Ch 19: Kinetic Theory of Ideal GassesSee all chapters

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Introduction to Ideal Gasses | 28 mins | 0 completed | Learn |

Intro to Kinetic Theory | 21 mins | 0 completed | Learn |

Kinetic Energy and Temperature | 34 mins | 0 completed | Learn |

Speed Distribution of Ideal Gasses | 18 mins | 0 completed | Learn |

Concept #1: Introduction to Kinetic Theory of Gasses

**Transcript**

Hey guys, in this video we're going to introduce the kinetic theory of gases, which really takes a look at the microscopic nature or the atomic nature of ideal gases in order to derive important equations about them like the equation of state like the average pressure on the walls of a container things like that it's all done with kinetic theory, so let's get to it. The ideal gas law takes a macroscopic view of the gases. It makes no relationship between the measurable, the measurable being the pressure the volume, the number sorry and the temperature and the particles, which are either N or capital N depending on which form of the ideal gas law you use kinetic theory takes a microscopic view of gases it does connect those measurables to the particles so we will be able to figure out how to find the temperature by looking at the particles how to find the pressure by looking at the particles there are several assumptions that you need to make in order for kinetic theory to really work. One, the average number of particles in the gas is large sorryy the number of particles in the gas is large that's typically a given you're almost always going to be dealing with gases that have moles in them, gases on the order of moles 5 moles 10 moles whatever a mole is 10 to the 23 particles which is very very large this needs to be true because the kinetic theory is going to take a statistical approach to looking at things and the more number of particles you have the larger the number of particles the better the statistics are going to be. Two, the average distance of particles is large compared to the size of particles this is actually an identical requirement to the 0 particle volume requirement for an ideal gas. Three the particles obey Newtonian mechanics, but move randomly. So if they collide they collide and they obey the momentum conservation if they collide with the wall, they collide with the wall and they obey momentum conservation but aside from that these not interacting particles are going to move in straight lines randomly inside of the gas. Four, the particles only interact through elastic collisions with each other or with the walls of the container and this is exactly the same as the non interacting particle requirement that we have for the ideal gas. Five, particles can collide elastically with the walls of the container of the gas I sort of included that in part five sorry part four and six all gas particles are identical, so that means that in your sample of gas you only have one type atomic hydrogen, molecular hydrogen, neon, Xenon, helium whatever but you don't have a combination. So for instance air which has a bunch of different gases it but it's mostly a combination of molecular nitrogen and molecular oxygen air would not be a good gas to use kinetic theory on because it doesn't obey part six requirement six, and I already talked about the two conditions for an ideal gas non interacting particles particles of 0 volume those are also conditions to apply kinetic theory, so kinetic theory works really really well when applied to ideal gases because the requirements for both overlap. Using kinetic theory we can do many things importantly we can justify the ideal gas law, we can compute the pressure of a gas on the walls of a container based on the average speed of the particles we can compute the average kinetic energy of the gas particles and then we can compute the temperature of the gas based on the average kinetic energy of the gas particles and all of this we can do by analyzing the motion of the particles inside of the gas and this is just a little diagram that shows a bunch of particles that are moving in random directions but one particle in particular that's making a collision with the wall alright and this is sort of a preview to the next thing that we're going to talk about which is how to figure out the pressure that collisions exert on the wall of a container so this particle is moving with an average speed V it has some perpendicular speed some speed perpendicular to the wall when it bounces elastically it's going to leave with the same speed so that change in momentum that impulse is going to be two times the mass of the particles times that perpendicular speed. That wraps up our introduction into the kinetic theory of gases a lot more to come on the kinetic theory later. Alright thanks for watching guys.

Concept #2: Kinetic Theory and Pressure

**Transcript**

Hey guys, in this video we're going to figure out how to calculate the pressure of the gas on the walls of the container that it's in using the kinetic theory, let's get to it remember that kinetic theory can be very useful for describing or explaining ideal gases at the microscopic level OK utilizing kinetic theory what we're going to do now is compute the pressure in terms of the average speed of gas particles. So this is taking a look at the particulate level so an actual feature of a particle the average speed that will end up telling us what the pressure of the gas is first we want to compute the impulse that a single particle receives from a container wall. If it has a perpendicular speed V perpendicular when it bounces elastically off of the wall, right the collision is going to be elastic then it's going to have that same perpendicular speed coming off of the wall if the particle has some random mass little m and some random speed V perpendicular then the momentum transfer that impulse is twice the mass times that perpendicular speed. Next we need to find the number of particles colliding with some area of the wall. Let me minimize myself so what we're taking a look at is we're taking a look at a very specific area of the wall just A, the only particles that are going to collide with the wall are those that are close enough to do so in some amount of time delta T. The furthest distance the furthest perpendicular distance that the particles can travel is this distance, so if we have particles over here moving at V perpendicular and there some extra distance away they're not going to make it to the wall within the time delta T there too far away only the particles within that distance within this distance are going to be close enough. That means the number is just going to be the density of the particles in the entire gas times the volume of this cylinder the cylinder that tells us which particles are close enough to collide with the wall. The density of the ideal gas Avogadro's law tells us is N divided by V and it's a constant. It's just the number divided by the volume and what's the volume of this cylinder well its the surface sorry the face area A times the length V perpendicular delta T. There is one assumption that we didn't make though this isn't quite the answer yet within this cylinder you have to remember that the particles are moving randomly one of the requirements of applying the kinetic theory sorry of applying the kinetic theory is that you always treat the particles as moving randomly so if some particles are moving towards the wall other particles have to be moving away from the wall and if it's truly random half will move towards half will move away so the actual number we have in this cylinder is half of this value so it's one half times the density times the volume of that cylinder so far so good. Now the total impulse that's going to be imparted on these particles is going to be the number of particles that collide in the time delta T times the impulse that each particle undergoes so the total number is one half times N divided by V time's A V perpendicular dela T and the impulse was 2 times M V perpendicular if you look at the figure, the first figure in this video, the impulse delivered to a particle by the wall was twice the mass times the perpendicular speed alright. So if we simplify this we're going to get N. A. M. V perpendicular squared divided by V something that's important to note this is speed. That V is speed this V is volume just so we don't get confused here alright and just in case you're wondering where the one half went it cancelled with the 2. Now what's the force on this surface going to be what's the force on that surface A going to be that we indicated it's going to be the impules delivered the total impules deliver divided by the duration of the impulse delta T Oh sorry I actually missed a delta T here, there should be a delta T multiplyed there. So it's what I wrote there divided by delta T so we just lose the delta T so there's N A M V perpendicular squared divided by V where once again V perpendicular the numerator is a speed V in the denominator is a volume the volume of the gas so the pressure due to these particle collisions is just the force divided by the area the force is being applied to which is the number times the mass times that perpendicular speed divided by the volume and don't forget the number divided by the volume is just the density, let's do an example 6 moles of gaseous hydrogen sorry gaseous molecular hydrogen fill a 0.005 cubic meter container if the pressure of the gas is just atmospheric pressure what is the perpendicular speed of the hydrogen molecules so remember that the pressure from the microscopic viewpoint is the number of particles times the mass of each particle times that perpendicular speed squared divided by the volume. If we want to solve for the perpendicular speed squared first we have to isolate sorry if we want to solve for the perpendicular speed first we have to isolate the perpendicular speed squared which is P V over capital N m and then we just take the square root so the perpendicular speed is the square root of P V over capital N little M. Now the trick with this problem is that they didn't tell us the number of particles they told us there were six moles so first, we have to figure out how many particles there are in 6 moles and if you remember this was our conversion equation Sorry no, this was our conversion equation it's just the number of moles times Avogadro's number and that equals 3.61 times 10 to the 24 particles alright now what we need is we need to find the mass of a molecule hyrdrogen because we don't know that either each hydrogen atom in molecule hyrdrogen there's two of them has one atomic unit of mass.So the mass of molecular hydrogen is going to be 2 atomic units and we need the conversion between atomic units and kilogrammes and it's 1.67 times 10 to the -27 kilogrammes per atomic unit. 1.67 times 10 to the -27 kilogrammes is just the mass of a proton or the mass of 1 hydrogen atom alright and if we multiply this it's 3.34 times 10 to the -27 kilogrammes. So we know the pressure we know the volume we know the number we know the mass so now we can find that perpendicular speed which is going to be atmospheric pressure 1 times 10 to the 5 the volume which we said was 0.005 cubic metres the number which we found was 3.61 times 10 to 24 and the mass which we found was 3.34 times 10 to the -27. All of that square rooted gives us 205 meters per second. So that is the average perpendicular speed of these hydrogen molecules in this gas at this particular pressure contained in a volume sorry contained in a container at this particular volume all of it derived using the kinetic theory alright no ideal gas law is used here. Alright guys that wraps up the kinetic theory in the pressure of gases. Thanks for watching.

Practice: A sample of 5 moles of and ideal atomic hydrogen gas is contained in a cylinder of 5 cm radius and 20 cm height. If the gas is moving at a perpendicular speed of 150 m/s, what is the force on the top face of the container? The pressure?

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