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# Kinetic molecular theory of gases

## Video transcript

so I want to talk to you a little more about the kinetic molecular theory of gases what this basically says is that the macroscopic properties of a gas like the pressure or the volume or the temperature are just a result of the microscopic properties of the gas molecules like the position and the speeds of these molecules so V down here is speed V appears volume the idea is this gas is made of molecules they're flying around in certain directions they have certain speeds and if you knew those speeds and you knew the distribution of speeds and the positions in here you could figure out these macroscopic properties what I want to basically do in this video is try to figure out what is the relationship if we know the microscopic properties how could we predict the macroscopic properties like if I knew the speed of all these molecules how could I figure out what pressure would be in there or vice versa if I knew the temperature of the gas could I say what the average speeds are of these molecules in this gas that's what we're going to do but first we have to make a few assumptions one assumption is that these molecules don't really interact and if they do interact it would only be because of a collision and if there is a collision between these molecules we have to assume it's elastic and kinetic energy will be conserved and momentum will be conserved similarly if one of these molecules strikes the wall of the container and has a collision there that should also be elastic there should be no kinetic energy lost so then let's get to it let me just clean that up get rid of that and start over up here I need to figure out how to relate a microscopic quantity to a macroscopic quantity let's just start with speed let's say you've got a particle in here a molecule moving this way with some speed I'll call it VX since I've drawn it in the X direction and it collides with this wall well that's going to impart a force on this wall and if you get a lot of these doing that in here you'll get a pressure on this wall but it's going to be an elastic collision so this particle is going to bounce backwards with the same velocity and let's try to figure out what force that would exert because if I can figure out the force on the wall I can figure out the pressure because pressure is just force per area so force ya it equals MA but it also you equals delta p the change in momentum over the change in time so this is an alternate way to write Newton's second law what would be the change in momentum so I'm going to try to find the force on this wall the change in momentum momentum is MV and if this mass doesn't change then change in momentum is just M Delta V where the V here is speed so mass times sorry excuse me velocity mass times change in velocity so what would be the change in velocity for this collision right here it struck the wall and bounced back with the same speed some people want to say zero because it comes in with the same speed that it goes out with but V is velocity and so the change in velocity is actually two times V because it came in with V and left with negative V so technically the change would be negative to V but I'm going to ignore negatives because I just want the size of this force on this wall so M times two times VX over delta T but I don't want delta T in here I want an equation of state that just has pressure and volume and speeds and stuff like that so how can I get rid of delta T well I know the distance in here let's just call the side lengths here L so you have a box of L by L by L to cube well the time it takes between collisions so there's an impulse here M Delta V right when this collision takes place and then this particle travels over here to the left bounces off of this wall then comes back over to here again hits it how long is it between those impacts well the time it would take to travel to the left and back I know speed is distance per time so the time the delta T is just going to be the distance per speed and the distance it's not just L because it's got to travel to this wall and then back I want to know the force on this wall over here I need to figure out how long is it between collisions with this wall so it's going to be 2 times L over the velocity in the x-direction that's what I can substitute in over here and I get that F is going to be M times 2 V x over delta T now is 2 L over V X but since I'm dividing by V X on the bottom I move that up top and look already got one here so I'm just going to square it I can cancel off the twos and I get that the force on the wall by this particle is mass times its velocity in the X Direction squared divided by L I should say this particle doesn't have to just be going in the X direction it might have some total velocity this way where the X component is just a part of it but if I just took the X component of the particle speed whatever whatever particle it is it had some velocity the X component I'd get the force contribution to the pressure on this wall over here so this is the force on this wall over here by one particle but I want to know the force from all the particles because I want to get the total pressure so how could I do that well if I want the total force I just need to add up the contributions from all the particles so let's say there was there were other particles well they're going to have the same mass M I'm assuming I got the same gas throughout all molecules have the same mass and L will be the same for all of them so the only difference in contribution will be that some may have a certain component of velocity in X direction I'll call this VX 1 squared plus there may be some other particle that has a different component too and there may be some particle that has a different component 3 you just have to add all these up so I'll have VX to the two references particle 2 squared plus the X 3 the X component of particle 3 is velocity squared plus I'd keep going to N many times I'll keep adding this up until I got what the X and the nth particle total amount there's n particles in there squared but this is looking like an average in fact if I just divide both sides by n the total number of particles look at what I get I get that force over N equals M divided by L times this whole thing divided by n is just the average value the average value of what the average value of VX squared and it's the average value of VX squares I'm going to put a bar over the top of this this is telling you that it's the average value of VX squared it's not the square of the average values of VX that's different if I took the averages of all the V X's VX one plus VX 2 plus VX 3 divided by N and then square it I'd get a different result so it's important to note first you square them all take the average and that's what you're doing here you're taking the average of the squares not the square of the average ok so moving on we get that F equals n times M over L the average value of the x squared but what do we want to do with this I promised you a relationship between speeds and pressures so let's turn this into pressure let's figure out what's the pressure on this area here so I have to turn this into a pressure that's not too hard pressures just force per area so I could just divide this by the area of this wall so I'll divide if I divide the left side by area I've got to divide the right-hand side by area as well what does that leave me with on the left hand side I get pressure that's good a macroscopic variable equals n times M times V x squared averaged over all the gas molecules divided by a times L but a what's a a is just l squared so I get l squared times L on the bottom that's just l cubed and look at what's going to happen l cubed that's just volume that's the volume of this cube and so I get n the number of molecules times M the mass of one of the molecules times the average value of the X component squared over all the gas molecule divided by V that's the volume we're getting close this is looking like the ideal gas law so this is really good let me just take this result actually and just put it in a new window so we can get a clean result and look at what I get I get that the pressure times the volume if I multiply both sides by V pressure times volume equals the number of gas molecules times M times the average squared X velocity in the gas so this is pretty cool if I went out and measured the pressure of a gas and the volume of the gas I can try to figure out now with this average squared X components of velocity are for the gas that's a mic scopic quantity we've got a relationship now but I mean I don't just care like I'm not trying to just single out X there's y and there's other directions in here why would we want an equation with just the X usually you just want a formula it'd be better if this just told us the total average squared velocity let's do that if this is in the X direction I had velocity in the X but these particles also have velocity in the Y direction and so the total we know the total V total would be V x squared plus V Y squared and there's also one more we live in three dimensions there we go t in the Z direction so this is the Pythagorean theorem in three dimensions it works in three just as well as it doesn't - but this equation works also if I average them all if it took all the averages of the x squared components of velocity and I took the average of all the V Y squared components of velocity so if I take these all and I averaged them well this equation is still true oops this should have been squared on my goodness this V total here should have been squared right there I take the average now I'm going to make a claim I'm going to claim that the particles in here are flying around randomly there's no direction that's singled out there's no preferred direction they have just as much velocity on average in every direction as any other direction so really V in the X Direction squared averaged over all the gas molecules has to be equal to V in the Y Direction squared because why would it be any different why would Y be preferred than X I mean on average if you had a lot of gas molecules these have got to basically statistically be even VZ it's got to be equal the average of those squares these have got to be equal so may as well write this down here as three times one of them so three times VX squared averaged because already have that one up here and now I can this is a way I can get V total in here I want V total not just one direction so I get V x squared averaged over all the gas molecules equals I'm just going to divide both sides by three here and I get V total squared averaged over all the gas molecules divided by three this is cool I can substitute this into there and I'll get a relationship that says that P times V P times V equals the total number of gas molecules times the mass of one gas molecule times the average of the total squared velocities divided by three and I'm going to rearrange this just a little bit more I'm going to say that let's multiply both sides by three I'll get that three times pv equals n times m times the average of V total squared for all the gas molecules and I'm going to do one more thing I'm going to multiply both sides by 1/2 you might think that's random but I've doing this for a reason check this out and now look what we got over here this whole term right there one-half M V squared this should look familiar this is just the average kinetic energy of one of the gas molecules this is awesome this says if I knew the pressure and the volume then I've got a way to figure out what's the average kinetic energy of one of these gas molecules this gives me a direct relationship between the kinetic energy of a gas molecule or the average kinetic energy and what the macroscopic pressure and volume are it's so important that I'm going to write it again what we found with the three halves times the pressure times the volume equals n times the average kinetic energy of a gas molecule what could we do with this we can do a few more things P times V I know what P times V is remember the ideal gas law PV equals capital n K T so I can substitute in and KT over here and I'll get that three-halves times capital n k t equals capital n average kinetic energy well these ends cancel and I get a direct formula that the average kinetic energy and a gas the average kinetic energy of one single gas molecule equals three-halves KBT this is nice it tells me that directly if I know the temperature I can directly figure out the average kinetic energy of one of these gas molecules no matter what kind of gas I have as long as it's an ideal gas that pretty cool something else that's useful is this is the average kinetic energy of one gas molecule this is n all of the gas molecules the total number of them so this whole thing right here is the total energy the total thermal energy of that gas if it's monatomic if the gas molecule isn't diatomic it's a single simple mono atomic gas all its god is kinetic energy that's the only energy it can have and so three-halves pv is the total energy of the gas or you can write it as three halves and KT would be the total internal energy or if you wanted to three halfs little an RT equals the total internal energy these are really useful to know but they're only true for a monatomic ideal gas a monatomic gas where the molecules that make up the gas are composed of only a single atom like helium or neon or any of the noble gases if you have a monatomic ideal gas these formulas give you a direct relationship between the macroscopic quantities and the total internal energy of that gas these are particularly useful and it's useful to note that by total internal energy for a monatomic ideal gas that's just a fancy word for the total kinetic energy see people used to think these were different energies remember people thought maybe there's thermal energy something new something different nope Boltzmann told us that's just kinetic energy in there for the most part and for a monoatomic ideal gas it's only kinetic energy in there so you total is just another word for the total kinetic energy but when we talk about thermal systems you'll often hear it referred to as the total internal energy of the gas
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