today we're going to be revising thermal physics again so this is part two all of ideal gases if you're looking for part one which is the rest of thermophysics please have a look at a link in the description of this video well let's get started the first aspect of revision for ideal gases would be what is a mole now contrary now contrary to part to popular belief this is not a small creature that lives on the ground but a mole is the amount of substance which contains the same number of particles as there are atoms in 12 grams of carbon 12. and this number is avogadro's number so let's call that an a which is about 6.02 multiplied by 10 to the power of 23 particles so for instance we could have one mole of water one mole of metal etc and each of them will contain the same number of particles 6.02 times 10 to the power of 23. okay next up is the concept of molar mass the molar mass will actually allow us to know how much there's a mole out of a particular substance way for instance the molar mass of water let's just say the molar mass of water is 18 grams per mole which is also commonly written as 18 grams divided by moles okay well this means that essentially if we had some water and if we had one more particles so remember one more particles is avogadro's number so if we had 18 grams of water exactly because that's the molar mass the number of particles contained within the container will be avogadros number or 6.02 times 10 to the power of 23 particles now if we were to have let's say 18 times twice is 36 grams of water we would have double that so we would have about 12 of times 10 to the power of 23 which is about 1.204 times 10 to the power of 24 particles and we can generalize this idea in a couple of different equations our first equation is that the number of moles i'm just going to write this as a word equation for once the number of moles typically given n is equal to your mass measured in grams of kilograms depending on the data we're given and that will be divided by the molar mass so the way i normally remember it is m over m mass over molar mass this equation is not typically given in your formula sheet and we can also kind of work it out just by thinking about units if we think about it right here on the top let's say we have grams or kilograms and then we're dividing that by grams per mole which is equal to one over one over mole which of course is just gives us moles so a number of moles is that once we have the number of moles what we can actually do is work out the number of particles number of individual particles so number of particles which will just be equal to our number of moles we can write this and this actually let's be consistent and let's write those number of moles multiplied by avogadro's number which is n a 6.02 times 10 to the power of 23. okay well let's have a look at a little example over here we have a glass of water containing 300 grams of water how many particles are contained within the glass the molar mass of water is 18 grams per mole well because we're curious we can start thinking about how many particles are we actually holding when we have a glass of water if the water weighs let's say 300 grams first off we need to work out the number of moles so remember n the number of moles is m over m so it's going to be mass over molar mass so that's going to be 300 grams divided by 18 grams per mole which is going to give us approximately 16.67 moles and what i'm going to do afterwards is i'm going to just multiply that number by avogadro's number to work out the number of particles because remember one of these moles is actually equal to an avogadro's number of particles so the total number of particles let's call that capital n so remember typically in physics it's lower case n is the number of moles capital n is the number of particles so the number of particles will then be equal to just 16.67 multiplied by 6.02 times 10 to the power of 23 and if we put that into a scientific calculator we are going to get about 1.0 times 10 to the power of 25 individual particles okay now that we've had a look at the number of moles let's look at ideal gases in a little bit more detail we can think about ideal gases using the kinetic theory model of the gases that tells us there are large number of particles now wrap random rapid motion this is actually the first of the assumptions of this model which would need to be remembered before an exam so this is our first assumption our second assumption tells us that the individual particles occupy negligible volume compared to the volume of the gas in other words the particles are very very small three all collisions are perfectly elastic remember in a perfectly elastic collision kinetic energy is conserved and the time spent is collision in collision is negligible compared to the time spent between collisions our fourth assumption is that there are negligible forces between the particles except during collision so this means the particles are basically not attracting each other unless uh or not interacting via any other forces unless they're colliding well this leads us to the ideal gas flow equation and just a couple of really important points to remember as well from the previous slide as we said in ideal gas the forces between the particles are assumed to be zero except during a collision because of that the potential energy of the gas is also assumed to be zero well there are two equations which govern the behavior of ideal gases and they're very very important pv is equal to nrt and pv is equal to n times k times t they're essentially the same equation p stands for pressure v stands for volume now in this equation the lowercase n is the number of moles so we can write that in the number of moles r here is the gas constant or the gas law constant uh which is just approximately equal to 8.31 so we can just write that in 8.31 and t is the temperature in kelvin so this is very very important quite a typical mistake would be to have to have a problem in which the temperature is given in degrees celsius remember we need to convert that to kelvin the lower equation is basically the same equation however rather than the number of moles we just converted to a number of particles so the upper case and in this case is the number of particles k is boltzmann's constant boltzmann's constant which is approximately equal to 1.38 times 10 to the power of minus 23 and t once again is the temperature in kelvin now looking at those two equations there's one more factor that i really like to mention and that is about boltzmann's constant k because pv is equal to nrt and pv is equal to nk times t then what we can do let me just make this a little bit more legible that's a t here you can do is essentially set those parts of the equation equal to one another n r will have to equal n multiplied by k well let's just rearrange for k k will be equal to n r divided by the number of particles once again in this equation lower case n is the number of moles uppercase n is the number of particles but how are they related remember the number of particles n is equal to the number of moles multiplied by avogadro's number 6 about 6 times 10 to the power of 23 okay well let's just sub that in here and what we're going to get is that boltzmann's constant k will be equal to n r over n times avogadro's number those guys are going to cancel and we're going to find that boltzmann's constant is equal to the gas law constant divided by avogadro's number now let's have a think about the ideal gas behavior if let's say the temperature is constant so first off let's look at uh well what will become known as boyle's law after we set the temperature to be a constant so let's say that our temperature is equal to a constant in practice for instance we could be increasing the pressure by decreasing the volume with a pump very very slowly to try and keep the temperature constant for instance this will be a physical scenario for this but if the temperature is a constant because pv is equal to nr t and let's say that the temperature is constant for a fixed volume so for a fixed amount of gas then the pressure could be varied the volume could be varied and the number of moles was a constant r is just a constant then the temperature is a constant so if that's the case the whole of the right hand side that i'm actually highlighting now is a constant so p times v will be a constant which would mean that p is going to be a constant divided by v which would mean that p is proportional to 1 over v in other words the pressure is inversely proportional to the volume at a constant temperature for a fixed amount of gas and this statement is known as boyle's law now let's think about what would happen if we kept v to be a constant so keep the volume constant maybe it's a metal container that we're dealing with or a glass container or something which doesn't really change its volume and let's say that also the amount of substance is constant so n is equal to a constant so no gas is is leaking out of um out of our container well if that's the case well then that would mean that p will be let's just rearrange pv is equal to rt so i'm going to start off with pv is equal to nrt here so pv is equal to nrt and because v is a constant now i am going to just rearrange p so i'm going to say that p is going to equal to n r divided by v times t well have a look at that n is a constant r is a constant and v is a constant so this whole highlighted area is equal to a constant so this means that the pressure is equal to some constant multiplied by t which means that the pressure is directly proportional to temperature for a fixed mass of a gas if the volume is kept constant this is known as our pressure law so we can write it down as our pressure pressure temperature law pressure temperature law now that we have had a good look at our two laws let's have a look at a an experiment for boyle's law so the way the way we're going to do this experiment to prove that boyle's law is correct we're going to take a co2 of a pressured gas with a volume scale typically this is filled with oil as well at least partially with oil so we're going to vary the pressure of the gas and we're going to do that with a foot pump or just a regular pump and we're going to have a pressure gauge and we also have a volume scaled we're going to measure the pressure with a with a pressure sensor or a pressure gauge once again so it might be worth just adding a list of measurements so let's do that really quickly so i'm going to say measurements let's say the volume with the scale with the tube and our pressure with the pressure gauge or a sensor as well if that's available afterwards we're going to plot a graph of the pressure against one over v and if the graph is a straight line through the origin this shows that our relationship is correct remember if we have a graph of the pressure against one over the volume we will be expecting a straight line relationship why is that remember p times v is equal to a constant which means that p is proportional to 1 over v so p will be equal to some constant let's call that k divided by v which means that p is equal to k times 1 over v now just underneath i'm going to write y is equal to mx plus c i'm going to have a plus 0 here so if p is on the y axis if one over v is on the x axis what's left for our gradient is is is our constant so in this case our gradient be equal to our constant which is actually equal to n r t and that is because pv is equal to nrt in general so we could if we were given the temperature for instance from the gradient of this of this graph we could work out our our our number of moles for instance or if we given both of those we could work out the constant r if we didn't know it etc so this is a really really important experiment now how could we make this this experiment um better well we need to ensure that the temperature is constant and this is really important so i'm just going to add that in in here so what i'm going to say is vary the pressure slowly so vary the pressure slowly so that the temperature remains constant for anyone that has pumped up some tires on a bike for instance you know that if you vary the uh the if you pump up the tires quite quickly that the pump even will get really really hot really quickly so we're going to try and keep that temperature constant now let's have a look at how we could actually investigate the pressure temperature law remember that says that pressure is proportional to temperature which means that the pressure is equal to some constant nrov multiplied by the temperature in order to investigate that we must really need to make sure that the volume of the gas in question is constant and additionally the amount of substance is constant so what we're going to do is we're actually going to use a a round bottom sealed flask of a fixed volume and we're going to submerge that in a water bath so this will be our first step of the method then we're going to place that water bath on top of a heater or or we could use a bunsen burner as well some heat pads to heat up the water and we can see my artistic impression of the flames over here afterwards we're going to vary the temperature and we're going to be measuring the pressure so the water will slowly heat up so we could start could even start with ice at zero degrees celsius and then slowly essentially heat up the ice melt it make sure it turns into water and then the water can sort of heat up we're going to take measurements regular measurements of the pressure with pressure gauge and of the temperature with a thermometer after we have taken our readings of the pressure and the temperature we're going to plot a graph of p against t now our graph will look something like this so we're going to have our pressure and we're going to have our temperature and the graph should be a straight line but won't be through the origin we could actually use we could actually extrapolate to figure out the um our absolute zero by drawing the line of best fit and extrapolating and if we do that at zero pressure we should be reaching theoretically the uh the absolute zero which is about minus 273 degrees celsius so let me just write this down because this is very very important so um let's write it right over here in blue so we're going to extrapolate or let's just say that we are going to draw a line of best fit so draw a line of best fit and extrapolate to find absolute zero zero like so and um at zero pressure so that'll be the fifth point at zero pressure you'll find absolute zero now remember at absolute zero the particles have minimum internal energy which would mean that their kinetic energy is starting to tend towards zero so they're hardly exerting any force on to the container there's hardly any change of momentum of the particles as they're colliding which means there's hardly going to be any pressure on the container so far we've talked quite a lot about pressure in terms of the kinetic model but let's actually explain it using the kinetic model what actually is pressure well let's imagine the side of a container and let's imagine that we have a molecule of gas which is going towards it then the molecule is going to collide with the container that will rebound back into the opposite direction if the initial part of the initial momentum of this particle was mv the uh the final amount of the particle is going to be mv in the other direction assuming assuming that no energy has been lost so we can just say that this is equal to minus mv so there's going to be a change of momentum and that change of momentum will be equal to final momentum which is minus mv minus the initial momentum which is mv which is equal to minus 2 m v but remember anytime there's a change of momentum there is a force that must be acting as well newton's second law says the net force acting on an object is proportional to the rate of change of momentum so that individual particle is going to experience a force of minus 2 m v divided by delta t the same force will be experienced by the container but in the opposite direction now because pressure is equal to force of area in general if we imagine there's so many of these particles which are constantly colliding with uh with uh with the container let's say the container is is here across all sides if we sum all of those forces and divide them by the area this is what pressure actually is so it's the sum of all forces of those those collision forces of the particles divided by the area so let's just write that mathematically as the sum of all the forces divided by the area this is what pressure actually is on a small scale okay now let's talk a little bit about the root mean squared speed so there's many many different molecules in a gas billions and billions of them and they all have different velocities in different directions so we need another method to have an idea of how much on average the the speeds of the particle are and we use the root mean square speed for that so imagine we have some particles over here let's say 320 meters per second one is 150 in this direction then the other one is 100 meters per second in the opposite direction then what we need to do would be to first off square the speeds then afterwards we're going to take the mean and then afterwards we are going to find the square root of our result okay well let's start off first off by squaring them so and then taking the mean so it's going to be 320 squared plus 150 squared plus 100 squared and because there's only three particles in our hypothetical scenario we're going to divide by three so let's do that on the calculator 320 squared plus 150 squared plus 100 squared divided by 3 and that's going to give me approximately let's just write it down over here as 4.4 multiplied actually let's just involve a little bit more precision because we're going to be square rooting the answer and i'm going to put it into the appropriate significant figures right at the very end so it can be 4.4967 times 10 to the power of 4 that's going to be meter squared s to the power of minus 2. okay well we have squared speed and we found the mean of of the squares now let's just square root our answer so it'll be the square root of 4.4967 times 10 to the power of 4 to get a root mean square speed of approximately 212 meters per second because we're using up to two significance in this we could also you leave that as about 210 meters per second now the mean squared speed is really important in this equation in particular which says that the pressure times the volume is equal to one-third times the number of particles multiplied by the mass and then multiplied by the um squared mean speed notice this is not the root mean square speed if it was there'll be a square root sign here but there is not so we've just squared the values and i've taken the mean of them a typical exam question might ask you to derive that this same equation up here can be written as the pressure is equal to one third multiplied by the density of the substance times uh times essentially the squared mean speed okay well how are we going to tackle this first off remember the density is equal to mass over volume now the total mass of of the gas will be equal to the number of particles multiplied by the by the individual mass of each of them then we're going to be dividing that by the volume so let's see how we can proceed one thing we could do is just rearrange for the volume then plug that into the original equation so we can say that the volume will be equal to the density i mean of course that the volume be equal to the number of particles multiplied by the mass divided by the density okay so now that we have v what i'm going to do is i'm just going to substitute that in here and what we're going to get is that p and rather than times v i'm going to say times n m divided by the density is equal to a third and m c squared we're basically there aren't we so and we can cancel out nm on either side and what we're going to get is that the pressure is equal to a third times the density times the squared mean speed okay so up next is the maxwell boltzmann distribution we've been talking about we've been revising about the speed of the particles the maxwell boltzmann distribution is a plot of the proportion of particles with a given speed against the speed of the particles so we can see that quite few of the particles a relatively small proportion have low speeds and we could just write this down so we can say that few of the particles have low speeds however the vast majority of the particles will have moderate speeds and this will be this region over here maybe i can just sort of highlight it like so it will be just this region over here with where the vast majority of the particles are you can see that those are the particles with the um with the greatest proportion in terms of numbers so we can say that most of the particles most of the particles have moderate so moderate speeds we can see that here that the speed is still not quite high but a very very small proportion of the particles will have some very high speeds and then we can also say that few of the particles will have very high speeds now what will happen to this curve if the temperature changes so let me draw with a different color what we expect to see if we increase the temperature first off the curve will become flatter and more of the particles are going to end up having higher speeds so the graph will suddenly appear to look something like this so let's see if i can use a larger highlighter so it will look something like this in which more of the particles are going to have larger speeds so we could just write that here so this orange graph here represents what would happen if we increase the temperature so this is a graph for a higher temperature where once again the there's a there's a greater proportion or there's more particles at higher speeds so there seems to be a link between how between how quickly a particle is going and what the temperature is let's have a look at that link the link is to be found within this extremely important equation which really tells us what temperature actually is and that is that the mean kinetic energy a half m c squared where c squared but a bar on top signifies that we're taking the mean squared speed is equal to three halves times k which is boltzmann's constants constant 1.38 times 10 to power minus 23 times the temperature in kelvin just please remember that this over here is the temperature in kelvin this over here is boltzmann's constant so baltimore's constant giving you formula booklet but 1.38 times 10 to power minus 23 this is just three halves representing the degrees of freedom this is something you only sort of worry about if you study physics at university c squared is the squared squared mean speed and m is the mass of a particle of the gas so this means that for instance if you know the the mass out of an air particle you will know that at a given temperature the speed should be something which you can actually calculate you'll be able to calculate the the squared mean speed or the root mean square speed of a particle of air at a given temperature which is really really exciting now how can we actually come up with how did we actually come up with this equation how can we derive this equation and this is something that we need to know for the exam so let's practice it so the way to derive this expression is uh first off let's look at the starting point so anytime we're deriving uh something in physics we need to look at some other equations and then finally come up with this equation so the derivation for this one is relatively straightforward we have two equations in our formula booklet one is that pv is equal to one third multiplied by the number of particles multiplied by the mass multiplied by c squared with a bar on top the other one is that pv is equal to n where n is the number of particles again k's boltzmann's constant and t is our temperature in kelvin well having a minute this equation is for pv and this version is for pv so we can just set those equal to one another so what we could really easily say is that one third n m c squared is equal to n k t okay so now we can cancel out the ends like so what we're left with is a third m c squared with a bar on top is equal to k times t now let's do a little mathematical trick so we could represent a third because what in the equations we're applying that we're trying to prove there's a factor of three over two but we could just represent this third as uh simply as simply two over thirds times one over two because the twos would uh would just cancel each other out and what we left with is one over one over three so m c squared that's equal to k times t we're basically there so what we're going to get is a half m c squared which is equal to three halves k t and we have successfully proven our equation okay now that we have done this so we derived this equation let's see whether we can apply to a problem so we know the molar mass of area is 29 grams but more estimate how quickly a molecule is moving at 20 degrees c room temperature as soon as i see this in question what i will do is just really quickly you know convert that to kelvin because my equation will only work in kelvin so we need to add 273 so this will be equal to 293 kelvin okay well we're going to need to know the mass of a single particle to estimate c squared from there because we know the temperature we know boltzmann's constant and the only thing that we don't know is m however we have the number of grams per mole so that's going to be 0.029 kilograms per mole however remember one more is avogadro's number 6.02 times 10 to the power of 23. so if we just divide that okay so if one more weighs 29 grams then one single particle will weigh 0.029 divided by 6.02 so 6.02 times 10 to the power of 23 which is going to give us about 4.8 times 10 to the power of minus 26 kilograms so we're expecting quite a small number which is which totally totally finds what exactly what we would expect okay so now that we have the mass what we can do is estimate the we could estimate the the speed squared so so let's do that using the equation that we derived earlier which is a half m c squared with a dash on top that's equal to three halves k t let's rearrange for c squared and this can cancel out so that's going to be equal to three k t and um we're gonna be dividing that by the mass so in other words c squared will be equal to 3 times 1.38 times 10 to the power of minus 23 multiplied by the temperature which was 293 kelvin we're going to divide that by the mass of an individual particle which is 4.8 times 10 to the power of minus 26 now this will give us an estimate for the squared mean speed how we're really interested in the root mean square speed as an estimate for the actual speed so i'm going to square root my answer and if we put that into a calculator we're going to get approximately 500 meters per second as an estimate for how fast particle is moving at room temperature okay folks well thank you very much for watching this very long revision lesson hopefully you found this useful in your revision thanks again for watching and i'll see you in the next video