Average Kinetic Energy of Gases - Video Tutorials & Practice Problems

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1

concept

Introduction to Kinetic Theory of Gasses

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3m

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Hey guys, so you may run across some problems. They ask you to calculate something called the average kinetic energy of an ideal gas. So what I'm gonna show you in this video is the equation for that and we're gonna see that this equation is actually one of the equations that make up what's called the kinetic molecular theory. So I'm gonna break it down for you, show you the equation and we'll do a quick example. Let's check this out. So remember that the kinetic molecular theory is really just a set of equations and they connect the variables for ideal gasses, macroscopic variables like pressure, volume and temperature to the other microscopic variables for individual particles, like velocities and energies and things like that. Now remember back to one of our earliest discussions on thermodynamics, we talked about temperature. We said that one of the definitions for temperature is a measure of how hot or cold something is, but it's not very useful. A more important definition or more sort of useful one is that it's related to the average kinetic energy more specifically the average kinetic energy per particle. So, in this video, I'm actually gonna show you now the equation that describes that relationship and it's that the average kinetic energy given by the capital letter K Is equal to 3/2ves KB times T. So, just be careful here because you're gonna have to kes the capital one is for the energy, the lower case one is for the bolts men constant. That's that KB and then your temperature has to be in kelvin's but this is it this is the equation. It's very straightforward and it's actually pretty fascinating because it tells us that if we know the temperature of a gas, it tells us the average amount of energy per particle Of that gas, which is pretty awesome. So let's go ahead and just take a look at our example here. So, we're gonna calculate the average kinetic energy of oxygen molecules if the gas is at 27°C. So, in part a all we have to do is just use our new equation. This is gonna be pretty straightforward, just gonna plug and chug. So this is our kinetic energy. This is gonna be three halves. This is gonna be the bolts one constant times the temperature. Now, just very quickly here. Remember that this temperature is given to us in Celsius and it has to be in kelvin's. The reason for that is that if you plug a negative number here, right, you can have negative Celsius, then you're going to get a negative energy. And that doesn't make any sense. So, we have to first convert this to kelvin's. So this is gonna be 2 73 plus 27. And you're gonna get 300 Kelvin and that's what we plug into this temperature over here. All right, so this is just gonna be equal to three halves. And this is gonna be 1.38 times 10 to the minus 23 then we're gonna multiply this by 300. If you work the sandwich you're gonna get is 6.21 times 10 to the minus 21. And that's gonna be in jewels. So if you have this gas here at 300 kelvin's, then obviously some of the molecules are going to have more energy, so we'll have less energy. But if you average them all, this is actually the energy that you're gonna get per particle. Alright, so now let's take a look at the second problem here. The second question is a more conceptual one that asks, would the answer be any different if we had a different type of gas? If it were nitrogen instead of oxygen. So if we just look at the equation, we can see the equation only. Just depends on a constant and temperature and nowhere in the equation do we actually have the type of gas, like the mass or something like that? So one important conceptual point to know about this average kinetic energy is that it depends only on the temperature of the gas And actually not the type of gas. So if you have nitrogen at 300 Kelvin, it would also have this amount of energy per particle would be the same exact number. So, the answer here is no. Alright, so that's it for this one. Guys, let's move on

2

Problem

Problem

In a sample of gas, you pick a particle at random. The mass of the particle is 1.67 × 10^{-27} kg and you measure its speed to be 1600 m/s. If that particle's kinetic energy is equal to the average kinetic energy of the gas particles, what is the temperature of the sample of gas?

A

232 K

B

0.065 K

C

103.3 K

D

206.5 K

3

example

Gas in a Balloon

Video duration:

3m

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Hey everybody. So let's work on this problem together here. So I've got a spherical balloon that I'm told some information about. So I'm gonna just draw this out real quickly. So I've got this spherical balloon and we're told that the volume is four times 10 to the third meters cubed and the pressure is 1.2 atmospheres. Now, remember whenever we're given units of atmospheres, we can always convert this to pascal's by using this conversion factor over here. Now we're told here also that the average kinetic energy of the particles inside the balloon is this number over here and ultimately we want to figure out is the number of moles of gas inside the balloon, basically how many how much gas do we have? So the variable for that number is going to be little end. So which equation do we start off with? Well remember we have one equation involving and it's going to be PV equals NRT. Alright, so let's go ahead and start off there. So we've got PV equals N. R. T. So if you want to figure out N then we just have to move everything over to the other side and we've seen this before. So basically what happens is you end up with PV divided by R. T. Is equal to N. Alright, so now what I'm gonna do is just gonna start plugging in some numbers here. So the pressure before I plug it in, I can't plug it in as 1.2. I'm gonna have to convert this really quickly here. So this 1.2 atmospheres, I can use this conversion factor to get it in terms of pascal's basically what I'm gonna do is I want to divide by units of atmospheres on the bottom so that it cancels. Well, the conversion factor is one atmosphere and this is 1.1 times 10 to the fifth pascal's. So this unit will cancel what you'll end up with here is 21.21 times 10 to the fifth pascal's. Alright, so, that's just the number that I'm gonna plug in here. 1.21 Times 10 to the 5th. Now, I've got the volume which is four times 10 to the minus three. Now, I'm gonna divide by the R which is the gas constant 8.314. And now, finally, I'm going to look at the temperature. So, what is the temperature? Well, actually, I'm not really told in this problem what the temperature is. So, I can't just go ahead right away and plug it in. So, I'm gonna have to figure this out. The only other information that I know about this problem is that the average kinetic energy is just this number over here. So, basically what they're giving me in this problem is they're actually giving me K average. Remember this equation here for k average is related to the temperature. So, that's how we figure out what T is equal to. All right. So, basically, I'm gonna go over here. So, I've got that K average is equal to three halves and this is gonna be K. B. T. Alright, so basically this just says that the average kinetic energy is related to the temperature. So that's the sort of relationship. But that's the link of how we get the temperature. Okay? So here's what we're gonna do, we're gonna do the uh this K average and I'm gonna just divide by this other stuff over here to get the temperature. So this is gonna be 7.2 times 10 to the minus 21 divided by uh This is gonna be three halves. And this is gonna be 1.38 times 10 to the minus 23. That's just the bolts, one constant, which is list right over there, That's equal to the temperature. And if you work this out, what you're gonna get here is you're gonna get 347.8 Kelvin. Alright, so this is the number here that we plug in for this over here. So now we're just gonna multiply by 347.8 or sorry, divide And that's going to give you your final answer for moles. And if you work this out, what you're gonna get is 1 0.1 Six moles. Actually it thinks this is a this is going to be seven or 168 moles. Alright, so that's it for this one. Guys, let me know if you have any questions

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