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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: Average Kinetic Energy and Temperature

The correct answer is 48 K.

**Transcript**

Hey guys, in this video we want to use kinetic theory to relate the average kinetic energy and the temperature which is something that we've been saying are related the entire time in our discussion of thermodynamics. Alright let's get to it. We can easily convert our equation for the pressure of a gas found by the kinetic theory by analysing collisions with the container wall into an equation of state. Remember that an equation of state just relates the pressure and the volume for an ideal gas, that's all that determines the state of an ideal gas and if you have your notes look at the pressure that we solved for in the video on kinetic theory and pressure and you'll see that this is the equation of state that you get. Something important to consider here is that when we calculated the impulse of a single particle along our process to find the pressure, we made one crucial assumption that we assume that each particle individually moved with the same perpendicular speed which is not a very good assumption because the kinetic theory at its basis is supposed to assume that these particles act randomly. A much better approach is to consider the average equation of states, the averaged PV for each particle then what we would do is we were to say that this is NMV perpendicular squared average. The number of particles, there's no average number of particles, the gas just has a number of particles so that's not included in the average. The mass of each particle remember one of the assumptions the last assumption of the kinetic theory was that the gas particles are all identical. You had a single gas so there is no average mass there's just the mass of each individual particle but the average speed squared is susceptible to being averaged and the order of this is really important this is the squared that's averaged, it's not the average of the square. It's not the average perpendicular speed squared, it's not that, it's this average of the square. Now something that's important is in three dimensional motion we always have this relationship because we can break up any total speed V squared into the components of the speed, the X, Y and Z components, using the Pythagorean theorem because the directions are arbitrary we can assign any direction to be the X direction or the Y direction or the Z direction. Just for convenience we'll say that the perpendicular direction happened to be the X direction so in this case if this is our wall and our particles are moving with their perpendicular speeds towards the wall I'm going to consider the direction towards the wall to be the X direction because those directions are arbitrarily chosen it doesn't matter if I call that the X direction. Now in truly random motion which, remember, is one of the fundamental assumptions of the kinetic theory the average speed should be isotropic. This should say isotropic not isotropically meaning it's the same in all directions. That the average speed should be just repeated the same in all directions. This is because particles are produced moving in every single direction equally. That is what truly randomness means. So if you plug this relationship into our equation for the average of the speed squared, you get that the average of the square is 1 third of the average of the square of the total speed. That's because you would say V squared average would be equivalent all 3 of these would be replaced by VX squared average. So there would be 3 of them, so if you divide the 3 over you get 1 third. Moving on. Now we're going to rewrite our equation of state to include the fact that we are to sorry replace that average perpendicular speed with 1 third of the average total speed so this is going to be 1 third NMV squared. And now we can use a little algebraic trick here to get the result to look a very specific way. Alright I'm going to pull a 2 out and that means I have to put a 1 half inside that 2 and that 1 half will cancel and will stay 1 third but this becomes, oh this by the way is average, MV squared average. Notice this though, what does this look like? This looks exactly like a kinetic energy. This is actually the average kinetic energy. For the average kinetic energy you don't use the average of the speed you actually use the average of the square of the speed. If I take the kinetic energy and I just average it 1 half doesn't average, it stays 1 half. The mass doesn't average because all the particles are identical the only thing that averages is V squared and it's very very important the order. Remember that this is not the average squared it's not that. It's not the square of the average, it's the average of the square. Substituting this into our equation of state we can write PV in terms of the average kinetic energy. This becomes 2 thirds N times that average kinetic energy or 3 halves PV over N is the average kinetic energy. And now if we use the ideal gas law. Where PV equals NKT and we solve for PV over N we can substitute PV over N for KT and we can write the average kinetic energy in terms of the temperature which is exactly what we're looking for and this is exactly the claim that we have been making this entire time, that we could relate the temperature to the average kinetic energy and it's just 3 halves, the Boltzmann constant times the temperature and that is the relationship for an ideal gas between the average kinetic energy and the temperature. Alright it's a very important relationship and it's one that we've been talking about since the beginning of thermodynamics but now we finally have the mathematical relationship between them. A gas of molecular nitrogen has an average kinetic energy per 100 molecules of 1 times 10 to the -19 joules. What's the temperature of the gas? It's average kinetic energy per 100 molecules so the average kinetic energy per 100 molecules is 1 times 10 to the -19 joules so each molecule has 1 100th of that total kinetic energy So that means that we have to drop that 10 to the -19 down by another 100th which is 2 decimal points. So this is 10 to the -21 joules alright and from our relationship between the average kinetic energy and the temperature of the gas, we can solve for that temperature which is exactly what the question is asking for. So the temperature is just 2 thirds, 1 times 10 to the -21, which is the kinetic energy per individual molecule not per 100 molecules divided by the Boltzmann constant which is 1.38 times 10 to the -23 and this whole thing comes out to 4831 Kelvin. So pretty hot, that's pretty hot like 2 thirds the temperature of the Sun. Alright guys that wraps up the relationship between kinetic energy and temperature we finally have a mathematical relationship and it came to us by analysing an ideal gas with kinetic theory. Alright thanks for watching guys.

Example #1: Atmosphere on the Planet X

K_{TH}=6.31e-21J

**Transcript**

Hey guys, let's do an example. Air has an average mass of 4.8 time 10 to the -6 kilograms per particle. If some planet X has an average surface temperature of 305 Kelvin could planet X support an atmosphere of air according to thermodynamics? Note that the mass of the planet is this number and the radius of the planet is this number. Hint: what according to thermodynamics means is check to see if the air on planet X would have enough thermal energy to escape gravity. What does it mean to escape gravity? Well here's the surface of X where X has some radius R of X and there's a particle that's going to have some kinetic energy. So at the surface of X it has some initial kinetic energy and it has some initial potential energy due to the gravitational interaction with planet X. Then what's going to happen is does that molecule escape? Well in order to know if it escapes, here's X, It has to get very, very, very far away, here's our molecule, from X. So far that there is no interaction. No interaction means that that final potential energy is zero. Infinitely far away, so far there is no gravitational interaction. Now in order to just have enough kinetic energy to escape, the kinetic energy final would also be zero. If you had more kinetic energy left over that means that when you started you had more than enough kinetic energy to overcome the gravitational attraction. If you end with zero it means you had just enough to overcome the gravitational interaction and if you have less than that, that means you don't have enough to overcome gravity and you're going to stop short when you run out of kinetic energy before you get infinitely far away, before you escape that interaction. So basically what kinetic energy can what energy conservation is going to say is that the initial kinetic energy plus the initial potential energy is going to equal zero. Normally it would be equal to the final kinetic and the final potential but as we said both of those are zero. So the initial kinetic energy, the gravitational potential energy is negative gravitational constant times the mass of the planet times the mass of your little gas molecule divided by the radius of the planet. So we can say that our initial kinetic energy just has to equal G capital MX little M over RX. If the kinetic energy equals at least this, it can escape gravity if it's greater than this it can escape gravity and then keep going forever if it's less than this it'll run out of kinetic energy short of escaping gravity. Now the gravitational potential energy we can easily solve for. This is 6.67 times 10 to the -11, we were told the mass of planet X 4.23 times 10 to the 23, we were told the mass of our air molecule 41. times 10 to the -26 and we were told the radius of planet X 1.25 times 10 to the 7 meters. So all of this tells us there our initial kinetic energy has to be at least 1.09 times 10 to the -19 jewels OK. So now the question is what is our thermal energy? That was the hint. The hint said check to see if air on planet X would have enough thermal energy to escape gravity. So we measure the thermal energy and if it's greater that this minimum escape kinetic energy then planet X can't have an atmosphere. If that thermal energy is less then planet X can have an atmosphere at least as far as thermodynamics is concerned. So the thermal kinetic energy is 3 halves K times the temperature. We were told that the average surface temperature was 305 five Kelvin, the Boltzmann constant is 1.38 times 10 to the -23 times 305 Kelvin. Which is going to equal 2.1 times 10 to the -21 joules. So that thermal kinetic energy is less than the required escape kinetic energy. That means atmosphere is possible at least according to thermodynamics an atmosphere of air on planet X is absolutely possible because the air molecules do not have enough thermal energy due to the surface temperature to escape so they would stay close to the planet and produce an atmosphere. Alright guys that wraps up this problem. Thanks for watching.

Concept #2: Kinetic Theory and RMS Speed

**Transcript**

Hey guys, in this video we're going to talk about the kinetic theory and a particular measurement called the RMS speed. So we need to figure out what the RMS speed is, how the kinetic theory can tell it to us and why it's useful. Let's get to it. Remember guys that our relationship between the average kinetic energy and the temperature is 3 halves KT. This we found in a previous video on the kinetic theory and temperature. Recall that the definition of the average kinetic energy is just 1 half times M times the average of the square of the speed. Not the square of the average right it's never the square of the average, it's the average of the square and those 2 are totally different. So we can relate this temperature conversion of kinetic energy into the definition of average kinetic energy, we can equate those 2 and we get this. That allows us to solve for the average of the square of the speed which is just 3 KT over M. There's an associated quantity to this average of the square called the root mean squared speed or the RMS speed and we can find it just by taking the square root of this value. So the RMS speed which is the square root of the average of the square is the square root of 3 KT over M. So this is what the RMS speed is. Now what exactly does the RMS speed mean? What does the root mean squared mean? It actually means the root of the mean of the squared. What that means is in the order of operations first we're taking the root of the mean of the squared. And you can find an RMS value for anything. The algorithm defining an RMS value is pretty easy. Start with X, your value, then you square it then you average the square and then you take the square root of the average of the square. Now RMS values are sometimes as good as it's going to get when it comes to some kind of average or statistical value. Some values, some variables, you can't find an average or the average is always zero and the RMS's then gonna become the closest possible thing to some sort of statistical average. The RMS is not the average, it's different but it's similar in the statistical applications. Let's see an example. Consider the graph above me where some value X is actually oscillating. It's oscillating above and below the horizontal axis. If this is a periodic function then the average of the X is just going to be zero and it's always going to be zero because it spins just as much time above the horizontal axis as it does below the horizontal axis. Whatever time it spends above, it spends below. Whatever time it spends above, it spends below. So that average is always going to be equal to zero. So you have no average X if X is a periodic variable or an oscillating variable but if you square X like we have up here now all of a sudden X squared is never negative. X can be negative but X squared can't. You can't have a square that results in a negative number. So X squared spends all of its time bouncing above the horizontal axis and it wouldn't take the average of X squared it is absolutely not equal to zero. Since the average of the square is not zero than the RMS value which is just the square root of this will also not be zero. So RMS values are incredibly important in oscillating functions. RMS values are not as important when talking about ideal gases because we can in fact find an average speed and the average speed will not be zero. The RMS value we use a lot because it's actually really really easy to find since we have that relationship between temperature and kinetic energy it's super easy to find the RMS speed it just comes out naturally. So let's do a couple examples. Gaseous atomic hydrogen has an RMS speed of 3 and a half kilometers per second. What's the temperature of this hydrogen gas? So we know the RMS speed is the square root of 3KT over M. If we square both sides, then the square of the RMS is 3KT over M and if we want to isolate T to solve for the temperature then we get MVRMS squared over 3 times the Boltzmann constant. Before we can plug everything in we need to know the mass of atomic hydrogen. Remember that atomic hydrogen is only 1 hydrogen atom which has a mass of 1 atomic mass unit and 1 atomic mass unit is 1.67 times 10 to the -27 kilograms. So we don't really have to do any converting if it was like molecular nitrogen which is N2 that has 28 atomic units, that we would have to actually multiply out to convert but if it's 1 atomic unit we know for every 1 atomic unit it's 1.67 times 10 to the -27. This is 3 and a half kilometers per second we need this in meters per second so it's 3500 meters per second and that thing is squared divided by 3 times the Boltzmann constant which is 1.38 times 10 to the -23 and so the temperature ends up being 494 Kelvin. And that's how you take an RMS speed and you find the temperature from it. Let's do one more example before wrapping up here. The gas is heated to 40 degrees Celsius. To what temperature must it be heated to double its RMS speed? Remember from the equation that the RMS speed is proportional to the square root of the temperature. So if we want to double the RMS speed, what do we have to do to the temperature? Do we double the temperature? No. Notice that according to this equation, doubling the temperature means that you take the square root of 2 times the temperature which is the square root of 2. That would lead to the square root of 2 because you have to take the square root. You actually need to multiply it by 4 because the square root of 4 times the temperature is going to lead to the square root of 4 times VRMS which is 2VRMS so to double the speed you have to quadruple the temperature. So the question is what is quadruple the temperature? Our new temperature which I called T prime this 4 times the temperature. What is quadruple the temperature though? Can we use forty degrees Celsius? No you can never use Celsius and Fahrenheit for standalone temperatures like this, you can only use Celsius in exchange with Kelvin for differences in temperature. So the first thing we need to do before even plugging anything in is convert this to Kelvin. This is 313 Kelvin. Now the new temperature is going to be 4 time that temperature. So this is going to be 4 times 313 Kelvin which turns out to be 1252 Kelvin. Since our original temperature was given in Celsius, we should probably give our resulted temperature in Celsius too. That means T prime in degrees Celsius is 1252 minus 273. Which is 979 degrees Celsius. That is what the new temperature has to be in degrees Celsius in order to quadruple the RMS speed. Alright guys this wraps up our discussion on kinetic theory and more importantly the RMS speed and how it relates to the temperature. Thanks for watching guys.

Practice: Deutrons, ^{2}H, are popular particles to produce plasmas with. A plasma can be thought of as an incredibly hot gas, with particles moving at incredibly large speeds. What temperature would a deutron plasma have to be so that the rms speed of the deuteron particles was 10% the speed of light? Note that the speed of light is 3.0 x 10^{8} m/s.

Example #2: Unknown Temperatures of Ideal Gasses

**Transcript**

Hey guys, let's do an example. Four different samples of ideal gases are kept in four different containers, A through D. The mass and the RMS speed of the gas in each container is shown in the following table. If the temperature of the gas in container A is 300 Kelvin, what is the temperature in containers B, C, and D? Assume that each container is insulated from the environment and each other. So pretty simple let's start with our equation for the RMS speed. And let's solve for the temperature. First we'll square. This is going to be 3KT over M, then we'll isolate the temperature over 3K and now if we consider M and VRMS to be the speed and mass in A, this equals 300 Kelvin. So let's look at B. The temperature for B is going to be 3 times the mass but the same RMS speed. So this is clearly just 3 times the temperature in A. Compare this equation to this equation, this question just has an extra 3, this is 3 times the temperature in A so it's 900 Kelvin. That's one answer. What about C, the temperature in C? C has 4 times the mass and twice the RMS speed so we're going to get 4 times the mass and 2 times the RMS speed but remember that the RMS speed is squared. That square has to go on the outside. So what factor do we have that's extra? We have a 4 here and we have a 2 here but the 2 is squared. So that 2 squared is 4 times the other 4 is 16 times that original temperature which is forty 48000 Kelvin. And lastly what's the temperature in box D? The mass is 2 times M and the speed is 4 times the RMS speed so we get a 2M here and we get a 4RMS. Once again the square applies to both the 4 and the VRMS. So we have a 2 here and we have a 4 here but that 4 is squared, what's 4 squared? 16. What's sixteen times 2? 32, times the original temperature which is just going to be 96000 Kelvin. Pretty straightforward problem, just an application of the equation for the RMS speed. Alright guys, that wraps up this problem. Thanks for watching.

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Suppose we have two monoatomic gasses, neon and argon. We cool the neon gas to 200 K in a 5 m 3 room, while the argon gas it put into a 0.25 m 3 balloon and left outside, where the temperature is 27 oC. Assuming the room is at atmospheric pressure, 1x10 5 Pa, if the balloon is brought into the room filled with argon and shrinks as it reaches equilibrium,
(a) What is the initial, average energy per particle of each gas?
(b) What is the final, average energy per particle of each gas?
(c) What is the final temperature of the room?

A popular material to make a plasma out of is deuterium, a hydrogen isotope with one proton and one neutron. What would the temperature of a deuterium plasma be if the rms speed of the deutrium was 10% the speed of light? Note the mass of deuterium is 3.34x10-27 kg, and the speed of light is 3x108 m/s.

A hydrogen plasma has an rms speed of v0 at a temperature of T0. What would be temperature be if the rms speed were to double?
(a) √2 T0
(b) 2T0
(c) 4T0
(d) T0/2

We have unknown amounts of 2 gases kept in separate containers but at the same temperature. Which of the following statements is definitely true?
1. Their average molecular velocities are the same.
2. Their pressures are the same.
3. Their average molecular kinetic energies are the same.

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