- Thermodynamics part 1: Molecular theory of gases
- Thermodynamics part 2: Ideal gas law
- Thermodynamics part 3: Kelvin scale and Ideal gas law example
- Thermodynamics part 4: Moles and the ideal gas law
- Thermodynamics part 5: Molar ideal gas law problem
- What is the ideal gas law?
- The Maxwell–Boltzmann distribution
- What is the Maxwell-Boltzmann distribution?
Intuition of how gases generate pressure in a container and why pressure x volume is proportional to the combined kinetic energy of the molecules in the volume. Created by Sal Khan.
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- I still didn't get why P.V is a constant. I mean I got what he said in video, but why is it a constant?(48 votes)
- While it is the case that the "constant" K in the video really does depend on the number of molecules in the gas and the absolute temperature, one can still easily understand why the product, PV, will be constant if the number of molecules and their average kinetic energy (measured by the temperature) doesn't change. Imagine inserting a partition in the volume, V, which just divides the volume into two equal parts of volume V/2 each. Because the N molecules will be uniformly distributed through the volume, V, before insertion of the partition, the insertion will result in N/2 molecules, on average, in each of the two subdivisions. But each of these two sets of N/2 molecules were exerting a pressure, P, on the walls of their separate subdivisions before the insertion and they will continue to do so after the insertion. Now imagine what would happen if the N/2 molecules in one of the subdivisions were now placed in the other subdivision. Clearly both sets of N/2 molecules would exert the pressure, P, on the walls of the V/2 volume they now share. Consequently, the total pressure would now be 2P. But the total occupied volume would now be V/2, and so the product (2P)(V/2) = PV, just as before. The argument generalizes to inserting (n - 1) partitions to create n subdivisions of volume, V/n, each, containing (on average) N/n molecules each generating equal pressure, P, in each subdivision. Now put all the molecules in just one of the subdivisions to produce a total pressure of nP in that one crowded subdivision. This time (nP)(V/n) = PV. Finally, consider that N/n of the molecules are contributing P/n to the pressure in the original volume,V. Now expand the volume to nV and realize that then only N/n molecules occupy each volume, V, and generate the pressure, P/n, within that volume as they did before the expansion. But then the pressure throughout the volume, nV, will just be that common pressure for each volume, V, P/n. So now we have (P/n)(nV) = PV.(1 vote)
- How is the momentum of the particles changing if in an ideal gas all collisions are inelastic and momentum is conserved? Thank you in advance.(19 votes)
- The collisions are ELASTIC. But it is the total momentum of the entire system that is conserved, not that of each particle. Conservation of momentum only applies to systems of particles, not individuals.
However, in a closed system momentum is conserved for both elastic and inelastic collisions.(23 votes)
- When Sal writes the formula for Force as "Change in momentum over Change in time", why is "momentum" P, and later, Pressure is also P? Are they both correct? Doesn't this get confusing?(5 votes)
- another fundamental question, if i heat up a liquid then the volume increases and the density decreases. If the volume is fixed, for example a liquid is in a bottle and I heat it up and the volume can't get even bigger so the pressure increases instead of the volume. Do the density decreases too in a fixed volume? I have a problem by imagineing this.
Thank's for help.(8 votes)
- Density is mass divided by volume so since the mass doesn't change if the volume doesn't change then the density doesn't change either.(15 votes)
- At2:56, why and how does the particle change its momentum after bouncing off the wall of the container ?(2 votes)
- The momentum changes because the particle is suffering an impulse as it collides. At this model, every collision is elastic, which makes sense, since in the macroscopical world non-elastic collisions lose energy in the form of heat and sound, two things that a microscopical collision can't create, by definition. Thus, when the particle hits the wall, it simply bounces back.(4 votes)
- is PV=k related to the ideal gas law PV=nRT?(1 vote)
- If CO2 is one of the main components of smoke and it is heavier than air, then why when there is a fire, people try to crawl on the floor so as not to suffocate in the smoke?
Won't you actually suffocate more quickly if you are down because its concentration near the floor will be greater than in then surrounding air?(2 votes)
- CO2 is not one of the main components of smoke. Also, while it may be present in smoke, it is not nearly as deadly for us to breathe as the other, major components, like carbon monoxide and other particulates, which are lighter than air and rise.(3 votes)
- How does this relate to the Venturi Effect? If Ihave a large diameter pipe flowing into another smaller pipe Bernoulli says the pressure will be greater in the larger pipe than in the smaller and the speed in the smaller pipe will increase and the pressure will reduce. This video however says larger volume smaller pressure. Does this mean only for a closed system ie. no flow in or out?(2 votes)
The video is talking about a gas in which there is ZERO BULK MOTION. in other words there is no flow of gas as a large body. The only kinetic energy in this ideal situation is the microscopic kinetic energy of the molecules.
The velocities are so many and so varied that they average out to zero
For the Bernoulli effect to occur, there needs to be a flow of gas. The flow of air outside an airplane for example: the pitot tube (small tube near the cockpit) measures the change in pressure due to change in air speed.
Your example of the venturi effect (in a carburetor) is another example(3 votes)
- i have heard that the momentum of the whole system is conserved but the momentum of individual particle is not conserved..........
i dont get it actually......
does it mean the product of mass and velocity of all the particles in the system remains constant.. but for and individual particle it is different? correct me if i am wrong...(1 vote)
- the law of conservation of matter and momentum says that momentum cannot be created or destroyed so in the case of collision between particles. the total momentum prior to the collision and after the collision will remain constant. the momentum is just exchanged between particles.(3 votes)
- at9:22you said that particle wont even hit the wall . hows that possible ,it always has velocity(2 votes)
- He says that it won't hit the wall in the same amount of time, meaning that it requires more time for the particle to get to the other wall when the walls are farther apart.(2 votes)
After all the work we've been doing with fluids, you probably have a pretty good sense of what pressure is. Now let's think a little bit about what it really means, especially when we think about it in terms of a gas in a volume. Remember, what was the difference between a gas and a liquid? They're both fluids, they both take the shape of their containers, but a gas is compressible, while a liquid is incompressible. Let's start focusing on gases. Let's say I have a container, and I have a bunch of gas in it. What is a gas made of? It's just made up of a whole bunch of the molecules of the gas itself, and I'll draw each of the molecules with a little dot-- it's just going to have a bunch of molecules in it. There's many, many, many more than what I've drawn, but that's indicative, and they'll all be going in random directions-- this one might be going really fast in that direction, and that one might be going a little bit slower in that direction. They all have their own little velocity vectors, and they're always constantly bumping into each other, and bumping into the sides of the container, and ricocheting here and there and changing velocity. In general, especially at this level of physics, we assume that this is an ideal gas, that all of the bumps that occur, there's no loss of energy. Or essentially that they're all elastic bumps between the different molecules. There's no loss of momentum. Let's keep that in mind, and everything you're going to see in high school and on the AP test is going to deal with ideal gases. Let's think about what pressure means in this context. A lot of what we think about pressure is something pushing on an area. If we think about pressure here-- let's pick an arbitrary area. Let's take this side. Let's take this surface of its container. Where's the pressure going to be generated onto this surface? It's going to be generated by just the millions and billions and trillions of little bumps every time-- let me draw a side view. If this is the side view of the container, that same side, every second there's always these little molecules of gas moving around. If we pick an arbitrary period of time, they're always ricocheting off of the side. We're looking at time over a super-small fraction of time. And over that period of time, this one might end up here, this one maybe bumped into it right after it ricocheted and came here, this one changes momentum and goes like that. This one might have already been going in that direction, and that one might ricochet. But what's happening is, at any given moment, since there's so many molecules, there's always going to be some molecules that are bumping into the side of the wall. When they bump, they have a change in momentum. All force is change in momentum over time. What I'm saying is that in any interval of time, over any period or any change in time, there's just going to be a bunch of particles that are changing their momentum on the side of this wall. That is going to generate force, and so if we think about how many on average-- because it's hard to keep track of each particle individually, and when we did kinematics and stuff, we'd keep track of the individual object at play. But when we're dealing with gases and things on a macro level, you can't keep track of any individual one, unless you have some kind of unbelievable supercomputer. We can say, on average, this many particles are changing momentum on this wall in this amount of time. And so the force exerted on this wall or this surface is going to be x. If we know what that force is, and we you know the area of the wall, we can figure out pressure, because pressure is equal to force divided by area. What does this help us with? I wanted to give you that intuition first, and now I'm just going to give you the one formula that you really just need to know in thermodynamics. And then as we go into the next few videos, I'll prove to you why it works, and hopefully give you more of an intuition. Now you understand, hopefully, what pressure means in the context of a gas in a container. With that out of the way, let me give you a formula. I hope by the end of this video you have the intuition for why this formula works. In general, if I have an ideal gas in a container, the pressure exerted on the gas-- on the side of the container, or actually even at any point within the gas, because it will all become homogeneous at some point-- and we'll talk about entropy in future videos-- but the pressure in the container and on its surface, times the volume of the container, is equal to some constant. We'll see in future videos that that constant is actually proportional to the average kinetic energy of the molecules bouncing around. That should make sense to you. If the molecules were moving around a lot faster, then you would have more kinetic energy, and then they would be changing momentum on the sides of the surface a lot more, so you would have more pressure. Let's see if we can get a little bit more intuition onto why pressure times volume is a constant. Let's say I have a container now, and it's got a bunch of molecules of gas in it. Just like I showed you in that last bit right before I erased, these are bouncing off of the sides at a certain rate. Each of the molecules might have a different kinetic energy-- it's always changing, because they're always transferring momentum to each other. But on average, they all have a given kinetic energy, they keep bumping at a certain rate into the wall, and that determines the pressure. What happens if I were able to squeeze the box, and if I were able to decrease the volume of the box? I just take that same box with the same number of molecules in it, but I squeeze. I make the volume of the box smaller-- what's going to happen? I have the same number of molecules in there, with the same kinetic energy, and on average, they're moving with the same velocities. So now what's going to happen? They're going to be hitting the sides more often-- at the same time here that this particle went bam, bam, now it could go bam, bam, bam. They're going to be hitting the sides more often, so you're going to have more changes in momentum, and so you're actually going to have each particle exert more force on each surface. Because it's going to be hitting them more often in a given amount of time. The surfaces themselves are smaller. You have more force on a surface, and on a smaller surface, you're going to have higher pressure. Hopefully, that gives you an intuition that if I had some amount of pressure in this situation-- if I squeeze the volume, the pressure increases. Another intuition-- if I have a balloon, what blows up a balloon? It's the internal air pressure of the helium, or your own exhales that you put into the balloon. The more and more you try to squeeze a balloon-- if you squeeze it from all directions, it gets harder and harder to do it, and that's because the pressure within the balloon increases as you decrease the volume. If volume goes down, pressure goes up, and that makes sense. That follows that when they multiply each other, you have to have a constant. Let's take the same example again, and what happens if you make the volume bigger? Let's say I have-- it's huge like that, and I should have done it more proportionally, but I think you get the idea. You have the same number of particles, and if I had a particle here, in some period of time it could have gone bam, bam, bam-- it could have hit the walls twice. Now, in this situation, with larger walls, it might just go bam, and in that same amount of time, it will maybe get here and won't even hit the other wall. The particles, on average, are going to be colliding with the wall less often, and the walls are going to have a larger area, as well. So in this case, when our volume goes up, the average pressure or the pressure in the container goes down. Hopefully, that gives you a little intuition, and so you'll never forget that pressure times volume is constant. And then we can use that to do some pretty common problems, which I'll do in the next video. I'm about to run out of time. See you soon.