Entropy change in irreversible processes — always - 0: Question bank

1.7.23 · D5 · Physics › Thermodynamics › Entropy change in irreversible processes — always - 0

Definition Symbol glossary — pehle yeh padho

Is page par use hone wale har symbol ka plain words mein matlab:

$S$ — entropy, ek number jo score karta hai ki ek state kitni spread-out / mixed-up hai (ek state function: sirf state par depend karta hai, is par nahin ki tum wahan kaise pahunche). Dekho Entropy as a state function.
$Δ S_{system}$ (yeh bhi $Δ S_{sys}$ likha jaata hai) — jis cheez ko hum study kar rahe hain (gas, ice) uski entropy ka change, start aur end states ke beech.
$Δ S_{surr}$ — surroundings ki entropy ka change (room, woh reservoir jise system touch karta hai).
$Δ S_{univ}$ — total: $Δ S_{univ} = Δ S_{system} + Δ S_{surr}$ . Yahi quantity hai jis par second law constraint lagata hai.
$δ Q$ — boundary cross karne wali heat ki ek tiny-si quantity (sign upar ke anusaar). $δ$ ( $d$ nahin) warn karta hai ki heat path-dependent hai, state function nahin.
$δ Q_{rev}$ — woh tiny heat jab process reversible hoti hai (infinitesimal gradient se drive hoti hai). Sirf yahi version entropy define karta hai: $d S = δ Q_{rev} / T$ .
$δ Q_{irr}$ — ek real (irreversible) process mein actual tiny heat; entropy-production equation mein yahi increment (reversible wala nahin) appear karta hai neeche.
$T$ — heat cross karne ke waqt heat absorb/release karne wale body ka absolute temperature (kelvin mein).
$T_{surr}$ — boundary par surroundings ka temperature; $T_{H}$ , $T_{C}$ — hot aur cold reservoir temperatures; $T_{m}$ — ice ka melting temperature ( $273 K$ ), woh phase change ke dauran fixed $T$ .
$σ$ (entropy production) — irreversibility se manufacture hone wali entropy, defined by $d S_{system} = \frac{δ Q _{irr}}{T _{surr}} + δ σ, δ σ \geq 0$ jahan $δ Q_{irr}$ boundary cross karne wali actual (irreversible) heat hai; $δ σ > 0$ exactly tab jab ek finite driving gradient present ho.

Figure — Entropy change in irreversible processes — always - 0

Formula Clausius-cycle proof, step by step draw kiya gaya

Neeche kai "Why" answers ke peechhe yahi reasoning hai, toh hum ise ek baar pura build karte hain. Figure s02 picture hai; coloured legs follow karo.

Step 1 — KYA: loop band karo. Koi bhi real irreversible process lo state $A$ se state $B$ tak (orange leg). Phir sochо ki $B$ se $A$ wapas reversible path ke along jaana (teal leg). KYUN: hum sirf ek hi law assume kar sakte hain $\oint δ Q / T_{surr} \leq 0$ , aur yeh sirf closed cycles ke baare mein bolta hai — toh loop band karna padega. DEKHNE MEIN: orange + teal legs diagram par ek closed loop form karte hain.

Step 2 — KYA: poore loop par Clausius lagao. $orange, irreversible \int_{A}^{B} \frac{δ Q _{irr}}{T _{surr}} + teal, reversible \int_{B}^{A} \frac{δ Q _{rev}}{T} \leq 0$ KYUN $\leq$ hai $=$ nahin: loop ka ek hissa irreversible tha, aur irreversibility hamesha loop integral ko zero se neeche karta hai.

Step 3 — KYA: teal leg ko $- Δ S$ pehchano. Definition se $d S = δ Q_{rev} / T$ , aur $S$ ek state function hai, toh $\int_{B}^{A} δ Q_{rev} / T = S_{A} - S_{B} = - Δ S_{system}$ . KYUN: reversible leg literally unhi do endpoints ke beech entropy difference ki definition hai.

Step 4 — KYA: rearrange karo. Step 3 ko Step 2 mein substitute karke: $\int_{A}^{B} \frac{δ Q _{irr}}{T _{surr}} - Δ S_{system} \leq 0 \Rightarrow Δ S_{system} \geq \int_{A}^{B} \frac{δ Q _{irr}}{T _{surr}} .$ Right-hand side $- Δ S_{surr}$ hai (woh heat jo surroundings lose karta hai, unke $T_{surr}$ par). Toh $Δ S_{system} + Δ S_{surr} \geq 0$ , yaani $Δ S_{univ} \geq 0$ .

True or false — justify karo

True or false: Har real process mein system ki entropy badhti hai.

False — sirf universe ki entropy

Δ S_{univ}

ko badhna chahiye; ek system entropy lose kar sakta hai (freezing mein

Δ S_{sys} = - Q / T_{m} < 0

hota hai) bas surroundings ko utna hi ya zyada gain karna chahiye.

True or false:

Δ S_{univ} = 0

wali process physically impossible hai.

False — equation

d S_{sys} = δ Q_{irr} / T_{surr} + δ σ

ko uski floor

δ σ = 0

par set karna exactly reversible limit deta hai; yeh idealized boundary hai jo har driving gradient

\to 0

hone par reach hoti hai, forbidden state nahin. Dekho Reversible vs irreversible processes.

True or false: Agar ek gas compress hoti hai aur uski entropy girti hai, toh second law violate hoti hai.

False — compression work heat

∣ Q ∣

surroundings mein push karta hai, toh

Δ S_{surr} = + ∣ Q ∣/ T_{surr}

utna hi badhta hai jitna

Δ S_{sys}

girta hai,

Δ S_{univ} \geq 0

maintain karte hue.

True or false: Adiabatic (

Q = 0

) process ke liye,

Δ S_{system} = 0

False — sirf ek reversible adiabat mein

Δ S_{sys} = 0

hota hai; ek irreversible adiabat (jaise free expansion, Figure s03 mein dashed path) mein

Q = 0

hota hai phir bhi

Δ S_{sys} = n R ln 2 > 0

hota hai.

True or false: Ek slow, quasi-static process hamesha reversible hoti hai.

False — slowness sirf pressure/temperature gradients ko zero tak shrink karti hai; friction phir bhi har step mein

δ σ > 0

inject karta hai, toh quasi-static-with-friction process irreversible hai.

True or false: Entropy production

σ

ek state function hai jaise

S

False —

S

sirf endpoints par depend karta hai, lekin

σ

(raaste mein manufacture hone wali entropy) path par depend karta hai, exactly jaise Figure s02 proof mein do alag loop-integrals dikhate hain.

True or false: Carnot cycle ke liye

\oint δ Q_{rev} / T = 0

hota hai.

True — Carnot cycle reversible hai, toh heat body ke apne

T

par cross karta hai (

T_{surr} = T

) aur Clausius inequality equality ke saath hold karta hai.

True or false: Agar kisi interval mein

Δ S_{univ} < 0

ho, toh process ko khud ko correct karne ke liye bas zyada time chahiye.

False —

Δ S_{univ} \geq 0

exact aur instantaneous hai (

δ σ \geq 0

har step par), toh ek real isolated system ke liye net decrease kabhi nahin hota; correct karne ke liye kuch bhi nahin hai.

True or false: Cold se hot heat flow karna hamesha second law violate karta hai.

False — yeh theek hai agar drive kiya jaaye (ek fridge); driving work kaafi extra entropy kahin aur dump karta hai ki

Δ S_{univ} \geq 0

overall rahe. Sirf spontaneous cold→hot flow, bina kisi input ke, forbidden hai.

Error dhundo

"Free expansion mein

Q = 0

hai, toh

Δ S = Q / T = 0

." — galti dhundo.

Formula

d S = δ Q_{rev} / T

reversible heat require karta hai; free expansion irreversible hai (dashed path, Figure s03), toh tum ek imagined reversible isothermal path (solid curve) ke through route karte ho jo same endpoints ke saath

Δ S = n R ln 2 > 0

deta hai.

"Finite

Δ T

ke across heat flow ke liye,

\int δ Q / T

mein dono bodies ke liye ek temperature

T

use karo." — galti dhundo.

Har body ko apne temperature par score kiya jaata hai: hot

Q / T_{H}

lose karta hai, cold

Q / T_{C}

gain karta hai; yeh mismatch exactly

σ = Q (1/ T_{C} - 1/ T_{H}) > 0

hai jo Figure s01 mein plot hai.

Δ S_{system} = \int_{A}^{B} δ Q_{irr} / T_{surr}

exactly." — galti dhundo.

Yeh

\geq

hona chahiye,

=

nahin; gap entropy production

δ σ \geq 0

hai (Figure s02 proof ka Step 4), aur equality sirf reversible limit mein hold hoti hai — yeh Clausius inequality hai.

"Kyunki entropy hamesha badhti hai, ek refrigerator apne contents ki entropy kam nahin kar sakta." — galti dhundo.

Contents (system) entropy lose kar sakta hai; compressor ka work heat plus extra entropy room mein dump karta hai (

Δ S_{surr} > 0

), toh total

Δ S_{univ}

phir bhi badhta hai.

"Ice melt hone se room thanda hota hai, toh

Δ S_{univ}

negative ho sakta hai." — galti dhundo.

Wahi heat

Q > 0

ice mein lower melting temperature

T_{m}

par enter karta hai (gain

+ Q / T_{m}

) jabki higher

T_{surr}

par room leave karta hai (loss

- Q / T_{surr}

); chhhote

T_{m}

se divide karne par gain bada ho jaata hai, toh

Δ S_{univ} > 0

hota hai.

"Ek reversible process mein

Δ S_{system} = 0

hota hai." — galti dhundo.

Reversible sirf

Δ S_{univ} = 0

force karta hai; system ki entropy freely change ho sakti hai (reversible isothermal expansion mein

Δ S_{sys} = n R ln 2 > 0

hota hai) jab tak surroundings exactly ulti amount se change ho.

Why questions

Kyun hum

Δ S

ek reversible path ke along evaluate karte hain, chahe process irreversible ho?

Kyunki

S

ek state function hai — uska change sirf endpoints par depend karta hai, aur definition

d S = δ Q_{rev} / T

sirf reversible paths par valid hai, toh hum ek convenient ek (jaise Figure s03 mein solid isotherm) borrow karte hain jo same endpoints wali ho.

Kyun irreversibility strict inequality

Δ S_{univ} > 0

force karta hai na ki

\geq

Kyunki ek real gradient measurable entropy banata hai: heat

Q

jo finite gap

T_{H} \to T_{C}

cross karta hai

σ = Q (1/ T_{C} - 1/ T_{H}) > 0

deta hai (Figure s01), jo strictly positive hota hai jab

T_{C} < T_{H}

hota hai aur sirf infinitesimal-gap limit mein vanish karta hai.

Kyun surroundings ko

T_{surr}

par score kiya jaata hai, system ke temperature par nahin?

Surroundings ek bada reservoir hai jo apne fixed

T_{surr}

par internally reversibly heat exchange karta hai, toh

Δ S_{surr} = - Q / T_{surr}

; system alag

T

par hai, aur dono ke beech finite gap extra

σ

ka source hai.

Kyun Clausius inequality

Δ S_{univ} \geq 0

prove karne ke liye ek cycle use karta hai?

Kyunki

\oint δ Q / T_{surr} \leq 0

sirf closed loops ke liye stated hai; irreversible

A \to B

(orange) ko reversible

B \to A

(teal) ke saath Figure s02 mein close karne par reversible leg

- Δ S_{sys}

supply karta hai (Step 3) jabki poora loop

\leq 0

rehta hai (Step 2),

Δ S_{sys} \geq \int δ Q_{irr} / T_{surr}

deta hai.

Kyun second law time ko ek direction deta hai?

Sirf

Δ S_{univ} \geq 0

allowed hai, toh mixed-up direction future hai; ek reversed movie (unmixing, doodh ka un-spill hona) ke liye

Δ S_{univ} < 0

chahiye hoga, jo kabhi nahin hota — yeh Arrow of time hai.

Kyun do identical macroscopic states reversible aur irreversible dono paths se reach kiye ja sakte hain phir bhi same

Δ S_{system}

share karte hain?

Kyunki

Δ S_{system}

sirf endpoints par depend karta hai; jo differ karta hai woh

Δ S_{surr}

hai aur isliye

Δ S_{univ}

, jo irreversible path ke liye exactly extra

σ

se bada hota hai.

Kyun statistical picture

S = k ln W

agree karta hai ki mixing entropy badhata hai?

Ek mixed/spread configuration vastly zyada microstates

W

se correspond karta hai, toh uska

ln W

(aur isliye

S

) bada hota hai — dekho Statistical interpretation of entropy (Boltzmann S = k ln W).

Edge cases

Edge case: do bodies ke beech heat flow hota hai equal temperature

T_{H} = T_{C}

par.

Δ S_{univ}

kya hai?

Zero — bracket

(1/ T_{C} - 1/ T_{H})

vanish ho jaata hai (Figure s01 ka right edge), reversible limit se match karta hai jahan koi driving gradient nahin hai.

Edge case: ek gas vacuum mein expand karta hai lekin container sirf half khali hoti hai (

V \to 1.5 V

). Kya

Δ S_{univ}

phir bhi positive hai?

Haan —

Δ S_{sys} = n R ln (1.5) > 0

surroundings unchanged ke saath, toh vacuum mein koi bhi real expansion, chahe kitni bhi chhoti ho, universe entropy badhata hai.

Edge case: infinitely slow, frictionless (reversible) process ki limit mein

Δ S_{univ}

kya hota hai?

Yeh exactly

0

ki taraf upar se approach karta hai; reversibility allowed entropy production ka mathematical infimum hai, jo kabhi actually beat nahin hota.

Edge case: ek isolated system jo pehle se maximum entropy (equilibrium) par hai. Aur kya change possible hai?

Spontaneously kuch nahin —

Δ S_{univ}

aur nahin badh sakta aur na hi gir sakta, toh state stable hai aur koi net macroscopic process nahin hota.

Edge case: ek finite-temperature body jo same

T

par infinite reservoir mein drop ki gayi hai. Reversible hai ya nahin?

Limit mein reversible — koi temperature gap nahin hone par, heat exchange koi entropy produce nahin karta; koi bhi nonzero gap use irreversible banata hai

Δ S_{univ} > 0

ke saath.

Edge case: kya

Δ S_{surr} = 0

ho sakta hai jabki

Δ S_{sys} > 0

ho?

Haan — exactly free-expansion case: surroundings koi heat exchange nahin karte (

Q = 0

) toh

Δ S_{surr} = 0

, phir bhi system ki entropy badhti hai,

Δ S_{univ} > 0

maintain karte hue.

Edge case: ek chemical reaction fixed

T

aur

p

par spontaneously run karti hai jabki heat absorb karti hai (endothermic). Kya yeh

Δ S_{univ} \geq 0

break karta hai?

Nahin — endothermic spontaneity ka matlab hai system ki apni entropy rise surroundings ke loss

- Q / T

ko overpower karne ke liye kaafi badi hai, toh

Δ S_{univ} = Δ S_{sys} - Q / T

phir bhi

\geq 0

hai (equivalently Gibbs free energy

Δ G \leq 0

Edge case: ek "negative absolute temperature" spin system (

T < 0

kelvin scale par) ek normal positive-

T

body ko heat deta hai. Heat kis direction mein flow karti hai, aur kya

Δ S_{univ} \geq 0

phir bhi obeyed hota hai?

Heat negative-

T

system se bahar positive-

T

wale mein flow karti hai, aur

Δ S_{univ} \geq 0

phir bhi hold karta hai kyunki negative-

T

state actually energy/entropy ordering mein saare positive-

T

states ke upar hoti hai, toh transfer phir bhi total entropy badhata hai.

Recall Lock karne ke liye ek-line summary

Law yeh hai: " $Δ S_{univ} \geq 0$ , real processes ke liye strict" — kabhi nahin " $S_{system}$ hamesha badhti hai," aur kabhi nahin " $Q / T$ kisi bhi path par kaam karta hai." Temperature ko body se match karo, $Q > 0$ = system mein rakho, aur $Δ S$ hamesha ek imagined reversible path ke through route karo.