Mean free path, mean speed, RMS speed — derivations: Visual walkthrough

1.7.11 · D2 · Physics › Thermodynamics › Mean free path, mean speed, RMS speed — derivations

Shuru karne se pehle, woh words jo hum baar baar use karenge — har ek neeche ek picture ke saath pinned:

Ek molecule = ek tiny hard ball. Iska cross-width diameter kehlata hai, jise $d$ likha jaata hai.
Number density $n$ = space ke har cubic metre mein kitne molecules baithe hain, yaani $n = N / V$ (unhe $N$ count karo, volume $V$ se divide karo jisme woh rehte hain).
Mean speed $\overset{v}{ˉ}$ = ek molecule ki everyday-average speed (parent note mein $\overset{v}{ˉ} = 8 k_{B} T / π m$ ke roop mein derive ki gayi hai). Hum ise ek single known number — metres per second mein ek speed — ki tarah treat karte hain — aur kuch nahin.
Kisi quantity ka average — upar bar ke saath likha jaata hai, jaise $\overline{v^{2}}$ — bas iska matlab hai: har molecule ke liye measure karo, sab add karo, jitne hain usse divide karo. Toh $\overline{v^{2}}$ hai "squared speeds ka average". Hum Step 6 se is notation par depend karenge.

Step 1 — Ek ball, uski width, aur "touching" ka matlab

KYA. Hum pehle decide karte hain ki do molecules kab collide hue maane jaate hain.

KYUN. Poori derivation crashes count karne ke baare mein hai. Hum crashes tab tak count nahin kar sakte jab tak hum agree na karein ki crash kya hota hai. Ek ball ka size hota hai, toh "touching" ka matlab "centres same point par" nahin hai — iska matlab hai "centres ek certain distance par".

PICTURE. Do balls har ek diameter $d$ ki (toh har ek ki radius $d /2$ hai). Woh tab just touch karte hain jab unke centres exactly $d$ apart hon — kyunki $d /2 + d /2 = d$ .

Figure — Mean free path, mean speed, RMS speed — derivations

Yeh single most-fumbled point hai, toh hum ise abhi frame karte hain: effective "target radius" $d$ hai.

Step 2 — Ek ball ko point tak shrink karo, target ko radius $d$ tak grow karo

KYA. Hum real picture (do fat balls) ko ek aasaan se replace karte hain (ek point molecule fat targets ka peecha kar raha hai).

KYUN. Do moving fat balls track karna mushkil hai. Lekin ek collision sirf centres ke beech ki distance par depend karta hai. Toh hum geometry identical rakh sakte hain jabki moving molecule ko ek dot bana dein aur target molecule ko combined radius $d$ dein. Same collisions, simpler bookkeeping.

PICTURE. Moving molecule ek dot ban jaata hai. Har doosra molecule radius $d$ ki disc ban jaata hai. Crash tab hota hai jab dot ka path in discs mein se kisi mein enter kare.

Step 3 — Dot ek tube drag karta hai: the collision cylinder

KYA. Jab dot ek straight line mein fly karta hai, toh woh ek lamba pata tube sweep karta hai. Koi bhi target jiska centre tube ke andar hai, use hit kiya jaata hai.

KYUN. Hum count karna chahte hain ki dot kitne targets se milta hai. Woh count barabar hai kitne target-centres swept region mein aate hain. Reach $d$ wala straight-flying dot radius $d$ ka cylinder sweep karta hai.

PICTURE. Maano $t$ woh time stretch hai jitne ke liye hum molecule ko dekh rahe hain (seconds mein measure kiya). Is time mein dot ek distance = (uski speed) × (woh time) = $\overset{v}{ˉ} t$ travel karta hai. Isliye tube ke paas hai:

radius $d$ (Step 1 se reach),
length $\overset{v}{ˉ} t$ (speed × time).

$π d^{2}$ kyun aur $π (d /2)^{2}$ kyun nahin? Kyunki tube radius Step 1 se reach $d$ hai, physical radius $d /2$ nahin. Circle ka area $π r^{2}$ hai jahan $r = d$ .

Step 4 — Tube ke andar targets count karo

KYA. Count karo ki swept tube mein kitne molecules baithe hain.

KYUN. Tube ke andar har target ek collision hai. Unhe count karna time $t$ mein crashes ki sankhya deta hai.

PICTURE. Surrounding gas ko $n$ molecules per cubic metre se bharo. Tube ka volume (face area)×(length) hai. Andar ka count paane ke liye $n$ se multiply karo.

$Volume of tube = face area σ \times length \overset{v}{ˉ} t = π d^{2} \overset{v}{ˉ} t$

$Number of collisions = targets per m^{3} n \times tube volume π d^{2} \overset{v}{ˉ} t = nπ d^{2} \overset{v}{ˉ} t$

Step 5 — Distance ÷ crashes = mean free path (pehla draft)

KYA. Total distance flown ko crashes ki sankhya se divide karo taaki crashes ke beech ki average distance mile. Hum us average distance ko ek naam dete hain: mean free path, jise $λ$ (Greek letter "lambda") likha jaata hai.

KYUN. "Mean free path" literally do collisions ke beech ki average straight run ka matlab hai. Agar aap total length $\overset{v}{ˉ} t$ fly karte hain aur $nπ d^{2} \overset{v}{ˉ} t$ baar cheezein hit karte hain, toh har hit par aapne cover kiya:

PICTURE. Ek zig-zag path collisions se segments mein chop kiya gaya; $λ$ average segment length hai.

$λ = \frac{total distance}{number of collisions} = \frac{v ˉ t}{nπ d ^{2} v ˉ t}$

Ab cancellation dekhein — yeh satisfying part hai. $\overset{v}{ˉ}$ cancel ho jaata hai (top aur bottom), aur $t$ cancel ho jaata hai (top aur bottom):

$λ = \frac{v ˉ t}{nπ d ^{2} v ˉ t} = \frac{1}{nπ d ^{2}}$

Yeh pehla draft hai — acha hai, lekin ek simplification chhupata hai jo hum aage fix karte hain.

Step 6 — Chhipi hui simplification: targets bhi move kar rahe hain

KYA. Step 3 mein humne targets ko frozen treat kiya. Woh hain nahin — woh hamari dot ki tarah hi zoom kar rahe hain. Hum iske liye correct karte hain.

KYUN. Collision rate is baat par depend karta hai ki dot target ke relative kitna fast approach karta hai, ground speed par nahin. Do moving molecules ek doosre ke paas zyada fast aate hain (average par) ek still wale ki taraf move karne wale se.

PICTURE. Dot ki velocity $v_{1}$ aur ek target ki velocity $v_{2}$ ko arrows ki tarah draw karo. Jo matter karta hai woh hai relative velocity $v_{rel} = v_{1} - v_{2}$ ( $v_{2}$ ki tip se $v_{1}$ ki tip tak ka arrow).

Yaad karo bar ka matlab hai "sab molecules par average" (upar defined). Average karne ke liye clean quantity squared relative speed hai, kyunki squaring vector subtraction ko ek dot product mein turn karta hai jise hum expand kar sakte hain:

$\overline{v_{rel}^{2}} = \overline{(v_{1} - v_{2})^{2}} = ball 1 \overline{v_{1}^{2}} + ball 2 \overline{v_{2}^{2}} - = 0 (random directions) 2 \overline{v_{1} \cdot v_{2}} = 2 \overline{v^{2}}$

Middle term $\overline{v_{1} \cdot v_{2}}$ zero par average hota hai: do molecules apni directions independently choose karte hain, toh unke arrows every-which-way point karte hain aur dot product utna hi often positive hota hai jitna negative. Dono balls ke same speed statistics hain, toh $\overline{v_{1}^{2}} = \overline{v_{2}^{2}} = \overline{v^{2}}$ , jisse $\overline{v_{rel}^{2}} = 2 \overline{v^{2}}$ milta hai.

Intuition

$\overline{v_{rel}^{2}} = 2 \overline{v^{2}}$ se honest $2$ tak Ek quick warning: root of the average square, $\overline{v_{rel}^{2}} = 2 \overline{v^{2}}$ , RMS relative speed hai — exactly mean relative speed $\overline{∣ v_{rel} ∣}$ nahin jo hume collision rate ke liye actually chahiye. Dono close hain lekin identical nahin. Precise calculation batana worth it hai: agar aap two independent Maxwell–Boltzmann velocities ko $∣ v_{1} - v_{2} ∣$ mein feed karein aur mean lein, toh do Maxwell distributions ka difference khud ek Maxwell distribution hai jisme twice the variance hai. Uski mean speed exactly $2$ times single-molecule mean speed hai. Toh honest, distribution-averaged result crude estimate se agree karta hai: $\overline{∣ v_{rel} ∣} = 2 \overset{v}{ˉ} .$ Isliye $2$ (aur nahin, koi $8/ π$ -flavoured mess) exact factor hai. "Difference of two Maxwells" fact ke liye Maxwell-Boltzmann distribution dekhein.

Step 7 — $2$ ko uss jagah rakhein jahan woh belong karta hai

KYA. Collision rate (crashes per time) mein jo speed matter karta hai woh mean relative speed $2 \overset{v}{ˉ}$ hai, $\overset{v}{ˉ}$ nahin. Lekin woh distance jo dot actually space mein travel karta hai phir bhi $\overset{v}{ˉ} t$ hai (uski apni ground speed).

KYUN. Distance flown = own speed × time. Collisions suffered = relative speed × time × (targets per volume × area). Different speeds alag alag jagah jaate hain.

PICTURE. Same tube jaise Step 5 mein, lekin crash-counter ab $2$ times faster tick karta hai kyunki targets approach karte hain.

$λ = \frac{distance dot flies}{crashes} = \frac{v ˉ t}{nπ d ^{2} ( 2 v ˉ ) t} = \frac{v ˉ}{2 nπ d ^{2} v ˉ}$

Ek baar phir $\overset{v}{ˉ}$ cancel karo:

Step 8 — Lab quantities mein rewrite karo (pressure aur temperature)

KYA. Microscopic $n$ (jo aap directly measure nahin kar sakte) ko pressure aur temperature (jo aap kar sakte hain, ek gauge aur thermometer se) se swap karo.

KYUN. $n$ invisible hai; gas ke pressure aur temperature har lab ki wall par hote hain. Swap karne ke liye hume teen aur physical symbols chahiye, jo yahaan plain words mein introduce kiye hain:

Pressure $P$ — woh push per unit area jo gas apne container walls par exert karta hai (pascals, Pa mein measure kiya).
Absolute temperature $T$ — gas ki hotness absolute zero se measure ki gayi (kelvin, K mein).
Boltzmann constant $k_{B}$ — ek fixed conversion number ( $1.38 \times 1 0^{- 23}$ joules per kelvin) jo "ek molecule worth temperature" ko energy mein convert karta hai. Yeh microscopic aur everyday ke beech ka bridge hai.

Ideal gas law inhe $P V = N k_{B} T$ ke roop mein tie karta hai. Dono sides ko $V k_{B} T$ se divide karne par exactly woh number density milti hai jो hume chahiye:

$n = \frac{N}{V} = \frac{P}{k _{B} T}$

Ise boxed result mein substitute karo:

$λ = \frac{1}{2 nπ d ^{2}} = \frac{1}{2 π d ^{2}} \cdot \frac{1}{n} = \frac{1}{2 π d ^{2}} \cdot \frac{k _{B} T}{P}$

Edge & degenerate cases (extremes par kya hota hai)

Intuition Har knob ko limit tak push karo

$P \to 0$ (vacuum): $n \to 0$ , toh $λ \to \infty$ . Ek molecule bina kisi crash ke poora container cross kar sakta hai — yeh free-molecular regime hai, aur isliye vacuum flasks (aur space) molecules ko forever travel karne dete hain. Diffusion and effusion dekhein.
$P \to \infty$ (bahut dense): $λ \to 0$ . Molecules hamesha touching hain; "crashes ke beech straight flight" ka idea toot jaata hai aur gas liquid-like behave karta hai. Hamara billiard-ball model quietly $λ ≫ d$ assume karta hai, yaani bahut zyada empty space.
$d \to 0$ (point particles): $σ = π d^{2} \to 0$ , toh $λ \to \infty$ — sizeless particles kabhi kuch hit nahin karte. Isliye size $d$ essential hai: no size, no collisions, no viscosity or heat conduction.
Single molecule ( $n \to 0$ again): hit karne ke liye kuch nahin, $λ$ undefined/infinite hai — $1/ n$ law ke saath consistent.

Ek-picture summary

Is page ki sab cheez ek frame mein: ek dot radius- $d$ target discs ke field se fly karta hai, $π d^{2}$ cross-section ka cylinder drag karte hue; targets bhi rush karte hain (woh $2$ ); distance ÷ crashes $λ$ deta hai.

Recall Feynman retelling — poora walkthrough ek 12-year-old ko explain karo

Imagine karo tum ek marble ho jo andhe hokar ek aise floor par roll kar raha ho jis par doosre marbles bikre hain. Pehla sawaal: tum kab kisi ko "hit" karte ho? Jab tumhare centres milte hain tab nahin — jab woh ek marble-width ke andar aa jaate hain, kyunki dono marbles fat hain. Toh hum ek trick karte hain: pretend karo ki tum ek tiny dot ho aur har doosra marble double size tak puff up ho gaya hai. Ab, jab tum roll karte ho, tum apne peeche ek invisible tube drag karte ho us puffed target ki width ka. Jo bhi marble jiska centre tumhari tube mein hai woh crash hai. Crashes count karo: yeh hai floor kitna crowded hai, times tube ka face size ( $π d^{2}$ ), times tum kitne door roll kiye. Jo distance roll kiye use crashes ki sankhya se divide karo aur — magic — tumhari speed aur time dono cancel ho jaate hain, chhodke $λ = 1/ (nπ d^{2})$ . Ek last honesty check: doosre marbles still nahin baithe hain, woh tumhare paas bhi roll kar rahe hain, toh tum unhe thoda zyada often milte ho — lagbhag $2$ times zyada. Us $2$ ko bottom mein daalo aur ho gaya: $λ = 1/ (2 nπ d^{2})$ . Bade marbles ya zyada crowded floor ⇒ bumps ke beech shorter runs. Simple.

Recall Quick self-test

Tube radius $d$ kyun hai aur $d /2$ kyun nahin? ::: Collision centre-separation $d$ par hota hai (do radii ka sum), toh reach ek full diameter hai.
$λ$ mein $\overset{v}{ˉ}$ aur $t$ kyun cancel ho jaate hain? ::: Faster/longer fly karo aur tum zyada distance cover karte ho aur same proportion mein zyada targets hit karte ho — ratio unchanged rehta hai.
$2$ kahan se aata hai? ::: Targets bhi move karte hain; do Maxwell distributions par $∣ v_{1} - v_{2} ∣$ average karna $2 \overset{v}{ˉ}$ deta hai.
$P \to 0$ par $λ$ ka kya hota hai? ::: $n \to 0$ , toh $λ \to \infty$ — free flight, no collisions.

Flashcards

Mean free path in terms of n and d

λ = \frac{1}{2 nπ d ^{2}}

Mean free path in terms of P and T

λ = \frac{k _{B} T}{2 π d ^{2} P}

Why collision cross-section is

π d^{2}

not

π (d /2)^{2}

Collision occurs at centre separation

d

, so effective target radius is

d

Origin of the

2

factor

Mean relative speed of two molecules is

2 \overset{v}{ˉ}

(difference of two Maxwell distributions)

Dependence of

λ

on pressure at fixed T

λ \propto 1/ P