Ch 19: Kinetic Theory of Ideal GassesSee all chapters

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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: Speed Distribution of an Ideal Gas

**Transcript**

Hey guys, in this video we're going to talk about the speed distribution of an ideal gas. Let's get to it. Remember guys that in an ideal gas not all particles move at the same speed. This was one of the assumptions that we had to make in kinetic theory in order to derive everything that we did using kinetic theory for ideal gases. The speeds of particles in an ideal gas can vary wildly, with some speeds occurring very frequently and some speeds occurring very rarely. You can think about it like this. What if the temperature was very low? Well then low speeds would occur very commonly. If the temperature is low, then the average kinetic energy is low. So the odds are that a particle having an extraordinarily high kinetic energy relative to the average existing is pretty rare so the odds of a particle having a really high speed at a low temperature is rare. It can happen but it's pretty rare for it to occur. Speeds of particles in an ideal gas form what's called a distribution. Technically a probability distribution. All that a distribution is is it's distributing the probability of finding any particle at each speed. It doesn't distribute it evenly, it distributes it according to a particular function. In this case the Maxwell-Boltzmann distribution. That's what governs the probability of finding a particle at any speed in an ideal gas at a given temperature. I plotted the Maxwell-Boltzmann probability distribution for three temperatures in this graph right here, the probability versus the speed. The hottest temperature T3, a medium temperature T2 and then a low temperature T1. As you can see these distributions are shifting from the high speeds to low speeds as the temperature decreases. This is exactly what should be expected from what I just explained about the probability of a particle having an extraordinarily high kinetic energy when the average kinetic energy of the gas is low. The most probable speed is the peak of the distribution. I marked the most probable speed here for the low temperature gas, here for mid-temperature gas and here for the high temperature gas. It's just the point of highest probability, the vertical axis is just probability so this point right here on the horizontal axis is the speed at which the highest probability occurs. Notice that as we expected, that most probable speed increases with temperature because as the temperature increases, the average kinetic energy increases, so on average particles will be traveling faster. In addition the most probable speed is a new statistical speed that we have. We already saw the RMS speed before, the most probable speed has a new statistically significant speed. So later on we're going to have an equation for the most probable speed as well. Now as the temperature decreases we can see that speeds nearer the most probable speed become more probable but speeds far from the most probable speed become less probable. So we can compare the high temperature gas to the low temperature gas. For the high temperature gas, look at all of these high probabilities for speeds that are very very close to the most probable speed, the dash line in the middle and now look at probabilities for speeds far away from that most probable speed. Those three dots that I just drew, the probability is zero for all of them. So far away from the most probable, you have an extremely low probability at low temperature but close to the most probable you have a very high probability. Now for the high temperature gas, it's actually the opposite. For the high temperature gas, look at the nearby points. The nearby points don't really have a very high probability they have a pretty low probability but far speeds away, very high speeds, speeds that are very far from the most probable speed they still have a probability. It's not zero. So as the temperature decreases, speeds near that peak speed become more probable but speeds far from it become less probable. This is called a narrowing of the distribution. And you can see that the distribution for T1 clearly looks like a more narrow peak than the distribution for T3. As temperature increases speeds become roughly the same probability. Maybe not exactly the same probability but they're on the order of the same probability. Maybe the most probable speed has a ten percent chance of being found but then far away you find a speed here that has a five percent chance of being found. Those are roughly the same probabilities, they are on the same order but here for the low temperature, the peak speed might have an eighty five percent chance of probability but if I go far away from it immediately that probability drops to zero. This is known as a widening of the distribution and T3 is clearly a more wide peak than T1. The the wider it gets the more equally probable all those speeds become and eventually at an infinite temperature all speeds become equally probable. Now we can write out our equations for the most probable speed of a particle in an ideal gas and the average speed for a particle in an ideal gas. The most probable speed is 2KT over M. Almost the exact same as the RMS speed just with a 2 inside the radical instead of a 3. The average speed on the other hand is 8KT over pi M. So it's 8 over pi instead of 3. A little bit different. Let's do an example before we get out here. You perform an experiment measuring individual particle speeds in gaseous atomic hydrogen and find that the average particle speed is 1.7 kilometers per second. If you were to take one more measurement, what is the likely speed you measure? So we have an average particle speed of 1.7 kilometers per second and this comes after many many many measurements and then if we take one measurement, V equals what? What do we think is going to equal? If you take one more measurement what's the most likely speed you're going to occur or you're going to measure? The most probable speed. So this is that crucial information to solving this problem. The average speed we know is the square root of a eight KT over M. That means that the temperature is MV average squared over sorry I'm missing a pi, pi M over 8 KB. So this temperature that this atomic hydrogen exists at, and remember atomic hydrogen has a mass of one atomic mass unit is pi, one atomic mass unit is 1.67 times 10 to the -27 the average speed we know is 1.7 kilometers per second but we need that in meters per second so this is 17000 meters per second divided by 8 times the Boltzmann constant which is 1.38 times 10 to the -23. And this leads us to finding that our gaseous atomic hydrogen has a temperature of a 137 Kelvin. So if our atomic hydrogen has a temperature of 137 Kelvin what is the most probable speed that would measure? Well the most probable speed is the square root of 2 times KT over M. So this is the square root of 2 times 1.38 times 10 to the -23 times 137 Kelvin divided by the mass, this is atomic hydrogen so once again it's got one atomic mass unit which is 1.67 times 10 to the -27 kilograms. Plugging all this into your calculator we get 1504 meters per second. That is the most likely speed that you would measure next. And that's one and a half kilometers per second, where the average is 1.7 per second. That wraps up our discussion on speed distributions in an ideal gas. Thanks for watching, guys.

Example #1: Data Collection on Particle Speeds in an Ideal Xenon Gas

Answer should be 299K

**Transcript**

Hey guys, let's do an example. The following data is recorded speeds of particles in an ideal Xenon 132 gas. Find A, the most probable speed, B, the average speed, C, the RMS speed and D, the temperature of the gas. So A, the most probable speed. What we have here is just the distribution of speeds. The most probable speed is always the peak of that distribution, is just the one that has the highest number of particles in it, 42 exists at this speed which is 200. So the most probable speed is 200 meters per second. Boom, that one's done. Super easy. B, what about that average speed though? To find the average speed, we need to average out all of these numbers. I'm not going to write them all down because it's huge you guys can write it down yourself and calculate it but I'll show you how to set it up and then give the answer. You have to write down each number times the speed. So first we have 7 times 50 plus 24 times 100 plus 38 times 150 etc. You have to keep writing that out. Next is 43 times 200, 37 times 250, 26 times 300, etc. All of this divided by the total number of particles. So that total number is going to be 7 plus 24, right? 7 plus 24 plus 38 plus 43 plus 38 plus 43 plus 37 plus 26, all of the number of particles and when you do the math for this you get the average speed is 219 meters per second so slightly higher than the most probable speed. Part C is the RMS speed that's the most difficult. In order to do the RMS speed you have to represent the data like this. The first column that I wrote is the number of each occurrence, the second column that I wrote is the speed, the third column is the square of the speed. Remember to find the RMS value for anything. To find RMS, you have to take X. X in this case is the speed. You have to square it so I already did that in the third column then you have to average it. That's what we're going to do next, we're going to take all of these values here in this third column and we're going to average them based on how frequently they occur and then finally you have to take the square roots that average. So first what's the average of V squared? Well what it's going to be is it's going to be the number times the corresponding probability sorry times the corresponding speed squared summed up over all of the possible V squares divided by the total number of particles exactly like we did before. So this is going to be 7 but it's now times 2500 not times 50 plus 24 now times 10000 plus 38 now times 22500, etc. You're doing this for all of the V squared's the next is 43 and 40000 and 37 and 62500 and 26 and 90000 etc. divided by the total number of particles which is just the sum of this column so 7 plus 24 plus 38 plus 43 plus 37, blah blah blah. If you do all of that, you get an average of the squared speed of 56219 meters per second. What this means is the RMS speed which is the square root of the average of the squared is the square root of 56219 and that's 237 meters per second. So those are the 3 statistically significant speeds, the most probable, the average and the RMS. Now part D is to find the temperature. We can use any one of those to find the temperature because all 3 of them have equations that relate them to the temperature. I'll just use the RMS speed. Solving for the temperature where the temperature is the mass times the RMS speed squared divided by 3 times the Boltzmann constant. So the temperature first of all the mass, it's Xenon 132. It's already telling you the atomic weight it's 132 atomic units. So it's 132 times the mass of one atomic mass unit which is 1.67 times 10 to the -27, the RMS speed we calculated was 237 meters per second squared divided by three times the Boltzmann constant and so the temperature is going to be somewhere around, sorry I can't find my notes. Right here 213 Kelvin. The temperature is gonna be somewhere in the neighborhood of 213 Kelvin but you could have also found the temperature from any of the other two statistically significant speeds because the most probable speed and the average speed each have an equation very similar to this that relate them to the temperature. So this problem shows you how to take raw data and find these values, the most probable speeded, the average speed, the RMS speed and then the temperature all from the data. Alright guys that wraps up this problem. Thanks for watching. You have to square it so I already did that in the third column then you have to average it.

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Concept #1: Speed Distribution of an Ideal Gas

Example #1: Data Collection on Particle Speeds in an Ideal X...

30 mol of monoatomic hydrogen gas is contained within a 30 cm x 40 cm x 60 cm box at a pressure of 5x105 Pa.
(a) What is the rms speed of the hydrogen gas? Note that the mass of hydrogen is 1.67x10 -27 kg.
(b) What would the rms speed be for a 30 mol sample of O 2 gas, considered to have a rigid inter-atomic bond, stored in the same container at the same pressure? Note that the mass of O2 is 5.34x10-26 kg.

What is the average kinetic energy of a molecule of oxygen at 303 K? Boltzmann’s constant is 1.38066 × 10−23 J/K.
1. 7.18632e-21
2. 9.2366e-21
3. 5.28102e-21
4. 6.79284e-21
5. 8.92595e-21
6. 6.27509e-21
7. 6.87568e-21
8. 7.7662e-21
9. 5.98515e-21
10. 6.97923e-21

If the average speed of a water molecule at 25°C is 640 m · s −1, what is the average speed at 100°C
1. 801 m/s
2. 320 m/s
3. 572 m/s
4. 716 m/s
5. 5120 m/s

If the average speed of a carbon dioxide molecule is 410 m • s −1 at 25°C, what is the average speed of a molecule of methane at the same temperature?
1. 679 m • s−1
2. 1000 m • s−1
3. 410 m • s−1
4. 1130 m • s−1
5. 247 m • s−1

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