Thermodynamic potentials — U (internal), H (enthalpy), F (Helmholtz), G (Gibbs): Exercises

2.4.1 · D4 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Thermodynamic potentials — U (internal), H (enthalpy), F (He

Intuition Is page ko kaise use karein

Yeh ek self-testing ladder hai. Har problem seedha likha gaya hai, phir uska poora solution ek collapsible [!recall]- callout ke andar chhupaaya gaya hai — problem padho, khud try karo, tab solution kholna. Difficulty paanch steps mein badhti hai:

L1 Recognition — kya tum sahi object ka naam le ya padhh sakte ho?
L2 Application — kya tum ek standard formula mein values daal sakte ho?
L3 Analysis — kya tum do-teen steps chain karke sahi tool choose kar sakte ho?
L4 Synthesis — kya tum khud koi result build kar sakte ho jo kisi ne tumhe diya nahi?
L5 Mastery — kya tum ek real, multi-part situation ko shuru se ant tak handle kar sakte ho?

Yahan sab kuch parent note par based hai. Shuru karne se pehle, ek reminder table taaki neeche har symbol pehle se samjha hua ho.

L1 — Recognition

Recall Solution 1.1

KYA karte hain: bas table padh lo.

$U (S, V)$ , $H (S, p)$ , $F (T, V)$ , $G (T, p)$ .
Open beaker ⇒ pressure $p$ atmosphere se fixed hai. Hot-plate/thermostat ⇒ temperature $T$ fixed hai.
Woh potential jiske natural variables exactly $(T, p)$ hain woh $== G ==$ hai. To Gibbs free energy $G$ minimize hoti hai.

YEH KYUN poora point hai: tum woh potential choose karte ho jiske natural variables un cheezoon se match karein jo physically constant hain. $(T, p)$ match karo $\to G$ .

Recall Solution 1.2

$d G = - S d T + V d p$ se, generic $d G = (\frac{\partial G}{\partial T})_{p} d T + (\frac{\partial G}{\partial p})_{T} d p$ se compare karo. $d T$ aur $d p$ ke coefficients match karne par: $- S = (\frac{\partial G}{\partial T})_{p}, V = (\frac{\partial G}{\partial p})_{T} .$ KYU: differential hi dictionary hai; $d T$ ke aage baitha coefficient hi $T$ direction mein fixed $p$ par $G$ ki slope hai.

L2 — Application

Recall Solution 2.1

Step 1 — tool choose karo. Hum $T$ fixed rakhhte hain, isliye $d G = - S d T + V d p$ use karo. $d T = 0$ hone se pehla term khatam ho jaata hai aur $d G = V d p$ bachta hai. Step 2 — equation of state substitute karo. $V = n R T / p$ , isliye $Δ G = \int_{p_{0}}^{p} \frac{n R T}{p} d p = n R T ln \frac{p}{p _{0}} .$ Step 3 — numbers daalo. $n R T = 1 \times 8.314 \times 300 = 2494.2 J$ , aur $ln (5/1) = 1.6094$ . $Δ G = 2494.2 \times 1.6094 \approx 4014 J .$ KYU positive: gas ko compress karna (pressure badhana) fixed $T$ par $G$ ko badhata hai, aur yahi $R T ln (p / p_{0})$ chemical potential $μ = μ^{\circ} + R T ln (p / p^{\circ})$ ki backbone hai.

Recall Solution 2.2

Step 1: $C_{p} - C_{V} = n R$ mein $n = 1$ dene par $C_{p} = C_{V} + R$ milta hai. Step 2: $C_{V} = \frac{3}{2} R$ , isliye $C_{p} = \frac{3}{2} R + R = \frac{5}{2} R$ . Step 3: $\frac{5}{2} \times 8.314 = 20.785 J mol^{- 1} K^{- 1}$ . KYU $C_{p} > C_{V}$ : constant pressure par gas garm karne se expand hoti hai, kuch energy atmosphere ko push back karne mein kharach hoti hai ( $p d V$ ), isliye per degree zyada heat supply karni padti hai. Dekho Heat capacities Cp and Cv.

L3 — Analysis

Recall Solution 3.1

Step 1 — $d U$ ko fixed $T$ par $d V$ se divide karo. $S = S (T, V)$ maano. $T$ constant rakh kar aur $d U = T d S - p d V$ ko $d V$ se divide karne par: $(\frac{\partial U}{\partial V})_{T} = T (\frac{\partial S}{\partial V})_{T} - p .$ Yeh step kyun? Hum $U$ ki volume par dependence fixed $T$ par chahte hain, isliye hum $T$ pin karke $V$ -direction mein differentiate karte hain. Step 2 — unmeasurable $\partial S / \partial V$ ko Maxwell relation se khatam karo. $d F = - S d T - p d V$ se, mixed partials ki equality deti hai $(\frac{\partial S}{\partial V})_{T} = (\frac{\partial p}{\partial T})_{V} .$ Yeh kyun? $F$ ke natural variables $(T, V)$ hain — exactly woh do jo hum hold/vary kar rahe hain — isliye iska Maxwell relation entropy-vs-volume ko pressure-vs-temperature mein convert karta hai, jo hum measure kar sakte hain. Step 3 — assemble karo. $(\frac{\partial U}{\partial V})_{T} = T (\frac{\partial p}{\partial T})_{V} - p . ✓$ Step 4 — ideal gas. $p = n R T / V \Rightarrow (\frac{\partial p}{\partial T})_{V} = n R / V = p / T$ . Tab $(\frac{\partial U}{\partial V})_{T} = T \cdot \frac{p}{T} - p = 0.$ Interpretation: ek ideal gas ki internal energy fixed $T$ par volume par depend nahi karti — energy sirf temperature mein rehti hai. (Yeh "Joule expansion" fact hai.)

Recall Solution 3.2

Step 1 — parent ke result se shuru karo. $C_{p} - C_{V} = T (\frac{\partial p}{\partial T})_{V} (\frac{\partial V}{\partial T})_{p}$ . Step 2 — triple product rule se $(\partial p / \partial T)_{V}$ ko rewrite karo. $(\frac{\partial p}{\partial T})_{V} = - \frac{( \partial V / \partial T ) _{p}}{( \partial V / \partial p ) _{T}} .$ Kyun? Yeh "cyclic" identity hume awkward constant- $V$ derivative ko do measurable constant- $p$ /constant- $T$ wale derivatives ke through express karne deti hai. Step 3 — substitute karo: $C_{p} - C_{V} = - T \frac{( \partial V / \partial T ) _{p}^{2}}{( \partial V / \partial p ) _{T}} . ✓$ Step 4 — sign. Numerator mein ek square (≥ 0) hai $T$ (> 0) se multiply. Stability ki demand hai ki $(\partial V / \partial p)_{T} < 0$ (zyada squeeze karo ⇒ chhhota volume). Ek positive number ko negative se divide karo, overall minus ke saath, to result ≥ 0 milta hai. Isliye $C_{p} \geq C_{V}$ hamesha. Step 5 — ideal-gas check. $(\partial V / \partial T)_{p} = n R / p$ , $(\partial V / \partial p)_{T} = - n R T / p^{2}$ . Tab $- T \cdot \frac{( n R / p ) ^{2}}{- n R T / p ^{2}} = - T \cdot \frac{n ^{2} R ^{2} / p ^{2}}{- n R T / p ^{2}} = - T \cdot (- \frac{n R}{T}) = n R . ✓$

L4 — Synthesis

Recall Solution 4.1

Step 1 — general formula. $d H = T d S + V d p$ se, fixed $T$ par $d p$ se divide karne par: $(\frac{\partial H}{\partial p})_{T} = T (\frac{\partial S}{\partial p})_{T} + V .$ Step 2 — Maxwell from $G$ . $d G = - S d T + V d p$ deta hai $(\frac{\partial S}{\partial p})_{T} = - (\frac{\partial V}{\partial T})_{p}$ . Isliye $(\frac{\partial H}{\partial p})_{T} = V - T (\frac{\partial V}{\partial T})_{p} .$ Yeh route kyun: hume sirf measurable $p, V, T$ derivatives se bana form chahiye. Step 3 — equation of state apply karo. $V$ ke liye solve karo: $V = \frac{n R T}{p} + nb$ . Fixed $p$ par, $(\frac{\partial V}{\partial T})_{p} = \frac{n R}{p} .$ Step 4 — assemble karo. $(\frac{\partial H}{\partial p})_{T} = (\frac{n R T}{p} + nb) - T \cdot \frac{n R}{p} = nb .$ Interpretation: ideal-gas cancellation ek residue $nb$ chhhodti hai. Is gas ki enthalpy pressure ke saath constant slope $nb$ par badhti hai — excluded volume kabhi dono terms ko poori tarah cancel nahi karne deta. (Compare karo ideal-gas parent result se, jahan yeh zero tha.)

Recall Solution 4.2

Step 1 — coexistence condition. Boundary par molar Gibbs energies equal hain, $g_{1} (T, p) = g_{2} (T, p)$ . Boundary ke saath ek tiny step lo: dono change karte hain lekin equal rehte hain, isliye $d g_{1} = d g_{2}$ . Step 2 — har $d g$ likhho. Har phase ke liye $d g = - s d T + v d p$ use karo (lower case = per mole): $- s_{1} d T + v_{1} d p = - s_{2} d T + v_{2} d p .$ Step 3 — collect karo. $(v_{1} - v_{2}) d p = (s_{1} - s_{2}) d T \Rightarrow \frac{d p}{d T} = \frac{s _{1} - s _{2}}{v _{1} - v _{2}} = \frac{Δ s}{Δ v} .$ Step 4 — latent heat. Fixed $T$ par reversible phase change $L = T Δ s$ heat absorb karta hai, isliye $Δ s = L / T$ aur $\frac{d p}{d T} = \frac{L}{T Δ v} .$ Yeh Clausius–Clapeyron hai: coexistence line ki slope sirf latent heat aur volume jump se decide hoti hai.

L5 — Mastery

Recall Solution 5.1

(a) $Δ U$ . Ideal gas ke liye $U$ sirf $T$ par depend karta hai (Problem 3.1). Process isothermal hai, isliye $Δ U = 0.$ (b) $Δ S$ . $d F = - S d T - p d V$ ke sibling ka use karo: fixed $T$ par, $d U = T d S - p d V$ aur $d U = 0$ se, $T d S = p d V \Rightarrow Δ S = \int_{V_{0}}^{V} \frac{p}{T} d V = \int_{V_{0}}^{V} \frac{n R}{V} d V = n R ln \frac{V}{V _{0}} .$ Numerically: $n R ln (30/10) = 8.314 \times ln 3 = 8.314 \times 1.0986 = 9.134 J K^{- 1}$ . (c) $Δ F$ . Fixed $T$ par, $d F = - p d V$ , isliye $Δ F = - \int_{V_{0}}^{V} p d V = - \int_{V_{0}}^{V} \frac{n R T}{V} d V = - n R T ln \frac{V}{V _{0}} .$ Numerically: $- 8.314 \times 400 \times 1.0986 = - 3653.5 J$ . Definition se cross-check: $F = U - T S$ , aur fixed $T$ par, $Δ F = Δ U - T Δ S = 0 - 400 \times 9.134 = - 3653.5 J$ . ✓ (Same number, do routes.) (d) Maximum work. Reversible isothermal expansion ke liye gas dwara kiya gaya work hai $W = \int_{V_{0}}^{V} p d V = n R T ln \frac{V}{V _{0}} = 3653.5 J .$ Aur wakai $- Δ F = + 3653.5 J = W_{m a x}$ . Yahi parent statement hai " $- Δ F$ at constant $T$ = maximum work." ✓

Figure — Thermodynamic potentials — U (internal), H (enthalpy), F (Helmholtz), G (Gibbs)

Figure padhna: red curve isotherm $p = n R T / V$ hai; iske neeche shaded area (jasmein $V_{0}$ se $V$ tak) exactly $\int p d V = W_{m a x} = - Δ F$ hai.

Recall Solution 5.2

Step 1 — energy. Free expansion: $W = 0$ (vacuum, push karne ke liye kuch nahi) aur $Q = 0$ (insulated). First law $Δ U = Q - W = 0$ . Step 2 — temperature. $U = \frac{3}{2} R T$ sirf $T$ par depend karta hai, aur $Δ U = 0 \Rightarrow Δ T = 0$ . Final $T = 400 K$ , unchanged. Step 3 — entropy. $S$ ek state function hai, isliye iska change sirf endpoints par depend karta hai, path par nahi. Endpoints ( $T = 400 K$ , $V : 10 \to 30 L$ ) Problem 5.1 ke identical hain. Isliye $Δ S = n R ln \frac{V}{V _{0}} = 9.134 J K^{- 1},$ bilkul pehle jaisa. Step 4 — lesson. Work aur heat path-dependent hain; entropy aur internal energy state functions hain. Reversible isotherm aur irreversible free expansion dono same do states ko connect karte hain, isliye $Δ U$ aur $Δ S$ match karte hain, lekin sirf reversible path actually $3653.5 J$ ka work extract karta hai. Free expansion us available work ko "waste" karta hai — woh khoi hui opportunity $- Δ F$ se measure hoti hai.

Recall Final self-audit (cover kar lo aur recall karo)

Jaane se pehle inhe answer karo. Fixed $T, p$ par kaun sa potential minimize hota hai? ::: $G$ Ideal gas ke liye $(\partial U / \partial V)_{T}$ kitna hota hai? ::: $0$ Kya $C_{p} - C_{V}$ kabhi negative ho sakta hai? ::: Nahi — yeh $- T (\partial V / \partial T)_{p}^{2} / (\partial V / \partial p)_{T} \geq 0$ hai Constant $T$ par $- Δ F$ kya measure karta hai? ::: obtainable maximum work Latent heat ke terms mein Clausius–Clapeyron slope? ::: $d p / d T = L / (T Δ v)$ Free aur reversible expansions same $Δ S$ kyun share karte hain? ::: $S$ ek state function hai; same endpoints

Potential	Definition	Natural vars	Differential
$U$	—	$S, V$	$d U = T d S - p d V$
$H$	$U + p V$	$S, p$	$d H = T d S + V d p$
$F$	$U - T S$	$T, V$	$d F = - S d T - p d V$
$G$	$H - T S$	$T, p$	$d G = - S d T + V d p$