Ideal gas law PV = nRT — derivation from kinetic theory: Visual walkthrough

1.7.8 · D2 · Physics › Thermodynamics › Ideal gas law PV = nRT — derivation from kinetic theory

Step 0 — The stage: balls ka ek box

Koi bhi maths se pehle, cast se milte hain.

Ek cube jiske har side ki length $L$ hai (toh uska volume $V = L^{3}$ hai — length times width times height, sab barabar).
Andar: $N$ chote balls (molecules), har ek ki mass $m$ hai (ek ball kitni heavy hai).
Har ball ek velocity ke saath zoom karti hai — ek arrow jo batata hai woh kitni fast aur kis direction mein move karti hai.

Hum us velocity arrow ko box ke teen edges ke saath teen pieces mein kaatate hain: $v_{x}$ (kitni fast left–right move karti hai), $v_{y}$ (front–back), $v_{z}$ (up–down). Yeh velocity components hain.

Figure — Ideal gas law PV = nRT — derivation from kinetic theory

Hum right wall ko study karenge, woh wali jo $x$ -axis se guzarti hai. Sirf $v_{x}$ hi us wall ko push karne ke liye matter karta hai.

Step 1 — Wall se ek bounce

KYA: ek single ball ko right wall se takraate aur seedha wapas bounce karte track karo.

KYUN: ek wall "temperature" feel nahi kar sakti. Woh sirf pushes — forces — feel karti hai. Newton ne bataya ki force kuch nahi balki momentum ka time ke saath badalna hai: $F = \frac{Δ p}{Δ t}$ jahaan $p = m v$ momentum hai (mass × velocity, "ek moving cheez mein kitna oomph hota hai") aur $Δ$ ka matlab hai "mein change". Toh koi bhi push find karne ke liye, pehle yeh dhundhna hoga ki ek ball har bounce mein kitna momentum transfer karti hai. Yahi yeh step hai. Dekho Pressure and Newton's Second Law.

PICTURE: ball right ki taraf aati hai $x$ -momentum $+ m v_{x}$ ke saath, aur left ki taraf jaati hai $x$ -momentum $- m v_{x}$ ke saath. Collision perfectly elastic hai — ball apni speed rakhti hai, bas direction flip ho jaata hai.

$x -momentum mein change Δ p_{x} = baad mein: left ki taraf (- m v_{x}) - pehle: right ki taraf (+ m v_{x}) = - 2 m v_{x}$

Ball ka momentum $- 2 m v_{x}$ se badla. Newton ke third law se ball wall ko opposite direction mein push karti hai, toh wall ko $2 m v_{x}$ size ka ek kick milta hai. Minus sign sirf "leftward" kehta hai; uski size hi wall ko push karti hai.

Step 2 — Woh wapas kitni baar aati hai?

KYA: usi ball ke usi wall par do hits ke beech ka time pata karo.

KYUN: ek kick steady push nahi hoti. Ek wall pressure feel karti hai kyunki kicks baar baar aate hain. Toh humein kicks ki rate chahiye, matlab unke beech ka time gap.

PICTURE: right wall se bounce karne ke baad, ball left wall tak poori tarah fly karti hai (distance $L$ ) aur wapas aati hai (aur $L$ ). Total round trip $= 2 L$ . $x$ -direction mein $v_{x}$ speed se travel karti hoi, time hai distance ÷ speed.

$Δ t = \frac{2 L round-trip distance}{v _{x} x -speed}$

Toh ek faster ball ( $v_{x}$ bada) jaldi wapas aati hai — woh zyada baar hit karti hai. Yeh yaad rakho: $v_{x}$ ab do jagah aaya hai — har kick ka size aur kicks kitni baar aate hain.

Step 3 — Ek ball se average push

KYA: Step 1 (kick size) aur Step 2 (kick rate) ko ek steady average force mein combine karo.

KYUN: kyunki ek wall jo 2,000 baar per second chote taps se hit ho rahi hai woh same feel karti hai jaise ek smooth push jo (tap size) × (taps per second) ke barabar ho. Woh product hi average force hai, seedha $F = Δ p /Δ t$ se.

$F_{1} = \frac{∣Δ p _{x} ∣}{Δ t} = \frac{2 m v _{x} har hit par kick (Step 1)}{har hit par time (Step 2) 2 L / v _{x}} = \frac{m v _{x}^{2}}{L}$

PICTURE: dekho $v_{x}$ do baar appear hota hai aur $v_{x}^{2}$ mein multiply ho jaata hai. Isliye final law mein speed-squared hota hai, speed nahi.

Step 4 — Saare $N$ balls ko add karo

KYA: one-ball force ko box ke har ball par sum karo.

KYUN: wall saari balls ko ek saath feel karti hai. Har ek independently contribute karti hai, toh total force sum hai. Aur $N$ cheezein ka sum $N$ times unke average ke barabar hota hai — exactly yahi "average" ka matlab hai.

$F = \frac{m}{L} \sum_{i = 1}^{N} v_{x, i}^{2} = \frac{m}{L} N v_{x}^{2} ka saari balls par mean ⟨ v_{x}^{2} ⟩$

Angle brackets $⟨ ⟩$ ka matlab hai "saare $N$ balls par average". Humne $N$ alag-alag numbers ka ek bada messy sum ek tidy quantity ke liye trade kar liya: mean square $x$ -speed.

Step 5 — Randomness se $\frac{1}{3}$ aata hai

KYA: wall-specific $⟨ v_{x}^{2} ⟩$ ko full speed $⟨ v^{2} ⟩$ se replace karo.

KYUN: humne $x$ -wall arbitrarily choose ki thi. Ek real gas mein, koi bhi direction special nahi hai — ise isotropy kehte hain (Greek: "sab directions mein same"). Toh average sideways-ness, average front-ness, aur average up-ness barabar honi chahiye: $⟨ v_{x}^{2} ⟩ = ⟨ v_{y}^{2} ⟩ = ⟨ v_{z}^{2} ⟩$

Kyunki $v^{2} = v_{x}^{2} + v_{y}^{2} + v_{z}^{2}$ hai, dono sides ka average lene par $⟨ v^{2} ⟩ = 3 ⟨ v_{x}^{2} ⟩$ milta hai, toh:

$⟨ v_{x}^{2} ⟩ = \frac{1}{3} ⟨ v^{2} ⟩$

PICTURE: speed-squared ek "pie" hai jo teen equal slices mein baanta hai, har direction ke liye ek. Har wall sirf apna slice feel karti hai — ek third.

Step 6 — Force se pressure tak

KYA: total force ko wall ke area se divide karke pressure mein badlo.

KYUN: pressure area par faili hui force hai — ek choti patch par badi force zyada press karti hai; usi force se ek badi wall mein barely press hota hai. Wall $L$ side ka ek square hai, toh uska area $L^{2}$ hai.

$P = \frac{F}{area} = \frac{F}{L ^{2}} = \frac{m N ⟨ v _{x}^{2} ⟩}{L ^{3}} = \frac{1}{3} \frac{m N ⟨ v ^{2} ⟩}{L ^{3}}$

Ab magic: $L^{3}$ bas volume $V$ hai. $L^{3} = V$ substitute karne par:

Step 7 — Bridge: temperature asal mein kya hai?

KYA: mechanical result ko temperature $T$ se connect karo.

KYUN: hamaara formula "speed" ki boli bolata hai; thermometers "temperature" ki. Humein ek translator chahiye. Nature ek deta hai: ek ideal gas ke liye, temperature ko average translational kinetic energy ke measure ke roop mein define kiya jaata hai (Equipartition Theorem ke zariye). Motion ke 3 directions mein se har ek $\frac{1}{2} k_{B} T$ energy store karta hai, toh teeno milke dete hain:

$avg kinetic energy ⟨ E_{k} ⟩ = \frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T$

Yahaan $k_{B}$ Boltzmann constant hai — woh chota number ( $1.38 \times 1 0^{- 23} J/K$ ) jo "degrees" ko "joules per molecule" mein convert karta hai. Dekho Boltzmann Constant and Gas Constant aur Temperature and Internal Energy.

PICTURE: ek dial jahan "hotness" (T) aur "average jiggle energy" ( $⟨ E_{k} ⟩$ ) do scales par padhi gayi same needle hain.

Ab Step 6 ke result ko us shape mein laao $\frac{1}{2}$ insert karke (2 se multiply aur divide karo):

$P V = \frac{1}{3} N m ⟨ v^{2} ⟩ = \frac{2}{3} N = ⟨ E_{k} ⟩ = \frac{3}{2} k_{B} T (\frac{1}{2} m ⟨ v^{2} ⟩) = \frac{2}{3} N \cdot \frac{3}{2} k_{B} T = N k_{B} T$

$\frac{2}{3}$ aur $\frac{3}{2}$ perfectly cancel ho jaate hain — ek sign ki yeh definition sahi key thi.

$P V = N k_{B} T$

Step 8 — Moles mein count karna

KYA: "balls ki sankhya $N$ " ko "moles ki sankhya $n$ " se swap karo.

KYUN: chemists gas ko moles mein count karte hain, individual molecules mein nahi. Ek mole mein $N_{A} = 6.022 \times 1 0^{23}$ molecules hote hain (Avogadro's number). Toh $N = n N_{A}$ .

$P V = N k_{B} T = (n N_{A}) k_{B} T = n ise R kahenge (N_{A} k_{B}) T$

Dono constants $N_{A} k_{B}$ ko ek single gas constant $R$ mein bundle karne par:

Real gases tab deviate karte hain jab "point particle, no forces" assumptions break hoti hain — woh hai Real Gases and Van der Waals Equation. $⟨ v^{2} ⟩$ ke peeche individual speeds ka spread Maxwell-Boltzmann Speed Distribution hai.

Ek-picture summary

Yeh raha poora climb ek slide par: bounce → kick → rate → force → sum → $\frac{1}{3}$ → pressure → temperature → moles.

Recall Feynman retelling — walkthrough plain words mein

Ek box imagine karo jo billions of tiny bouncy balls se bhara hua hai. Ek ball follow karo. Woh right wall se takraati hai aur seedha wapas bounce karti hai — uski sideways motion flip ho jaati hai, aur woh flip wall ko $2 m v_{x}$ ki ek kick deta hai. Phir woh across fly karti hai, far wall se hit hoti hai, wapas aati hai, aur phir kick karti hai — har $2 L / v_{x}$ seconds mein. Is size ki kick itni baar aana ek steady push jaise feel hota hai jitna $m v_{x}^{2} / L$ ho. Saare $N$ balls ka push add karo aur milega $N$ times average of $v_{x}^{2}$ . Lekin koi bhi direction special nahi hai, toh sideways jiggle bilkul total jiggle ka ek-third hai — wahan se woh famous $\frac{1}{3}$ aata hai. Total force ko wall ke area se divide karo aur pressure nikalti hai: $P = \frac{1}{3} N m ⟨ v^{2} ⟩ / V$ . Abhi tak yeh pure Newton hai — kahin bhi heat nahi. Last trick yeh realize karna hai ki "temperature" sirf nature ka word hai "average jiggle energy" ke liye: $\frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T$ . Woh slot in karo aur fractions cancel hokar $P V = N k_{B} T$ dete hain. Finally, molecules ki jagah moles mein count karo, dono constants ko $R$ mein roll karo, aur tumne $P V = n R T$ ek single bouncing ball se banaa liya.

Recall Rapid self-test

Force $v_{x}$ kyun nahi, $v_{x}^{2}$ se kyun scale karti hai? ::: Kick size $\propto v_{x}$ aur hit rate $\propto v_{x}$ ; woh multiply karte hain.
$\frac{1}{3}$ kahaan se aata hai? ::: Isotropy — $x$ teen equal directions mein se ek hai, $⟨ v_{x}^{2} ⟩ = \frac{1}{3} ⟨ v^{2} ⟩$ .
Kaunsa physical input mechanics ko thermodynamics mein badalta hai? ::: $\frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T$ (temperature = average kinetic energy).
$R$ kya hai? ::: $R = N_{A} k_{B}$ — Boltzmann's constant per-molecule se per-mole tak scale kiya gaya.

Parent: 1.7.08 Ideal gas law PV = nRT — derivation from kinetic theory (Hinglish)