Average Velocity of gas molecules calculated with a Boltzmann distribution

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The average velocity of gas molecules calculated using the Boltzmann distribution reflects the average speed of a large sample of molecules, assuming they are at thermal equilibrium. Individual gas molecules do not have the same average velocity, as it varies with molecular mass and species. The average kinetic energy of gas molecules is uniform at a given temperature, but the average velocity scales inversely with the square root of molecular mass. The discussion also touches on the relationship between instantaneous and time-averaged speeds for individual molecules, suggesting they may be equivalent. Overall, the Boltzmann distribution provides a statistical approach to understanding molecular velocities in gases.
Viona
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What is the meaning of the average velocity of gas molecules calculated by Boltzmann distribution (in kinetic theory of gases)?
Does all molecules have the same average velocity?
 
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Viona said:
What is the meaning of the average velocity of gas molecules calculated by Boltzmann distribution (in kinetic theory of gases)?
I guess you mean speed, because the average velocity is normally 0.

The meaning is the same as for any average. Were you to sample a bunch of molecules, the average speed of these molecules would be equal to the average speed calculated by the Boltzmann distribution, provided that the number of sampled molecules is high enough (for smaller numbers, you will get fluctuations around the average value).

Viona said:
Does all molecules have the same average velocity?
It depends. Check the parameters that enter into the calculation of the average speed. Provided that these parameters are equal, then the average speed will be the same.
 
Viona said:
Does all molecules have the same average velocity?
No.

I don't understand your question. Using an average calculation in any circumstance is needed only when the values are not all equal.
 
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The average thermal speed is useful for a number of things, but I don't really understand what the question is here. As for your question about all molecules, are you asking if different molecular species have the same average velocity? If so, then generally no. In a gas volume at a given temperature, the average kinetic energy of all molecules is the same (assuming it has reached equilibrium), so the average velocity of any gas species will scale with the inverse square root of the molecular mass.
 
anorlunda said:
No.

I don't understand your question. Using an average calculation in any circumstance is needed only when the values are not all equal.

cjl said:
The average thermal speed is useful for a number of things, but I don't really understand what the question is here. As for your question about all molecules, are you asking if different molecular species have the same average velocity? If so, then generally no. In a gas volume at a given temperature, the average kinetic energy of all molecules is the same (assuming it has reached equilibrium), so the average velocity of any gas species will scale with the inverse square root of the molecular mass.
I think I am confused: is the average velocity calculated using Boltzmann distribution is similar to, for example, measuring the velocity for one molecule many times and take the average?
 
Viona said:
I think I am confused: is the average velocity calculated using Boltzmann distribution is similar to, for example, measuring the velocity for one molecule many times and take the average?

First answer @cjl 's question. Are you asking about a gas with one molecule specie. or multiple species ? (such as air with O2, and N2)

If the average kinetic energy is E, then E=mv2, and the corresponding velocity is ##v=\sqrt{E/m}## . But m is different for each specie.

If you are just asking about the definition of average for multibody cases, try to answer the question yourself and post your conclusion.
 
I believe the instantaneous average speed over all the molecules is equal to the time-average for a single molecule, but I'm not sure I could prove it.
 
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