Maxwell-Boltzmann distribution of speeds — most probable, mean, rms

1. Building the distribution from scratch

HOW we derive it (Feynman-style, two ingredients):

Ingredient 1 — Boltzmann factor. The probability a molecule has kinetic energy $\varepsilon=\tfrac12 mv^2$ falls off like $e^{-\varepsilon/kT}=e^{-mv^2/2kT}$. WHY: fewer molecules can afford large energy at temperature $T$.

Ingredient 2 — Volume of "velocity shell". Velocity is 3D: $(v_x,v_y,v_z)$. All velocity vectors with speed near $v$ live on a spherical shell of radius $v$ and thickness $dv$. Its volume $\propto 4\pi v^2\,dv$. WHY the $v^2$: bigger spheres have more surface, so more ways to have a large speed — this pushes the peak away from $v=0$.

Multiply the two and add a normalizing constant $A$: $f(v) = A\,\underbrace{4\pi v^2}_{\text{how many ways}}\;\underbrace{e^{-mv^2/2kT}}_{\text{how likely}}$

Finding $A$ (normalization). Require $\int_0^\infty f(v)\,dv=1$. Using the Gaussian integral $\int_0^\infty v^2 e^{-\alpha v^2}dv=\tfrac14\sqrt{\pi/\alpha^3}$ with $\alpha=m/2kT$, you get $A=\left(\dfrac{m}{2\pi kT}\right)^{3/2}$.

2. The three characteristic speeds — each derived

Let $\alpha=\dfrac{m}{2kT}$ so $f(v)\propto v^2 e^{-\alpha v^2}$. Note $\dfrac1\alpha=\dfrac{2kT}{m}$.

(a) Most probable speed $v_{mp}$ — the PEAK

WHAT: the speed at which $f(v)$ is maximum (most molecules near here). HOW: set $\dfrac{df}{dv}=0$. $\frac{d}{dv}\Big(v^2 e^{-\alpha v^2}\Big)=\big(2v-2\alpha v^3\big)e^{-\alpha v^2}=0$ Why this step? Product rule; the exponential is never zero, so the bracket must vanish. $2v(1-\alpha v^2)=0\Rightarrow v^2=\frac1\alpha=\frac{2kT}{m}$ $\boxed{v_{mp}=\sqrt{\frac{2kT}{m}}=\sqrt{\frac{2RT}{M}}}$

(b) Mean (average) speed $\bar v$ — the BALANCE point

WHAT: average of speed itself, $\bar v=\int_0^\infty v\,f(v)\,dv$. HOW: need $\int_0^\infty v^3 e^{-\alpha v^2}dv=\dfrac{1}{2\alpha^2}$. $\bar v = 4\pi\Big(\frac{\alpha}{\pi}\Big)^{3/2}\cdot\frac{1}{2\alpha^2}=\frac{2}{\sqrt{\pi\alpha}}$ Why this step? Now substitute $\dfrac1\alpha=\dfrac{2kT}{m}$, so $\dfrac{1}{\pi\alpha}=\dfrac{2kT}{\pi m}$ and $\dfrac{2}{\sqrt{\pi\alpha}}=2\sqrt{\dfrac{2kT}{\pi m}}=\sqrt{\dfrac{4\cdot 2kT}{\pi m}}=\sqrt{\dfrac{8kT}{\pi m}}$ (the factor $2$ becomes $4$ inside the root, giving $8$, not $4$). $\boxed{\bar v=\sqrt{\frac{8kT}{\pi m}}=\sqrt{\frac{8RT}{\pi M}}}$

(c) Root-mean-square speed $v_{rms}$ — the ENERGY average

WHAT: $v_{rms}=\sqrt{\overline{v^2}}$, where $\overline{v^2}=\int_0^\infty v^2 f(v)\,dv$. This is the one tied to kinetic energy: $\overline{KE}=\tfrac12 m\overline{v^2}=\tfrac32 kT$. HOW: need $\int_0^\infty v^4 e^{-\alpha v^2}dv=\dfrac{3}{8}\sqrt{\dfrac{\pi}{\alpha^5}}$. $\overline{v^2}=4\pi\Big(\frac{\alpha}{\pi}\Big)^{3/2}\cdot\frac{3}{8}\sqrt{\frac{\pi}{\alpha^5}}=\frac{3}{2\alpha}=\frac{3kT}{m}$ $\boxed{v_{rms}=\sqrt{\frac{3kT}{m}}=\sqrt{\frac{3RT}{M}}}$

3. Effect of temperature and mass

4. Worked examples

5. Common mistakes (Steel-man + fix)

6. Flashcards

Concept Map

Hinglish (regional understanding)