Exercises — Entropy change in irreversible processes — always - 0

Master rule yaad karo: $\Delta S_{\text{univ}}\ge 0$ hamesha, $=0$ sirf reversible processes ke liye. Akele system pe aisi koi restriction nahi hai.

• (a) Free expansion irreversible hai → $\Delta S_{\text{univ}}>0$. Yahaan $\Delta S_{\text{sys}}>0$ bhi hai (gas phail jaati hai).
• (b) Freezing real hai (paani aur fridge coils ke beech finite $\Delta T$) → $\Delta S_{\text{univ}}>0$. Lekin $\Delta S_{\text{sys}}<0$: liquid → ordered crystal entropy kho deta hai. Surroundings zyada gain karte hain, isliye universe fir bhi badhta hai. Yeh woh key "system drop kar sakta hai" wala case hai.
• (c) Coffee ka thanda hona: irreversible heat flow → $\Delta S_{\text{univ}}>0$; $\Delta S_{\text{sys}}<0$ (coffee thandi hoti hai).
• (d) Ek true Carnot cycle reversible hai → $\Delta S_{\text{univ}}=0$. Yeh reference (boundary) case hai.

First law setup. Thermodynamics ka first law kehta hai $\Delta U=Q-W$, jahan $\Delta U$ gas ki internal energy ka change hai, $Q$ gas mein add ki gayi heat hai, aur $W$ gas dwara kiya gaya kaam hai. Free expansion mein gas kuch nahi push karta (vacuum), isliye $W=0$; aur yeh isolated hai, isliye $Q=0$. Atah $\Delta U=0$, aur ideal gas ki internal energy sirf temperature pe depend karti hai, isliye $T$ unchanged hai.

Hum $\Delta S=Q/T=0$ nahi likh sakte — woh formula reversible path ke liye chahiye. Kyunki $S$ ek state function hai, real path ko same endpoints ke beech ek reversible isothermal expansion se replace karo: $\Delta S_{\text{sys}}=nR\ln \frac{V_f}{V_i}=2\times 8.314\times \ln 3=18.26\text{J/K}.$ Surroundings ne koi heat exchange nahi ki: $\Delta S_{\text{surr}}=0$. Isliye $\Delta S_{\text{univ}}=18.26\text{J/K}>0.✓$

Har reservoir bahut bada hai, isliye har ek apne temperature pe reversibly heat exchange karta hai. Hot wala $Q$ kho deta hai, cold wala $Q$ gain karta hai: $\Delta S_{\text{univ}}=-\frac{Q}{T_H}+\frac{Q}{T_C}=1000\left(\frac{1}{300}-\frac{1}{500}\right)=1000\times (0.003333-0.002)=1.333\text{J/K}>0.✓$ Bracket positive hai kyunki $T_C<T_H$: same heat lower temperature pe entropy ke liye zyada "count" hoti hai.

Heat absorbed: $Q=mL=0.036\times 334000=12024$ J. Ice ise $273$ K pe absorb karti hai (phase change ke dauran constant $T$); room ise $300$ K pe kho deta hai: $\Delta S_{\text{sys}}=\frac{Q}{273}=44.04\text{J/K},\Delta S_{\text{surr}}=-\frac{Q}{300}=-40.08\text{J/K}.$ $\Delta S_{\text{univ}}=44.04-40.08=3.96\text{J/K}>0.✓$

Har gas ka available volume double ho jaata hai ($V\to 2V$) jabki uski temperature aur pressure conditions hume har ek ko uss species ke independent free-into-vacuum-like expansion ki tarah treat karne deti hain. Isliye har gas $nR\ln 2$ gain karti hai: $\Delta S_{\text{univ}}=nR\ln 2+nR\ln 2=2nR\ln 2=2\times 1\times 8.314\times 0.693=11.53\text{J/K}>0.✓$ Insulated boundary ke paas koi heat cross nahi hoti, isliye $\Delta S_{\text{surr}}=0$ aur yeh saari system entropy hai. Mixing irreversible hai: tum kabhi nahi dekhoge ki do mixed gases spontaneously alag halves mein un-mix ho jaayein — yeh free expansion jaisa hi "arrow" hai.

Ab integral kyun? Block ka temperature constant nahi hai — woh $400$ K se gir ke $300$ K tak aata hai. Entropy change $\int \delta Q_{\text{rev}}/T$ hai, aur yahaan integral ke neeche $T$ block ka apna current temperature hai, isliye hum ise bahar nahi nikal sakte.

Neeche ka figure temperature (vertical axis, kelvin mein) ko time (horizontal axis) ke against plot karta hai. Coral curve block ka temperature hai jo $400$ K se gir ke lake temperature pe level off hoti hai; dashed lavender line lake hai, jo apne $300$ K pe flat pinned hai kyunki woh bahut badi hai. Butter-shaded area unke beech ka temperature mismatch hai jo heat flow drive karta hai — woh gap jitna bada, utni zyada entropy manufacture hoti hai. Dekho ki coral curve lake line se hamesha upar hai: heat sirf block → lake ki taraf flow karti hai.

Block ke liye, $\delta Q=mcdT$ (reservoirs ki ek ladder ke through imaginarily socha gaya reversible warming/cooling): $\Delta S_{\text{block}}={\int }_{400}^{300}\frac{mcdT}{T}=mc\ln \frac{300}{400}=385\ln (0.75)=-110.75\text{J/K}.$ Negative — block thanda hota hai, entropy kho deta hai.

Lake bahut badi hai, isliye woh $300$ K pe rehti hai aur block ne jo total heat release ki woh absorb karti hai: $Q=mc(T_i-T_L)=385\times (400-300)=38500\text{J},$ $\Delta S_{\text{lake}}=+\frac{Q}{300}=\frac{38500}{300}=+128.33\text{J/K}.$ $\Delta S_{\text{univ}}=-110.75+128.33=17.58\text{J/K}>0.✓$

Final temperature (energy conservation, equal heat capacities → arithmetic mean): $T_f=\frac{T_1+T_2}{2}=300\text{K}.$ Hot block $400\to 300$ thanda hota hai; cold block $200\to 300$ garam hota hai. Har ek $C\ln (T_f/T_{\text{start}})$ contribute karta hai: $\Delta S=C\ln \frac{T_f}{T_1}+C\ln \frac{T_f}{T_2}=400\left(\ln \frac{300}{400}+\ln \frac{300}{200}\right).$ $=400\ln \frac{300\times 300}{400\times 200}=400\ln (1.125)=47.14\text{J/K}>0.✓$ Product form $\ln \frac{T_f^2}{T_1{T}_2}$ note karo: kyunki arithmetic mean $T_f$ geometric mean $\sqrt{T_1{T}_2}$ se zyada hai, argument $>1$ hai aur $\Delta S>0$ — hamesha, jab tak blocks ki temperature equal na ho.

Step 0 — Kya saari ice pighlegii? Pehle edge case check karo. Garam paani zyada se zyada itna release kar sakta hai (saara $273$ K tak cool ho jaaye): $Q_{\text{avail}}=m_{2}c(293-273)=0.20\times 4186\times 20=16744\text{J}.$ Saari ice pighlaane ke liye chahiye $Q_{\text{melt}}=m_{1}L=0.05\times 334000=16700$ J. Kyunki $Q_{\text{avail}}=16744>16700=Q_{\text{melt}}$, paani saari ice pighla sakta hai — lekin barely: $273$ K se upar temperature badhane ke liye sirf $16744-16700=44$ J bache hain. Isliye $T_f$ $273$ K se bahut thoda upar hai. (Agar $Q_{\text{avail}}$ $Q_{\text{melt}}$ se kam hota, toh system exactly $273$ K pe settle karta kuch unmelted ice ke saath — ek alag calculation.)

Step 1 — Final temperature. Ice pighlaane ki heat + melt ko $273$ se $T_f$ tak garam karne ki heat = garam paani ka $293$ se $T_f$ tak thanda hone mein khoyi heat: $m_{1}L+m_{1}c(T_f-273)=m_{2}c(293-T_f).$ $16700+209.3(T_f-273)=837.2(293-T_f).$ Expand karo: $16700+209.3T_f-57139=245300-837.2T_f$. (Yahaan $837.2\times 293=245300$ J aur $209.3\times 273=57139$ J.) Collect karo: $1046.5T_f=245300+57139-16700=285739\Rightarrow T_f=\frac{285739}{1046.5}=273.04\text{K}.$ Toh $T_f=273.04$ K — melting point se sirf $0.04$ K upar, exactly woh "tiny excess" jo bache hue $44$ J khareedte hain. Step 0 se consistent hai. ✔

Step 2 — Har piece ki Entropy.

• $273$ K pe ice pighalna: $\Delta S_1=\frac{m_{1}L}{273}=\frac{16700}{273}=+61.17\text{J/K}.$
• Pighli ice ko $273$ se $T_f=273.04$ K tak garam karna: $\Delta S_2=m_{1}c\ln \frac{T_f}{273}=209.3\times \ln \frac{273.04}{273}=209.3\times (1.466\times 10^{-4})=+0.0307\text{J/K}.$
• Garam paani ko $293$ se $T_f$ tak thanda karna: $\Delta S_3=m_{2}c\ln \frac{T_f}{293}=837.2\times \ln \frac{273.04}{293}=837.2\times (-0.07053)=-59.05\text{J/K}.$

Step 3 — Sum. $\Delta S_{\text{univ}}=\Delta S_1+\Delta S_2+\Delta S_3=61.17+0.03-59.05=+2.15\text{J/K}>0.✓$ Positive, jaisa hona chahiye — mixing/melting irreversible hai.

Microscopic count. $N$ molecules, har ek ke paas independent roop se do gune zyada jagahein hain, isliye arrangements ki sankhya $2^N$ se multiply ho jaati hai: $\Delta S=k\ln (W\cdot 2^N)-k\ln W=k\ln 2^N=Nk\ln 2.$ Macroscopic (Clausius). L2 se, ek mole double volume tak free-expand hota hai: $\Delta S=R\ln 2$. Yeh match karte hain, kyunki $R=N_{A}k$: $Nk\ln 2=N_{A}k\ln 2=R\ln 2=8.314\times 0.693=5.76\text{J/K}.✓$ Ek akela positive number $R\ln 2$ simultaneously "reversible path pe heat divided by temperature" bhi hai aur "gas khud ko kitne zyada tarike se arrange kar sakti hai uska log" bhi. Woh equivalence hi woh gehri wajah hai jis se time ek direction mein jaata hai: mixed-up macrostate mein overwhelmingly zyada microstates hain, isliye duniya uski taraf drift karti hai aur kabhi wapas nahi aati.

System (half volume tak isothermal compression): $\Delta S_{\text{sys}}=nR\ln \frac{V}{2V}=-nR\ln 2=-2\times 8.314\times 0.693=-11.53\text{J/K}.$ Claimed $\Delta S_{\text{surr}}=+6$ J/K. Toh $\Delta S_{\text{univ}}=-11.53+6=-5.53\text{J/K}<0.$ Yeh second law violate karta hai ($\Delta S_{\text{univ}}\ge 0$) → impossible. Process ke barely allowed hone ke liye, surroundings ko kam se kam $11.53$ J/K gain karna hoga (reversible limit, equality). $✓$

• Free expansion $\Delta S$ formula → $nR\ln (V_f/V_i)$ ek kalpnik reversible isothermal path ke through.
• Finite $\Delta T$ ke paas heat → $\Delta S_{\text{univ}}=Q(1/T_C-1/T_H)>0$.
• Entropy of mixing (1 mol + 1 mol, equal halves) → $2nR\ln 2$.
• Finite body cooling → $\Delta S=mc\ln (T_f/T_i)$; temperature integral ke andar hai.
• Two-block mixing sign → positive kyunki arithmetic mean $>$ geometric mean.
• Boltzmann ↔ Clausius bridge → $R=N_{A}k$, isliye $N_{A}k\ln 2=R\ln 2$.
• Forbidden direction → koi bhi process jo $\Delta S_{\text{univ}}<0$ de, impossible hai.