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? Apr 14, 2016

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 ${\upsilon}_{\text{mp}}$, $\left\langle\upsilon\right\rangle$, and ${\upsilon}_{\text{rms}}$ be the most-probable, average, and root-mean-square speeds, respectively.

• ${\upsilon}_{\text{mp}} = \sqrt{\frac{2 {k}_{B} T}{m}}$
• $\left\langle\upsilon\right\rangle = \sqrt{\frac{8 {k}_{B} T}{\pi m}}$
• ${\upsilon}_{\text{rms}} = \sqrt{\frac{3 {k}_{B} T}{m}}$

From factoring out everything that is not $\sqrt{\frac{{k}_{B} T}{m}}$, you get the order $\sqrt{3} > 2 \sqrt{\frac{2}{\pi}} > \sqrt{2}$. That corresponds to the horizontal location of each type of speed on the graph. That is, ${\upsilon}_{\text{rms" > << upsilon >> > upsilon_"mp}}$, 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 ${\upsilon}_{\text{mp}}$ 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 $\frac{\mathrm{df} \left(\upsilon\right)}{\mathrm{du} \psi l o n}$ of the speed distribution function $f \left(\upsilon\right)$ (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 ${\upsilon}_{\text{mp}}$ 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)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 $\upsilon$, we would get:

$\textcolor{g r e e n}{\frac{\mathrm{df} \left(\upsilon\right)}{\mathrm{du} \psi l o n}}$

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

Using the product rule, we have $\frac{d}{\mathrm{du} \psi l o n} \left[g \left(\upsilon\right) h \left(k \left(\upsilon\right)\right)\right] = \left[g \left(\upsilon\right) h ' \left(k \left(\upsilon\right)\right) \cdot k ' \left(\upsilon\right) + h \left(k \left(\upsilon\right)\right) g ' \left(\upsilon\right)\right]$, 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 $g \left(\upsilon\right) = {\upsilon}^{2}$, $h \left(\upsilon\right) = {e}^{- m {\upsilon}^{2} \text{/} 2 {k}_{B} T}$, and $k \left(\upsilon\right) = - \frac{m {\upsilon}^{2}}{2 {k}_{B} T}$.

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

$= {e}^{- m {\upsilon}^{2} \text{/} 2 {k}_{B} T} \left[2 \upsilon - {\upsilon}^{3} \left(\frac{m}{{k}_{B} T}\right)\right] = 0$

Of course, ${e}^{x} \ne 0$, so the only thing that can be $0$ is:

$0 = 2 \upsilon - {\upsilon}^{3} \left(\frac{m}{{k}_{B} T}\right)$

So we get, given that speeds are always positive:

${\upsilon}^{{\cancel{3}}^{2}} \left(\frac{m}{{k}_{B} T}\right) = 2 \cancel{\upsilon}$

${\upsilon}^{2} = \frac{2 {k}_{B} T}{m} \implies \textcolor{b l u e}{{\upsilon}_{\text{mp}}} = \textcolor{b l u e}{\sqrt{\frac{2 {k}_{B} T}{m}}}$

AVERAGE SPEED

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

$\textcolor{g r e e n}{\left\langle\upsilon\right\rangle = {\int}_{0}^{\infty} \upsilon f \left(\upsilon\right) \mathrm{du} \psi l o n}$

= 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 $\alpha = \frac{m}{2 {k}_{B} T}$ to get:

$= 4 \pi {\left(\frac{m}{2 \pi {k}_{B} T}\right)}^{\text{3/2}} \cdot \frac{1}{2 {\left(\frac{m}{2 {k}_{B} T}\right)}^{1 + 1}}$

$= 4 \pi {\left(\frac{m}{2 \pi {k}_{B} T}\right)}^{\text{3/2}} \cdot \frac{2 {\left({k}_{B} T\right)}^{2}}{{m}^{2}}$

= 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"))

$= 8 \pi {\left(\frac{1}{2 \pi}\right)}^{\text{2/2"cdot(1/(2pi))^"1/2" cdot ((k_BT)/(m))^"1/2}}$

$= 4 \cdot {\left(\frac{{k}_{B} T}{2 \pi m}\right)}^{\text{1/2}}$

$\implies \textcolor{b l u e}{\left\langle\upsilon\right\rangle = \sqrt{\frac{8 {k}_{B} T}{\pi m}}}$

ROOT-MEAN-SQUARE SPEED

Now, for the root-mean-square speed!

By definition, ${\upsilon}_{\text{rms}} = \sqrt{\left\langle{\upsilon}^{2}\right\rangle}$. Back to the Maxwell-Boltzmann distribution! We update our previous formula for the average using the substitution $\upsilon \to {\upsilon}^{2}$:

$\textcolor{g r e e n}{\left\langle{\upsilon}^{2}\right\rangle = {\int}_{0}^{\infty} {\upsilon}^{2} f \left(\upsilon\right) \mathrm{du} \psi l o n}$

(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 $2 n + 1 \ne 4$:

${\int}_{0}^{\infty} {x}^{2 n} {e}^{- \alpha {x}^{2}} \mathrm{dx} = \frac{1 \cdot 3 \cdot 5 \cdots \left(2 n - 1\right)}{{2}^{n + 1} {\alpha}^{n}} {\left(\frac{\pi}{\alpha}\right)}^{\text{1/2}}$

For this, $\alpha = \frac{m}{2 {k}_{B} T}$, $x = \upsilon$, 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")]

$= \frac{3}{\cancel{2}} \frac{\cancel{\pi} m}{\cancel{2} \cancel{\pi} {k}_{B} T} {\left(\frac{\cancel{2} {k}_{B} T}{m}\right)}^{2}$

$= 3 \cancel{\frac{m}{{k}_{B} T}} {\left(\frac{{k}_{B} T}{m}\right)}^{\cancel{2}}$

$= \frac{3 {k}_{B} T}{m}$

Therefore:

$\textcolor{b l u e}{{\upsilon}_{\text{rms}}} = \sqrt{\left\langle{\upsilon}^{2}\right\rangle} = \textcolor{b l u e}{\sqrt{\frac{3 {k}_{B} T}{m}}}$

NOW WHAT?

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

• ${\upsilon}_{\text{mp}} = \sqrt{\frac{2 {k}_{B} T}{m}}$
• $\left\langle\upsilon\right\rangle = \sqrt{\frac{8 {k}_{B} T}{\pi m}}$
• ${\upsilon}_{\text{rms}} = \sqrt{\frac{3 {k}_{B} T}{m}}$

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

$\sqrt{3} \sqrt{\frac{{k}_{B} T}{m}} > 2 \sqrt{\frac{2}{\pi}} \sqrt{\frac{{k}_{B} T}{m}} > \sqrt{2} \cdot \sqrt{\frac{{k}_{B} T}{m}}$

That is, $\textcolor{b l u e}{{\upsilon}_{\text{rms" > << upsilon >> > upsilon_"mp}}}$

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.