What kinds of speeds can be found from a Maxwell-Boltzmann distribution?

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Answer 1 · 2017-12-23T05:02:34

See below.

Concerning the velocity distribution for a Maxwellian gas:

Hyperphysics

Most probable speed

The most probable speed corresponds to the maximum of the velocity distribution, where the slope is zero. One solves the equation

$(dbar(f)(nu))/(dnu)=sqrt(2/pi)(m/(kT))^(3/2)[2nu+((-mnu)/(kT))nu^2]e^((-mnu^2)//(2kT))=0$

where $bar(f)(nu)$ is the Maxwell velocity distribution (probability distribution for a molecule's velocity) as a function of velocity $nu$ .

From this, the most probable speed, denoted $nu_"m.p."$ emerges as:

$color(blue)(nu_"m.p."=sqrt((2kT)/m))$

Mean speed

An average or mean speed $< nu >$ is computed by weighting the speed $nu$ with its probability of occurrence $bar(f)(nu)dnu$ and then integrating:

$< nu > = int_0^(oo)nubar(f)(nu)dnu=int_0^(oo)e^((-mnu^2)//(2kT))sqrt(2/pi)(m/(kT))^(3/2)nu^3dnu$

$=> color(blue)(< nu > = sqrt(8/pi)sqrt((kT)/m))$

Root mean square speed

$< nu^2 > = int_0^(oo)nu^2bar(f)(nu)dnu=3(kT)/m$

$=>1/2m< nu^2 > = 3/2kT$

$=>color(blue)(nu_"r.m.s"=sqrt((3kT)/m))$

Note:

The mean speed $< nu >$ is $13%$ larger than $nu_"m.p."$ and $nu_"r.m.s"$ is $22%$ larger.
The common proportionality to $sqrt(kT//m)$ has two immediate implications: higher temperature implies higher speed, and larger mass implies lower speed.

**The equivalent expressions in terms of the universal/ideal gas constant $R$ are given in the figure above.