so if I just multiply n times that this is this statement right here is the kinetic energy of the system kinetic energy of the system now we're almost there in fact we are there we just established that the kinetic energy of the system is equal to three-halves times the pressure times the volume of the system now what is the kinetic energy of the system it's the internal energy because we said all the energy in the system because it's a simple ideal monoatomic gas all of the energy in the system is in kinetic energy right so we could say the internal energy of the system the internal energy of the system is equal to that's just the total kinetic energy of the system it's equal to three-halves times our total pressure times our total volume now you might say hey Sal you just figured out the pressure on this side what about the pressure on that side and that side and that side or on every side of the cube well the pressure on every side of the cube is the same value so all we have to do is find in terms of the pressure on one side and that's essentially the pressure of the system so what else can we do with that well we know that PV is equal to NRT our ideal gas formula P V is equal to n RT or this is the number of moles of gas this is the ideal gas constant this is our temperature in Kelvin so if we make that replacement will say that internal energy can also be written as 3 halves times the number of moles we have times the ideal gas constant times our temperature now I did a lot of work and a little bit of math E but these results are one interesting because now you have a direct relationship if you know the pressure and the volume you know what the actual kinetic you know the actual antennal internal energy or the total kinetic energy of the system is or if you know what the temperature and the number of molecules you have are you also know what the internal energy of the system is and there's a couple of key takeaways I want you to have if the temperature does not change in our ideal situation here if the temperature does not change if delta T is equal to zero if this doesn't change the number of particles are going to change then our internal energy does not change as well and so if we know so if we say that there is some change in internal energy and I'll use this in future proofs we could say that that's equal to three-halves three-halves times n R times well the only thing that can change not the number of molecules of the ideal gas constant times the change in T or it could also be written as three halves times the change in PV we don't know if either of these are constant so we have to say the change in the product anyway this was a little bit mathy and I apologize for it but hopefully it gets you it gives you a little bit more sense that this is really is just the sum of all of the kinetic energy we're related it to some of these macro state variables like pressure volume and time and now since I've done the video on it we can actually use this result in future proofs or at least you won't complain too much if I do anyway see in the next video