Entropy — Clausius definition dS = dQ_rev - T: Exercises

1.7.22 · D4 · Physics › Thermodynamics › Entropy — Clausius definition dS = dQ_rev - T

Yeh page ek ladder hai. Har rung pehle se zyada mushkil hai, aur har problem ka complete solution ek collapsible callout mein chhupa hua hai taaki tum pehle khud try kar sako. Yahan sab kuch parent note Entropy — Clausius definition par build karta hai.

Shuru karne se pehle, yeh teen tools paas rakh lo — inhe hum use karte karte kamayenge.

Har symbol ka matlab ek quick reminder, kyunki hum koi bhi symbol bina clarity ke use nahi karenge:

$n$ = moles ki sankhya (kitna gas hai).
$T$ = absolute temperature kelvin (K) mein — kabhi celsius nahi, kyunki $1/ T$ kabhi blow up ya negative nahi hona chahiye.
$V$ = volume; $P$ = pressure; $Q$ = heat; $c$ = specific heat (1 kg ko 1 K garm karne ki energy); $C_{V}$ = molar heat capacity at constant volume.
$ln$ = natural logarithm — number $e$ ka "undo" button. Yeh isliye aata hai kyunki $\int \frac{d x}{x} = ln x$ , aur heat ko temperature se divide karne par hamesha kuch $\frac{1}{x} d x$ jaisa integrate karna padta hai.

Level 1 — Recognition

Goal: kya tum spot kar sakte ho ki kaun sa formula apply hoga aur sign kya hoga?

Problem 1.1

Entropy change ki Clausius definition aur uski SI unit batao.

Recall Solution 1.1

$d S = \frac{d Q _{r e v}}{T}, Δ S = \int_{A}^{B} \frac{d Q _{r e v}}{T} .$ Subscript rev ka matlab hai ki heat ko ek reversible path ke along flow hote imagin karna hoga. Unit: $J K^{- 1}$ (joules per kelvin), kyunki hum energy (J) ko temperature (K) se divide karte hain.

Problem 1.2

Ek gas expand karti hai. Bina kisi number ke, kya uski entropy badhegi ya ghateggi, aur kyun?

Recall Solution 1.2

Yeh badhegi. Zyada volume matlab molecules ke paas rehne ki zyada jagah — zyada microscopic arrangements (microstates). Formula $n R ln \frac{V _{2}}{V _{1}}$ mein, expansion matlab $V_{2} > V_{1}$ , toh $\frac{V _{2}}{V _{1}} > 1$ , aur 1 se bade number ka $ln$ positive hota hai. Figure dekho: ek wide box, zyada dots spread out.

Figure — Entropy — Clausius definition dS = dQ_rev - T

Problem 1.3

Ek reversible adiabatic process mein $d Q_{r e v} = 0$ poore time. $Δ S$ kya hai?

Recall Solution 1.3

$Δ S = \int \frac{0}{T} = 0.$ Zero heat reversibly exchange hua ⇒ zero entropy change. Is special case ko isentropic kehte hain ("same entropy").

Level 2 — Application

Goal: ek formula mein numbers sahi se plug karo.

Problem 2.1

$2 mol$ ideal gas isothermally aur reversibly expand karke apne original volume ka teen guna ho jaata hai. $Δ S$ nikalo.

Recall Solution 2.1

Isothermal ⇒ sirf volume term bachti hai: $Δ S = n R ln \frac{V _{2}}{V _{1}} = 2 \times 8.314 \times ln 3.$ $ln 3 = 1.0986$ , toh $Δ S = 2 \times 8.314 \times 1.0986 = 18.27 J K^{- 1} .$ Positive, jaisa expansion ke liye expected tha. ✔

Problem 2.2

$0.5 kg$ paani ( $c = 4186 J kg^{- 1} K^{- 1}$ ) ko $290 K$ se $340 K$ tak heat karo. Paani ka $Δ S$ nikalo.

Recall Solution 2.2

Temperature change ho raha hai, toh single $Q / T$ use nahi kar sakte; local $T$ ke saath integrate karte hain: $Δ S = m c ln \frac{T _{2}}{T _{1}} = 0.5 \times 4186 \times ln \frac{340}{290} .$ $ln (340/290) = ln (1.1724) = 0.15906$ , toh $Δ S = 2093 \times 0.15906 = 332.9 J K^{- 1} .$

Problem 2.3

$1 mol$ ideal gas ( $C_{V} = \frac{3}{2} R$ ) ko constant volume par $300 K$ se $600 K$ tak heat kiya jaata hai. $Δ S$ nikalo.

Recall Solution 2.3

Constant volume ⇒ volume term vanish ho jaati hai ( $ln 1 = 0$ ): $Δ S = n C_{V} ln \frac{T _{2}}{T _{1}} = 1 \times \frac{3}{2} \times 8.314 \times ln 2.$ $\frac{3}{2} \times 8.314 = 12.471$ , aur $ln 2 = 0.6931$ , toh $Δ S = 12.471 \times 0.6931 = 8.644 J K^{- 1} .$

Level 3 — Analysis

Goal: ek saath do effects, ya ek irreversible process ke liye reversible stand-in dhundho.

Problem 3.1

$1 mol$ ideal gas ( $C_{V} = \frac{3}{2} R$ ) $(T_{1}, V_{1}) = (300 K, 1 L)$ se $(T_{2}, V_{2}) = (600 K, 4 L)$ tak jaata hai. $Δ S$ nikalo.

Recall Solution 3.1

Temperature aur volume dono change ho rahe hain, toh full ideal-gas formula use karo: $Δ S = n C_{V} ln \frac{T _{2}}{T _{1}} + n R ln \frac{V _{2}}{V _{1}} .$ Term 1: $\frac{3}{2} \times 8.314 \times ln 2 = 12.471 \times 0.6931 = 8.644 J K^{- 1}$ . Term 2: $8.314 \times ln 4 = 8.314 \times 1.3863 = 11.526 J K^{- 1}$ . $Δ S = 8.644 + 11.526 = 20.17 J K^{- 1} .$ Kyunki $S$ ek state function hai, yeh answer is baat ki parwah nahi karta ki gas pehle heat ki gayi phir expand ki, ya pehle expand ki phir heat ki — sirf endpoints matter karte hain.

Problem 3.2 (Free expansion)

$3 mol$ ideal gas ek evacuated, insulated chamber mein rush karke apna volume double kar leta hai. Gas ka entropy change nikalo.

Recall Solution 3.2

Real process irreversible hai, toh hum actual heat ( $Q_{a c t u a l} = 0$ ) use nahi kar sakte. Iske bajaay:

Insulated + no external pressure ⇒ $Q = 0$ , $W = 0$ , toh $Δ U = 0$ ⇒ $T$ unchanged (ideal gas $U$ sirf $T$ par depend karta hai).
Same endpoints jaise ek reversible isothermal doubling ke hain. Woh substitute path use karo: $Δ S = n R ln \frac{V _{2}}{V _{1}} = 3 \times 8.314 \times ln 2 = 24.94 \times 0.6931 = 17.29 J K^{- 1} .$ Positive — irreversibility created entropy ke roop mein dikhti hai. Figure mein trick dikhi hai: tedhi real path ko smooth reversible path se replace kiya gaya hai jo same do dots ko join karti hai.

Problem 3.3 (Cycle mein heat dono directions mein)

Ek Carnot engine $T_{H} = 400 K$ par $Q_{H} = 500 J$ absorb karta hai aur $T_{C} = 300 K$ par heat reject karta hai. $Q_{C}$ aur do entropy pieces nikal ke verify karo ki $\oint d Q_{r e v} / T = 0$ .

Recall Solution 3.3

Reversible Carnot cycle ke liye, $\frac{Q _{H}}{T _{H}} = \frac{Q _{C}}{T _{C}}$ , toh $Q_{C} = Q_{H} \frac{T _{C}}{T _{H}} = 500 \times \frac{300}{400} = 375 J .$ Gas dwara heat lene par entropy gain: $+ \frac{500}{400} = + 1.25 J K^{- 1}$ . Heat reject karne par entropy loss: $- \frac{375}{300} = - 1.25 J K^{- 1}$ . Loop ke around sum: $1.25 - 1.25 = 0$ . ✔ Yeh ek concrete cycle mein Clausius theorem hai.

Level 4 — Synthesis

Goal: kai bodies ko combine karo; system aur surroundings ki entropy track karo.

Problem 4.1 (Mixing / thermal contact)

$400 K$ par copper ka $1 kg$ block ( $c = 385 J kg^{- 1} K^{- 1}$ ) ek huge reservoir mein $300 K$ par daala jaata hai. $Δ S_{b l oc k}$ , $Δ S_{r eser v o i r}$ , aur $Δ S_{u ni v er se}$ nikalo.

Recall Solution 4.1

Block (uska $T$ $400 \to 300$ girta hai, integrate karo): $Δ S_{b l oc k} = m c ln \frac{T _{2}}{T _{1}} = 385 \times ln \frac{300}{400} = 385 \times (- 0.28768) = - 110.8 J K^{- 1} .$ Negative — cooling entropy lose karti hai, sahi hai.

Reservoir (itna bada hai ki uska $T$ $300 K$ par rehta hai; woh block ki di hui heat absorb karta hai): $Q_{in t o r es} = m c (T_{1} - T_{2}) = 385 \times (400 - 300) = 38500 J, Δ S_{r es} = \frac{+ 38500}{300} = + 128.3 J K^{- 1} .$ (Yahan hum single reservoir temperature se divide karte hain kyunki reservoir ka temperature kabhi change nahi hota.)

Universe: $Δ S_{u ni v} = - 110.8 + 128.3 = + 17.5 J K^{- 1} > 0.$ Hot se cold ki taraf heat flow irreversible hai, toh universe ki entropy badhti hai. ✔

Problem 4.2 (Do finite blocks ek common temperature par pahunchte hain)

Do identical blocks, har ek mass $m$ , specific heat $c$ , $T_{1} = 400 K$ aur $T_{2} = 200 K$ par, contact mein rakh ke isolate kar diye jaate hain. Dikhao ki final temperature arithmetic mean hai aur $m c = 100 J K^{- 1}$ ke liye $Δ S_{u ni v}$ nikalo.

Recall Solution 4.2

Final temperature: equal masses/heat capacities, energy conserved: $m c (T_{f} - T_{1}) + m c (T_{f} - T_{2}) = 0 \Rightarrow T_{f} = \frac{T _{1} + T _{2}}{2} = \frac{400 + 200}{2} = 300 K .$ Entropy har block ki (integrate karo, uska $T$ vary karta hai): $Δ S = m c ln \frac{T _{f}}{T _{1}} + m c ln \frac{T _{f}}{T _{2}} = m c ln \frac{T _{f}^{2}}{T _{1} T _{2}} .$ $= 100 \times ln \frac{30 0 ^{2}}{400 \times 200} = 100 \times ln \frac{90000}{80000} = 100 \times ln 1.125 = 100 \times 0.11778 = 11.78 J K^{- 1} .$ Phir se positive. Note karo $T_{f}^{2} = 90000 > T_{1} T_{2} = 80000$ : arithmetic mean hamesha geometric mean se bada hota hai, aur yahi wajah hai ki $Δ S_{u ni v} > 0$ kisi bhi do unequal temperatures ke liye. Woh inequality plain algebra mein chhupa second law hai.

Level 5 — Mastery

Goal: ek general result prove karo, ya zero entropy change wala process design karo.

Problem 5.1 (Isentropic condition — ek jaana-maana law derive karna)

Reversible adiabatic ideal-gas process ke liye, $Δ S = 0$ . Full formula se start karke $T$ aur $V$ ke beech ka relation derive karo. ( $γ = C_{P} / C_{V}$ lo aur $C_{P} - C_{V} = R$ use karo.)

Recall Solution 5.1

Total entropy change zero set karo: $n C_{V} ln \frac{T _{2}}{T _{1}} + n R ln \frac{V _{2}}{V _{1}} = 0.$ $n C_{V}$ se divide karo aur $R / C_{V} = (C_{P} - C_{V}) / C_{V} = γ - 1$ use karo: $ln \frac{T _{2}}{T _{1}} + (γ - 1) ln \frac{V _{2}}{V _{1}} = 0 \Rightarrow ln [\frac{T _{2}}{T _{1}} (\frac{V _{2}}{V _{1}})^{γ - 1}] = 0.$ Exponentiate karo ( $ln$ undo karo): $\frac{T _{2}}{T _{1}} (\frac{V _{2}}{V _{1}})^{γ - 1} = 1 \Rightarrow T V^{γ - 1} = const .$ Toh jaana-maana adiabatic law exactly yeh statement hai ki "entropy constant rehti hai." Numerically, $γ = 5/3$ ke liye, volume double karne par ( $V_{2} / V_{1} = 2$ ) $\frac{T _{2}}{T _{1}} = 2^{- (γ - 1)} = 2^{- 2/3} = 0.6300,$ yaani gas apne start temperature ka $63.0%$ ho jaata hai. Check karo: term 1 $= C_{V} ln 0.6300$ aur term 2 $= R ln 2$ cancel ho jaate hain.

Problem 5.2 (Ek reversible heat exchange design karna)

Tumhe $1 mol$ gas ( $C_{V} = \frac{5}{2} R$ ) ko constant volume par $300 K$ se $600 K$ tak heat karna hai, lekin tum chahte ho ki universe ka entropy change as small as possible ho. Compare karo (a) single $600 K$ reservoir use karna versus (b) reservoirs ki infinite ladder use karna jo har baar infinitesimally zyada hot ho. Dono cases mein $Δ S_{u ni v}$ nikalo.

Recall Solution 5.2

Gas entropy change dono tarafon se same hai (state function): $Δ S_{g a s} = n C_{V} ln \frac{T _{2}}{T _{1}} = \frac{5}{2} (8.314) ln 2 = 20.785 \times 0.6931 = 14.41 J K^{- 1} .$ Heat needed: $Q = n C_{V} Δ T = 20.785 \times 300 = 6235.5 J$ .

(a) Single hot reservoir at $600 K$ saara $6235.5 J$ $600 K$ par deta hai: $Δ S_{r es} = - \frac{6235.5}{600} = - 10.39 J K^{- 1}, Δ S_{u ni v} = 14.41 - 10.39 = + 4.02 J K^{- 1} .$

(b) Reservoirs ki ladder, har ek gas temperature $T$ se match karta hai jaise woh badhti hai, $d Q = n C_{V} d T$ us hi $T$ par deta hai: $Δ S_{r es} = - \int_{300}^{600} \frac{n C _{V} d T}{T} = - n C_{V} ln 2 = - 14.41 J K^{- 1}, Δ S_{u ni v} = 14.41 - 14.41 = 0.$ Ladder reversible limit hai: heat hamesha (almost) equal temperature wali bodies ke beech cross karti hai, toh koi entropy create nahi hoti. "Reversible" ka yahi physical matlab hai — har step par temperatures match karna.

Connections

Carnot Cycle and Efficiency — Problem 3.3 $Q_{H} / T_{H} = Q_{C} / T_{C}$ use karta hai.
First Law of Thermodynamics — energy conservation jo har track kiye $Q$ ke peeche hai.
Second Law of Thermodynamics — har $Δ S_{u ni v} > 0$ result iska action mein example hai.
Reversible vs Irreversible Processes — L5 ladder reversible limit hai.
Exact and Inexact Differentials — kyun $1/ T$ $d Q$ ko state function bana deta hai.
Clausius Inequality — L3 free-expansion trap.
Statistical Entropy — Boltzmann S = k ln W — Problem 1.2 ke peeche microstate picture.