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QUESTION

# Most probable velocity&lt;average velocity&lt;root mean square velocity so in the given figure why the slope of most probable velocity at the peak?

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.