# Most probable molecular speed?

## 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?

## Answers and Replies

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?

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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