[Araconan]

Hi!

I've been reading about the Maxwell-Boltzmann Distribution Law, and according to my textbook, there's essentially three ways to describe the velocity for a set of particles in an ideal gas.
1. Root Mean Square Velocity
2. Average Velocity
3. Most Probable Velocity

I was confused on the meaning and usage of the root mean square velocity and why it was important when I could just calculate the average velocity directly without squaring anything. So I searched it up, and what I found out was that when considering a large amount of particles, the particles are generally isotropic. So since the directions of the velocities are all balanced, velocities of opposite directions would cancel each other out, leaving an average velocity of 0. Hence a root mean square velocity would actually provide an actual number to work with, instead of 0.

But what I don't understand, is that if this was true, why is there even an equation for calculating the average velocity? v = √(8RT)/(pi)(M). Wouldn't it just always be 0?
If the statement about the velocities cancelling each other out is wrong, and that the value for the average velocity isn't 0, then why would I need to know the root mean square velocity? Aren't they essentially measuring the same thing, and wouldn't the average velocity give me the necessary information regarding the average speed of the particles?

I'm basically having a hard time understanding the difference between these two velocities, and when to use which one. Also, my textbook also states that both effusion and diffusion rates depends on the average velocity. Why is it dependant on the average velocity instead of the root mean square velocity?

Thank you in advance!

[alexmahone]

Are you sure the formula you have isn't for average speed?

[Araconan]

Well, according to my textbook, it says that the formula is for the average velocity (Average Velocity = √(8RT)/(pi)(M) )
It also says that this formula is derived by solving Maxwell-Boltzmann's Distribution Law, which describes the distribution of particle velocities. It also keeps on mentioning the average velocity when deriving expressions for the rate of diffusion/effusion, and the rate of collisions of gas particles on container walls.

However, I've actually just looked up the Maxwell-Boltzmann's Distribution Law on Wikipedia, and the equation for average velocity, was actually stated as average speed on Wikipedia, and the equation for the most probable velocity, was labelled as the equation for most probable speed, by Wikipedia. So in that case, can I think of the average velocity and the most probable velocity as instead the average speed and most probable speed?

[alexmahone]

Your textbook is wrong; those are speeds, not velocities.

[Araconan]

Ahh okay, that would make a lot more sense. Thank you!

[juanrga]

Hi!

I've been reading about the Maxwell-Boltzmann Distribution Law, and according to my textbook, there's essentially three ways to describe the velocity for a set of particles in an ideal gas.
1. Root Mean Square Velocity
2. Average Velocity
3. Most Probable Velocity

I was confused on the meaning and usage of the root mean square velocity and why it was important when I could just calculate the average velocity directly without squaring anything. So I searched it up, and what I found out was that when considering a large amount of particles, the particles are generally isotropic. So since the directions of the velocities are all balanced, velocities of opposite directions would cancel each other out, leaving an average velocity of 0. Hence a root mean square velocity would actually provide an actual number to work with, instead of 0.

Effectively, the average velocity can be zero in situations where there positive velocities compensate the negative velocities. For instance if 10 particles have velocity v and other 10 velocity -v the average is zero. The average velocity is zero for a gas that is at rest as a whole!

The Root Mean Square Speed (not velocity) is a speed and it is nonzero (unless all particles are at rest) because the speeds cannot cancel. For instance the 10 above particles all have speed v.

I'm basically having a hard time understanding the difference between these two velocities, and when to use which one. Also, my textbook also states that both effusion and diffusion rates depends on the average velocity. Why is it dependant on the average velocity instead of the root mean square velocity?

Thank you in advance!

It is not the same a flow of particle diffusing at the left than a flow of particles diffusing at the right although the root mean square velocity could be the same in both cases.

[Araconan]

Wait, so if the root mean square value is a speed value, then why can't I take all the velocity values, take the absolute value of them, and then calculate the average that way? Why do I have to square all of the values? When compared to the average speed, √(8RT)/(pi)(M), doesn't the root mean square value then become redundant? Like why do I even need it if it's just another speed value? (When I can just calculate the average speed instead)

[alexmahone]

Wait, so if the root mean square value is a speed value, then why can't I take all the velocity values, take the absolute value of them, and then calculate the average that way?

That's how the average speed is calculated.

Why do I have to square all of the values? When compared to the average speed, √(8RT)/(pi)(M), doesn't the root mean square value then become redundant? Like why do I even need it if it's just another speed value? (When I can just calculate the average speed instead)

Root mean square speed is a different kind of average speed. Wikipedia says that "Root-mean-square speed is the measure of the speed of particles in a gas that is most convenient for problem solving within the kinetic theory of gases."

[Araconan]

I know that the root mean square speed pops up, when trying to prove Kinetic Molecular Theory in terms of the Ideal Gas Law. But so can I then assume that when I'm doing any calculations/intuitive understandings outside of proving the KMT, I will always use average speed? Is there any other situations in which I will use the root mean square speed instead of the average speed?

[Jorriss]

I know that the root mean square speed pops up, when trying to prove Kinetic Molecular Theory in terms of the Ideal Gas Law. But so can I then assume that when I'm doing any calculations/intuitive understandings outside of proving the KMT, I will always use average speed? Is there any other situations in which I will use the root mean square speed instead of the average speed?

Root mean square quantities show up all the time. They show up in calculating variances, diffusion, brownian motion and more places.

No one can say will you always use average speed or root-mean-square speed. You just need to take it problem by problem.

[juanrga]

Wait, so if the root mean square value is a speed value, then why can't I take all the velocity values, take the absolute value of them, and then calculate the average that way? Why do I have to square all of the values? When compared to the average speed, √(8RT)/(pi)(M), doesn't the root mean square value then become redundant? Like why do I even need it if it's just another speed value? (When I can just calculate the average speed instead)

Speed, a scalar, is the magnitude or norm of the vector velocity. The definition is
$$v \equiv \sqrt{\mathbf{v}\mathbf{v}}$$
You can take the average ##\langle v\rangle##, but this quantity is not really important to characterize the population of particles. A more important expression is ##\langle v^2\rangle##, where
$$v^2 = \mathbf{v}\mathbf{v}$$
For example, as stated in Stochastic scientific quantities and states, you need know both ##\langle \mathbf{v} \rangle## and ##\langle v^2 \rangle## for obtaining the average kinetic energy of a gas of particles, because ##\langle K(\mathbf{v}) \rangle \neq K(\langle \mathbf{v} \rangle)##.