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Learn Advanced Placement Chemistry: Gases 1 Video
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 Learn Advanced Placement Chemistry: Gases 1 Video
Learn about Advanced Placement Chemistry, Gases 1, in this comprehensive video by bannanaiscool.
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Rob Lederer: To describe ideal gas behavior, we have to understand the kinetic molecular theory postulates in order to do that. Now, here is the first one. Individually, gas molecules are so small that they occupy very little space in the container volume that they are in, so much so, that we just say that their volume is zero, we kind of cheat. So, the volume of the container is the volume of the container and we don't have to worry about the size of the molecules that are involved. Okay, that's the first postulate. Second one would be that we believe that all molecules are moving about and colliding with each other in a random fashion constantly in motion and the pressure is actually determined by them, those molecules reacting or at least bumping into the container walls. Okay, that's the second. Now, the third one would be that, we take for granted although it's not true in all cases that a molecule does not have any type of attracting or repelling force with other molecules in the gas sample in order to treat them ideally, and then the last one, which is kind of obvious is that temperature is proportional to the average kinetic energy of the molecules. So, as the temperature goes up the molecules collide with more vigor and they also move faster. Speeds involve too because kinetic energy and speed have a relationship, don't they? Oh! Here it comes. EK=1/2 mv2. you know that from grade nine physics probably. Okay, did you also know? Probably not. That Ek=3/2 RT. Well that's true for gases where R here however, is not the usual 8.314 Kilopascal liters per Kelvin mole, but actually 8.314 joules per Kelvin more. It's got the same number, but it's just different unit. Now, so if these two both are equal to EK here then these two equal each other Na. What do I mean by that? Hey, listen. If we can calculate use this formula to calculate the about cutting in the mass of a gas molecule and its speed to calculate its EK. Well, if we multiply that by 6.02 two times 1023 will get how much energy is a way or is being exemplified by the collisions of one mole of those molecules. So, hey numbered to Avogadro that's how we write it, that's Avogadro's number and multiplied by 6.02 two time 1023 atoms or molecules per mole, right? So, take that down here that's what we get it here. Hey, I'm isolating for v2 here and so it get this, the three halfs divided one half gives me 3RT/mNa. Hey, mass is in kilograms because that's what it is up here in that EK formula to get your joules here, but joules is kilograms times, meters, squared, divided by second square, kilograms meters squared per second square. So this is kilogram here and that's per mole. When you multiply those two numbers together, you get kilograms per mole. That's pretty close to grams per mole, or it's pretty close to mole or mass it's a molar mass in kilograms per moles instead of grams per mole. Look at this that means then that this if I take the square route of each side, I get now  because a lot of text books have replaced the V velocity with µ okay, but that just stands for route mean square velocity or the average speed of the molecules in a sample equals the square root of three times, R times the temperature divided by the molar mass. Hey that's pretty cool. So that means, if you know the molar mass of a gas and you know its temperature, you can calculate how fast the particles are going on average in that sample. That's very impressive.