What Fraction of Ideal Gas Molecules Have Speeds Around 2v?

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Homework Help Overview

The discussion revolves around determining the fraction of molecules in an ideal gas that have speeds within a specific range, using the Maxwell velocity distribution at a given temperature. The original poster presents a problem involving the calculation of this fraction, denoted as dN/N, where dN represents the number of molecules in the speed range and N is the total number of molecules.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between dN and N, questioning whether it is necessary to calculate N to find the fraction dN/N. There is discussion about using probability distribution functions to derive the fraction directly.

Discussion Status

Some participants have provided guidance on using the probability distribution function for the ideal gas to find the fraction of molecules without needing to calculate the total number of molecules. There is ongoing exploration of the relationship between the Maxwell-Boltzmann distribution and the most probable speed, with some participants attempting to clarify the connections between different theoretical concepts.

Contextual Notes

Participants note the lack of specific details such as the mass of the gas molecules and the temperature, which may affect the calculations. The discussion also touches on the equipartition theorem and its relevance to the problem, indicating that assumptions about the system are being examined.

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Homework Statement


An ideal gas is in a container at temperture 300 degrees. What fraction of its moelcules will have speeds in the range 1.95v and 2.05v where v is the most probable molecular speed?

Assume dN is constant in the range around 2v


Homework Equations


Maxwell velocity distribution (MVD)


The Attempt at a Solution


The answer should be in the form of dN/N
dN/(.1v)=MVD so dN can be worked out. But how do you work out N?
 
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You don't need to work out N. dN/N is the fraction of molecules- the number having that speed out of the total.
 
I don't understand. I can only seem to work out dN only by itself which is dN=.1v(MVD) but what is N?

Are you saying it is possible to work out dN/N in one go? If so how?
 
Yes, it's possible.

1) Again, dN/N is the fraction of molecules having that speed. It's possible to work out the fraction from a probability distribution- even if you don't know the total number N.

2) Consider a probability distribution function f(x). The probability of obtaining a result between x and x+dx is defined to be equal to

Prob=f(x)dx/ Int [f(x)dx]

where Int[] is the integral over all x. Usually, f(x) is 'normalized' such that Int[f(x)dx]=1, so that the probability is given by Prob=f(x)dx (when f(x) is normalized).

3) In your case, dN/N is the fraction of molecules having a velocity between 1.95 <v> and 2.05 <v>. Thus, we have

dN=f(v)dV * N

or dN/N=f(v) dV.

dV=0.1 <v>.

4) You need to find the probability distribution function for your problem. (It's the Maxwell Boltzmann distribution)

5) Then you need to work out the average velocity, <v>.

Let's go from there.
 
How do I work out <v>. Is it needed? The details like m is not given either. I don't think that to work the answer out, T is needed either.
 
I worked it out. You need to include one extra thing which is use the equipartition theorem. The equation relating the molecule's velocity and temperture. If you sub it into the (MVD) then you will get the fraction decimal.
 
Showing that <v^2>=3/2 kT can be done from an integral over the Maxwell Boltzmann distribution.

Equipartition is essentially a consequence of the MB distribution.
 
But the MB is different to the MVD so some new information is needed. Although I will try to see how that relation is derived from MB.
 
No they're not. How are they different?
 
  • #10
I see, it is possible to derive the relationship I described with an integral. Although one doesn't use the MVD but a part of it to calculate the probability of the mean squared velocity. Basically use the MVD except with number particles and a 4pi factor missing.

In my textbook however, they derived the velocity probability density function using the relationship .5mv^2=3kT/2 which they derived from basic kinetic theory and Newton's laws. So it is not surprising that by using the probaility density function .5mv^2=3kT/2 was able to come out from an integral. It was made so that this relation can come out!
 
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