Intuition The big pictureA thermodynamic potential is an energy-like quantity whose natural variables you can control in a lab, and whose minimum tells you the equilibrium state. Just as a ball rolls to the bottom of a valley (minimum of potential energy), a thermodynamic system "rolls" to the minimum of the right potential — but which potential depends on what you hold fixed (entropy? temperature? pressure? volume?).
WHY four of them? Because in different experiments different things are held constant. You don't control entropy easily, but you can control temperature (thermostat) and pressure (open beaker). Each potential is engineered to make a particular pair of variables the "natural" ones.
Definition Internal energyU U U U U U combined 1st + 2nd law for a reversible process (closed system, no other work) is
d U = T d S − p d V dU = T\,dS - p\,dV d U = T d S − p d V
WHY this form? First law: d U = δ Q − δ W dU = \delta Q - \delta W d U = δ Q − δ W δ Q = T d S \delta Q = T\,dS δ Q = T d S δ W = p d V \delta W = p\,dV δ W = p d V d U = T d S − p d V . dU = T\,dS - p\,dV. d U = T d S − p d V .
This equation is the mother of all potentials. Reading it as U = U ( S , V ) U=U(S,V) U = U ( S , V ) natural variables of U U U = = S and V = = ==S \text{ and } V== == S and V == d U dU d U d S dS d S d V dV d V
Intuition What is a Legendre transform (Feynman style)?Suppose your energy is written in a variable you can't control (like S S S partner T T T can control with a thermostat). A Legendre transform swaps a variable for its conjugate slope by subtracting off the product .
HOW it works. We have T = ( ∂ U / ∂ S ) V T = (\partial U/\partial S)_V T = ( ∂ U / ∂ S ) V S S S T T T T S TS T S F = U − T S . F = U - TS. F = U − T S . d F = d U − T d S − S d T = ( T d S − p d V ) − T d S − S d T = − S d T − p d V . dF = dU - T\,dS - S\,dT = (T\,dS - p\,dV) - T\,dS - S\,dT = -S\,dT - p\,dV. d F = d U − T d S − S d T = ( T d S − p d V ) − T d S − S d T = − S d T − p d V . T d S T\,dS T d S cancel , leaving d F dF d F d T dT d T d V dV d V
The same trick swapping − p d V → + V d p -p\,dV \to +V\,dp − p d V → + V d p + p V +pV + p V
U (internal energy) U \quad\text{(internal energy)} U (internal energy) H = U + p V (enthalpy) H = U + pV \quad\text{(enthalpy)} H = U + p V (enthalpy) F = U − T S (Helmholtz free energy) F = U - TS \quad\text{(Helmholtz free energy)} F = U − T S (Helmholtz free energy) G = H − T S = U + p V − T S (Gibbs free energy) G = H - TS = U + pV - TS \quad\text{(Gibbs free energy)} G = H − T S = U + p V − T S (Gibbs free energy)
Derivation of d H dH d H H = U + p V ⇒ d H = d U + p d V + V d p = ( T d S − p d V ) + p d V + V d p = T d S + V d p . H=U+pV \Rightarrow dH = dU + p\,dV + V\,dp = (T\,dS - p\,dV)+p\,dV+V\,dp = T\,dS+V\,dp. H = U + p V ⇒ d H = d U + p d V + V d p = ( T d S − p d V ) + p d V + V d p = T d S + V d p . Why this step? The − p d V -p\,dV − p d V d U dU d U + p d V +p\,dV + p d V d ( p V ) d(pV) d ( p V )
Derivation of d F dF d F done above. Why? Subtracting T S TS T S T d S T\,dS T d S T T T
Derivation of d G dG d G G = U + p V − T S G=U+pV-TS G = U + p V − T S d G = d U + p d V + V d p − T d S − S d T = − S d T + V d p . dG = dU + p\,dV+V\,dp - T\,dS - S\,dT = -S\,dT + V\,dp. d G = d U + p d V + V d p − T d S − S d T = − S d T + V d p . Why? Both T d S T\,dS T d S p d V p\,dV p d V ( T , p ) (T,p) ( T , p )
Intuition Which potential is "the boss"?The 2nd law says the total entropy of the universe never decreases . Translating that constraint to the system alone under different lab conditions turns "entropy maximized" into "the right potential minimized."
Held constant
Minimized potential
Lab situation
S , V S,V S , V U U U isolated, rigid (rare)
S , p S,p S , p H H H adiabatic, constant pressure
T , V T,V T , V F F F thermostat, rigid box
T , p T,p T , p G G G thermostat, open to atmosphere (chemistry! )
Intuition Why "free energy"?F = U − T S F=U-TS F = U − T S free to do work at constant T T T T S TS T S Δ G < 0 \Delta G<0 Δ G < 0
Worked example Example 1 — Enthalpy = heat at constant pressureShow δ Q p = d H \delta Q_p = dH δ Q p = d H Step 1: d H = T d S + V d p dH = T\,dS + V\,dp d H = T d S + V d p Why? Use the differential we derived.
Step 2: At constant pressure d p = 0 dp=0 d p = 0 d H = T d S = δ Q rev dH = T\,dS = \delta Q_{\text{rev}} d H = T d S = δ Q rev Why? δ Q = T d S \delta Q=T\,dS δ Q = T d S Result: heat absorbed at constant p p p Δ H \Delta H Δ H C p = ( ∂ H / ∂ T ) p C_p=(\partial H/\partial T)_p C p = ( ∂ H / ∂ T ) p
Worked example Example 2 — Get an entropy fromG G G G ( T , p ) G(T,p) G ( T , p ) S S S Step 1: d G = − S d T + V d p dG=-S\,dT+V\,dp d G = − S d T + V d p Why? Natural-variable differential.
Step 2: At constant p p p ( ∂ G ∂ T ) p = − S \left(\frac{\partial G}{\partial T}\right)_p=-S ( ∂ T ∂ G ) p = − S Why? Match the d T dT d T Result: S = − ( ∂ G ∂ T ) p S=-\left(\dfrac{\partial G}{\partial T}\right)_p S = − ( ∂ T ∂ G ) p V = ( ∂ G ∂ p ) T V=\left(\dfrac{\partial G}{\partial p}\right)_T V = ( ∂ p ∂ G ) T
Worked example Example 3 — A Maxwell relation fromF F F ( ∂ S ∂ V ) T = ( ∂ p ∂ T ) V \left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial p}{\partial T}\right)_V ( ∂ V ∂ S ) T = ( ∂ T ∂ p ) V Step 1: d F = − S d T − p d V dF=-S\,dT-p\,dV d F = − S d T − p d V S = − ( ∂ F / ∂ T ) V S=-(\partial F/\partial T)_V S = − ( ∂ F / ∂ T ) V p = − ( ∂ F / ∂ V ) T p=-(\partial F/\partial V)_T p = − ( ∂ F / ∂ V ) T Step 2: Mixed second derivatives of F F F ∂ 2 F ∂ V ∂ T = ∂ 2 F ∂ T ∂ V \frac{\partial^2 F}{\partial V\,\partial T}=\frac{\partial^2 F}{\partial T\,\partial V} ∂ V ∂ T ∂ 2 F = ∂ T ∂ V ∂ 2 F Why? F F F Step 3: That gives − ( ∂ S ∂ V ) T = − ( ∂ p ∂ T ) V -\left(\frac{\partial S}{\partial V}\right)_T = -\left(\frac{\partial p}{\partial T}\right)_V − ( ∂ V ∂ S ) T = − ( ∂ T ∂ p ) V Result: ( ∂ S ∂ V ) T = ( ∂ p ∂ T ) V \boxed{\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial p}{\partial T}\right)_V} ( ∂ V ∂ S ) T = ( ∂ T ∂ p ) V
Common mistake "Free energy means energy that's free / costs nothing."Why it feels right: the word "free." Truth: F F F G G G maximum useful work extractable at fixed T T T p p p G G G T S TS T S locked away by the second law. Fix: read "free" as "available to do work," not "free of charge."
Common mistake "All potentials are minimized at equilibrium, always."Why it feels right: symmetry of the four definitions. Truth: minimization holds only for the matching constraints (e.g. G G G T , p T,p T , p Fix: match potential ⇄ its natural variables held fixed.
Common mistake Wrong sign ind H dH d H d H = T d S − V d p dH=T\,dS-V\,dp d H = T d S − V d p Why it feels right: d U dU d U Fix: the p V pV p V + p d V +p\,dV + p d V d ( p V ) d(pV) d ( p V ) − p d V -p\,dV − p d V + V d p +V\,dp + V d p
Recall Feynman: explain to a 12-year-oldEnergy is like money in your wallet. But how much you can actually spend depends on the rules of where you are. In a hot room (fixed temperature) a "tax" called T S TS T S free energy F F F p V pV p V G G G U , H , F , G U,H,F,G U , H , F , G after accounting for which taxes apply in your situation , and the system always settles where its relevant wallet is smallest.
Mnemonic Remembering the four & their variables"Good Physicists Have Studied Under Very Fine Teachers" isn't it — use the square :
Corners clockwise: U U U H H H G G G F F F Add p V pV p V T S TS T S
Sign rule: "S follows T, V follows p." When T T T − S d T -S\,dT − S d T p p p + V d p +V\,dp + V d p
What is the fundamental differential of U U U d U = T d S − p d V dU = T\,dS - p\,dV d U = T d S − p d V (natural variables
S , V S,V S , V )
Define enthalpy H H H d H dH d H H = U + p V H=U+pV H = U + p V ;
d H = T d S + V d p dH=T\,dS+V\,dp d H = T d S + V d p (natural variables
S , p S,p S , p )
Define Helmholtz free energy and give d F dF d F F = U − T S F=U-TS F = U − T S ;
d F = − S d T − p d V dF=-S\,dT-p\,dV d F = − S d T − p d V (natural variables
T , V T,V T , V )
Define Gibbs free energy and give d G dG d G G = U + p V − T S G=U+pV-TS G = U + p V − T S ;
d G = − S d T + V d p dG=-S\,dT+V\,dp d G = − S d T + V d p (natural variables
T , p T,p T , p )
Which potential is minimized at constant T , p T,p T , p Gibbs free energy
G G G (the chemistry case)
Which potential is minimized at constant T , V T,V T , V Helmholtz free energy
F F F Why does enthalpy matter for constant-pressure heat? At
d p = 0 dp=0 d p = 0 ,
d H = T d S = δ Q dH=T\,dS=\delta Q d H = T d S = δ Q , so
Δ H \Delta H Δ H is heat absorbed
Express S S S V V V G ( T , p ) G(T,p) G ( T , p ) S = − ( ∂ G / ∂ T ) p S=-(\partial G/\partial T)_p S = − ( ∂ G / ∂ T ) p ,
V = ( ∂ G / ∂ p ) T V=(\partial G/\partial p)_T V = ( ∂ G / ∂ p ) T What is a Legendre transform doing here? Swapping a variable for its conjugate by subtracting their product, to change natural variables
State the Maxwell relation from F F F ( ∂ S / ∂ V ) T = ( ∂ p / ∂ T ) V (\partial S/\partial V)_T=(\partial p/\partial T)_V ( ∂ S / ∂ V ) T = ( ∂ p / ∂ T ) V What does "free" in free energy mean? Maximum work extractable at fixed
T T T (and
p p p for
G G G ); not "costless"
First Law of Thermodynamics — source of d U = T d S − p d V dU=TdS-pdV d U = T d S − p d V Entropy and the Second Law — why minimization replaces maximizationMaxwell Relations — direct consequence of exact differentials of U , H , F , G U,H,F,G U , H , F , G Legendre Transform — the mathematical engineHeat Capacities Cp and Cv — C p = ( ∂ H / ∂ T ) p C_p=(\partial H/\partial T)_p C p = ( ∂ H / ∂ T ) p C v = ( ∂ U / ∂ T ) V C_v=(\partial U/\partial T)_V C v = ( ∂ U / ∂ T ) V Gibbs Free Energy and Chemical Equilibrium — Δ G < 0 \Delta G<0 Δ G < 0
swaps variable for conjugate
Intuition Hinglish mein samjhoDekho, thermodynamic potential ka matlab hai ek energy-jaisa quantity (U , H , F , G U, H, F, G U , H , F , G minimum batata hai ki system equilibrium me kahan settle hoga — bilkul jaise ball valley ke neeche tak roll karke ruk jaati hai. Lekin twist ye hai: kaunsa potential minimize hoga ye depend karta hai ki tum lab me kya constant rakh rahe ho. Agar temperature aur pressure fix hain (jaise open beaker, room temperature — yani chemistry), to Gibbs free energy G G G boss hai.
Sab kuch nikalta hai ek hi master equation se: d U = T d S − p d V dU = T\,dS - p\,dV d U = T d S − p d V S S S T T T Legendre transform ka trick use karte hain — variable ko uske partner se swap karne ke liye unka product subtract/add kar dete hain. T S TS T S F = U − T S F = U - TS F = U − T S T d S T\,dS T d S d F = − S d T − p d V dF = -S\,dT - p\,dV d F = − S d T − p d V p V pV p V H H H G G G
"Free energy" me "free" ka matlab muft nahi hai bhai — iska matlab hai kitna useful kaam nikaal sakte ho fixed temperature pe. T S TS T S Δ G < 0 \Delta G < 0 Δ G < 0 G G G ∂ S / ∂ V \partial S/\partial V ∂ S / ∂ V ∂ p / ∂ T \partial p/\partial T ∂ p / ∂ T
