Visual walkthrough — Second law — Kelvin-Planck statement, Clausius statement

1.7.18 · D2 · Physics › Thermodynamics › Second law — Kelvin-Planck statement, Clausius statement

Step 1 — Wo do boxes jo hum draw kar sakte hain

KYA. Neeche har machine in dono walls ke beech mein hogi. Wall se bahar jaane wale arrows = usse heat bahar ja rahi hai; andar jaane wale arrows = heat andar aa rahi hai.

KYUN. Poora Second Law in dono boxes ke beech arrows ke baare mein ek statement hai. Agar hum abhi picture-language fix kar lein, toh proof sirf arrows ki bookkeeping ban jaata hai.

PICTURE. Orange bar hot hai, blue bar cold hai, aur unke beech ka khaali space wahan hai jahan machines baithenge.

Step 2 — Engine, teen arrows mein draw ki gayi

KYA. Teen arrows: ek hot se andar aata hai, ek work ke roop mein bahar jaata hai, ek cold wall mein jaata hai.

TEEN KYUN? Kyunki $\Delta U=0$ incoming energy $Q_H$ ko exactly do bahar jaane wale pieces mein split karne pe majboor karta hai, $W$ aur $Q_C$. Do arrows bahar, ek andar.

PICTURE. Neeche arrow ki chaurai scale mein draw ki gayi hai: $Q_H$ sabse moti hai, aur aankhon se dekh sakte hain ki $Q_H=W+Q_C$.

Step 3 — Refrigerator, engine ulta chalaya gaya

KYA. Step 2 se compare karein: har arrow opposite direction mein point kar raha hai, aur work ab ek input hai.

PROOF KE LIYE KYUN ZAROORI HAI. Ek fridge hamara "ordinary, legal" uphill mover hai — isko $W>0$ chahiye. Baad mein aane wali forbidden magic machines fridges (ya engines) hongi jinmein kuch jagah zero hoga.

PICTURE. Step 2 aur Step 3 ko side by side rakho aur har arrowhead ki direction badal gayi hai.

Step 4 — Do forbidden machines draw karo

Ab hum exactly woh cheezein draw karte hain jo har statement ban karti hai. Forbidden machine ki picture woh hai jo hum secretly dobara banana chahenge.

KYA. Do impossible lagte boxes: perfect engine mein koi cold arrow nahi; free fridge mein koi work arrow nahi.

KYUN. Parent statements exactly yahi hain ki "yeh do boxes exist nahi kar sakte." Hamara kaam: dikhao ki agar ek exist karta hai, toh doosra bhi karna chahiye.

PICTURE. Neeche red arrows woh missing/forbidden pieces hain — ek perfect engine "ek engine hai jisme cold arrow delete ho gayi hai," ek free fridge "ek fridge hai jisme work arrow delete ho gayi hai."

Step 5 — Clausius todo ⟹ Kelvin–Planck bhi toot gaya

Yeh equivalence ka pehla aadha hai, pictures mein.

Ordinary engine exactly $Q_C$ reject kare (free fridge ki uphill heat se match karte hue). Cold wall track karo:

$\underset{\text{engine dumps in}}{\underbrace{+Q_C}}\underset{\text{free fridge sucks out}}{\underbrace{-Q_C}}=0$

Har symbol yahan: $+Q_C$ engine ka waste arrow cold mein; $-Q_C$ free fridge ka arrow cold se bahar; yeh hamare tuning ki wajah se equal hain, isliye sum zero hai — cold wall untouched rahti hai.

Ab poora combo track karo:

$W=Q_H-Q_C,\text{net heat from hot}=Q_H-Q_C=W$

toh combined box sirf hot wall se heat leta hai aur work $W$ produce karta hai, kuch nahi — yeh exactly ek K–P violator hai. ∎

KYA. Humne do legal-lagte parts fuse kiye aur cold reservoir existence se bahar cancel ho gaya.

$Q_C$ ko match karne ke liye tune KYUN? Taaki do cold arrows equal-aur-opposite hon. Koi aur choice cold wall ka leftover arrow chhodti hai aur "single reservoir" conclusion kharab ho jaata hai.

PICTURE. Neeche cold wall par do blue arrows dekho — same thickness, opposite direction — ek doosre ko mita dete hain, ek single-reservoir engine bachta hai.

Step 6 — Kelvin–Planck todo ⟹ Clausius bhi toot gaya

Mirror image.

Perfect engine $W=Q_H$ deta hai. Ise ek fridge ko do jo cold se $Q_C$ khich raha hai aur $Q_C+W$ hot mein dump kar raha hai. Hot wall track karo:

$\underset{\text{engine takes out}}{\underbrace{-Q_H}}\underset{\text{fridge dumps in}}{\underbrace{+(Q_C+W)}}=Q_C(\text{since }W=Q_H)$

Term by term: $-Q_H$ perfect engine hot wall ko drain kar raha hai; $+(Q_C+W)$ fridge use refill kar raha hai; kyunki $W=Q_H$ yeh cancel hokar net $+Q_C$ hot mein deliver ho jaata hai. Iske saath cold wall ne sirf $Q_C$ khoya.

$\text{net result: }{Q}_C\text{ cold}\to \text{hot move hua, }{W}_{\text{external}}=0.$

Bahar se koi work nahi li gayi, koi aur change nahi — yeh ek Clausius violation hai. ∎

KYA. Perfect engine ki work poori fridge mein loop ho jaati hai; baahri duniya ko sirf heat bina cost ke upar chadhte dikha.

Work KYUN cancel hoti hai? Kyunki $W=Q_H$: engine ka output exactly fridge ka input hai, isliye combo ke bahar koi bhi work meter kabhi nahi hilta.

PICTURE. Neeche, green work arrow ek internal loop hai (engine → fridge) — yeh box se kabhi bahar nahi jaata, isliye bahar se yeh box ek free fridge lagta hai.

Step 7 — Degenerate aur edge cases (taaki koi scenario surprise na kare)

KYA. Har "zero" jo hum plug karte hain exactly kisi ek forbidden machine par ya ek harmless do-nothing par land karta hai.

INHE SABB KYUN DIKHAO. Second Law boundaries ke baare mein ek statement hai. Sirf $Q_C$, $W$, aur $T_H-T_C$ ko zero tak push karke hi hum exactly dekh sakte hain kahan "legal" "forbidden" mein badal jaata hai.

PICTURE. Neeche number line efficiency $\eta$ ko $0$ se $1$ tak mark karti hai: strictly 1 se kam sab kuch allowed hai; ek akela point $\eta =1$ forbidden wall hai.

Ek picture mein summary

Proof ke dono hisse, ek frame mein: left par, free fridge + ordinary engine ⟹ cold wall cancel ⟹ perfect engine. Right par, perfect engine + ordinary fridge ⟹ work internally loop hoti hai ⟹ free fridge. Do forbidden machines ek doosre ko banate hain.

Recall