Average Kinetic Energy of Gases - Online Tutor, Practice Problems & Exam Prep

Introduction to Kinetic Theory of Gasses

Hey, guys. So you may run across some problems that ask you to calculate something called the average kinetic energy of an ideal gas. So what I'm going to show you in this video is the equation for that, and we're going to see that this equation is actually one of the equations that make up what's called the kinetic molecular theory. So I'm going to 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 gases, 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 that'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 going to 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 three halves, k_B times T. So just be careful here because you're going to have two k's, the capital one is for the energy, the lowercase one is for the Boltzmann constant, that's that k_B, and then your temperature has to be in Kelvins. But this is it, this is the equation, it's very straightforward. 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 going to calculate the average kinetic energy of oxygen molecules, if the gas is at 27 degrees Celsius. So in part a, all we have to do is just use our new equation. This is going to be pretty straightforward, just going to plug and chug. So this is our kinetic energy, this is going to be three halves. This is going to be the Boltzmann 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 Kelvins. The reason for that is that if you plugged 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 Kelvins. So this is going to be, 273 plus 27, and you're going to get 300 Kelvin, and that's what we plug into this temperature over here. Alright. So this is just going to be equal to three halves, and this is going to be 1.38 × 10^-2323-23, and then we're going to multiply this by 300. If you work this out, what you're going to get is 6.21 × 10^-2121-21 joules. So if you have this gas here at 300 Kelvins, then obviously some of the molecules are going to have more energy, some will have less energy. But if you average them all, this is actually the energy that you're going to 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 that 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 had nitrogen at 300 Kelvin, it would also have this amount of energy per particle. It would be the same exact numbers. The answer here is no. Alright, so that's it for this one, guys. Let's move on.

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

Gas in a Balloon

Hey, everybody. So let's work on this problem together here. I've got a spherical balloon that I'm told some information about. I'm going to just draw this out real quickly. I've got this spherical balloon and we're told that the volume is 43 meters cubed, and the pressure is 1.2 atmospheres. Now remember, whenever we are given units of atmospheres, we can always convert this to pascals 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 much gas we have. So the variable for that, remember, is going to be little n. So which equation do we start off with? Well, remember we have one equation involving n. It's going to be PV= nRT. Alright? So let's go ahead and start off there. So we've got PV=nRT. 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/RT=n. Alright? So now what I'm going to do is just 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 going to have to convert this really quickly here. So this 1.2 atmospheres, I can use this conversion factor to get it in terms of pascals. Basically what I'm going to do is I want to divide by units of the atmospheres on the bottom so that it cancels. Well, the conversion factor is 1 atmosphere and this is 1.015 Pascals. So this unit will cancel. What you'll end up with here is 1.215 Pascals. Alright? So that's just the number that I'm going to plug in here, 1.215 Pascals. Now I've got the volume, which is 4-3. Now I'm going to 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 going to 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. And remember, this equation here for k average is related to the temperature. So that's how we figure out what T is equal to. Alright? So basically, I'm going to go over here. So I've got that k average is equal to 32kbT. 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 going to do. We're going to do this k average, and then I'm going to just divide by this other stuff over here to get the temperature. So this is going to be 7.2×10-21 divided by, this is going to be 32, and this is going to be 1.38×10-23. That's just the Boltzmann constant, which is listed right over there. That's equal to the temperature, and if you work this out, what you're going to get here is you're going to get 347.8 Kelvin. Alright? So this is the number here that we plug in for this over here. So now we're just going to 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 going to get is 167 moles. Alright. So that's it for this one, guys. Let me know if you have any questions.