Mean free path, mean speed, RMS speed — derivations

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1. Setting up: the Maxwell–Boltzmann speed distribution

Molecules don't all move at the same speed. The fraction with speed in $[v, v+dv]$ is:

$f(v)\,dv = 4\pi N \left(\frac{m}{2\pi k_B T}\right)^{3/2} v^2 e^{-\frac{mv^2}{2k_BT}}\,dv$

WHY three different numbers? Because $\langle v\rangle \ne \sqrt{\langle v^2\rangle}$ for any spread-out distribution. Squaring before averaging gives extra weight to fast molecules, so $v_{rms} > \bar v > v_p$ always.

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2. Deriving RMS speed — from pressure first principles

WHAT: Get $v_{rms}$ without even needing the full distribution, using the kinetic theory of pressure.

HOW (derivation from scratch):

Consider one molecule of mass $m$ in a cube of side $L$, moving with $x$-velocity $v_x$. It bounces off a wall.

1. Momentum change per collision with one wall: $\Delta p = mv_x - (-mv_x) = 2mv_x$ Why this step? The wall reverses $v_x$; momentum is a vector so the change is twice the magnitude.

2. Time between hits on that wall: it must travel $2L$ (there and back): $\Delta t = \frac{2L}{v_x}$

3. Force from one molecule = rate of momentum transfer: $F = \frac{\Delta p}{\Delta t} = \frac{2mv_x}{2L/v_x} = \frac{mv_x^2}{L}$

4. Pressure from $N$ molecules on area $L^2$: $P = \frac{\sum F}{L^2} = \frac{m}{L^3}\sum v_x^2 = \frac{Nm}{V}\,\overline{v_x^2}$ Why this step? $L^3 = V$, and we average $v_x^2$ over all molecules.

5. Isotropy: no direction is special, so $\overline{v_x^2}=\overline{v_y^2}=\overline{v_z^2}$, and $\overline{v^2}=\overline{v_x^2}+\overline{v_y^2}+\overline{v_z^2}=3\,\overline{v_x^2}$ Thus $\overline{v_x^2} = \tfrac13 \overline{v^2}$ and:

6. Compare with ideal gas $PV = Nk_BT$: $\frac{1}{3}m\overline{v^2}=k_BT \;\Rightarrow\; \overline{v^2}=\frac{3k_BT}{m}$

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3. Deriving mean speed $\bar v$ from the distribution

$\bar v = \frac{\int_0^\infty v\, f(v)\,dv}{\int_0^\infty f(v)\,dv}$

Let $a=\dfrac{m}{2k_BT}$. The shape is $v^2 e^{-av^2}$ (constants cancel in the ratio).

• Numerator $\int_0^\infty v\cdot v^2 e^{-av^2}dv = \int_0^\infty v^3 e^{-av^2}dv = \dfrac{1}{2a^2}$.
• Denominator $\int_0^\infty v^2 e^{-av^2}dv = \dfrac{\sqrt\pi}{4a^{3/2}}$.

Why these integrals? They're standard Gaussian moments; the odd one ($v^3$) is elementary (sub $u=v^2$), the even one uses $\int_0^\infty e^{-av^2}dv = \tfrac12\sqrt{\pi/a}$ differentiated w.r.t. $a$.

$\bar v = \frac{1/(2a^2)}{\sqrt\pi/(4a^{3/2})}=\frac{4a^{3/2}}{2a^2\sqrt\pi}=\frac{2}{\sqrt{\pi a}}=\sqrt{\frac{4}{\pi a}}$

Substitute $a=\frac{m}{2k_BT}$:

Similarly the peak gives $v_p=\sqrt{\dfrac{2k_BT}{m}}$.

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4. Deriving mean free path $\lambda$

WHAT: Average distance a molecule travels between collisions.

HOW:

1. Model molecules as spheres of diameter $d$. Two molecules collide when their centres come within $d$. So the moving molecule sweeps a collision cylinder of radius $d$, i.e. cross-section: $\sigma = \pi d^2$ Why $d$ not $d/2$? Collision happens when centre-to-centre distance $= d$ (both have radius $d/2$). Effective target radius is $d$.

2. In time $t$ a molecule sweeps volume $\sigma\,\bar v\, t$. If $n=N/V$ is number density, number of collisions $= n\sigma\bar v t$.

3. Mean free path = total distance / number of collisions: $\lambda = \frac{\bar v\, t}{n\sigma\bar v\, t}=\frac{1}{n\sigma}=\frac{1}{n\pi d^2}$

4. Correction: other molecules also move. The relative speed between two molecules averages $\sqrt2\,\bar v$ (vector difference of two independent velocities of equal mean speed). So replace $\bar v$ with $\sqrt2\,\bar v$ in the collision rate:

Reading the formula: $\lambda \propto 1/P$ (squeeze the gas → more crowding → shorter flights) and $\lambda\propto T$ at fixed $P$.

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5. Worked examples

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6. Common mistakes (Steel-man + fix)

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7. Active recall

Recall Self-test (hide answers)

• Why is $v_{rms}>\bar v$?
• Where does the $\sqrt2$ in $\lambda$ come from?
• Why is $\overline{v_x^2}=\tfrac13\overline{v^2}$?
• What is the temperature dependence of $\lambda$ at fixed $P$? Of $v_{rms}$?

Recall Feynman (explain to a 12-year-old)

Imagine a room full of bouncing super-balls. If you ask "how fast is a typical ball?" you can answer three ways. The fairest energy-answer squares each speed, averages, then square-roots (RMS) — that's because faster balls carry way more punch. The plain average ($\bar v$) is a touch slower. And "how far does a ball zip before banging into another?" depends on how crowded the room is: pack twice as many balls in, and each one travels half as far before a crash. That crash distance is the mean free path.

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Flashcards

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Connections

• Maxwell-Boltzmann distribution — parent of all three speeds
• Kinetic theory of pressure — source of $PV=\tfrac13Nm\overline{v^2}$
• Ideal gas law — closes the loop to get $T$
• Equipartition theorem — explains $\tfrac32 k_BT$ per molecule
• Diffusion and effusion — uses $\bar v$ and $\lambda$
• Viscosity and thermal conductivity of gases — both $\propto \lambda \bar v$
• Speed of sound in gases — comparable to $v_{rms}$

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, gas ke andar molecules billiard balls ki tarah idhar-udhar bhaag rahe hote hain, alag-alag speeds par. Inko describe karne ke liye teen numbers use karte hain. RMS speed ($\sqrt{3RT/M}$) wo speed hai jo energy carry karti hai, kyunki kinetic energy $v^2$ par depend karti hai — isliye square karke average lete hain phir root. Mean speed ($\sqrt{8RT/\pi M}$) simple average hai, jo collision rate aur diffusion control karti hai. Aur ek rule yaad rakho: hamesha $v_p < \bar v < v_{rms}$, ratio approx $1.41:1.60:1.73$.

RMS ka derivation pressure se aata hai. Ek molecule wall se takrata hai, momentum change $2mv_x$, do takkaron ke beech time $2L/v_x$, force = $mv_x^2/L$. Sab molecules jodo to $PV=\tfrac13 Nm\overline{v^2}$. Isko ideal gas $PV=Nk_BT$ se compare karo, mil jata hai $\overline{v^2}=3k_BT/m$. Bas yahi $v_{rms}$ hai. Yahan trick hai isotropy: koi direction special nahi, isliye $\overline{v_x^2}=\tfrac13\overline{v^2}$.

Mean free path $\lambda$ matlab ek molecule kitni doori chalta hai do takkaron ke beech. Molecule ek cylinder sweep karta hai jiska cross-section $\pi d^2$ hai (yaad rakho, $d$ — diameter, kyunki collision centre-to-centre distance $d$ par hoti hai, $d/2$ nahi). Toh $\lambda = 1/(n\pi d^2)$. Par ek important correction: doosre molecules bhi move kar rahe hain, isliye relative speed average $\sqrt2\,\bar v$ ho jati hai, aur ek $\sqrt2$ neeche aa jata hai: $\lambda = 1/(\sqrt2\,n\pi d^2)=k_BT/(\sqrt2\,\pi d^2 P)$.

Yeh sab kyun zaroori? Kyunki yeh invisible micro-world ko measurable cheezon se jodta hai — pressure, temperature, sound speed, diffusion. Common galti: $M$ ko grams me daal dena (kg/mol use karo!), ya $\sqrt2$ bhool jaana. Bas formula ratta mat maaro, derivation samjho — exam me kuch bhi pucha jaye nikaal loge.