What kinds of speeds can be found from a MaxwellBoltzmann distribution?
1 Answer
See below.
Explanation:
Concerning the velocity distribution for a Maxwellian gas:
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
From this, the most probable speed, denoted
#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
 A calculation of
#< nu^2 ># proceeds as:
#< 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