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.

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##, #Hows does hubris play into the Odyssey? ,English,,

Get a Unique Answer HERE or Order a Similar Paper

Our website has a team of professional writers who can help you write any of your homework. They will write your papers from scratch. We also have a team of editors just to make sure all papers are of HIGH QUALITY & PLAGIARISM FREE. To make an Order you only need to click Ask A Question and we will direct you to our Order Page at WriteDemy. Then fill Our Order Form with all your assignment instructions. Select your deadline and pay for your paper. You will get it few hours before your set deadline.