Thermodynamics · Upper Division

Phase Transitions

Phase transitions — melting, boiling, ferromagnetism, superconductivity — are collective phenomena where macroscopic properties change discontinuously. They are among the most beautiful applications of thermodynamics and statistical mechanics.

PrerequisitesLawsofthermodynamics(Ch.12)Statisticalmechanics(Ch.S)BasiccalculusLaws of thermodynamics (Ch. 12) \cdot Statistical mechanics (Ch. S) \cdot Basic calculus
Learning Goals
  • Apply the Gibbs phase rule to determine degrees of freedom for single and multi-component systems.
  • Use the Clausius-Clapeyron equation to find coexistence curve slopes from latent heat and volume change.
  • Distinguish first-order and second-order phase transitions by their signatures in entropy and heat capacity.
  • Minimize the Landau free energy to find the order parameter and its temperature dependence.
  • Explain the concept of universality and why critical exponents are independent of microscopic details.

P.1 Phase Diagrams and Coexistence

A phase diagram maps the equilibrium phase of a substance in the (T, P) plane. The boundaries between phases are coexistence curves where two phases can exist simultaneously. Three curves meet at the triple point— the unique (T, P) where all three phases coexist. The liquid–gas curve ends at thecritical point (Tc, Pc) beyond which liquid and gas become indistinguishable.

Definition P.1Gibbs Phase Rule
For a system with C chemical components and P phases in thermodynamic equilibrium, the number of independent intensive variables (degrees of freedom) is:F=CP+2F = C - P + 2Forpurewater(C=1):singlephaseF=2(TandPfree);twophasecoexistenceF=1(acuFor pure water (C=1): single phase \to F=2 (T and P free); two-phase coexistence \to F=1 (a curveintheTPplane);triplepointF=0(afixedpointrve in the T-P plane); triple point \to F=0 (a fixed point.

P.2 The Clausius–Clapeyron Equation

Along a coexistence curve, two phases α and β have equal Gibbs free energy: G_α(T, P) = G_β(T, P). Differentiating this equality with respect to T along the curve:

dPdT=LTΔV(Clausius-Clapeyron)\frac{dP}{dT}=\frac{L}{T\Delta V} \qquad \text{(Clausius-Clapeyron)}(P.1)

Here L is the latent heat (heat absorbed per mole at the transition) and ΔV = V_β − V_α is the molar volume change. This equation determines the slope of every coexistence curve:

Liquid–gas: ΔV ≈ RT/P (ideal gas), so dP/dT = LP/(RT²), giving P = P₀ exp(−L/RT) — the vapor pressure increases exponentially with temperature.

Solid–liquid:ΔV is small (liquids and solids have similar densities). For water, ΔV < 0 (ice is less dense than water), so dP/dT < 0 — increasing pressure lowers the melting point. Ice melts under pressure, which is why ice skating works (though the effect is only 0.007°C/atm, too small to matter for skating — surface friction actually explains that).

Example P.1Vapor Pressure of Water

Use ClausiusClapeyrontoestimatetheboilingpointofwaterat0.5atm(highaltitude).Given:L=us–Clapeyron to estimate the boiling point of water at 0.5 atm (high altitude). Given: L =ol at 100°C.

Integrated form:ln(P2/P1)=(L/R)(1/T21/T1ln(P_{2}/P_{1}) = -(L/R)(1/T_{2} - 1/T_{1}
Known point:P1=1atmatT1=373KP_{1} = 1 atm at T_{1} = 373 K
Solve for T₂:1/T2=1/T1(R/L)ln(P2/P1)=1/373(8.314/40700)ln(0.51/T_{2} = 1/T_{1} - (R/L)ln(P_{2}/P_{1}) = 1/373 - (8.314/40700)ln(0.5
Compute:= 0.002681(2.04×104)(0.693)=0.002681+0.000141=0.002822K10.002681 - (2.04\times10^{-4})(-0.693) = 0.002681 + 0.000141 = 0.002822 K^{-1}
T₂:T2=1/0.002822=354K=81°C.Waterboilsat 81°CataltitudeyourpastatakeslongerT_{2} = 1/0.002822 = 354 K = 81°C. Water boils at ~81°C at altitude — your pasta takes longer to cook.

P.3 First and Second Order Phase Transitions

Definition P.2Order of a Phase Transition
First-order transition: ThefirstderivativeofG(entropyS=\partialG/\partialTandvolumeV=\partialG/\partialP)isdiscontinuous.TherThe first derivative of G (entropy S = -\partialG/\partialT and volume V = \partialG/\partialP) is discontinuous. Ther is latent heat. Examples: melting, boiling, liquid crystal transitions.Second-order (continuous) transition: ThefirstderivativeofGiscontinuousbutthesecondderivative(heatcapacityCp=T2The first derivative of G is continuous but the second derivative (heat capacity Cp = -T\partial^{2}/\partialT2)diverges.Nolatentheat.Examples:ferromagneticCuriepoint,superconductingtransi\partialT^{2}) diverges. No latent heat. Examples: ferromagnetic Curie point, superconducting transiion, liquid He superfluidity, critical point.

At a second-order transition, the system develops an order parameterthat grows continuously from zero. For a ferromagnet, this is the spontaneous magnetization M; for a superconductor, the amplitude of the Cooper pair wavefunction.

P.4 Landau Theory

Near a continuous phase transition, Landau (1937) proposed expanding the Gibbs free energy in powers of the order parameter φ:

G(T,ϕ)=G0+a(T)ϕ2+bϕ4+(b>0)G(T,\phi)=G_0+a(T)\phi^2+b\phi^4+\cdots \qquad (b>0)(P.2)

where a(T) = a₀(T − Tc) changes sign at the critical temperature Tc. For T > Tc, the minimum is at φ = 0 (disordered phase). For T < Tc, two minima appear at φ = ±√(−a/2b) — spontaneous symmetry breaking.

ϕ(TcT)ββ=12 (Landau mean-field)|\phi|\propto (T_c-T)^\beta \qquad \beta=\frac{1}{2}\ \text{(Landau mean-field)}(P.3)

The exponent β = ½ is the mean-field critical exponent. Real systems near the critical point have β ≈ 0.326 (Ising model in 3D) — a universal value that depends only on the symmetry of the order parameter and the dimensionality, not on microscopic details. This universality is explained by the renormalization group.

Example P.2Ferromagnetic Phase Transition

TheIsingmodelinmeanfieldapproximationgivestheselfconsistencyequationm=tanh(JThe Ising model in mean-field approximation gives the self-consistency equation m = tanh(Jzm/kBT),wheremisthemagnetizationpersite,ziscoordinationnumber,Jisexchangeezm/k_{BT}), where m is the magnetization per site, z is coordination number, J is exchange energy. Find Tc and m(T) near Tc.

At Tc:Forsmallm,tanh(x)xx3/3.m=(Jz/kBT)m(Jz/kBT)3m3/3For small m, tanh(x) \approx x - x^{3}/3. m = (Jz/k_{BT})m - (Jz/k_{BT})^{3} m^{3}/3
Critical condition:NontrivialsolutionrequiresJz/kBTc=1Tc=Jz/kBNon-trivial solution requires Jz/k_{BTc} = 1 \to Tc = Jz/k_{B}
Near Tc:Letτ=(TTc)/Tc1.Thenm23(1T/Tc)/(Jz/kBT)3(TcTLet \tau = (T-Tc)/Tc ≪ 1. Then m^{2} \approx 3(1 - T/Tc)/(Jz/k_{BT})^{3} \propto (Tc-T
Order parameter:m(TcT)1/2themeanfieldβ=1/2exponentm \propto (Tc - T)^{1/2} — the mean-field \beta = 1/2 exponent.
Heat capacity:C jumps discontinuously at Tc but has no divergence — a characteristicofmeanfieldtheory(realIsing:Cdivergeslogarithmicallyin2D,as(TTcharacteristic of mean-field theory (real Ising: C diverges logarithmically in 2D, as (T-T
Definition P.3Common Traps
  • Phase-rule P means number of phases: donotconfuseitwithpressureinF=CP+2do not confuse it with pressure in F = C - P + 2
  • Latent heat marks first-order transitions: continuous transitions have no latent heat even though heat capacity can diverge.
  • Thesignof\DeltaVmattersThe sign of \DeltaV matters: water's melting curve slopes backward because ice has larger molar volume than liquid water.
  • Mean-field exponents are not universal truth: real critical exponents depend on symmetry and dimensionality.
Exercises — P.1–P.4 Phase Transitions
0.
State the triple point and critical point conditions for water. Why is the triple point used to define the Kelvin temperature scale? What is a supercritical fluid?
°C
Straightforward
1.State the triple point and critical point conditions for water. Why is the triple point used to define the Kelvin temperature scale? What is a supercritical fluid?
Straightforward
2.Calculate dP/dT for the ice–water coexistence curve. How much does pressure lower the melting point? Does this explain ice skating?
Intermediate
3.MinimizetheLandaufreeenergyG=a(TTc)ϕ2+bϕ4tofindtheorderparameterandheatcMinimize the Landau free energy G = a(T-Tc)\phi^{2} + b\phi^{4} to find the order parameter and heat capacity on both sides of Tc.
Intermediate
4.Find the critical point (Tc, Pc,Vc)ofthevanderWaalsgas.Showthatthereducedequationofstate(intermsofTr=c, Vc) of the van der Waals gas. Show that the reduced equation of state (in terms of Tr =e law of corresponding states.
Challenging
Key Takeaways
  • Coexistencecurvesseparatephases;triplepoint(allthreephases)hasF=0;criticalpoinCoexistence curves separate phases; triple point (all three phases) has F=0; critical point ends liquid–gas curve.
  • ClausiusClapeyron:dP/dT=L/(T\DeltaV)slopeofcoexistencecurvefromlatentheatandvoluClausius–Clapeyron: dP/dT = L/(T\DeltaV) — slope of coexistence curve from latent heat and volume change.
  • First-order: latent heat, discontinuous S and V. Second-order: continuous S, divergent Cp, no latent heat.
  • OrderparameterϕappearsatTc;growsas(TcT)β.Meanfield:β=12Order parameter \phi appears at Tc; grows as (Tc-T)^\beta. Mean-field: \beta = \frac{1}{2}.
  • Landautheory:G=aϕ2+bϕ4;spontaneoussymmetrybreakingwhenachangessignLandau theory: G = a\phi^{2} + b\phi^{4}; spontaneous symmetry breaking when a changes sign.
  • Critical exponents are universal — depend on symmetry and dimension, not microscopic details.