Home Uncategorized Most probable velocity<average velocity<root mean square velocity so in the given figure why the slope of most probable velocity at the peak?

Most probable velocity<average velocity<root mean square velocity so in the given figure why the slope of most probable velocity at the peak?

Author:

February 20th, 2023

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 ##upsilon_”mp”##, ##<< upsilon >>##, and ##upsilon_”rms”## be the 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 ##sqrt((k_BT)/m)##, you get the order ##sqrt3 > 2sqrt(2/pi) > sqrt2##. That corresponds to the horizontal location of each type of speed on the graph. That is, ##upsilon_”rms” > << upsilon >> > upsilon_”mp”##, since molecular speed 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 ##upsilon_”mp”## is the 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 ##(dF(upsilon))/(dupsilon)## of the speed distribution function ##F(upsilon)## (with respect to speed ##upsilon##) is ##0##.
The ##y## axis is the fraction of molecules with that speed, as the graph states, so it says that ##upsilon_”mp”## is the speed that most molecules are likely to have (which is the intuitive interpretation of “most-probable” speed).
The Maxwell-Boltzmann speed distribution function (Physical Chemistry: A Molecular Approach, McQuarrie) is given as
##mathbf(F(upsilon) = 4pi(m/(2pik_BT))^”3/2″ upsilon^2 e^(-m upsilon^2″/”2k_BT))##
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 ##upsilon##, we would get:
##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 ##d/(dupsilon)[f(upsilon)g(h(upsilon))] = [f(upsilon)g'(h(upsilon))*h'(upsilon) + g(h(upsilon))f'(upsilon)]##, as follows:
##= 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 ##f(upsilon) = upsilon^2##, ##g(upsilon) = e^(-m upsilon^2″/”2k_BT)##, and ##h(upsilon) = -(m upsilon^2)/(2k_BT)##.
Now simply note that the constants can never be ##0##, so they can be divided out to leave:
##= 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, ##e^x ne 0##, so the only thing that can be ##0## is:
##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 ##x = upsilon##, ##n = 1##, and ##alpha = m/(2k_BT)## to get:
##= 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, ##upsilon_”rms” = sqrt(<< upsilon^2 >>)##. Back to the Maxwell-Boltzmann distribution! We update our previous formula for the average using the substitution ##upsilon -> upsilon^2##:
##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 ##2n+1 ne 4##:
##int_(0)^(oo) x^(2n)e^(-alphax^2)dx = (1cdot3cdot5cdots(2n-1))/(2^(n+1)alpha^n)(pi/(alpha))^”1/2″##
For this, ##alpha = m/(2k_BT)##, ##x = upsilon##, and ##n = 2##.
##=> 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, ##color(blue)(upsilon_”rms” > << upsilon >> > upsilon_”mp”)##
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.