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

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1. The master equation (everything starts here)

WHY this form? First law: $dU = \delta Q - \delta W$. For a reversible path $\delta Q = TdS$ (definition of entropy) and $\delta W = p\,dV$. Because $U$ is a state function, $dU$ is the same for any path between two states — so this exact differential holds always, even though we derived it reversibly.

The key reading: $U = U(S,V)$. The variables $S$ and $V$ are the natural variables of $U$ because $dU$ is naturally written in terms of $dS$ and $dV$. From $dU = TdS - pdV$ we read off: $T = \left(\frac{\partial U}{\partial S}\right)_V, \qquad -p = \left(\frac{\partial U}{\partial V}\right)_S$

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2. Building the other three by Legendre transforms

Enthalpy $H$ — swap $V \to p$

$H \equiv U + pV$ HOW the differential follows: $dH = dU + p\,dV + V\,dp = (TdS - pdV) + pdV + Vdp = TdS + Vdp$ So $H = H(S,p)$, with $T = (\partial H/\partial S)_p$ and $V=(\partial H/\partial p)_S$.

Helmholtz free energy $F$ — swap $S \to T$

$F \equiv U - TS$ $dF = dU - TdS - SdT = -SdT - pdV$ So $F = F(T,V)$, with $-S = (\partial F/\partial T)_V$ and $-p = (\partial F/\partial V)_T$.

Gibbs free energy $G$ — swap both

$G \equiv U - TS + pV = H - TS = F + pV$ $dG = -SdT + Vdp$ So $G = G(T,p)$, with $-S = (\partial G/\partial T)_p$ and $V = (\partial G/\partial p)_T$.

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3. WHY each potential is the "useful" one

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4. Maxwell relations (free bonus from exactness)

From $dF = -SdT - pdV$: $\frac{\partial}{\partial V}\Big|_T(-S) = \frac{\partial}{\partial T}\Big|_V(-p)\ \Rightarrow\ \boxed{\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial p}{\partial T}\right)_V}$ The other three (from $U,H,G$):

\left(\frac{\partial T}{\partial p}\right)_S = \left(\frac{\partial V}{\partial S}\right)_p,\quad \left(\frac{\partial S}{\partial p}\right)_T = -\left(\frac{\partial V}{\partial T}\right)_p$$ --- ## 5. Worked examples > [!example] (a) Enthalpy of an ideal gas depends only on $T$ > Show $(\partial H/\partial p)_T = 0$ for an ideal gas. > **Step 1:** Use Maxwell from $G$: $(\partial H/\partial p)_T = V - T(\partial V/\partial T)_p$. > *Why this step?* We need a measurable form; start from $dH=TdS+Vdp \Rightarrow (\partial H/\partial p)_T = T(\partial S/\partial p)_T + V$, then sub the Maxwell $(\partial S/\partial p)_T=-(\partial V/\partial T)_p$. > **Step 2:** Ideal gas $pV=nRT \Rightarrow (\partial V/\partial T)_p = nR/p = V/T$. > *Why?* Just differentiate the equation of state at fixed $p$. > **Step 3:** $(\partial H/\partial p)_T = V - T\cdot(V/T) = 0$. ✓ So $H=H(T)$ only. > [!example] (b) Gibbs energy of an ideal gas vs pressure (isothermal) > Find $G(T,p)-G(T,p_0)$ at fixed $T$. > **Step 1:** $dG = -SdT + Vdp$; at constant $T$, $dG = Vdp$. *Why?* $dT=0$ kills the first term. > **Step 2:** $V = nRT/p$, so $\Delta G = \int_{p_0}^{p} \frac{nRT}{p}dp = nRT\ln(p/p_0)$. > *Why this step?* Substitute equation of state and integrate. This $RT\ln(p/p_0)$ is the workhorse behind chemical potential $\mu(T,p)=\mu^\circ + RT\ln(p/p^\circ)$. > [!example] (c) $C_p - C_V$ from potentials > **Step 1:** $C_V = T(\partial S/\partial T)_V$, $C_p = T(\partial S/\partial T)_p$. *Why?* $\delta Q = TdS$ at const $V$ or $p$. > **Step 2:** Expanding $S(T,V)$ and using Maxwell gives > $$C_p - C_V = T\left(\frac{\partial p}{\partial T}\right)_V\left(\frac{\partial V}{\partial T}\right)_p$$ > For ideal gas: $(\partial p/\partial T)_V = nR/V$, $(\partial V/\partial T)_p = nR/p$, product $\times T = nR$. So $C_p-C_V = nR$. ✓ --- ## 6. Common mistakes (Steel-manned) > [!mistake] "$F$ and $G$ are the same — both are 'free energy'." > **Why it feels right:** both are energies you subtract $TS$ from, both are minimized at equilibrium, both called "free energy." **The fix:** they differ by $pV$ ($G=F+pV$) and have *different natural variables*. Use $F$ when **volume** is fixed (rigid box), $G$ when **pressure** is fixed (open beaker). Choosing the wrong one means your "minimize at equilibrium" criterion is just false for your constraints. > [!mistake] "Legendre transform means just *add* $pV$ randomly." > **Why it feels right:** $H=U+pV$ adds, $F=U-TS$ subtracts — looks inconsistent. **The fix:** the rule is *subtract the conjugate-pair product to remove a variable, add it to bring its slope in.* $H=U-(-p)V$: we swap $V$ for $p$. Sign comes from which variable carries a minus in $dU=TdS-pdV$. Track the signs from $dU$, never memorize. > [!mistake] "Maxwell relation $(\partial S/\partial V)_T=(\partial p/\partial T)_V$ — I'll guess the signs." > **Why it feels right:** four relations, two have minus signs, easy to mix. **The fix:** derive each from the matching potential's $d\Phi$ using equality of mixed partials. The signs are *forced* by the signs in $d\Phi$ — never a guess. --- ## 7. Active recall > [!recall]- Quick self-test (cover the answers) > - What are the natural variables of $G$? → $T,p$. > - Write $dF$. → $-SdT - pdV$. > - At constant $T,p$ which potential is minimized? → $G$. > - $-\Delta F$ at constant $T$ equals what? → maximum work obtainable. > - State the Maxwell relation from $G$. → $(\partial S/\partial p)_T = -(\partial V/\partial T)_p$. > [!recall]- Feynman: explain to a 12-year-old > Imagine energy is money in a piggy bank. $U$ is the total money inside. But sometimes you can't use *all* of it — the universe charges a "messiness tax" ($TS$) that you can't spend. What's left after the tax is the **free** money you can actually use to do something — that's $F$. If you're also playing the game outdoors where the air pushes on you, you must rent space from the atmosphere ($pV$); adding that gives $H$, and after both the tax and the rent you get $G$. Depending on whether your jar is sealed (fixed volume) or open to the air (fixed pressure), a different "money" number tells you whether you've settled down (reached equilibrium) — nature always slides toward the *smallest* one allowed. > [!mnemonic] Remembering the potentials > **"Good Physicists Have Studied Under Very Fine Teachers"** — square of state functions: > Corners $U, H, G, F$ with variables $S, V, T, p$ on the **Thermodynamic Square**. > For the **swaps**: *H = U **H**oists pV* (add $pV$); *F = U **F**rees by losing TS* (subtract $TS$); *G gets **both**.* --- ## #flashcards/physics Definition of enthalpy $H$ ::: $H \equiv U + pV$ Definition of Helmholtz free energy $F$ ::: $F \equiv U - TS$ Definition of Gibbs free energy $G$ ::: $G \equiv U - TS + pV = H - TS$ Master differential of $U$ ::: $dU = TdS - pdV$ Differential of $H$ and its natural variables ::: $dH = TdS + Vdp$; natural vars $S,p$ Differential of $F$ and its natural variables ::: $dF = -SdT - pdV$; natural vars $T,V$ Differential of $G$ and its natural variables ::: $dG = -SdT + Vdp$; natural vars $T,p$ What is a Legendre transform doing physically ::: swapping a variable for its conjugate slope (e.g. $V\leftrightarrow p$) by subtracting/adding the product Which potential is minimized at constant $T,p$ ::: Gibbs free energy $G$ Which potential is minimized at constant $S,V$ ::: internal energy $U$ Physical meaning of $-\Delta F$ at constant $T$ ::: maximum total work the system can do Physical meaning of $\Delta H$ at constant $p$ ::: heat exchanged (heat of reaction) Maxwell relation from $F$ ::: $(\partial S/\partial V)_T = (\partial p/\partial T)_V$ Maxwell relation from $G$ ::: $(\partial S/\partial p)_T = -(\partial V/\partial T)_p$ General $C_p - C_V$ formula ::: $T(\partial p/\partial T)_V(\partial V/\partial T)_p$ $G$ change for ideal gas isothermal pressure change ::: $\Delta G = nRT\ln(p/p_0)$ Why are Maxwell relations true ::: $d\Phi$ is exact, so mixed second partials are equal --- ## Connections - [[Legendre transforms]] — the mathematical machine generating all four potentials - [[Maxwell relations]] — direct consequence of exactness of $d\Phi$ - [[Entropy and the Second Law]] — supplies the $TdS$ term - [[Chemical potential and the grand potential]] — extends to variable $N$: add $\mu\,dN$ - [[Phase equilibria & Clausius–Clapeyron]] — uses $G$ minimization and the $F$-Maxwell relation - [[Heat capacities Cp and Cv]] — derived via potentials in Example (c) ## 🖼️ Concept Map ```mermaid flowchart TD U[Internal energy U] MASTER[dU = TdS - pdV] NAT[Natural vars S and V] LEG[Legendre transform] H[Enthalpy H] F[Helmholtz F] G[Gibbs G] EOS[Equations of state] EQ[Minimized at equilibrium] U -->|state function gives| MASTER MASTER -->|read off| NAT LEG -->|swaps var for slope| U U -->|add pV, swap V to p| H U -->|subtract TS, swap S to T| F U -->|subtract TS add pV| G H -->|dH = TdS + Vdp| EOS F -->|dF = -SdT - pdV| EOS G -->|dG = -SdT + Vdp| EOS F -->|at fixed T,V| EQ G -->|at fixed T,p| EQ ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Dekho, thermodynamic potentials ka funda simple hai: ek hi system ki energy ko hum alag-alag tareeke se likhte hain, depending on ki lab mein hum kya control kar sakte hain. Internal energy $U$ ke natural variables $S$ aur $V$ hain, lekin entropy ko directly clamp karna possible nahi hota. Real life mein hum **temperature** (thermostat se) aur **pressure** (khuli atmosphere se) control karte hain. Isliye hum $U$ ko transform karke $H$, $F$, aur $G$ banate hain — same physics, bas convenient variables mein. > > Trick ka naam hai **Legendre transform**: ek variable ko uske conjugate (slope) se swap karne ke liye, unka product subtract/add karte ho. $H=U+pV$ (volume ko pressure se swap), $F=U-TS$ (entropy ko temperature se swap), aur $G=U-TS+pV$ (dono swap). Sign hamesha $dU=TdS-pdV$ se nikaalo — ratto mat, derive karo. > > Har potential ka apna physical use hai: $F$ ka $-\Delta F$ constant $T$ par **maximum work** deta hai; $H$ ka $\Delta H$ constant pressure par **heat of reaction** hai; aur sabse important — $G$ constant $T,p$ par **minimize** hota hai equilibrium par, jo chemistry aur phase transitions ke liye perfect hai. Bonus: kyunki har $d\Phi$ exact differential hai, mixed partials barabar hote hain, jisse **Maxwell relations** free mein mil jaate hain — jinse hum un cheezon ko measure kar paate hain jo seedhe naapna mushkil hota hai (jaise $\partial S/\partial V$). Yahi advanced thermodynamics ki reedh ki haddi hai. ![[audio/2.4.01-Thermodynamic-potentials-—-U-(internal),-H-(enthalpy),-F-(Helmholtz),-G-(Gibbs).mp3]]