Entropy change in irreversible processes — always - 0: Worked examples

1.7.23 · D3 · Physics › Thermodynamics › Entropy change in irreversible processes — always - 0

Yeh page parent topic ka drill hall hai. Hum ek hi law $Δ S_{univ} \geq 0$ ko har angle se attack karte hain: system entropy upar, system entropy neeche, zero (reversible), degenerate, real-world, aur ek exam twist. Shuru karne se pehle, kuch symbols saral shabdon mein.

Definition Har symbol jo hum use karte hain, ek baar define kiya

$Q$ = heat: energy jo ek boundary cross karti hai temperature difference ki wajah se. Sign convention (is poori page par use hoga): $Q > 0$ matlab heat us object mein andar aa rahi hai jise hum dekh rahe hain; $Q < 0$ matlab heat us object se bahar ja rahi hai. Toh agar ek hot body apni heat $Q$ ek cold body ko deti hai, toh hot body ko $- Q$ dikhta hai aur cold body ko $+ Q$ .
$W$ = system dwara surroundings par kiya gaya work (jaise gas piston ko baahir dhakelta hai). First law: $Δ U = Q - W$ — internal energy $U$ heat add hone se badhti hai aur system jo work bahar deta hai usse ghatti hai.
$n$ = gas ke moles ki sankhya (ek mole $= 6.02 \times 1 0^{23}$ molecules); yeh sirf count karta hai ki kitni gas hai.
$R = 8.314$ J·mol⁻¹·K⁻¹ = ideal-gas constant, $p V = n R T$ mein fixed conversion factor. Yeh "moles aur kelvin" ko "joules" mein convert karta hai.
$c$ = specific heat capacity: ek kilogram ko ek kelvin garam karne ke liye joules (units J·kg⁻¹·K⁻¹). Mass $m$ ko $d T$ garam karne mein $δ Q = m c d T$ heat lagti hai.
$C \equiv m c$ = poore object ki heat capacity (units J·K⁻¹): us object ko ek kelvin garam karne ke joules. Jab problem seedha $C$ deta hai toh humein $m$ aur $c$ alag-alag nahi chahiye.

Do tools jo hum baar-baar use karenge. Dono sirf definition $d S = δ Q_{rev} / T$ hain jo kisi chosen reversible path par evaluate ki gayi hain (kyunki $S$ ek state function hai — path matter nahi karta, sirf endpoints matter karte hain).

The scenario matrix

Cell	Kya special hai	Example
A $Δ S_{sys} > 0$ , koi surroundings nahi	isolated, irreversible	Ex 1 (free expansion)
B do reservoirs, finite $Δ T$	classic "hot→cold"	Ex 2 (heat conduction)
C $Δ S_{sys} < 0$ lekin $Δ S_{univ} > 0$	system entropy girta hai	Ex 3 (water freezing)
D temperature changes → $ln$ chahiye	thermal equilibration	Ex 4 (two blocks)
E limiting case $\to$ reversible	$Δ S_{univ} \to 0$	Ex 5 (shrink the gap)
F degenerate / zero input	$Δ T = 0$ , ya $V$ unchanged	Ex 6 (equal-temp mixing)
G real-world word problem	numbers, units, phase change	Ex 7 (coffee cooling)
H exam twist — mixing counting	Boltzmann view cross-check	Ex 8 (gas mixing)

Ab hum saatonon cover karte hain.

Ex 1 — Cell A: free expansion (isolated, entropy up)

Worked example Statement

$n = 2$ mol ideal gas $T = 300$ K par $V$ se $3 V$ mein vacuum mein expand karta hai. Rigid insulated box. $Δ S_{univ}$ find karo.

Forecast: $Q = 0$ aur $W = 0$ hai. Aage padhne se pehle guess karo ki $Δ S_{univ}$ $0$ hai ya $> 0$ .

Steps.

$Δ U = Q - W = 0 - 0 = 0$ , toh ideal gas ke liye $T$ $300$ K par hi rehta hai. Yeh step kyun? First law endpoint state fix karta hai; state function use karne se pehle humein dono endpoints chahiye.
Ek hi endpoints ( $V \to 3 V$ at $300$ K) ke beech ek reversible isothermal path choose karo. Uske along $δ Q_{rev} = p d V$ aur $p = n R T / V$ (yahan $n = 2$ mol aur $R = 8.314$ J·mol⁻¹·K⁻¹, dono upar define hain): $Δ S_{sys} = \int_{V}^{3 V} \frac{p d V}{T} = n R ln \frac{3 V}{V} = n R ln 3.$ Yeh step kyun? Real path irreversible hai toh $δ Q / T$ directly use nahi ho sakta; lekin $S$ sirf endpoints ki parwah karta hai, toh hum ek reversible route invent karte hain.
Numbers: $2 \times 8.314 \times ln 3 = 18.27$ J/K.
Surroundings: boundary par kuch cross nahi hua, $Δ S_{surr} = 0$ . Toh $Δ S_{univ} = 18.27$ J/K $> 0$ .

Verify: positive, jaisa ek irreversible process hona chahiye. Units: $mol \cdot J mol^{- 1} K^{- 1} = J/K$ ✔. Dekho Free expansion of an ideal gas.

Figure neeche padhein: left box mein gas $V$ mein packed hai (red dots left mein crowded); right box mein same molecules partition hatnے ke baad poore $3 V$ mein ghoom rahe hain. Yeh picture hi entropy rise hai — zyada room matlab arrange hone ke zyada tarike, aur kala "expand" arrow ek-taraf ki street mark karta hai: dots kabhi spontaneously wapas left mein crowd nahi honge.

Figure — Entropy change in irreversible processes — always - 0

Ex 2 — Cell B: finite temperature gap par heat flow

Worked example Statement

$Q = 1000$ J heat ek hot reservoir $T_{H} = 400$ K se ek cold reservoir $T_{C} = 300$ K mein conduct karti hai. $Δ S_{univ}$ find karo.

Forecast: hot body entropy khota hai, cold body entropy paata hai. Kaun jeetega? (Hint: same $Q$ , cold side par chhota $T$ .)

Steps.

Hot reservoir apne temperature par $Q$ khota hai (heat jaati hai, toh uska entropy change $- Q$ use karta hai): $Δ S_{H} = - \frac{Q}{T _{H}} = - \frac{1000}{400} = - 2.500 J/K .$ Yeh step kyun? Reservoir bahut bada hai; uska temperature nahi hilta, toh fixed- $T$ recipe exactly apply hoti hai.
Cold reservoir same $Q$ apne temperature par paata hai: $Δ S_{C} = + \frac{Q}{T _{C}} = + \frac{1000}{300} = + 3.333 J/K .$ Yeh step kyun? Energy conservation: hot side se jo bhi joule gaya woh cold side par pahunch gaya, toh same $Q$ yahan enter karta hai — lekin cold temperature se divide hota hai, jo chhoti hai.
Add karo: $Δ S_{univ} = - 2.500 + 3.333 = + 0.833 J/K$ . Yeh step kyun? Universe bas system + surroundings hai, toh uska entropy change dono reservoirs ke changes ka plain sum hai.

Verify: $Δ S_{univ} = Q (\frac{1}{T _{C}} - \frac{1}{T _{H}})$ , aur kyunki $T_{C} < T_{H}$ hai toh bracket positive hai ✔.

Figure neeche padhein: red arrow $1000$ J heat-packet ka hot box se cold box mein crossing hai. Same packet wahan se jaate waqt $Q / T_{H}$ "cost" karta hai lekin wahan land hone par $Q / T_{C}$ "khareedata" hai; kyunki landing temperature kam hai, purchase cost se zyada hai — woh surplus hai jo universe manufacture karta hai.

Ex 3 — Cell C: system entropy NEECHE jaata hai (water freezing)

Worked example Statement

$m = 0.100$ kg paani $0^{\circ}$ C ( $T_{m} = 273$ K) par ek freezer mein freeze hota hai jiska air $T_{env} = 263$ K par hai. Latent heat of fusion $L = 3.34 \times 1 0^{5}$ J/kg. Dikhao ki $Δ S_{sys} < 0$ hai phir bhi $Δ S_{univ} > 0$ .

Forecast: freezing = molecules line up ho rahe hain = zyada ordered = system entropy girta hai. Kya law toota? Padhne se pehle guess karo.

Steps.

Paani freeze hone par release hua heat: $Q = m L = 0.100 \times 3.34 \times 1 0^{5} = 3.34 \times 1 0^{4}$ J. Yeh paani se bahar jaati hai (toh paani ke liye $Q$ negative hai). Yeh step kyun? Freezing latent heat dump karti hai; wahi heat hai jo surroundings receive karte hain.
Paani (system) constant $273$ K par $Q$ khota hai: $Δ S_{sys} = - \frac{Q}{273} = - 122.3 J/K (< 0) .$ Yeh step kyun? Phase change ek fixed temperature ( $273$ K) par hota hai, toh fixed- $T$ recipe apply hoti hai; minus sign isliye hai kyunki heat paani se jaati hai.
Freezer air $263$ K par $Q$ paata hai: $Δ S_{surr} = + \frac{Q}{263} = + 127.0 J/K .$ Yeh step kyun? Same $Q$ ab freezer ki cold air (fixed $263$ K par ek reservoir) mein enter karta hai; plus sign kyunki heat receive ho rahi hai.
Total: $Δ S_{univ} = - 122.3 + 127.0 = + 4.66 J/K > 0$ . Yeh step kyun? System aur surroundings ko jodne se universe milta hai; cold air ka gain (neeche $263$ K par heat bank karke) paani ke loss ko outweigh karta hai.

Verify: system entropy negative hai (order badha) lekin universe phir bhi badha ✔. Yeh textbook proof hai ki "entropy hamesha badhti hai" sirf universe ke baare mein hai, system ke nahi. Compare karo Reversible vs irreversible processes.

Ex 4 — Cell D: temperatures CHANGE hote hain → $ln$ tool chahiye

Worked example Statement

Do identical copper blocks, har ek ki heat capacity $C = m c = 500$ J/K (jaise $m = 0.176$ kg copper with $c = 2840$ J·kg⁻¹·K⁻¹ — lekin yahan sirf product $C$ matter karta hai). Ek $T_{1} = 400$ K se start karta hai, doosra $T_{2} = 200$ K se. Woh touch karte hain aur common final temperature $T_{f}$ par pahunchte hain. $T_{f}$ aur $Δ S_{univ}$ find karo.

Forecast: $T_{f}$ guess karo (kya yeh plain average hai?) aur $Δ S_{univ}$ ka sign guess karo.

Steps.

Equal heat capacities aur bahar koi heat loss nahi ⇒ energy balance $C (T_{f} - T_{1}) + C (T_{f} - T_{2}) = 0$ , jisse $T_{f} = \frac{T _{1} + T _{2}}{2} = 300$ K milta hai. Yeh step kyun? Kisi bhi entropy formula se pehle humein endpoint temperature chahiye; equal $C$ ise arithmetic mean banata hai.
Har block ki temperature move karti hai, toh $Δ S = C ln (T_{f} / T_{i})$ use karna hoga, $Q / T$ nahi. Yeh step kyun? Heat temperatures ki range par enter/exit hoti hai; sirf $ln$ recipe slices ko sahi tarike se sum karti hai.
Hot block cools $400 \to 300$ : $Δ S_{1} = 500 ln \frac{300}{400} = - 143.8$ J/K. Cold block warms $200 \to 300$ : $Δ S_{2} = 500 ln \frac{300}{200} = + 202.7$ J/K. Yeh step kyun? Ab hum har block ke liye $ln$ tool mein actual numbers dalte hain; ratio $< 1$ (cooling) negative log deta hai, ratio $> 1$ (warming) positive log deta hai — signs isi tarah aane chahiye.
$Δ S_{univ} = - 143.8 + 202.7 = + 58.9 J/K > 0$ . Yeh step kyun? Dono block changes jodne se universe milta hai; cold block tab garam hua jab woh thanda tha (chhote $T$ se divide kiya), toh usne hot block ke loss se zyada entropy gain ki — isliye total positive hai.

Verify: general form $C ln \frac{T _{f}^{2}}{T _{1} T _{2}} = 500 ln \frac{30 0 ^{2}}{400 \cdot 200} > 0$ kyunki $T_{f}^{2} = 90000 > T_{1} T_{2} = 80000$ (arithmetic mean geometric mean ko beat karta hai) ✔.

Figure neeche padhein: black curve $Δ S = 500 ln (T /300)$ hai; do red dots blocks ki starting temperatures mark karte hain. Hot dot curve ke shallow part par $300$ K tak neeche slide karta hai (chhota entropy loss), jabki cold dot steep part par $300$ K tak upar chadta hai (bada entropy gain). $ln$ curve ki yeh lopsided steepness exactly wahi reason hai kyun net positive hai.

Ex 5 — Cell E: limiting case, gap shrink karo → reversible

Worked example Statement

Ex 2 dobara karo lekin cold reservoir ko $T_{C} = 399$ K par rakho ( $T_{H} = 400$ K se sirf $1$ K neeche), abhi bhi $Q = 1000$ J. Phir $T_{C} \to 400$ K imagine karo.

Forecast: jaise gap band hota hai, kya $Δ S_{univ}$ $0$ ki taraf shrink karta hai ya blow up hota hai?

Steps.

$Δ S_{univ} = Q (\frac{1}{399} - \frac{1}{400}) = 1000 \times (0.0025063 - 0.0025000) = + 0.00627 J/K$ . Yeh step kyun? Same recipe jaise Ex 2 mein; hum sirf number dekh rahe hain jaise driving gradient shrink hota hai.
Limit $T_{C} \to T_{H}$ lo: bracket $\to 0$ , toh $Δ S_{univ} \to 0$ . Yeh step kyun? Vanishing temperature difference reversible heat exchange ki definition hai — koi wasted gradient nahi, koi entropy produce nahi.

Verify: tiny positive ( $0.00627 ≪ 0.833$ of Ex 2), $0$ ki taraf ja raha hai ✔. Yeh woh $Δ S_{univ} = 0$ boundary hai jis par reversible processes rehte hain — dekho Carnot cycle and efficiency.

Ex 6 — Cell F: degenerate input, zero gradient

Worked example Statement

$Q = 1000$ J do reservoirs ke beech flow karta hai dono $T = 300$ K par. Aur alag se: $n = 1$ mol gas ek reversible isothermal step undergo karta hai jisme uska volume change nahi hota ( $V \to V$ ). Dono cases mein teeno entropies find karo.

Forecast: koi gradient nahi aur koi change nahi, toh har entropy kya hona chahiye?

Steps.

Reservoirs: hot side $- Q /300$ , cold side $+ Q /300$ , toh $Δ S_{univ} = Q (\frac{1}{300} - \frac{1}{300}) = 0.$ Yeh step kyun? Equal temperatures = heat ke liye koi direction prefer karne ki zaroorat nahi = zero entropy production (reversible limit).
Gas with $V \to V$ ( $n = 1$ mol ke saath): system ki entropy hai $Δ S_{sys} = n R ln \frac{V}{V} = n R ln 1 = 0.$ Yeh step kyun? $ln 1 = 0$ : identical endpoints ka matlab state function $S$ mein zero change.
Kyunki volume kabhi nahi badla, koi heat boundary cross nahi ki, toh surroundings bhi kuch nahi dekhte: $Δ S_{surr} = 0, Δ S_{univ} = Δ S_{sys} + Δ S_{surr} = 0.$ Yeh step kyun? Teeno state karne zaroori hain; law ka equality case $Δ S_{univ} = 0$ exactly wahi hai jo ek degenerate (no-change) input produce karta hai.

Verify: har entropy exactly $0$ hai ✔. Degenerate inputs precisely reversible boundary par land karte hain — koi violation nahi, bas equality case.

Ex 7 — Cell G: real-world word problem (coffee cooling)

Worked example Statement

$0.250$ kg coffee ka cup (paani maano, $c = 4186$ J·kg⁻¹·K⁻¹) $T_{1} = 350$ K se room temperature $T_{room} = 293$ K tak cool hota hai. $Δ S_{sys}$ , $Δ S_{surr}$ , $Δ S_{univ}$ find karo.

Forecast: coffee ki temperature change hogi ( $ln$ use karo), lekin room ek giant reservoir hai fixed $293$ K par ( $Q / T$ use karo). Mismatch ka kaun sa side entropy produce karta hai?

Steps.

Coffee ki heat capacity: $C = m c = 0.250 \times 4186 = 1046.5$ J/K. Coffee cool hone par nikli heat: $Q = C (T_{1} - T_{room}) = 1046.5 \times (350 - 293) = 5.965 \times 1 0^{4}$ J. Yeh step kyun? Yahi total $Q$ hai jo room receive karta hai; humein reservoir side ke liye yeh chahiye.
Coffee ek range par cool hoti hai ⇒ $ln$ tool: $Δ S_{sys} = C ln \frac{T _{room}}{T _{1}} = 1046.5 ln \frac{293}{350} = - 185.3 J/K .$ Yeh step kyun? Coffee ki temperature continuously $350$ se $293$ K tak slide karti hai, toh heat ka har scoop alag $T$ par nikalta hai; sirf $ln$ tool un slices ko sahi tarike se jodta hai (fixed- $T$ recipe yahan galat hogi).
Room (fixed $T$ ) saara $Q$ absorb karta hai: $Δ S_{surr} = + \frac{Q}{293} = + 203.6 J/K .$ Yeh step kyun? Reservoir temperature kabhi nahi hilti, toh fixed- $T$ recipe exact hai.
$Δ S_{univ} = - 185.3 + 203.6 = + 18.3 J/K > 0$ . Yeh step kyun? Jodne se universe milta hai; room har joule ko single low temperature $293$ K par bank karta hai, jabki coffee ne woh joules $350$ K tak ke temperatures par shed kiye — wahi mismatch produced entropy hai.

Verify: system down (cooling = kam spread), universe up ✔. Room heat "collect" karta hai jo $350$ K tak jaate waqt nikli thi lekin sab ko $293$ K par bank karta hai — temperature mismatch exactly woh irreversibility hai.

Ex 8 — Cell H: exam twist, mixing (Boltzmann cross-check)

Worked example Statement

Ek box do equal halves mein split hai. Left half: $n = 1$ mol gas A. Right half: $n = 1$ mol gas B, same $T$ , same $p$ . Partition hataao; gases poore box mein interdiffuse ho jaate hain. $Δ S_{univ}$ find karo, aur counting se cross-check karo.

Forecast: kuch garam ya thanda nahi hota, koi heat flow nahi — toh kya $Δ S_{univ}$ zero hai? (Trap!)

Steps.

Har gas $V /2$ se $V$ tak free-expand karta hai (har species ab poore box mein hai). Ex 1 ka tool har gas ke liye use karte hain (har ek mein $n = 1$ mol): $Δ S = n R ln \frac{V}{V /2} + n R ln \frac{V}{V /2} = 2 n R ln 2.$ Yeh step kyun? Mixing do simultaneous free expansions hain; har ek apna available volume double karta hai, toh $ln (2)$ do baar aata hai.
$n = 1$ mol har gas ke liye: $Δ S_{univ} = 2 R ln 2 = 11.53$ J/K $> 0$ (surroundings unchanged, kyunki koi heat kisi boundary se cross nahi ki). Yeh step kyun? Hum ek concrete number dalte hain ( $n = 1$ mol per gas, $R = 8.314$ ) effect ka size dekhne ke liye aur confirm karne ke liye ki yeh strictly positive hai — mixing ne really entropy produce ki zero heat flow ke bawajood.
Boltzmann cross-check (Statistical interpretation of entropy (Boltzmann S = k ln W)): entropy of mixing hai $- R n_{tot} \sum x_{i} ln x_{i}$ . Mole fractions $x_{A} = x_{B} = \frac{1}{2}$ aur $n_{tot} = 2$ mol ke saath: $Δ S = - 2 R (\frac{1}{2} ln \frac{1}{2} + \frac{1}{2} ln \frac{1}{2}) = 2 R ln 2 = 11.53 J/K .$ Yeh step kyun? Macroscopic "reversible path" answer microscopic counting answer ke barabar hona chahiye — do raaste, ek number.

Verify: dono routes $2 R ln 2 = 11.53$ J/K dete hain ✔. Mixing zero heat flow ke saath entropy badhata hai — classic exam trap jisme " $Q = 0 \Rightarrow Δ S = 0$ " galti phans jaati hai.

Recall Har cell ke liye one-line self-test

Ex 1 free expansion ::: $Δ S = n R ln 3 = 18.27$ J/K, surroundings $0$ . Ex 2 heat flow ::: $Q (1/ T_{C} - 1/ T_{H}) = + 0.833$ J/K. Ex 3 freezing ::: $Δ S_{sys} < 0$ ( $- 122.3$ ) lekin $Δ S_{univ} = + 4.66$ J/K. Ex 4 two blocks ::: $T_{f} = 300$ K, $Δ S_{univ} = + 58.9$ J/K. Ex 5 tiny gap ::: $+ 0.00627$ J/K $\to 0$ jaise gap band hota hai. Ex 6 degenerate ::: teeno entropies $= 0$ . Ex 7 coffee ::: $+ 18.3$ J/K. Ex 8 mixing ::: $2 R ln 2 = 11.53$ J/K, counting se match karta hai.