Second law — Kelvin-Planck statement, Clausius statement

1.7.18 · Physics › Thermodynamics

1. Stage set karna: heat engines aur refrigerators

Pehle yeh WHY define karna? Kyunki Second Law ke dono statements iske baare mein hain ki yeh do machines kya nahi kar sakti.

2. Kelvin–Planck statement (engine statement)

YEH kya forbid karta hai: ek "single-reservoir engine" jisme efficiency $η = 1$ ho ( $Q_{C} = 0$ ).

YEH pehle wrong kyun lagta hai? Tum kar sakte ho single step mein heat ko poori tarah work mein convert — jaise gas ki isothermal expansion saari absorbed heat ko work mein badal deti hai. Catch yeh hai "in a cycle." Gas expand ho gayi; dobara karne ke liye, use compress karna padega, jisme work lagti hai aur heat dump hoti hai. Ek poore cycle mein tum hamesha kuch heat reject karte ho. (Mistake 1 dekho.)

3. Clausius statement (refrigerator statement)

YEH kya forbid karta hai: ek refrigerator jo heat ko cold se hot move karne ke liye zero work ( $W = 0$ ) use kare.

WHY yeh wrong lagta hai: heat actually tumhare fridge mein cold se hot ki taraf flow karti hai! Lekin tumhara fridge wall mein plug hota hai — woh work $W > 0$ consume karta hai. Clausius sirf spontaneous, work-free version ko forbid karta hai.

4. Dono statements EQUIVALENT hain (derive karo!)

Yeh topic ka core hai. Hum prove karte hain: ek violate karo ⟹ tum dusra violate kar sakte ho. Isliye dono exactly same physics forbid karte hain.

Figure — Second law — Kelvin-Planck statement, Clausius statement

4a. Clausius violate karo ⟹ Kelvin–Planck violate hoga

Maan lo ek magic Clausius-violator heat $Q_{C}$ ko cold se hot ki taraf bina kisi work ke move karta hai.
Ab usi reservoirs ke beech ek ordinary engine chalao: woh hot se $Q_{H}$ absorb karta hai, work $W$ karta hai, cold ko $Q_{C}$ reject karta hai. (Iska rejected heat magic device ke $Q_{C}$ ke barabar choose karo.)

Yeh choice kyun? Taaki cold reservoir ka net heat exactly zero ho aur cancel ho jaaye.
Cold reservoir par net effect: $+ Q_{C}$ (engine se) $- Q_{C}$ (magic device ne kheeencha) $= 0$ .
Overall net effect: hot reservoir $Q_{H} - Q_{C}$ lose karta hai, aur combination work $W = Q_{H} - Q_{C}$ deliver karta hai, cold reservoir untouched.
Woh combination ek single (hot) reservoir se heat lekar work produce karta hai — yeh ek Kelvin–Planck violation hai. ∎

4b. Kelvin–Planck violate karo ⟹ Clausius violate hoga

Maan lo ek magic K–P engine hot reservoir se $Q_{H}$ leta hai aur use poori tarah work mein convert karta hai $W = Q_{H}$ (koi heat reject nahi).
Woh work ek ordinary refrigerator mein daalo: woh $W$ use karke cold se $Q_{C}$ kheenchta hai aur hot mein $Q_{C} + W$ dump karta hai.

Isme kyun daalo? Poora point yahi hai: free work fridge chalata hai.
Hot reservoir par net effect: $- Q_{H}$ (engine ko diya) $+ (Q_{C} + W)$ (fridge ne dump kiya) $= Q_{C}$ (kyunki $W = Q_{H}$ ).
Net effect: heat $Q_{C}$ cold se hot ki taraf move hui bina kisi external work ke aur bina kisi aur change ke — yeh ek Clausius violation hai. ∎

5. Worked examples

6. Common mistakes (steel-manned)

Recall Feynman: 12-year-old ko explain karo

Socho heat paani ki tarah hai. Paani khud se neechay (hot se cold) bahta hai — aasaan aur free. Paani ko upar push karne ke liye (cold se hot, jaise ek fridge) tumhe ek pump chahiye jo electricity par chale; yeh kabhi free nahi. Yehi Clausius idea hai. Ab socho girte paani se chalti ek chakki: tumhe paani ki saari motion work mein nahi badalti — kuch hamesha neechay splash ho jaata hai. Isliye ek aisi machine jo heat ko completely useful work mein, baar baar, bina kuch waste kiye convert kare, exist nahi kar sakti. Yehi Kelvin–Planck hai. Aur cool part: yeh do rules secretly ek hi rule hain jo do costumes pehne hain.

7. Flashcards

Kelvin–Planck statement batao.

Koi bhi cyclic device single reservoir se heat lekar use poori tarah work mein convert nahi kar sakti bina kisi aur effect ke (

η = 1

impossible).

Clausius statement batao.

Koi bhi cyclic device heat ko colder se hotter body mein sirf ek sole effect ke roop mein transfer nahi kar sakti (yaani zero work input ke saath).

Ek isothermal expansion mein poori heat se work conversion K–P violation kyun nahi hai?

Yeh cycle nahi hai; gas restore karne ke liye compression chahiye jo heat reject karta hai, isliye net

Q_{C} > 0

Engine cycle ke liye First-law relation.

W = Q_{H} - Q_{C}

, isliye

η = 1 - Q_{C} / Q_{H}

Hum K–P aur Clausius ko equivalent kaise prove karte hain?

Contradiction se: ek ka violator ek ordinary device ke saath combine hota hai doosre ka violator banane ke liye.

Clausius-violator ko ek ordinary engine ke saath combine karne se kya banta hai?

Ek single-reservoir engine jo work kare = Kelvin–Planck violator.

K–P-violator ko ordinary refrigerator chalane par kya banta hai?

Heat cold se hot move ho bina net work ke = Clausius violator.

Fridge ko Clausius violate na karne ke liye kya chahiye?

Ek nonzero work input

W > 0

(jaise electricity se).

Kisi bhi single-reservoir cyclic engine ki maximum efficiency?

Strictly 1 se kam; kabhi 100% nahi pa sakte.

Connections

Carnot engine and Carnot theorem — maximum $η = 1 - T_{C} / T_{H}$ quantify karta hai jo in statements se allowed hai.
Entropy and the Clausius inequality — quantitative form: $\oint \frac{d Q}{T} \leq 0$ .
First law of thermodynamics — $W = Q_{H} - Q_{C}$ supply karta hai jo har proof mein use hota hai.
Reversible and irreversible processes — Second Law irreversibility ki direction define karta hai.
Arrow of time — kyun macroscopic processes one-way hote hain.