Phase Rule and Phase Diagrams: Worked Problems

Introduction

Phase-diagram problems often combine three distinct calculations: a phase-rule count, a material-balance lever rule and a boundary-slope equation. Each has different inputs and answers. The phase rule counts possible intensive coordinates, the lever rule allocates a known overall composition between phases, and Clapeyron gives a local relationship between equilibrium pressure and temperature. Solving them in that order prevents the common mistake of using one formula to answer another formula's question.

Core explanation

For an ordinary bulk equilibrium with C independent components and P coexisting phases, F=C−P+2. Count phases as distinct homogeneous equilibrium regions, not separate pieces of the same phase. Count components after accounting for independent reactions or restrictions. If pressure is specified externally and remains fixed, the remaining variable count is F'=C−P+1. This is a constrained count, not a claim that pressure disappeared from thermodynamics. For a pure liquid–vapour pair, C=1 and P=2, so F=1. At a chosen pressure, the saturation temperature is fixed. For a binary liquid–vapour pair, C=2 and P=2, so F=2; fixing pressure leaves one intensive freedom such as temperature.

The lever rule begins with a total and component balance. On a binary tie line at one temperature and pressure, let phase α have composition x α, phase β have x β, and the overall mixture have z. With phase amount fractions f α and f β, z=f α x α+f β x β and f α+f β=1. Therefore f β=(z−x α)/(x β−x α). These are fractions of phase amounts on the same composition basis. Tie-line endpoints come from equilibrium measurements or calculations; the phase rule does not provide them. If z lies outside the interval between endpoints, the proposed two-phase split is invalid under those conditions.

For a pure-substance coexistence boundary, the Clapeyron equation follows from equal chemical potentials: dP/dT=ΔS/ΔV=ΔH/(TΔV). Use molar ΔH and molar ΔV consistently, so units are (J mol⁻¹)/[(K)(m³ mol⁻¹)]=Pa K⁻¹. A positive vaporisation ΔH and positive vaporisation ΔV give a positive slope. Fusion ΔH is usually positive, but fusion ΔV can be negative, as for ordinary ice melting, giving a negative slope. Clapeyron gives a tangent at a point, not automatically a global straight-line equation.

Numerical problems require assumptions about changes in ΔH and ΔV over the temperature interval. The Clausius–Clapeyron approximation neglects liquid volume relative to ideal vapour volume and treats vaporisation enthalpy as constant, yielding ln(P₂/P₁)=−ΔH vap/R(1/T₂−1/T₁). It is convenient over moderate ranges but is less reliable near critical conditions or over a wide interval where enthalpy changes appreciably. State these assumptions before applying the logarithmic relation.

Consider the distinction between amount and freedom. At a pure three-phase triple point, F=0. The relative masses of solid, liquid and vapour may still vary. Conversely, a lever-rule fraction cannot be found from phase count alone: it needs overall composition and endpoint compositions. A diagram shows where phase assemblages are possible, but material balance and thermodynamic property data supply quantitative answers.

Finally, check physical range and sign. Phase fractions must lie from zero to one and sum to one. A slope sign must match ΔH/ΔV for a positive absolute T. A pressure ratio from Clausius–Clapeyron should increase with temperature for ordinary vaporisation. These simple checks often catch misplaced signs, endpoint swaps or Celsius temperatures inserted where Kelvin is required.

Step-by-step reasoning

Read the diagram's fixed variables and phase labels. Count C and P, and calculate F or F' with assumptions stated. For an amount question, identify a single valid tie line and solve a component balance. For a coexistence slope, use absolute temperature and consistent molar property units. Finish by checking signs, ranges and physical interpretation.

Visual explanation

Imagine a P–T boundary with a tangent arrow; Clapeyron gives that arrow's slope. Beside it, imagine a T–composition tie line with overall z between endpoints; the lever rule divides amounts along that segment. The phase rule tells the dimension of the region or line containing the state. The three pictures answer different questions.

Real-world analogy

A railway map can tell how many directions a train may continue, a passenger count can tell how many people ride each branch, and a gradient can tell how steep one track is. These correspond loosely to freedom, phase amounts and boundary slope. None of the three quantities substitutes for the others, even though they describe one journey.

Real-world example

In a flash evaporator, an engineer chooses pressure and temperature, uses VLE to determine liquid and vapour compositions, then applies overall feed balance to calculate their flow rates. A phase-rule count confirms how many intensive choices are available. In a different problem, latent heat and volume change predict how saturation pressure shifts with temperature through Clapeyron.

Why?

Why does the lever rule need z but the phase rule does not? Phase equilibrium fixes which phase compositions can coexist at selected intensive conditions. The total feed composition determines how much of each phase is needed to conserve every component. The two calculations act on equilibrium geometry and material quantity, respectively.

Common misconception

F=0 does not mean there are zero moles or that no phase transformation can continue. It means no independently adjustable intensive coordinate while all stated phases coexist. Likewise, applying a lever rule between endpoints from different temperatures or pressures is invalid because they are not one equilibrium tie line.

Worked example

At fixed pressure, a binary mixture has liquid x A=0.20, vapour y A=0.70 and overall z A=0.40. With C=2 and P=2, F'=2−2+1=1: choosing the equilibrium temperature determines endpoints. The vapour amount fraction is (0.40−0.20)/(0.70−0.20)=0.40. Now for an unrelated pure fusion boundary with ΔH=4.0 kJ mol⁻¹, ΔV=−2.0×10⁻⁶ m³ mol⁻¹ and T=300 K, dP/dT=4000/[300(−2.0×10⁻⁶)]=−6.67×10⁶ Pa K⁻¹. The negative slope follows the contraction on melting.

Quick check

1. Which equation determines phase amounts after x, y and overall z are known? Answer: The lever rule derived from z=f L x+f V y and f L+f V=1; phase-rule F does not determine amounts.

Exam focus

Separate the answers by quantity and unit: F has no unit, phase fractions are dimensionless, and dP/dT has pressure-per-temperature units. Use Kelvin in Clapeyron. State whether pressure is imposed before using F', and show that a phase-fraction result lies in [0,1].

Advanced insight

At a critical endpoint, ΔH and ΔV both tend toward zero, so their ratio must be understood as a limiting coexistence slope rather than evaluated by naive zero-over-zero substitution. Near complex multiphase points, stability and independent constraints matter as much as algebraic counts. Thermodynamic models couple all three tools but do not erase their distinct roles.

Summary

The phase rule counts intensive freedoms, the lever rule uses conservation to determine phase fractions and Clapeyron relates the slope of a pure-substance boundary to enthalpy and volume changes. Correct phase and component identification, one valid tie line, consistent units and sign checks make combined phase-diagram problems manageable.

Practice questions

1. Pure water has ice, liquid and vapour at equilibrium. What is F? Answer: With C=1 and P=3, F=1−3+2=0; this is an invariant triple-point assemblage. 2. On a tie line x=0.10, y=0.60, z=0.35, what is the vapour fraction? Answer: (0.35−0.10)/(0.60−0.10)=0.50, so half the total moles are vapour. 3. A transition has ΔH>0 and ΔV<0. What is the Clapeyron slope sign? Answer: Negative, because dP/dT=ΔH/(TΔV) and absolute T is positive. 4. Why should a Clapeyron calculation use K rather than °C? Answer: T is absolute thermodynamic temperature in ΔH/(TΔV); Celsius values would give incorrect magnitude and could even give nonsensical signs.