Entropy change in irreversible processes — always - 0: Foundations

1.7.23 · D1 · Physics › Thermodynamics › Entropy change in irreversible processes — always - 0

Is page par assume kiya gaya hai ki aapne parent note Entropy change in irreversible processes ke koi bhi symbols pehle nahi dekhe hain. Hum har ek ko ek picture se build karenge, pehle equation mein aane se.

1. System, surroundings, universe

Kisi bhi symbol se pehle, hume jaanna hai ki hum dekh kya rahe hain.

Figure — Entropy change in irreversible processes — always - 0

Ye split kyun zaroori hai. Poora topic ek subtle point par tika hai: system akele orderliness kho sakta hai, phir bhi sum system + surroundings sirf use gain kar sakta hai. Agar hum kabhi boundary nahi kheeche, toh hum nahi keh sakte ki hum kis side ka score kar rahe hain.

2. "Change" symbol $Δ$ (pehle define karte hain)

Ise pehle kyun chahiye. Is topic ka har claim — aur neeche ki pictures — is baare mein baat karti hain ki koi quantity kaise badi. $Δ$ ko abhi naam dene ka matlab hai ki pehli equation jo aap milte hain wo already readable hai.

3. Heat aur symbols $Q$ , $δ Q$

Curly $δ$ kyun, na ki usual " $d$ "? Plain $d$ (jaise $d x$ mein) tab use hota hai jab koi cheez jo body own karti hai — ek state — usme tiny change ho. Lekin ek body heat own nahi karti; heat sirf in transit hoti hai. Curly $δ$ ek flag hai jo kehta hai: "ye ek flow ka sliver hai, kisi stored quantity ka change nahi." Aap ye nahi pooch sakte ki "gas mein kitni heat hai" — sirf ye ki "wall ke across kitni cross hui."

4. Temperature aur symbol $T$

Yahan Kelvin kyun zaroori hai, Celsius kyun nahi? Kyunki $T$ $δ Q / T$ ke denominator mein aayega. Agar hum $0^{\circ} C$ ya negative Celsius numbers allow karte, toh hum zero se ya negative se divide kar rahe hote — physically nonsense. Kelvin scale guarantee karta hai ki $T > 0$ hamesha, isliye ratio hamesha well-behaved hai.

5. Ek "state" aur ek state function

Picture ko words mein. Ek pahaad par altitude ke baare mein socho. Samundar ke level se aapki height sirf aap kahan khade hain par depend karti hai, na ki aapne kaunsa trail climb kiya. Summit par milne waale do hikers same altitude share karte hain chahe ek ne spiral path li ho aur doosre ne seedha upar gaya ho.

Topic is par kyun tika hai. Entropy $S$ (agla section) ek state function hai. Ye single fact free-expansion example ke peeche secret trick hai: hum entropy change ko kisi bhi convenient imaginary path ke along compute kar sakte hain same do states ke beech, chahe real process ne messier route li ho.

6. Entropy $S$ aur definition $d S = δ Q_{rev} / T$

Heat ko temperature se divide kyun karte hain? Kyunki heat ka wahi sliver zyada matter karta hai jab wo kisi thandi jagah land kare rather than garam jagah. Ek cup garam paani ek thandi lake mein daalo aur lake ki order barely hilaayegi; ek already ubalte hue pot mein daalo aur wo aur bhi kam hilaayegi. Ratio $δ Q / T$ exactly capture karta hai "ye heat kitna extra disorder khareed sakti hai," aur ye tab barhta hai jab $T$ ghatta hai — isliye $T$ denominator mein.

Subscript "rev" kyun? Kyunki ye clean equation sirf ek idealised, gradient-free path ke along exact hai. Real (irreversible) paths ke liye hum $Δ S$ compute karte hain ek reversible path imagine karke jiske same endpoints hon — exactly isliye legal kyunki $S$ ek state function hai (Section 5). Dekho Entropy as a state function.

7. Reversible vs irreversible, aur subscripts "rev" / "irr"

Topic ko is split ki zaroorat kyun hai. Parent ka poora claim hai $Δ S_{univ} \geq 0$ : " $=$ " reversible ideal ka hai, " $>$ " har real process ka. Dono ko naam liye bina, inequality ke do sides hi nahi hain. Dekho Reversible vs irreversible processes.

8. Surroundings ko hamesha $T_{surr}$ par kyun score kiya ja sakta hai

Ye point subtle enough hai ki iska apna section milta hai, kyunki Section 10 isi par lean karta hai.

Intuition Ek reservoir apne aap ke liye hamesha reversibly heat exchange karta hai

Surroundings ek huge reservoir hai — ek kamra, ek ocean — itna vast ki ek moderate $Q$ lena ya dena uska temperature bilkul nahi badalta. Uska temperature poore exchange mein $T_{surr}$ par pinned rehta hai. Kyunki exchange ke dauran uska apna temperature kabhi gradient develop nahi karta, reservoir ka entropy change clean formula se compute hota hai chahe overall transfer irreversible ho: $Δ S_{surr} = \frac{Q _{surr}}{T _{surr}} = \int \frac{δ Q _{surr}}{T _{surr}} .$ Irreversibility system aur surroundings ke beech finite gap mein rehti hai (figure s03, right), reservoir ke andar nahi. Isliye hum hamesha surroundings ko $T_{surr}$ par aur system ko apne $T$ par score karte hain; mismatch exactly wahi hai jo extra entropy produce karta hai.

9. Cyclic-integral symbol $\oint$ aur Clausius inequality

Yahan $T_{surr}$ (surroundings' temperature) kyun use hoti hai, system ka $T$ kyun nahi? Kyunki real transfer mein dono finite gap se differ karte hain. Har sliver ko us temperature par score karna hoga jahan bahar se koi heat hand over kar raha hai — exactly Section 8 ka reservoir reasoning. Dono ko mix karna parent ka mistake #3 hai.

10. Pieces ko saath laana: kyun $Δ S_{univ} \geq 0$

Ab hamare paas har wo symbol hai jo central claim ko assemble karne ke liye chahiye, sirf state karne ke liye nahi.

Formula Clausius se

$Δ S_{univ} \geq 0$ tak Step 1 — Real process se ek cycle banao. Koi bhi irreversible change state $A$ se state $B$ tak lo. Phir imagine karo $B \to A$ ek reversible path ke along wapas aana. Saath mein ye ek closed loop banate hain, isliye loop tool $\oint$ apply hota hai. Ye kaisa dikhta hai: real leg figure s02 ki messy pink route hai; return leg same do dots ke beech ek tidy blue route hai.

Step 2 — Loop par Clausius inequality apply karo. Real leg $δ Q_{irr}$ carry karta hai (irreversible slivers, Section 7); gentle return leg $δ Q_{rev}$ carry karta hai: $\int_{A}^{B} \frac{δ Q _{irr}}{T _{surr}} + \int_{B}^{A} \frac{δ Q _{rev}}{T} \leq 0.$ Kyun: $\oint = \int_{A}^{B} + \int_{B}^{A}$ ; ye " $\leq$ " hai kyunki ek leg irreversible tha.

Step 3 — Reversible leg ko naam do. Exactly definition $d S = δ Q_{rev} / T$ se, $\int_{B}^{A} \frac{δ Q _{rev}}{T} = S_{A} - S_{B} = - Δ S_{sys} .$ Kyun: $S$ ek state function hai, isliye ye integral exactly endpoints ke beech entropy difference hai.

Step 4 — Rearrange karo: $Δ S_{sys} \geq \int_{A}^{B} \frac{δ Q _{irr}}{T _{surr}} .$

Step 5 — Right-hand side ko surroundings ke taur par padho. Do facts, dono already build ho chuke hain:

Energy conservation (Section 3): system jo heat swallow karta hai wo heat surroundings hand over karte hain, $δ Q_{irr}^{sys} = - δ Q^{surr}$ .
Reservoir scoring (Section 8): surroundings ka apna entropy change $Δ S_{surr} = \int δ Q^{surr} / T_{surr}$ hai, yahan bhi valid kyunki reservoir $T_{surr}$ par rehta hai.

Combine karne par, right-hand integral $- Δ S_{surr}$ ke barabar hai: $\int_{A}^{B} \frac{δ Q _{irr}^{sys}}{T _{surr}} = \int \frac{- δ Q ^{surr}}{T _{surr}} = - Δ S_{surr} .$ Toh $Δ S_{sys} \geq - Δ S_{surr} .$

Step 6 — Parts add karo (entropy is extensive!). $Δ S_{surr}$ ko across laao aur $Δ S_{univ} = Δ S_{sys} + Δ S_{surr}$ use karo (legal kyunki entropy parts mein add hoti hai, Section 6): $Δ S_{sys} + Δ S_{surr} = Δ S_{univ} \geq 0$ Irreversible leg ke liye $> 0$ , reversible limit mein hi $= 0$ . Ye poora topic hai.

Prerequisite map

Equipment checklist

Δ X

ka matlab kya hai?

Final value minus initial value —

X

mein net change; positive = badha, negative = ghata.

δ Q

mein

δ

aapko kya warn karta hai?

Heat in transit hai, stored/owned quantity nahi — ye ek state function nahi hai.

Q

T

, aur

S

ki SI units kya hain?

Q

joules (J) mein,

T

kelvin (K) mein,

S

joules per kelvin (J/K) mein.

δ Q / T

ke andar

T

Kelvin mein kyun hona chahiye?

T

denominator mein baitha hai; Kelvin ise strictly positive rakhta hai taaki ratio hamesha physical rahe.

Celsius-to-Kelvin exact conversion kya hai?

T_{K} = T_{^{\circ} C} + 273.15

(classwork often

+ 273

round karta hai).

Entropy ko state function kya banata hai?

Uska change sirf do endpoint states par depend karta hai, path par nahi — isliye unke beech koi bhi reversible path same

Δ S

deta hai.

"Entropy extensive hai" ka kya matlab hai aur ye kyun matter karta hai?

Entropy parts mein add hoti hai (

S_{total} = S_{1} + S_{2}

), isliye

Δ S_{univ} = Δ S_{sys} + Δ S_{surr}

likhna legal hai.

δ Q_{rev}

aur

δ Q_{irr}

mein kya fark hai?

Same heat, same units — subscript record karta hai ki path reversible (gentle) tha ya irreversible (finite gap).

Heat

Q

ka sign convention kya hai?

System mein jaane par positive, system se bahar jaane par negative.

δ Q_{sys} = - δ Q_{surr}

kyun hota hai?

Energy conservation — heat transit mein conserved hoti hai, isliye jo ek body gain kare wo doosri khoye; minus sirf arrow ka direction flip hai.

Irreversible transfer ke liye bhi

Δ S_{surr} = \int δ Q_{surr} / T_{surr}

kyun ho sakta hai?

Surroundings ek huge reservoir hain jinka temperature

T_{surr}

par rehta hai; irreversibility bodies ke beech gap mein hai, reservoir ke andar nahi.

\int

aur

\oint

mein kya fark hai?

\int_{A}^{B}

ek open path ke along slivers summate karta hai;

\oint

unhe start tak wapas ek closed loop ke around summate karta hai.

\oint δ Q / T_{surr}

positive kyun nahi ho sakta?

Positive loop allow karta single reservoir se heat ko fully work mein convert karna bina kisi aur effect ke, jo second law forbid karta hai.

Clausius equality kab milti hai?

Sirf ek fully reversible cycle ke liye.

Clausius inequality mein kaun sa temperature jaata hai?

Surroundings ka temperature

T_{surr}

jahan har heat sliver boundary cross karta hai.

Kya

δ Q = 0

hone par bhi

Δ S > 0

ho sakta hai?

Haan — ek adiabatic irreversible process (jaise free expansion) mein

Q = 0

hone par bhi

Δ S_{univ} = n R ln 2 > 0

hota hai.

Clausius se

Δ S_{univ} \geq 0

tak ke six steps outline karo.

Ek cycle banao (irrev + rev return);

\oint \leq 0

apply karo; reversible leg ko

- Δ S_{sys}

naam do; rearrange karo; RHS ko

- Δ S_{surr}

padhо; parts add karo

Δ S_{univ} \geq 0

paane ke liye.