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.

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)?

Area under Maxwell distribution curve represents the number of molecules in the system

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.

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.

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.

Can V(rms) be equal to V(avg)?

yes when all the velocity data values are same. eg. -3,-3,-3,3,3,3 then V aver.: 3 and V rms : 3 .

So, why are there three kinds of mean speeds/velocities? i mean why not just one kind but three. Any reason?

It's kind of like how you'd have a mode, mean and median in statistics. They're all averages, but represent different things. For example, here, the most probable speed is equivalent to the mode, and the average speed is equivalent to the mean. Root mean square is a type of generalised mean, I believe.

Main content

Course: Physics library > Unit 10

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

What is the Maxwell-Boltzmann distribution?

Google Classroom

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.

Yes, good point. The number of gas molecules moving with a speed of exactly

231.46191432032804 \frac{m}{s}

is close to zero, since the odds of a gas molecule having that exact speed is pretty low. For this reason, you might think a graph of "number of molecules" on the y axis should just be zero for all exact speeds on the x-axis.

We can avoid this issue by just treating the x-axis as composed of "bins" of width

Δ v = 1 \frac{m}{s}

, so that we just graph the number of molecules with speeds between

v

and

v + 1 \frac{m}{s}

. This will give us non-zero values for numbers of molecules. Or we could change the "bin" size to be as small as we want, to have the desired level of precision.

While we're talking about it, to avoid this issue entirely, people often use something called probability density on the vertical axis instead of using # of molecules. The probability density is the probability, per unit speed, of finding a particle with a speed near

v

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.

OK, here's the equation for the Maxwell-Boltzmann distribution.

f (v) = \sqrt{[\frac{m}{2 π k T}]^{3}} 4 π v^{2} e^{(- \frac{m v^{2}}{2 k T})}

The quantity

f (v)

gives the probability density as a function of the speed

v

. This equation says that the probability density depends on the temperature

T

of the gas, the mass

m

of the molecules, and Boltzmann's constant

k

. Probability density is an alternate choice for the vertical axis, as opposed to # of molecules. The probability density is the probability, per unit speed, of finding a particle with a speed near

v

You don't really need to fully understand this mathematical equation to grasp many of the conceptual ideas associated with the Maxwell-Boltzmann distribution. So don't worry too much if the above equation looks like mathematical witchcraft to you.

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

most probable speed v_{p}

, since it is the speed that is most likely to be found for a molecule in a gas.

The

average speed v_{a v g}

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

root-mean-square speed v_{r m s}

. 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_{r m s} = \sqrt{\frac{1}{N} (v_{1}^{2} + v_{2}^{2} + v_{3}^{2} + …)}

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).

Root-mean-square speeds are useful to know because the kinetic energy of a molecule is proportional to the square of the speed (

K = \frac{1}{2} m v^{2}

). So to find the average kinetic energy

K_{a v g}

, we have to take the average of the squares of the velocities.

K_{a v g} = \frac{1}{N} (K_{1} + K_{2} + K_{3} + …)

K_{a v g} = \frac{1}{N} (\frac{1}{2} m v_{1}^{2} + \frac{1}{2} m v_{2}^{2} + \frac{1}{2} m v_{3}^{2} + …)

K_{a v g} = \frac{1}{2} m [\frac{1}{N} (v_{1}^{2} + v_{2}^{2} + v_{3}^{2} + …)]

K_{a v g} = \frac{1}{2} m v_{r m s}^{2}

So the average molecular kinetic energy of a gas is proportional to the square of the root-mean-square (rms) speed. It seems strange that the average kinetic energy is proportional to the rms speed instead of the average speed, but to find average kinetic energy we have to square first and then find the average, as opposed to finding the average and then taking a square.

It should be noted that all three of these quantities (

v_{p}

v_{a v g}

, and

v_{r m s}

) are quite large, even for a gas at room temperature. For example, Neon gas at room temperature (

293 K

) has a most probable speed, average speed, and root-mean-square speed of about,

v_{p} = 491 \frac{m}{s}

(or

1100 \frac{mi}{hr})

v_{a v g} = 554 \frac{m}{s}

(or

1240 \frac{mi}{hr})

v_{r m s} = 602 \frac{m}{s}

(or

1350 \frac{mi}{hr})

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)

(Choice A)
The root-mean-square molecular speed of the gas increases
(Choice B)
The graph of number of molecules vs. speed becomes more narrow
(Choice C)
The average molecular speed of the gas decreases
(Choice D)
The peak height of the graph of number of molecules vs. speed gets shorter

As a gas cools, the average speed gets smaller, since a cold gas has slower moving molecules.

Also, the graph of number of molecules vs. speed will get more narrow as the gas cools. Since the number of molecules remains fixed for a sealed container, the area under the curve will be constant. As the peak of the graph shifts left, the peak height will increase to maintain the same total area under the curve, thereby making the curve taller and more narrow as it cools.

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?

(Choice A)
Hotter gas molecules are allowed to enter and come into equilibrium with the initial gas.
(Choice B)
Some of the gas molecules are allowed to leave, and the remaining gas sample is heated in a sealed container.
(Choice C)
Some of the gas molecules are allowed to leave the sample, and the remaining gas sample is cooled in a sealed container.
(Choice D)
No molecules are allowed to enter or exit, but the gas is cooled in a sealed container

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Nitin
Posted 8 years ago. Direct link to Nitin's post “Why isn't V(rms) equal to...”
Why isn't V(rms) equal to V(avg) ? Here V(avg) is avg speed of the molecules.
Button navigates to signup pageComment on Nitin's post “Why isn't V(rms) equal to...”
(19 votes)
Answer
- Sean Meadows
  Posted 8 years ago. Direct link to Sean Meadows's post “Hey Nitin, I think I can ...”
  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.
  Comment on Sean Meadows's post “Hey Nitin, I think I can ...”
  (73 votes)
Toni Krmek
Posted 8 years ago. Direct link to Toni Krmek's post “Hello! Why is it that th...”
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.
Button navigates to signup pageComment on Toni Krmek's post “Hello! Why is it that th...”
(4 votes)
Answer
- Thom.Wescott
  Posted 8 years ago. Direct link to Thom.Wescott's post “This is a very good quest...”
  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.
  Button navigates to signup page
  (22 votes)
ricky chen
Posted 8 years ago. Direct link to ricky chen's post “Hi, I don't quite underst...”
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!
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- C Hart
  Posted 8 years ago. Direct link to C Hart's post “The wording is a little c...”
  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.
  Button navigates to signup page
  (9 votes)
Jake Savell
Posted 8 years ago. Direct link to Jake Savell's post “Just above the section "W...”
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)?
Button navigates to signup pageComment on Jake Savell's post “Just above the section "W...”
(5 votes)
Answer
- Prakriti
  Posted 6 years ago. Direct link to Prakriti's post “Area under Maxwell distri...”
  Area under Maxwell distribution curve represents the number of molecules in the system
  Button navigates to signup page
  (2 votes)
Abhijeeta
Posted 8 years ago. Direct link to Abhijeeta's post “Can V(rms) be equal to V(...”
Can V(rms) be equal to V(avg)?
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(1 vote)
Answer
- ms15135iiserm
  Posted 7 years ago. Direct link to ms15135iiserm's post “yes when all the velocity...”
  yes when all the velocity data values are same.
  eg. -3,-3,-3,3,3,3 then V aver.: 3 and V rms : 3 .
  Button navigates to signup page
  (4 votes)
Katie
Posted 5 years ago. Direct link to Katie's post “If we're plotting speed o...”
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?
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(3 votes)
Answer
- NoahTheCuber
  Posted 3 years ago. Direct link to NoahTheCuber's post “Technically yes, but it i...”
  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.
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  (3 votes)
Tatpong
Posted 7 years ago. Direct link to Tatpong's post “How did Boltzmann measure...”
How did Boltzmann measure gas particle speed?
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(4 votes)
Answer
- Shourya M. K.
  Posted 4 years ago. Direct link to Shourya M. K.'s post “It doesn't. It is just a ...”
  It doesn't. It is just a representation of particle speed and the number of particles with that speed. In other words, average particle speed is directly proportional to temperature, so you could measure the temperature, but that would be unpractical.
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  (1 vote)
overtakou
Posted 7 years ago. Direct link to overtakou's post “So what is the interpreta...”
So what is the interpretation behind each speed? I read you well explained how the most probable speed is different than the average speed, which are different than the rms speed, and I understood it. But what do they mean physical, for each particle or the gas?
Thanks!!
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(3 votes)
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Sukrit J. Baruah
Posted 8 years ago. Direct link to Sukrit J. Baruah's post “So, why are there three k...”
So, why are there three kinds of mean speeds/velocities? i mean why not just one kind but three. Any reason?
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(1 vote)
Answer
- beckydurrant97
  Posted 8 years ago. Direct link to beckydurrant97's post “It's kind of like how you...”
  It's kind of like how you'd have a mode, mean and median in statistics. They're all averages, but represent different things. For example, here, the most probable speed is equivalent to the mode, and the average speed is equivalent to the mean. Root mean square is a type of generalised mean, I believe.
  Button navigates to signup page
  (4 votes)
aebelshajan
Posted 4 years ago. Direct link to aebelshajan's post “Why cant we take the abso...”
Why cant we take the absolute value for velocity and then use this in our calculations?
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(2 votes)
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