Adiabatic relations — PV^γ = const, TV^(γ−1) = const (derivation): Visual walkthrough

1.7.16 · D2 · Physics › Thermodynamics › Adiabatic relations — PV^γ = const, TV^(γ−1) = const (deriva

Pehle koi bhi symbol aane se pehle, aao is natak ke har character ka naam lete hain.

Hume do "chhota change" wale symbols bhi chahiye. Jab main $d T$ likhta hun, ise "T mein ek bahut chhota change" padhte hain. Letter $d$ koi aisa number nahin hai jise tum multiply karo — yeh ek label hai jiska matlab hai ek infinitesimal nudge. Hum tiny nudges isliye use karte hain kyunki pressure aur volume process ke dauran saath-saath aur continuously change hote hain, isliye hum unhe ek waqt mein ek whisker track karte hain aur end mein whiskers ko jod dete hain (yahi jodna hai jo integral sign $\int$ Step 6 mein karega).

Step 1 — Box band karo: $Q = 0$

KYA. Hum gas ko ek bilkul insulated cylinder mein ek piston ke saath rakhte hain. Koi bhi heat walls se leak nahin hoti. Symbols mein poori process is condition ko follow karti hai: $heat through walls Q = 0.$

KYUN. Yeh akeli condition adiabatic aur baaki sab ke beech poora farak hai. Ek Isothermal Process mein heat andar aati rahti hai taaki $T$ fixed rahe; yahan hum us flow ko forbid karte hain. To gas ko apna piston push karne ke liye jis bhi energy ki zarurat hai, woh use apne andar se dhundni hogi.

PICTURE. Neeche, mothi chalk wall par hatching hai — woh hatching "insulated, koi heat cross nahin karta" ka universal engineering symbol hai. Andar aane ki koshish karte hue blue wiggly arrows sab rok diye jaate hain.

Figure — Adiabatic relations — PV^γ = const, TV^(γ−1) = const (derivation)

Step 2 — Energy ledger: First Law bina heat ke

KYA. Energy kabhi kho nahin jaati, sirf move hoti hai. First Law of Thermodynamics accountant ki equation hai: $heat in δ Q = change in stored energy d U + work pushed out δ W .$ $δ Q = 0$ set karna (hamara Step 1) deta hai: $0 = d U + δ W .$

KYUN. Yeh ek bookkeeping identity hai: andar aane wali heat ya to andar rehti hai ( $U$ badhati hai) ya work $W$ ke roop mein bahar jaati hai. Koi heat andar na aane se, right side ke dono terms cancel hone chahiye — agar gas work kharach karti hai, to uski stored energy bilkul utni hi girni chahiye.

PICTURE. $U$ ko piggy bank mein paise socho aur $δ W$ ko kharcha hua paisa. Koi allowance nahin ( $δ Q = 0$ ), spending bank ko ek-se-ek drain karta hai.

Step 3 — Dono terms ko gas quantities se naam do

KYA. Abstract $d U$ aur $δ W$ ko un chezon se replace karo jo hum measure kar sakte hain. $d U = n C_{V} d T and δ W = P d V .$ Dono ko $0 = d U + δ W$ mein plug karo: $0 = energy jo molecules gain/lose karte hain n C_{V} d T + piston push karne ka kaam P d V .$

Yahan $n$ moles ki sankhya hai (kitna gas hai), aur $C_{V}$ heat capacity at constant volume hai — ek mole ko ek degree warm karne mein kitne joules lagte hain jab volume fixed rakhte hain.

HEP kyon?

$d U = n C_{V} d T$ : ek ideal gas ke liye stored energy sirf temperature par depend karti hai. To $U$ mein koi bhi change $n C_{V}$ times temperature change hai — tab bhi jab volume change ho raha ho. (Yeh woh subtle point hai jo parent note steel-man karta hai: $C_{V}$ mein " $V$ " batata hai yeh kaise measure kiya gaya, na ki formula kab allowed hai.)
$δ W = P d V$ : piston ko volume ke ek sliver $d V$ se bahar push karne ke liye, gas pressure $P$ se press karta hai; force $\times$ distance $P d V$ ban jaata hai. Is geometry ke liye Work Done in Thermodynamic Processes dekho.

PICTURE. Piston ke neeche shaded sliver ka area $P d V$ hai — woh thin strip hi work hai. Chhota thermometer $d T$ dikhata hai, woh temperature nudge jo uske saath jaata hai.

Step 4 — Ideal-gas law use karke $P$ nikalo

KYA. Hamare paas teen variables ( $P$ , $V$ , $T$ ) hain lekin sirf do chahiye. Ideal-gas law $P V = n R T$ hume ek ko doosre se trade karne deta hai. Ise $P$ ke liye solve karo: $P = \frac{n R T}{V},$ aur $0 = n C_{V} d T + P d V$ mein substitute karo: $0 = n C_{V} d T + this is P \frac{n R T}{V} d V .$

Yahan $R$ universal gas constant hai (ek fixed number jo moles-and-degrees ko energy mein convert karta hai).

KYUN. Ek equation mein tangled teen unknowns integrate nahin ho sakte. $P$ eliminate karke hum choose kar rahe hain ki pehle $T$ aur $V$ ke beech ek relation pe pahunchein. Hum utni hi aasaani se $T$ eliminate karke seedha $P V^{γ}$ ki taraf bhi ja sakte the — same destination, alag raasta.

PICTURE. Ideal-gas law teen quantities ko connect karne wala bridge hai. Hum isse cross karke equation mein se $P$ hatate hain.

Step 5 — Variables alag karo aur $γ$ appear hone do

KYA. Har term ko $n C_{V} T$ se divide karo taaki har variable apni jagah baithe: $\frac{d T}{T} + \frac{R}{C _{V}} \frac{d V}{V} = 0.$ Ab akela messy coefficient $R / C_{V}$ hai. Mayer's Relation $C_{P} - C_{V} = R$ use karke: $\frac{R}{C _{V}} = \frac{C _{P} - C _{V}}{C _{V}} = = γ \frac{C _{P}}{C _{V}} - 1 = γ - 1.$ To equation ban jaati hai: $\frac{d T}{T} + (γ - 1) \frac{d V}{V} = 0.$

Symbol $γ = C_{P} / C_{V}$ ratio of specific heats hai, hamesha $1$ se bada. Yeh secretly encode karta hai ki molecules energy store karne ke kitne tarike rakhte hain — Degrees of Freedom and $\gamma$ dekho.

KYUN. "Variables separate karna" matlab hai saare $T$ wale ek term ki side aur saare $V$ wale doosri side, taaki har term ko akele integrate kiya ja sake. Aur $γ$ bilkul isliye aata hai kyunki Mayer ka substitution hota hai — woh exact moment hai jab gas ki inner nature ( $γ$ ) derivation mein enter karti hai.

PICTURE. Dekho tangled coefficient $R / C_{V}$ kaise clean $γ - 1$ mein collapse hota hai.

Step 6 — Saare whiskers jodo: integrate karo

KYA. Har term $\frac{d T}{T}$ aur $\frac{d V}{V}$ ek "fractional nudge" hai. Fractional nudges ka sum ek logarithm ( $ln$ ) deta hai, kyunki $ln$ precisely woh running total hai "ab tak yeh kitne percent badha hai." Integrate karne par: $ln T + (γ - 1) ln V = const .$

Logarithm kyun aur kuch nahin? Hume woh tool chahiye jo answer kare: "agar ek quantity har step mein fraction $d x / x$ se change hoti hai, to uska accumulated change kya hai?" Woh unique function jiska tiny change $d x / x$ hai woh $ln x$ hai. Koi aur tool us shape mein fit nahin hota — isliye $ln$ , na ki, maan lo, square root.

Ise product mein fold karo. Log rule $ln a + ln b = ln (ab)$ , aur $c ln V = ln V^{c}$ use karo: $ln (T V^{γ - 1}) = const ⟹ T V^{γ - 1} = const .$ Dono sides ka $ln$ lena trick tha; logs ka sum constant ke barabar hone ka matlab hai andar ka product constant hai.

PICTURE. $T_{1}$ se $T_{2}$ tak curve $1/ T$ ke neeche area hi $ln T_{2} - ln T_{1}$ hai. Figure un dono shaded areas ko balance hote dikhata hai taaki unka weighted sum fixed rahe.

Step 7 — $P V^{γ}$ aur $T^{γ} P^{1 - γ}$ mein convert karo

KYA. Hamare paas $T V^{γ - 1} = const$ hai. Pressure form paane ke liye, $T$ ko $T = \frac{P V}{n R}$ se replace karo: $\frac{P V}{n R} V^{γ - 1} = const ⟹ P V^{1 + (γ - 1)} = P V^{γ} = const .$ $V$ powers add hote hain: $V^{1} \cdot V^{γ - 1} = V^{γ}$ . $n R$ ek constant hai, isliye woh right side ke naye constant mein fold ho jaata hai.

Teesre relation ke liye, $P V^{γ} = const$ se shuru karo aur $V = \frac{n R T}{P}$ replace karo: $P (\frac{n R T}{P})^{γ} = const ⟹ P^{1 - γ} T^{γ} = const .$

KYUN. Teeno relations usi curve ko describe karte hain; inhe sirf re-express kiya gaya hai taaki jo bhi do variables tumhari problem deti hai unse match kare. (V-ratio diya aur $T$ chahiye? $T V^{γ - 1}$ use karo. $P$ diya? $P V^{γ}$ use karo.)

PICTURE. Ek curve, teen labels — $P$ , $V$ , $T$ ka ek triangle jahan ideal-gas law kisi bhi do faces ke beech convert karta hai.

Step 8 — Edge cases: pakde mat jaana

KYA AUR KYUN. Ek achhi picture corners mein bhi survive karni chahiye.

$γ \to 1$ (kalpnik "heat-hungry" gas): tab $γ - 1 \to 0$ , isliye $T V^{γ - 1} = T V^{0} = T = const$ aur $P V^{γ} \to P V = const$ . Adiabat isotherm mein degenerate ho jaata hai. Sanity check: $γ = 1$ ka matlab hoga $C_{P} = C_{V}$ , yaani constant pressure par heat add karne mein kuch extra nahin lagta — ek aisa gas jo kabhi temperature nahin badalta. To isotherm $γ \to 1$ limit hai. Real gases mein $γ > 1$ hota hai.
Compression ( $d V < 0$ ): work term $P d V$ ab negative hai, isliye $d U > 0$ : internal energy badhta hai, temperature badhta hai. Compression garm karta hai — bicycle pump. Same equation, $d V$ ka opposite sign.
Expansion ( $d V > 0$ ): $P d V > 0$ , isliye $d U < 0$ : expansion thanda karta hai — upar uthti cloud air.
$V \to 0$ ya $T \to 0$ : logs $ln V$ , $ln T$ $- \infty$ tak blow up ho jaate hain; physically ideal-gas model toot jaata hai (real gases liquify ho jaate hain). Yeh relation ek high-temperature, dilute-gas statement hai — iska domain respect karo.

PICTURE. Do adiabats (steeper, blue) ek isotherm (flatter, pink) ko same $P$ – $V$ axes par bracket karte hue, arrows ke saath jo dikhate hain "compress $\Rightarrow$ hotter, expand $\Rightarrow$ cooler." Isotherm $γ = 1$ degenerate adiabat hai.

Ek-picture summary

Upar sab kuch ek chain hai: box seal karo → energy ledger → terms ko naam do → $P$ hatao → $γ$ appear hone do → integrate karo → rename karo. Final figure poora flow compress karta hai, $Q = 0$ se teen boxed relations tak.

Recall Feynman retelling — saara walkthrough saral shabdon mein

Ek thermos socho jisme ek sliding lid hai aur hawa bhari hai. Hum ise seal karte hain taaki koi bhi garmahat andar ya bahar na ghus sake (Step 1: $Q = 0$ ). Ab accountant ka rule kehta hai: lid push karne mein air jo bhi energy kharach kare woh apni khud ki inner garmahat se churani hogi (Steps 2–3), kyunki piggy bank refill karne ke liye koi heat delivery nahin hai. Hum "kharche" ko $P d V$ aur "khoyi garmahat" ko $n C_{V} d T$ likhte hain, phir gas law use karke pressure symbol hatate hain (Step 4). Saaf divide karne par, messy factor $R / C_{V}$ gas ke personal number $γ - 1$ mein badal jaata hai (Step 5) — wahin gas ki character enter karti hai. Saare tiny fractional changes jodne par logarithms milte hain, aur constant rehne wale logs ka sum matlab ek product constant hai: $T V^{γ - 1} = const$ (Step 6). Temperature ko pressure se swap karo aur milta hai $P V^{γ} = const$ (Step 7). Thermos dabao aur trapped garmahat concentrate hoti hai — woh garm hoti hai; phailne do aur garmahat patli ho jaati hai — woh thandi hoti hai (Step 8). Yahi poori kahani hai, aur $P V^{γ} = const$ sirf uski rulebook hai.

Connections

First Law of Thermodynamics — Step 2 ka ledger $δ Q = d U + δ W$ .
Isothermal Process — Step 8 mein $γ \to 1$ degenerate case.
Mayer's Relation — Step 5 mein $R / C_{V}$ ko $γ - 1$ mein badalta hai.
Degrees of Freedom and $\gamma$ — $γ$ secretly kya count karta hai.
Work Done in Thermodynamic Processes — Step 3 ka $P d V$ sliver.
Speed of Sound in Gases — jahan adiabatic slope $γ P$ dobara aata hai.