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## Physics library

### Course: Physics library>Unit 10

Lesson 1: Temperature, kinetic theory, and the ideal gas law

# What is the Maxwell-Boltzmann distribution?

In a gas, there are lots of molecules traveling at lots of different speeds. Here's a framework for thinking about that.

## What is the Maxwell-Boltzmann distribution?

The air molecules surrounding us are not all traveling at the same speed, even if the air is all at a single temperature. Some of the air molecules will be moving extremely fast, some will be moving with moderate speeds, and some of the air molecules will hardly be moving at all. Because of this, we can't ask questions like "What is the speed of an air molecule in a gas?" since a molecule in a gas could have any one of a huge number of possible speeds.
So instead of asking about any one particular gas molecule, we ask questions like, "What is the distribution of speeds in a gas at a certain temperature?" In the mid to late 1800s, James Clerk Maxwell and Ludwig Boltzmann figured out the answer to this question. Their result is referred to as the Maxwell-Boltzmann distribution , because it shows how the speeds of molecules are distributed for an ideal gas. The Maxwell-Boltzmann distribution is often represented with the following graph.
The y-axis of the Maxwell-Boltzmann graph can be thought of as giving the number of molecules per unit speed. So, if the graph is higher in a given region, it means that there are more gas molecules moving with those speeds.
Notice that the graph is not symmetrical. There is a longer "tail" on the high speed right end of the graph. The graph continues to the right to extremely large speeds, but to the left the graph must end at zero (since a molecule can't have a speed less than zero).
The actual mathematical equation for the Maxwell-Boltzmann distribution is a little intimidating and not typically needed for many introductory algebra classes.

## What does root-mean-square speed mean?

You might think that the speed located directly under the peak of the Maxwell-Boltzmann graph is the average speed of a molecule in the gas, but that's not true. The speed located directly under the peak is the start color #e84d39, start text, m, o, s, t, space, p, r, o, b, a, b, l, e, space, s, p, e, e, d, space, end text, v, start subscript, p, end subscript, end color #e84d39, since it is the speed that is most likely to be found for a molecule in a gas.
The start color #11accd, start text, a, v, e, r, a, g, e, space, s, p, e, e, d, space, end text, v, start subscript, a, v, g, end subscript, end color #11accd of a molecule in the gas is actually located a bit to the right of the peak. The reason the average speed is located to the right of the peak is due to the longer "tail" on the right side of the Maxwell-Boltzmann distribution graph. This longer tail pulls the average speed slightly to the right of the peak of the graph.
Another useful quantity is known as the start color #1fab54, start text, r, o, o, t, negative, m, e, a, n, negative, s, q, u, a, r, e, space, s, p, e, e, d, space, end text, v, start subscript, r, m, s, end subscript, end color #1fab54. This quantity is interesting because the definition is hidden in the name itself. The root-mean-square speed is the square root of the mean of the squares of the velocities. Mean is just another word for average here. We can write the root-mean-square speed mathematically as,
v, start subscript, r, m, s, end subscript, equals, square root of, start fraction, 1, divided by, N, end fraction, left parenthesis, v, start subscript, 1, end subscript, squared, plus, v, start subscript, 2, end subscript, squared, plus, v, start subscript, 3, end subscript, squared, plus, point, point, point, right parenthesis, end square root
It might seem like this technique of finding an average value is unnecessarily complicated since we squared all the velocities, only to later take a square root. You might wonder, ""Why not just take an average of the velocities?" But remember that velocity is a vector and has a direction. The average gas molecule velocity is zero, since there are just as many gas molecules going right (+ velocity) as there are going left (- velocity). This is why we square the velocities first, making them all positive. This ensures that taking the mean (i.e. average value) will not give us zero. Physicists use this trick often to find average values over quantities that can take positive and negative values (e.g. voltage and current in an alternating current circuit).
It should be noted that all three of these quantities (start color #e84d39, v, start subscript, p, end subscript, end color #e84d39, start color #11accd, v, start subscript, a, v, g, end subscript, end color #11accd, and start color #1fab54, v, start subscript, r, m, s, end subscript, end color #1fab54) are quite large, even for a gas at room temperature. For example, Neon gas at room temperature (293, start text, space, K, end text) has a most probable speed, average speed, and root-mean-square speed of about,
start color #e84d39, v, start subscript, p, end subscript, end color #e84d39, equals, 491, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction (or 1100, start fraction, start text, m, i, end text, divided by, start text, h, r, end text, end fraction, right parenthesis
start color #11accd, v, start subscript, a, v, g, end subscript, end color #11accd, equals, 554, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction (or 1240, start fraction, start text, m, i, end text, divided by, start text, h, r, end text, end fraction, right parenthesis
start color #1fab54, v, start subscript, r, m, s, end subscript, end color #1fab54, equals, 602, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction (or 1350, start fraction, start text, m, i, end text, divided by, start text, h, r, end text, end fraction, right parenthesis

## What does the area under a Maxwell-Boltzmann distribution represent?

The y-axis of the Maxwell-Boltzmann distribution graph gives the number of molecules per unit speed. The total area under the entire curve is equal to the total number of molecules in the gas.
If we heat the gas to a higher temperature, the peak of the graph will shift to the right (since the average molecular speed will increase). As the graph shifts to the right, the height of the graph has to decrease in order to maintain the same total area under the curve. Similarly, as a gas cools to a lower temperature, the peak of the graph shifts to the left. As the graph shifts to the left, the height of the graph has to increase in order to maintain the same area under the curve. This can be seen in the curves below which represent a sample of gas (with a constant amount of molecules) at different temperatures.
As the gas gets colder, the graph becomes taller and more narrow. Similarly, as the gas gets hotter the graph becomes shorter and wider. This is required for the area under the curve (i.e. total number of molecules) to stay constant.
If molecules enter the sample, the total area under the curve would increase. Similarly, if molecules were to leave the sample, the total area under the curve would decrease.

## What do solved examples involving the Maxwell-Boltzmann distribution look like?

### Example 1: Cooling a gas

A gas of diatomic nitrogen is in a sealed container. The sealed container is then placed in an ice bath and reaches a lower equilibrium temperature with the ice bath.
What happens to the following quantities as the gas cools? (select two correct statements)

### Example 2: Change in the gas

A gas has a distribution of speeds that looks like the following.
Which one of the following series of actions could cause the distribution graph to change from curve 1 to curve 2, as seen below?

## Want to join the conversation?

• Why isn't V(rms) equal to V(avg) ? Here V(avg) is avg speed of the molecules. •   Hey Nitin, I think I can answer your question. Let's say there are 6 molecules in a sample of gas. These 6 molecules have the following velocities, respectively: -1 m/s, -2 m/s, -3 m/s, +1 m/s, +2 m/s, and +3 m/s. If we want to find the average speed of a molecule in this sample of gas, we need to convert these 6 vector quantities (i.e. velocities) into 6 scalar quantities (i.e. speeds) by removing the signs. After performing this conversion, we have the following speeds: 1 m/s, 2 m/s, 3 m/s, 1 m/s, 2 m/s, and 3 m/s. To find the average speed, v(avg), we need to add these 6 speeds together and then divide the sum by the total number of speeds:

1 + 2 + 3 + 1 + 2 + 3 = 12
12 / 6 = 2

So the average speed is 2 m/s. On the other hand, if we want to find the root-mean-square speed of a molecule in this sample of gas, we don't need to convert those 6 velocities into 6 speeds. Instead, we need to find the square root of the average of the squares of the velocities:

(-1)^2 + (-2)^2 + (-3)^2 + 1^2 + 2^2 + 3^2 = 28
28 / 6 = 4.67
The square root of 4.67 is 2.16.

So the root-mean-square speed is 2.16 m/s, which is greater than the average speed (2 m/s). This is why the root-mean-square speed does not equal the average speed. I hope this helps. If my reasoning is incorrect, please let me know.
• Hello!
Why is it that the average kinetic energy is proportional to the rms speed instead of the average speed .I understood about squares and roots, but still if i have the average speed of particles it would be normal for me to take the average speed for the average kin,energy?
Thank you. • This is a very good question that gets to the root (pun not intended) of something that is apparently so subtle that it was never explicitly stated to me all the way through second year college physics. Massive bodies in motion have two parameters that diverge the greater their velocity becomes. Momentum is simply mass times velocity, while kinetic energy is one half of the product of the mass and the square of the velocity. So when we look at the simple mode and mean of the Maxwell-Boltzmann distribution (which assumes a uniform mass) we are looking at momentum, but when we consider the rms value we are now looking at a description of kinetic energy.
I'm pretty sure you've already understood that, but are still as unsure as I am about why that matters. First off, in the context of this lesson, all we need is to be aware that there is a difference, and that at some point in the future it will become significant.
Looking beyond the scope of this lesson, I expect the implications of this follow from the two facts that a) both momentum and kinetic energy must be conserved, in any non-relativistic event at least, and b) they begin to scale at radically different rates as the velocity increases.
At the human scale, if my car hits another car at 1 mph the momentum is conserved as we both continue at 1/2 mph until various frictions bring us to a stop, and pretty much all of the excess kinetic energy is distributed without too much trauma. However, if I hit the same stationary car at 100 mph and momentum is conserved as we both continue down the road at 50 mph, there is now 10,000 times as much energy to dissipate. My car retains one half times m times 50 squared in its residual velocity, the second car absorbs the same amount as it accelerates, leaving one half m times 10,000 - 2500 - 2500, that is fully half the original kinetic energy to be dissipated as a bunch of highway ugliness.
Just how that translates into behaviours at the molecular scale, I am not sure. Hopefully, I'll get a better handle on this as these lessons progress.
• Hi, I don't quite understand the idea of "number of molecules per unit speed" on the y-axis. What is this idea of "unit speed"? Which "unit speed" are we dividing the number of molecules by? Could you explain? Thanks! • The wording is a little confusing. In order to get a distribution, you are dividing up the molecules into bins according to their speed (like a histogram). In this case, the bins are so thin that instead of a lot of bars, you get a smooth curve that represents the top of each thin bar. So when they say "number of molecules per unit speed" they aren't mathematically dividing, it's more like "number of molecules in each bin of speed" with the bins on the x-axis. But remember that the bins are so thin that we actually refer to them just by the x-value (speed). So there is a number of molecules at 600.000 m/s, and a number at 600.001 m/s, and a number at 600.002 m/s, and thus you get a continuous curve of number of molecules vs speed.
• Just above the section "What does the area under a Maxwell-Boltzmann distribution represent?", should the 605 m/s quantity be v(rms) instead of v(p)? • Can V(rms) be equal to V(avg)?
(1 vote) • If we're plotting speed on the x axis, is there then a maximum (the speed of light) the individual particles can't go beyond? So the graph doesn't tail off to infinity? • Technically yes, but it is a bit more complex. As particles approach the speed of light they are called relativistic particles. At that point, special relativity must be taken into account. Look up the Maxwell–Jüttner distribution. It describes the distribution of speeds for relativistic particles.
If any of this is incorrect please let me know and I will correct it.    