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

2.4.1 · D3 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Thermodynamic potentials — U (internal), H (enthalpy), F (He

Yeh page ek drill room hai. Parent note ne chaar potentials, unke differentials, aur Maxwell relations banaye the. Yahan hum har tarah ke problem un tools par throw karte hain — har constraint jo tum laga sako, har degenerate case, aur exam-style traps — aur har ek ko bilkul shuru se work out karte hain.

Definition Is page par use kiye gaye symbols

Poori jagah, $n$ = number of moles of gas (mole count), $R$ = gas constant $= 8.314 J mol^{- 1} K^{- 1}$ , $T$ = temperature, $p$ = pressure, $V$ = volume, $S$ = entropy. Potentials: $U$ (internal energy), $H$ (enthalpy), $F$ (Helmholtz free energy), $G$ (Gibbs free energy). Heat capacities: $C_{V}$ fixed volume par, $C_{p}$ fixed pressure par.

First law mein do small-change symbols aate hain jinhe dhyan se samajhna hai. $δ Q$ = a tiny bit of heat added system mein, aur $δ W$ = a tiny bit of work done by the system. Hum inhe $δ$ se likhte hain (na ki $d$ se) kyunki heat aur work state functions nahin hain — total heat li gayi path par depend karti hai, sirf endpoints par nahin. Iske ulat $d U$ , $d S$ , $d V$ mein $d$ use hota hai kyunki $U, S, V$ state functions hain. Ek reversible path par $δ Q = T d S$ aur $δ W = p d V$ .

Scenario matrix

Is topic ke har problem ka ghar in cells mein se kisi ek mein hai. Neeche ke worked examples un cell(s) se tagged hain jo woh cover karte hain, isliye end tak koi bhi cell andheri nahin rahi.

#	Cell class	Kya clamp hai	Sahi tool	Example
C1	Constant $V$ (rigid box)	$T, V$	$F$ , $C_{V}$	Ex. 1
C2	Constant $p$ (open beaker)	$T, p$	$G$ , $H$ , $C_{p}$	Ex. 2
C3	Constant $S$ (adiabatic + reversible)	$S, V$ ya $S, p$	$U$ , $H$	Ex. 3
C4	Isothermal work / free energy	$T$	$Δ F = - W_{m a x}$	Ex. 4
C5	Spontaneity ki sign / direction	$T, p$	$Δ G$ sign	Ex. 5
C6	Degenerate input: ideal gas ( $H, U$ sirf $T$ par depend karte hain)	—	Maxwell + EoS	Ex. 6
C7	Limiting case: $p \to 0$ , $T \to 0$ , $V \to \infty$	limits	asymptotics	Ex. 7
C8	Real-world word problem (phase change)	$T, p$	$G$ equality	Ex. 8
C9	Exam twist: kaun sa potential? wrong-tool trap	mixed	$Φ$ chuno	Ex. 9

Poori jagah, ek monatomic ideal gas ke liye hum $C_{V} = \frac{3}{2} n R$ , $C_{p} = \frac{5}{2} n R$ , aur gas constant $R = 8.314 J mol^{- 1} K^{- 1}$ use karte hain. Dekho Heat capacities Cp and Cv.

Example 1 — Cell C1: ek rigid sealed box mein gas ko garam karna

Worked example Constant volume, constant

$N$ $2.0$ mol monatomic ideal gas ek rigid sealed tank mein baitha hai (volume badal nahin sakta). Hum ise $T_{1} = 300 K$ se $T_{2} = 400 K$ tak garam karte hain. Absorbed heat $Q$ aur internal-energy change $Δ U$ nikalo.

Forecast: Pehle guess karo — kya yahan $Q$ equals $Δ U$ hoga, ya kuch energy kaam ke roop mein nikal jaayegi?

Step 1. Clamp ka naam lo. Volume fixed hai $\Rightarrow$ $d V = 0$ . Yeh step kyun? Rigid walls ka matlab hai gas kuch bhi push nahin kar sakta jo move kare, isliye $p V$ -work $= p d V = 0$ .

Step 2. Master equation $d U = T d S - p d V$ mein $d V = 0$ lagao: $d U = T d S = δ Q .$ Yeh step kyun? Jab koi work nahin, to first law $d U = δ Q - δ W$ collapse hokar $d U = δ Q$ ban jaata hai. Isliye constant volume par, heat hi internal-energy change hai.

Step 3. $C_{V} = (\partial U / \partial T)_{V} = \frac{3}{2} n R$ use karke integrate karo: $Δ U = Q = \int_{T_{1}}^{T_{2}} C_{V} d T = \frac{3}{2} n R (T_{2} - T_{1}) .$ Yeh step kyun? $C_{V}$ precisely woh heat hai jo fixed volume par per degree chahiye — yahi iska definition hai.

Step 4. Numbers: $Q = \frac{3}{2} (2.0) (8.314) (400 - 300) = 2494.2 J \approx 2.49 kJ .$

Verify: Units: $mol \cdot J mol^{- 1} K^{- 1} \cdot K = J$ ✓. Kyunki koi work nahin hua, $Q = Δ U$ exactly — forecast se match karta hai agar tumne predict kiya tha "saari heat internal energy ban jaayegi." Garam karne par positive $Q$ ✓.

Example 2 — Cell C2: usi gas ko ek open beaker mein garam karna

Worked example Constant pressure, constant

$N$ Wohi $2.0$ mol monatomic ideal gas, wohi $300 \to 400 K$ , lekin ab ek cylinder mein frictionless piston ke saath jo atmosphere ke liye khula hai (pressure fixed). Heat $Q$ nikalo aur Example 1 se compare karo.

Forecast: Kya zaroorat ki heat Example 1 ke $2.49 kJ$ se zyada hogi ya kam?

Step 1. Clamp ka naam lo: pressure fixed $\Rightarrow d p = 0$ . Yeh step kyun? Constant $p$ par natural potential enthalpy hai, jiska differential $d H = T d S + V d p$ hai.

Step 2. $d p = 0$ set karo: $d H = T d S = δ Q .$ Yeh step kyun? Constant pressure par, heat enthalpy change ke barabar hoti hai — isliye chemists $Δ H$ ko "heat of reaction" kehte hain.

Step 3. $C_{p} = (\partial H / \partial T)_{p} = \frac{5}{2} n R$ se integrate karo: $Q = Δ H = \frac{5}{2} n R (T_{2} - T_{1}) = \frac{5}{2} (2.0) (8.314) (100) = 4157.0 J \approx 4.16 kJ .$

Step 4. Example 1 se extra heat hai: $Q_{p} - Q_{V} = (C_{p} - C_{V}) Δ T = n R Δ T = (2.0) (8.314) (100) = 1662.8 J .$ Yeh step kyun? Constant $p$ par gas expand hoti hai aur atmosphere ko push karne mein energy lagaati hai ( $p Δ V$ ); woh push-out work hi extra heat hai jo tumhe supply karni padti hai. Yahi parent note ke Example (c) ka $C_{p} - C_{V} = n R$ hai.

Verify: $Q_{p} > Q_{V}$ ✓ (constant- $p$ heating hamesha zyada costly hoti hai). $p Δ V = n R Δ T = 1662.8 J$ , $Q_{p} - Q_{V}$ se match karta hai ✓. Units sab Joules ✓.

Example 3 — Cell C3: reversible adiabatic compression (constant entropy)

Worked example Constant

$S$ (adiabatic + reversible) $1.0$ mol monatomic ideal gas ko reversibly aur adiabatically $V_{1} = 22.4 L$ se $V_{2} = 11.2 L$ tak compress kiya jaata hai, starting at $T_{1} = 273 K$ . $T_{2}$ aur $Δ U$ nikalo.

Forecast: Adiabatic matlab koi heat andar ya bahar nahin. Jab tum ise squeeze karte ho to temperature badhti hai, ghatti hai, ya same rehti hai?

Step 1. Reversible + adiabatic $\Rightarrow$ $δ Q = 0$ aur reversible $\Rightarrow δ Q = T d S$ , isliye $d S = 0$ : entropy constant hai. Yeh step kyun? Constant $S$ se $U (S, V)$ natural potential ban jaata hai; $d U = T d S - p d V = - p d V$ .

Step 2. Ideal gas ke liye constant $S$ par, $T V^{γ - 1} = const$ jahan $γ = C_{p} / C_{V} = 5/3$ . Yeh step kyun? $d S = \frac{C _{V}}{T} d T + \frac{n R}{V} d V$ mein $d S = 0$ set karke integrate karne par exactly yahi relation milta hai. Dekho Entropy and the Second Law.

Step 3. $T_{2}$ nikalo: $T_{2} = T_{1} (\frac{V _{1}}{V _{2}})^{γ - 1} = 273 (\frac{22.4}{11.2})^{2/3} = 273 \cdot 2^{2/3} = 433.4 K .$

Step 4. Kyunki $δ Q = 0$ , saari internal-energy change work se aayi: $Δ U = \int C_{V} d T = \frac{3}{2} n R (T_{2} - T_{1}) = \frac{3}{2} (1.0) (8.314) (433.4 - 273) = 2000.4 J .$

Verify: Temperature badhi (compression gas ko garam karta hai) — forecast se match karta hai agar tumne "badhegi" predict kiya tha ✓. $Δ U > 0$ aur gas par kiye gaye work ke barabar hai kyunki $Q = 0$ ✓. $2^{2/3} \approx 1.587$ , isliye $273 \times 1.587 = 433.4$ ✓.

Example 4 — Cell C4: isothermal expansion se maximum work

Worked example Constant

$T$ , free-energy = available work $1.0$ mol ideal gas isothermally $T = 300 K$ par $V_{1} = 10 L$ se $V_{2} = 30 L$ tak expand hoti hai. $Δ F$ aur maximum obtainable work nikalo.

Forecast: Parent note kehta hai constant $T$ par $- Δ F$ maximum work hai. Jab gas expand hoti hai to kya tumhe $Δ F$ positive ya negative expect hai?

Step 1. Constant $T$ $\Rightarrow d T = 0$ . $d F = - S d T - p d V = - p d V$ use karo. Yeh step kyun? Helmholtz $F (T, V)$ ke natural variables $T, V$ hain — exactly jo ek isothermal process control karta hai.

Step 2. $p = n R T / V$ se integrate karo: $Δ F = - \int_{V_{1}}^{V_{2}} p d V = - n R T ln \frac{V _{2}}{V _{1}} .$ Yeh step kyun? Equation of state substitute karo aur $1/ V$ integrate karo.

Step 3. Numbers: $Δ F = - (1.0) (8.314) (300) ln \frac{30}{10} = - (2494.2) (1.0986) = - 2740.2 J .$

Step 4. Maximum work $W_{m a x} = - Δ F = 2740.2 J$ . Yeh step kyun? Constant $T$ par, $d F = - δ W$ , isliye $F$ mein jo drop hoti hai woh gas ki deliver ki jaane wali work hai.

Verify: Expansion ke liye $Δ F < 0$ (gas kar sakti hai work), forecast se match karta hai ✓. Reversible isothermal work $W = n R T ln (V_{2} / V_{1}) = + 2740.2 J = - Δ F$ se compare karo ✓ — donon computations agree karte hain. Units: J ✓.

Example 5 — Cell C5: $Δ G$ ki sign spontaneity decide karti hai

Worked example Yeh kis direction mein jaayega?

$T = 298 K$ aur constant pressure par, ek reaction ka $Δ H = - 50.0 kJ$ aur $Δ S = - 100.0 J K^{- 1}$ hai. Kya yeh spontaneous hai? Kis temperature par yeh flip hoga?

Forecast: Exothermic ( $Δ H < 0$ ) lekin entropy-lowering ( $Δ S < 0$ ) — ek tug of war. Room temperature par kaun jeetta hai?

Step 1. Constant $T, p$ par, spontaneity $G$ se govern hoti hai: spontaneous $⟺ Δ G < 0$ . Yeh step kyun? Parent note: fixed $T, p$ par, $G$ equilibrium par minimize hoti hai, isliye systems lower $G$ ki taraf slide karte hain.

Step 2. $Δ G = Δ H - T Δ S$ use karo (constant $T$ par valid): $Δ G = - 50000 - (298) (- 100) = - 50000 + 29800 = - 20200 J = - 20.2 kJ .$ Yeh step kyun? $G = H - T S$ , isliye fixed $T$ par, $Δ G = Δ H - T Δ S$ .

Step 3. $Δ G < 0 \Rightarrow$ spontaneous $298 K$ par. Enthalpy term jeet jaata hai.

Step 4. Flip temperature: $Δ G = 0$ set karo: $T^{*} = \frac{Δ H}{Δ S} = \frac{- 50000}{- 100} = 500 K .$ Yeh step kyun? $T^{*}$ se upar $- T Δ S = + 100 T$ term (positive, kyunki $Δ S < 0$ ) $Δ H$ ko overwhelm kar deta hai, aur $Δ G > 0$ ban jaata hai.

Verify: $T = 500$ check karo: $Δ G = - 50000 - 500 (- 100) = 0$ ✓. $T = 600 > T^{*}$ par: $Δ G = - 50000 + 60000 = + 10000 > 0$ — non-spontaneous ✓, flip confirm hota hai.

Example 6 — Cell C6: degenerate input, ideal-gas ki special structure

Worked example Ideal gas ke liye

$(\partial U / \partial V)_{T} = 0$ dikhao (Joule's law) Prove karo ki ideal gas ki internal energy fixed temperature par volume change hone se nahin badlti.

Forecast: Ek gas ko constant $T$ par squeeze karo. Kya uski internal energy shift hogi?

Step 1. $d U = T d S - p d V$ se shuru karo aur $(\partial / \partial V)_{T}$ lo: $(\frac{\partial U}{\partial V})_{T} = T (\frac{\partial S}{\partial V})_{T} - p .$ Yeh step kyun? Hum ek measurable expression chahte hain; $(\partial S / \partial V)_{T}$ directly measurable nahin hai, isliye hum ise Maxwell se swap karenge.

Step 2. $F$ wala Maxwell relation apply karo: $(\partial S / \partial V)_{T} = (\partial p / \partial T)_{V}$ : $(\frac{\partial U}{\partial V})_{T} = T (\frac{\partial p}{\partial T})_{V} - p .$ Yeh step kyun? Yeh parent note ka boxed Maxwell relation hai — yeh ek entropy derivative ko ek pressure derivative se trade karta hai jo hum equation of state se compute kar sakte hain.

Step 3. Ideal gas $p = n R T / V \Rightarrow (\partial p / \partial T)_{V} = n R / V = p / T$ : $(\frac{\partial U}{\partial V})_{T} = T \cdot \frac{p}{T} - p = 0.$

Verify: Yeh parent ke Example (a) ka mirror hai jisne $(\partial H / \partial p)_{T} = 0$ dikhaya tha; yahan $(\partial U / \partial V)_{T} = 0$ uska twin hai. Dono kehte hain ideal-gas $U$ aur $H$ sirf $T$ par depend karte hain — ideal gas ki degenerate structure ✓.

Example 7 — Cell C7: teen limiting cases $p \to 0$ , $T \to 0$ , $V \to \infty$

Worked example Isothermal Gibbs energy apni edges par

$Δ G = n R T ln (p / p_{0})$ use karo (parent Example b), jahan $p_{0}$ ek freely chosen reference pressure hai — ek fixed baseline (typically standard $p_{0} = 1 bar = 1 0^{5} Pa$ ) jiske against hum pressure $p$ measure karte hain. Kyunki sirf ratio $p / p_{0}$ logarithm mein aata hai, $p_{0}$ sirf $G$ ka zero set karta hai; physics is baat mein hai ki $p$ usse kaise compare karta hai. Teen limits examine karo: $p \to 0^{+}$ , $T \to 0^{+}$ , aur $V \to \infty$ .

Forecast: Jab tum ek gas ko fixed $T$ par expand hone do, to kya uski Gibbs energy blow up hoti hai, vanish hoti hai, ya $- \infty$ ki taraf jaati hai? Aur jab tum sab kuch absolute zero ki taraf cool karo to kya hota hai?

Step 1. Degenerate anchor $p = p_{0}$ : $ln (1) = 0 \Rightarrow Δ G = 0$ . Yeh step kyun? Sanity check — reference state ko zero change dena chahiye, yeh confirm karta hai ki $p_{0}$ sirf baseline fix karta hai.

Step 2. (Limit $p \to 0^{+}$ .) $ln (p / p_{0}) \to - \infty$ , isliye $Δ G \to - \infty$ . Yeh step kyun? Ek shrinking positive argument ka $ln$ $- \infty$ tak diverge karta hai; ek gas fixed $T$ par zyada volume (lower pressure) mein spread hokar hamesha apni Gibbs energy lower karti hai. Concrete point, $n = 1$ , $T = 300 K$ , $p / p_{0} = 0.01$ : $Δ G = (1) (8.314) (300) ln (0.01) = 2494.2 \times (- 4.6052) = - 11487.5 J .$

Step 3. (Limit $V \to \infty$ .) Fixed $T$ par, ideal gas $p = n R T / V$ , isliye $p \to 0$ jab $V \to \infty$ . $p / p_{0} = n R T / (p_{0} V)$ substitute karo: $Δ G = n R T ln \frac{n R T}{p _{0} V} V \to \infty - \infty.$ Yeh step kyun? $V \to \infty$ wahi physical limit hai jaise $p \to 0$ equation of state se padhna — infinite dilution $G$ ko $- \infty$ tak drive karta hai, isliye ek gas kabhi spontaneously re-compress nahin hoti.

Step 4. (Limit $T \to 0^{+}$ .) Prefactor $n R T \to 0$ . Chahe $ln (p / p_{0})$ bada ho, product $T ln (p / p_{0}) \to 0$ kisi bhi fixed finite $p / p_{0}$ ke liye: $lim_{T \to 0^{+}} n R T ln (p / p_{0}) = 0.$ Yeh step kyun? Bahar wala $T$ (bounded, fixed) logarithm ko hara deta hai, isliye $G$ ki isothermal pressure-dependence absolute zero par switch off ho jaati hai — Third Law se consistent hai ki entropy (aur $S = - (\partial G / \partial T)_{p}$ -type responses) $T \to 0$ par freeze out ho jaate hain. Dekho Entropy and the Second Law.

Verify: $Δ G (p = p_{0}) = 0$ ✓. Sign negative $p < p_{0}$ ke liye ✓. $p \to 0$ point $- 11487.5 J$ already bada aur negative hai, divergence se consistent ✓. $V \to \infty$ $p \to 0$ limit reproduce karta hai ✓. $T \to 0$ par $0$ milta hai ✓.

Example 8 — Cell C8: real-world phase change (Clausius–Clapeyron)

Worked example Water–vapour equilibrium boiling point par

Phase boundary par, do phases coexist karte hain. Paani $T = 373.15 K$ par boil hota hai, latent heat $L = 40.66 kJ mol^{- 1}$ aur molar volume change $Δ V = 3.03 \times 1 0^{- 2} m^{3} mol^{- 1}$ (vapour minus liquid) ke saath. Boiling curve ka slope $d p / d T$ estimate karo.

Forecast: Boiling ke paas $d p / d T$ ka order of magnitude guess karo — hazaaron Pa per K? Daas?

Step 1. Phase boundary par molar Gibbs energies equal hain: $G_{liq} = G_{vap}$ , aur boundary par rehne ka matlab hai $d G_{liq} = d G_{vap}$ . Yeh step kyun? Same $T, p$ par coexisting do phases ka $G$ equal hona chahiye (warna saari matter lower- $G$ phase ki taraf flow ho jaayega). Dekho Phase equilibria & Clausius–Clapeyron.

Step 2. Har phase ke liye $d G = - S d T + V d p$ insert karo aur equal set karo: $- S_{liq} d T + V_{liq} d p = - S_{vap} d T + V_{vap} d p \Rightarrow \frac{d p}{d T} = \frac{S _{vap} - S _{liq}}{V _{vap} - V _{liq}} = \frac{Δ S}{Δ V} .$ Yeh step kyun? $G$ ka natural differential $S$ aur $V$ carry karta hai; donon phases ko equal karne se slope isolate ho jaata hai.

Step 3. $Δ S$ ko $L / T$ se replace karo. Yeh step kyun? Phase change reversibly fixed $T$ par hoti hai (dono phases us ek temperature par coexist karte hain), isliye entropy jump reversible heat over temperature hai: $Δ S = δ Q_{rev} / T = L / T$ , jahan $L$ liquid $\to$ vapour mein absorbed molar latent heat hai. Yeh crucial substitution hai jo un-measurable entropy difference ko measured latent heat mein convert karti hai. $\frac{d p}{d T} = \frac{Δ S}{Δ V} = \frac{L}{T Δ V} .$

Step 4. Numbers: $\frac{d p}{d T} = \frac{40660}{( 373.15 ) ( 3.03 \times 1 0 ^{- 2} )} = 3596.3 Pa K^{- 1} .$

Verify: Units: $\frac{J mol ^{- 1}}{K \cdot m ^{3} mol ^{- 1}} = \frac{J}{K m ^{3}} = \frac{Pa}{K}$ ✓ (kyunki $1 J = 1 Pa \cdot m^{3}$ ). Result $\approx 3.6 kPa/K$ paani ke boiling curve ke jaane-maane real slope se match karta hai $10 0^{\circ} C$ ke paas (approximately $3.6 kPa/K$ ) ✓, aur positive sign ✓ kehta hai ki zyada pressure boiling point badhata hai — exactly isliye pressure cooker zyada garam pakkata hai. Dekho figure.

Example 9 — Cell C9: exam twist, "kaun sa potential?" trap

Worked example Wrong-tool trap

Ek gas ek rigid insulated box mein band hai aur andar ek vacuum partition mein expand hoti hai (free expansion). Ek student likhta hai " $Δ G = 0$ isliye kuch nahin hota." Error diagnose karo aur ideal gas ke liye sahi invariant nikalo.

Forecast: Kya $G$ sahi potential hai jab na $T$ control hai na $p$ ?

Step 1. Identify karo ki actually kya clamp hai. Rigid box $\Rightarrow$ total $V$ fixed. Insulated $\Rightarrow Q = 0$ . Free expansion koi external work nahin karta $\Rightarrow W = 0$ , isliye $Δ U = Q - W = 0$ : internal energy conserved hai. Yeh step kyun? Yahan controlled/conserved quantity $U$ hai with $V$ fixed — na ki $T, p$ . Isliye $G$ (natural vars $T, p$ ) galat tool hai; free expansion mein na $T$ na $p$ hold kiya jaata hai.

Step 2. Ideal gas ke liye, $U$ sirf $T$ par depend karta hai (Example 6), isliye $Δ U = 0 \Rightarrow Δ T = 0$ : temperature unchanged. Yeh step kyun? Constant $U$ + $U = U (T)$ forces constant $T$ ideal gas ke liye.

Step 3. Entropy, lekin, badhti hai. $T$ fixed aur $V : V_{1} \to V_{2}$ ke saath, $Δ S = n R ln \frac{V _{2}}{V _{1}} > 0.$ Yeh step kyun? $(\partial S / \partial V)_{T} = (\partial p / \partial T)_{V} = n R / V$ ; volume increase par integrate karo. Process irreversible hai, isliye entropy badhti hai chahe $Q = 0$ ho.

Step 4. $n = 1$ ke liye numbers, volume double karte hue ( $V_{2} / V_{1} = 2$ ): $Δ S = (1) (8.314) ln 2 = 5.763 J K^{- 1} .$

Verify: Student ka " $Δ G = 0$ " meaningless hai kyunki $G$ ka natural clamp active nahin hai. Sahi facts: $Δ U = 0$ , $Δ T = 0$ , $Δ S = + 5.763 J K^{- 1} > 0$ ✓ — Second Law se consistent hai ek irreversible, isolated process ke liye ✓.

Active recall

Recall Har scenario ko uske tool se match karo (answers cover karo)

Constant volume, sealed tank — kaun sa heat capacity aur potential? ::: $C_{V}$ aur $F$ ; $Q = Δ U$ . Fixed pressure par open beaker — heat kiske change ke barabar hai? ::: $Δ H$ (constant- $p$ heat), $C_{p}$ ke through. Reversible adiabatic — kaun si quantity hold hoti hai? ::: entropy $S$ ( $d S = 0$ ). Isothermal — $- Δ F$ physically kya hai? ::: maximum obtainable work. Constant $T, p$ par spontaneous matlab kisi ki sign kya hai? ::: $Δ G < 0$ . Phase coexistence boundary condition? ::: donon phases ka molar $G$ equal hona chahiye. Insulated rigid box mein free expansion kya conserve karta hai? ::: internal energy $U$ (aur ideal gas ke liye, $T$ ).