Worked examples — Thermodynamic potentials — U, H, F, G (preview)

1.7.26 · D3 · Physics › Thermodynamics › Thermodynamic potentials — U, H, F, G (preview)

Yeh page potentials note ka "drill ground" hai. Hum chaar differentials dobara derive nahi karenge — hum assume karte hain ki aapne yeh dekha hai: $dU=TdS-pdV,dH=TdS+Vdp,dF=-SdT-pdV,dG=-SdT+Vdp.$ Hamaara goal yahan alag hai: in equations ko har tarah ki situation mein use karna, including weird corners (ek sign flip, ek zero, ek ideal gas, ek phase change, ek exam trap). Har example aapko bataata hai ki woh matrix ka kaun sa cell fill karta hai.

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The scenario matrix

Neeche har cell ko kam se kam ek worked example ne cover kiya hai (uska label brackets mein hai). Neeche ka map figure dikhata hai ki potentials aur unke slopes aapas mein kaise connected hain — kaam karte waqt isko refer karte raho.

Case class	Specific scenario	Example
Read a slope	positive slope (get $T$ or $V$)	Ex 1
Sign trap	negative slope (get $-S$, $-p$)	Ex 2
Hold-fixed + integrate	$dp=0$: enthalpy = heat	Ex 3
Second derivative	Maxwell relation, both signs	Ex 4
Real-world number	ideal-gas $\Delta H$ heating	Ex 5
Zero / degenerate input	$\Delta (pV)=0$ so $\Delta H=\Delta U$	Ex 6
Limiting behaviour	$T\to 0$: free energy $\to$ internal energy	Ex 7
Spontaneity call	sign of $\Delta G$, all three sign cases	Ex 8
Exam-style twist	which potential? wrong-constraint trap	Ex 9

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Ex 1 — Positive slope padhna (cell: read a slope)

Forecast: padhne se pehle andaza lagao — kya $T$, $U$ ka $S$-slope hai ya $V$-slope, aur kya yeh negative ho sakta hai?

1. Coefficients match karo. Chain rule $dU={\left(\frac{\partial U}{\partial S}\right)}_{V}dS+{\left(\frac{\partial U}{\partial V}\right)}_{S}dV$ ko $dU=TdS-pdV$ ke saath likho. Yeh step kyun? Ek hi $dU$ ke liye do expressions ke $dS$ aur $dV$ ke aage equal coefficients hone chahiye — yeh is topic ka poora engine hai.
2. $dS$ coefficient pick karo: ${\left(\frac{\partial U}{\partial S}\right)}_V=T$. Yeh step kyun? Physics side par $dS$ ka coefficient exactly $+T$ hai.
3. Sign ka matlab: $T$ absolute temperature hai, isliye $T>0$ hamesha. Isliye internal energy hamesha badhti hai jab aap fixed volume par entropy add karte ho — $U$-vs-$S$ curve ka slope strictly positive hai. Yeh step kyun? Sign check nonsense pakadta hai: yahan negative $T$ unphysical hoga.

Verify: Ek ideal monatomic gas ke liye $U=\frac{3}{2}N{k}_{B}T$ aur (constants tak) $S=N{k}_B\ln (V{T}^{3/2})+\text{const}$, toh fixed $V$ par, $T\propto e^{2S/3N{k}_B}>0$ — hamesha positive, slope hamesha upar. ✓

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Ex 2 — Negative-slope trap (cell: sign trap)

Forecast: kya entropy plus ke saath nikalti hai ya minus ke saath?

1. $dT$ coefficient match karo: physics side par yeh $-S$ hai, toh ${\left(\frac{\partial F}{\partial T}\right)}_V=-S$. Yeh step kyun? Wahi coefficient-matching, lekin ab coefficient khud minus sign carry karta hai.
2. $S$ ke liye solve karo: $S=-{\left(\frac{\partial F}{\partial T}\right)}_V$. Student ne minus drop kar diya. Yeh step kyun? $S$ positive hona chahiye (third law), jabki $F$ badhte $T$ ke saath decrease karta hai, isliye $\partial F/\partial T<0$; extra minus use positive banata hai. Sign aur physics sirf minus ke saath agree karte hain.
3. Pressure ke liye bhi wahi: ${\left(\frac{\partial F}{\partial V}\right)}_T=-p$, isliye $p=-{\left(\frac{\partial F}{\partial V}\right)}_T$. Yeh step kyun? Box squeeze karne par ($V\downarrow$) $F$ badhta hai, toh $\partial F/\partial V<0$, aur minus $p>0$ banata hai. ✓

Verify: Standard ideal-gas Helmholtz energy $F=-N{k}_{B}T\ln (V)+f(T)$ lo, jahan $f(T)$ volume-independent piece hai (yeh saare purely temperature-dependent terms collect karta hai — thermal/kinetic aur constant parts — jo $V$-derivative ke liye matter nahi karte). Toh $-\partial F/\partial V=N{k}_{B}T/V=p$, ideal-gas law — positive, jaisa required hai. ✓

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Ex 3 — Pressure fixed rakho, integrate karo (cell: hold-fixed + integrate)

Forecast: kya $Q$ depend karega ki volume kaise badla, ya sirf endpoints par?

1. $dH=TdS+Vdp$ se shuru karo. $dp=0$ set karo: $dH=TdS$. Yeh step kyun? "Constant pressure" literally instruction $dp=0$ hai, jo doosra term delete kar deta hai.
2. Heat pehchano. Reversible path ke liye $\delta Q=TdS$, toh $\delta Q_p=dH$. Yeh step kyun? Yeh entropy ki definition hai jo $TdS$ ko reversible heat se jodhti hai (dekho First Law of Thermodynamics).
3. $C_p$ laao. Definition se $C_p={\left(\frac{\partial H}{\partial T}\right)}_p$, toh constant $p$ par, $dH=C_{p}dT$. Yeh step kyun? Hum answer measurable temperatures mein chahte hain, aur $C_p$ exactly $H$ ka $T$-slope hai (dekho Heat Capacities Cp and Cv).
4. Integrate karo: $Q_p=\Delta H={\int }_{T_1}^{T_2}{C}_{p}dT=C_p(T_2-T_1)$ constant $C_p$ ke liye. Yeh step kyun? Constant $C_p$ integral se bahar aa jaata hai, sirf plain difference bachta hai.

Verify (numbers): $C_p=29.1{\text{J K}}^{-1}{\text{mol}}^{-1}$ (diatomic gas), $1$ mol ko $T_1=300$ K se $T_2=350$ K tak heat karo: $Q_p=29.1\times (350-300)=1455\text{J}.$ Units: ${\text{J K}}^{-1}\times \text{K}=\text{J}$. ✓ Sirf endpoints par depend karta hai — sahi hai, $H$ ek state function hai.

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Ex 4 — Maxwell relation ke dono signs (cell: second derivative)

Forecast: kya dono mixed partials same sign ke saath niklenge ya opposite signs ke saath?

Neeche wala figure Maxwell square for $G$ hai: dono arrows same mixed second derivative ${\partial }^{2}G/\partial T\partial p$ tak paahunchne ke dono raaste hain, aur unhe follow karna neeche di gayi relation force karta hai.

1. Dono pehle slopes padho. $dG$ se: ${\left(\frac{\partial G}{\partial T}\right)}_p=-S$ aur ${\left(\frac{\partial G}{\partial p}\right)}_T=+V$. Yeh step kyun? $S$ follows $T$ (minus), $V$ follows $p$ (plus) — exactly $dG$ mein jo signs hain; yeh square ke top par dono labels hain.
2. Cross-differentiate karo. Kyunki $G$ ek state function hai, uski mixed second derivative path-independent hai: $\frac{\partial }{\partial p}{\left(\frac{\partial G}{\partial T}\right)}_p=\frac{\partial }{\partial T}{\left(\frac{\partial G}{\partial p}\right)}_T.$ Yeh step kyun? Mixed partials ki equality (exact differential) woh mathematical fact hai jo har Maxwell relation banata hai; figure mein yeh statement hai ki dono arrows same corner value tak paahunchte hain.
3. Step 1 ke slopes substitute karo: left side hai $\frac{\partial (-S)}{\partial p}=-{\left(\frac{\partial S}{\partial p}\right)}_T$; right side hai $\frac{\partial V}{\partial T}={\left(\frac{\partial V}{\partial T}\right)}_p$. Yeh step kyun? Figure mein amber arrow dekho: $-S$ label ko $p$ se differentiate karna $+V$ label ko $T$ se differentiate karne ke equal hona chahiye.
4. Result: $\boxed{{\left(\frac{\partial S}{\partial p}\right)}_T=-{\left(\frac{\partial V}{\partial T}\right)}_p}$. Minus yahan survive karta hai (contrast karo $F$-relation se jisme koi leftover minus nahi tha). Yeh step kyun? Akela minus $-S$ se aaya lekin $+V$ se nahi — asymmetric signs asymmetric result dete hain.

Verify: Ideal gas $V=nRT/p$. Toh ${\left(\frac{\partial V}{\partial T}\right)}_p=nR/p$, isliye relation predict karta hai ${\left(\frac{\partial S}{\partial p}\right)}_T=-nR/p$. $S=nR\ln (V)+g(T)=nR\ln (nRT/p)+g(T)$ se, jahan $g(T)$ entropy ka volume-independent (purely temperature-dependent) part hai jo $\partial /\partial p$ ke under drop out ho jaata hai, hum actually $\partial S/\partial p=-nR/p$ paate hain. ✓

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Ex 5 — Ek real-world number: gas ko garam karna (cell: real-world number)

Forecast: kaun sa bada hai, aur kitna — gap $\Delta (pV)$ ke equal hona chahiye.

1. Capacities chunno. Monatomic ideal gas: $C_v=\frac{3}{2}R$, $C_p=\frac{5}{2}R$ per mole, jahan $R=8.314{\text{J K}}^{-1}{\text{mol}}^{-1}$. Yeh step kyun? $\Delta H$ ke liye $C_p$ use hota hai, $\Delta U$ ke liye $C_v$; $H$ vs $U$ ka poora point yahi pair hai.
2. $\Delta H=n{C}_p\Delta T$ compute karo: $2.0\times \frac{5}{2}(8.314)\times 50=2078.5\text{J}$. Yeh step kyun? Ex 3 ne diya $\Delta H=\int C_{p}dT$; constant $C_p$ ise $n{C}_p\Delta T$ banata hai.
3. $\Delta U=n{C}_v\Delta T$ compute karo: $2.0\times \frac{3}{2}(8.314)\times 50=1247.1\text{J}$. Yeh step kyun? Fixed volume par $U$ ka natural heat measure $C_v$ hai; ideal gas ke liye $\Delta U$ sirf $T$ par depend karta hai.
4. Gap check karo: $\Delta H-\Delta U=831.4\text{J}$, jo $\Delta (pV)=nR\Delta T=2.0\times 8.314\times 50=831.4\text{J}$ ke equal hona chahiye. Yeh step kyun? $H=U+pV$, isliye $\Delta H-\Delta U=\Delta (pV)$; match hona confirm karta hai ki koi arithmetic slip nahi hai.

Verify: $2078.5-1247.1=831.4$ aur $nR\Delta T=831.4$. Units sab joules. ✓ $\Delta H>\Delta U$ kyunki expanding gas atmosphere par kaam karta hai. ✓

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Ex 6 — Ek degenerate case: $\Delta (pV)=0$ (cell: zero / degenerate input)

Forecast: agar $pV$ apni purani value par wapas aata hai, toh kya enthalpy ka "extra term" gayab ho jaata hai?

1. Definition change likho. $\Delta H=\Delta U+\Delta (pV)$. Yeh step kyun? Enthalpy define hoti hai $U+pV$ ke roop mein, isliye uska change cleanly split hota hai.
2. Degenerate condition daalo. $\Delta (pV)=p_2{V}_2-p_1{V}_1=0$ hypothesis se. Yeh step kyun? Yahan "zero input" $\Delta (pV)=0$ hai; yeh corner case hai jahan added $pV$ term kuch contribute nahi karta.
3. Conclude karo: $\Delta H=\Delta U+0=\Delta U$. Yeh step kyun? Correction term gone hone par, dono potentials exactly ek saath track karte hain.

Verify (numbers): ideal gas isothermal $T=300$ K par, $n=1$: $pV=nRT=2494.2$ J dono endpoints par, isliye $\Delta (pV)=0$ aur $\Delta U=0$ (ideal gas, constant $T$) — isliye $\Delta H=0$ bhi. ✓ Consistent.

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Ex 7 — $T\to 0$ par limiting behaviour (cell: limiting behaviour)

Forecast: jab $T$ zero hit karta hai toh "tax" term $TS$ ka kya hota hai?

1. Definition ko limit tak le jaao. $F=U-TS$. Jab $T\to 0$, ek perfect crystal ki entropy zero tend karti hai (third law), aur even ek general system ke liye $S$ bounded rehti hai, isliye product $TS\to 0$. Hence $F\to U$. Yeh step kyun? Entropic tax $TS$, $T$ ke saath scale karta hai; $T\to 0$ aur $S$ finite (actually perfect crystal ke liye $S\to 0$) ke saath, tax nichod ke kuch nahi reh jaata.
2. Differentials compare karo. $dF=-SdT-pdV$. Ek change consider karo jo $T=0$ par (ya infinitesimally near) kiya gaya ho, jahan $S\to 0$: term $-SdT$ gayab ho jaata hai kyunki uska coefficient $S$ zero ho jaata hai, $dF\to -pdV$ bachta hai. Yeh step kyun? Hum dekhna chahte hain ki absolute zero par $dF$ ka kaun sa term survive karta hai; kyunki $S\to 0$ $dT$ ke coefficient ko kill karta hai, sirf mechanical $-pdV$ term bachta hai — yahi $F$ ko wahan $U$ jaisa behave karaata hai.
3. $U$ ka slope recover karo. Fixed $S$ par, $dU=-pdV$ bhi, isliye $-pdV$ mechanical part shared hai: ${\left(\frac{\partial F}{\partial V}\right)}_T\to -p={\left(\frac{\partial U}{\partial V}\right)}_S$ is limit mein. Yeh step kyun? Dono potentials pure mechanical work par reduce ho jaate hain jab thermal terms khatam ho jaate hain.

Verify: Perfect crystal ke liye $S\to 0$ jab $T\to 0$ (third law), isliye $U=U_0$ finite ke saath, $T=0$ par: $F=U_0-0\cdot 0=U_0=U$. Numerically, $U_0=100$ J aur residual $S=0$ J/K lo: $F=100-0\times 0=100$ J $=U$. ✓ (Hypothetical finite $S$ bhi $TS\to 0$ deta hai, isliye conclusion robust hai.) Dekho Entropy and the Second Law third-law statement $S\to 0$ ke liye.

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Ex 8 — Spontaneity: $\Delta G$ ke teeno signs (cell: spontaneity call)

Forecast: compute karne se pehle har ek ke liye $\Delta G$ ka sign guess karo.

1. Sahi potential identify karo. Lab $T$ aur $p$ fixed rakhta hai (open flask + thermostat) — boss hai Gibbs: $\Delta G=\Delta H-T\Delta S$, aur $\Delta G<0$ matlab spontaneous. Yeh step kyun? Matching-constraint rule: $T,p$ fixed ⇒ $G$ minimise karo (dekho Gibbs Free Energy and Chemical Equilibrium).
2. Case (a): $\Delta G=-40000-298(50)=-40000-14900=-54900\text{J}=-54.9\text{kJ}<0$. Spontaneous (dono terms help karte hain). Yeh step kyun? Exothermic aur entropy-increasing — easy always-go case.
3. Case (b): $\Delta G=+30000-298(-20)=30000+5960=+35960\text{J}=+35.96\text{kJ}>0$. Spontaneous nahi (dono terms oppose karte hain). Yeh step kyun? Endothermic aur entropy-decreasing — always-stuck case.
4. Case (c): $\Delta G=+50000-298(100)=50000-29800=+20200\text{J}=+20.2\text{kJ}>0$ at 298 K — abhi spontaneous nahi, lekin $T^{\ast }=\Delta H/\Delta S=50000/100=500\text{K}$ se upar spontaneous ho jaata hai. Yeh step kyun? Jab $\Delta H>0$ aur $\Delta S>0$ ho, entropy term sirf tab jeetta hai jab $T$ kaafi bada ho — temperature-triggered case.

Verify: (a) $-54900$ J, (b) $+35960$ J, (c) $+20200$ J, crossover $T^{\ast }=500$ K. ✓ Signs: (a) −, (b) +, (c) + below 500 K. ✓

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Ex 9 — Exam twist: wrong-constraint trap (cell: exam-style twist)

Forecast: kya $F$ is case ko judge karne ke liye sahi potential bhi hai?

1. Constraints identify karo. Lab $T$ aur $p$ fixed rakhta hai (open flask + thermostat). Yeh step kyun? Kisi bhi potential problem mein pehla move: kya fixed hai yeh decide karta hai ki boss kaun hai.
2. Matching potential chuno. Constant $T,p$ ⇒ spontaneity $\Delta G$ se govern hoti hai, $\Delta F$ se nahi. $F$ sirf constant $T,V$ par minimise hota hai. Yeh step kyun? Yahan $F$ use karna classic wrong-constraint error hai; jab $V$ free ho tab $F$ zaruri nahi decrease kare.
3. Dikhao ki woh disagree kar sakte hain. $G=F+pV$, isliye $\Delta G=\Delta F+\Delta (pV)$. Agar volume itna shrink ho ki $\Delta (pV)=-8$ kJ, toh $\Delta G=+5-8=-3\text{kJ}<0$ — spontaneous $\Delta F>0$ hone ke bawajood. Yeh step kyun? Ek concrete number prove karta hai ki student ki logic galat hai: wrong potential ne galat verdict diya.

Verify: $\Delta G=\Delta F+\Delta (pV)=5+(-8)=-3$ kJ $<0$. ✓ Sahi verdict: yeh proceed kar sakta hai, kyunki constant $T,p$ par sirf $\Delta G$ ka sign matter karta hai.

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Recall Quick self-test

Gas constant $p$ par heat hota hai; kya $Q=\Delta H$ hai ya $\Delta U$? ::: $Q=\Delta H$ (constant pressure, Ex 3) Sign of $S$ from $F$? ::: $S=-(\partial F/\partial T{)}_V$, minus essential hai (Ex 2) $\Delta H=\Delta U$ kab hota hai? ::: Jab $\Delta (pV)=0$ ho (Ex 6) Reaction with $\Delta H>0,\Delta S>0$ turns spontaneous above which $T$? ::: $T^{\ast }=\Delta H/\Delta S$ (Ex 8c, 500 K) Constant-$T,p$ spontaneity kaun sa potential judge karta hai? ::: Gibbs $G$, $F$ nahi (Ex 9)

Related builders: Legendre Transform (potentials kaise bante hain), Maxwell Relations (Ex 4), Heat Capacities Cp and Cv (Ex 3, 5).