# Why is most-probable speed correspond to the top of the Maxwell-Boltzmann distribution curve, even though the root-mean-square speed is the highest? Why are average and root-mean-square to the right?

##### 1 Answer

*Because most-probable speed is most likely, i.e. a greater fraction of molecules will have that speed. It does NOT indicate that it is the highest in magnitude---only likelihood.*

Let **most-probable**, **average**, and **root-mean-square speeds**, respectively.

#upsilon_"mp" = sqrt((2k_BT)/m)# #<< upsilon >> = sqrt((8k_BT)/(pim))# #upsilon_"rms" = sqrt((3k_BT)/m)#

From factoring out everything that is not **increases from left to right** on the x-axis.

*The height on the y-axis does not indicate a faster speed.*

I derive these equations below using the Maxwell-Boltzmann distribution (you would need to know how to perform derivatives, but the integrals used are tabled).

**MOST PROBABLE SPEED**

Given that *most-probable* speed, then on the Maxwell-Boltzmann distribution plot, which is a **probability density** **plot**, it must be found at a local maximum, i.e. when the derivative

The **the fraction of molecules with that speed**, as the graph states, so it says that

The **Maxwell-Boltzmann speed distribution function** (*Physical Chemistry: A Molecular Approach, McQuarrie*) is given as

#\mathbf(f(upsilon)dupsilon = 4pi(m/(2pik_BT))^"3/2" upsilon^2 e^(-m upsilon^2"/"2k_BT)dupsilon)# where

#k_B# is the Boltzmann constant,#T # is temperature,#m# is the mass of the gas, and#upsilon# is speed.

Taking the derivative with respect to

#color(green)((df(upsilon))/(dupsilon))#

#= color(green)(4pi(m/(2pik_BT))^"3/2" d/(dupsilon)[upsilon^2e^(-m upsilon^2"/"2k_BT)])#

Using the product rule, we have

#= 4pi(m/(2pik_BT))^"3/2" [upsilon^2cdot(-cancel(2)upsilon*m/(cancel(2)k_BT))e^(-m upsilon^2"/"2k_BT) + 2upsilone^(-m upsilon^2"/"2k_BT)]# where

#g(upsilon) = upsilon^2# ,#h(upsilon) = e^(-m upsilon^2"/"2k_BT)# , and#k(upsilon) = -(m upsilon^2)/(2k_BT)# .

Now simply note that the constants can never be

#= cancel(4pi(m/(2pik_BT))^"3/2") [e^(-m upsilon^2"/"2k_BT)(2upsilon - upsilon^3(m/(k_BT)))] = 0#

#= e^(-m upsilon^2"/"2k_BT)[2upsilon - upsilon^3(m/(k_BT))] = 0#

Of course,

#0 = 2upsilon - upsilon^3(m/(k_BT))#

So we get, given that speeds are always *positive*:

#upsilon^(cancel(3)^(2)) (m/(k_BT)) = 2cancel(upsilon)#

#upsilon^2 = (2k_BT)/(m) => color(blue)(upsilon_"mp") = color(blue)(sqrt((2k_BT)/m))#

**AVERAGE SPEED**

The average speed can be gotten from the integral formula for averages, using the Maxwell-Boltzmann distribution from before:

#color(green)(<< upsilon >> = int_(0)^(oo) upsilonf(upsilon)dupsilon)#

#= 4pi(m/(2pik_BT))^"3/2" int_(0)^(oo) upsilon^3 e^(-m upsilon^2"/"2k_BT)#

Using this tabled integral:

#int_(0)^(oo) x^(2n+1)e^(-alphax^2)dx = (n!)/(2alpha^(n+1))#

we utilize

#= 4pi(m/(2pik_BT))^"3/2" cdot 1/(2(m/(2k_BT))^(1+1))#

#= 4pi(m/(2pik_BT))^"3/2" cdot (2(k_BT)^2)/(m^2)#

#= 8pi(cancel(m)/(2pi))^"3/2"cancel((1/(k_BT))^"3/2") cdot ((k_BT)^(cancel("4/2")^"1/2"))/(m^(cancel("4/2")^"1/2"))#

#= 8pi(1/(2pi))^"2/2"cdot(1/(2pi))^"1/2" cdot ((k_BT)/(m))^"1/2"#

#= 4cdot ((k_BT)/(2pim))^"1/2"#

#=> color(blue)(<< upsilon >> = sqrt((8k_BT)/(pim)))#

**ROOT-MEAN-SQUARE SPEED**

Now, for the root-mean-square speed!

By definition,

#color(green)(<< upsilon^2 >> = int_(0)^(oo) upsilon^2f(upsilon)dupsilon)#

(You may want to compare this back to the average speed integral to see the difference.)

#= 4pi(m/(2pik_BT))^"3/2" int_(0)^(oo) upsilon^4 e^(-m upsilon^2"/"2k_BT)#

We use another tabled integral, since

#int_(0)^(oo) x^(2n)e^(-alphax^2)dx = (1cdot3cdot5cdots(2n-1))/(2^(n+1)alpha^n)(pi/(alpha))^"1/2"#

For this,

#=> 4pi(m/(2pik_BT))^"3/2"[(1cdot(2*2-1))/(2^(2+1)(m/(2k_BT))^2)(pi/(m/(2k_BT)))^"1/2"]#

#= 4pi(m/(2pik_BT))^"3/2"[(3)/(8(m/(2k_BT))^2)(pi/(m/(2k_BT)))^"1/2"]#

#= 4pi(m/(2pik_BT))^cancel("3/2")[(3/8) * ((2k_BT)/m)^2cancel(((2pik_BT)/m)^"1/2")]#

#= 3/cancel(2)(cancel(pi)m)/(cancel(2)cancel(pi)k_BT) ((cancel(2)k_BT)/m)^2#

#= 3cancel((m)/(k_BT)) ((k_BT)/m)^cancel(2)#

#= (3k_BT)/m#

Therefore:

#color(blue)(upsilon_"rms") = sqrt(<< upsilon^2 >>) = color(blue)(sqrt((3k_BT)/m))#

**NOW WHAT?**

Now why the heck did we do all that? To get the comparable formulas, of course! We have:

#upsilon_"mp" = sqrt((2k_BT)/m)# #<< upsilon >> = sqrt((8k_BT)/(pim))# #upsilon_"rms" = sqrt((3k_BT)/m)#

Notice that you can factor these out to get the following magnitude orders:

#sqrt3sqrt((k_BT)/m) > 2sqrt(2/pi)sqrt((k_BT)/(m)) > sqrt2*sqrt((k_BT)/m)#

That is,

*As your Maxwell-Boltzmann distribution plot shows, the root-mean square speed is farthest to the right on the graph, meaning that it is largest, and the most-probable speed is farthest to the left, meaning that it is smallest, as predicted in the comparison above.*