Which one of the following is an intensive property?

The chemical potential of a component in a solution

The enthalpy H of a 1 kg sample of the metal at 300 K

The entropy S of a given single crystal of the metal

The chemical potential of a component in a solution

The heat capacity of a whole casting of the alloy

On the pressure–temperature diagram of pure water the solid–liquid boundary slopes backwards (dP/dT < 0), unlike that of most substances. Using the Clapeyron equation dP/dT = ΔH/(TΔV) for melting, the reason is that

the volume decreases on melting, since ice is less dense than liquid water

the enthalpy of fusion of ice is negative because heat is released when ice melts

the volume decreases on melting, since ice is less dense than liquid water

the entropy of liquid water is lower than that of ice at the melting point

the vapour pressure of ice is higher than that of liquid water at 0 °C

Two phases α and β of a binary A–B system are in equilibrium at fixed T and P. On a plot of molar Gibbs energy against composition, the equilibrium compositions of α and β are fixed by the condition that

one straight line is tangent to both curves, so μ_A and μ_B are each the same in α and in β

the two curves have equal Gibbs energies at the two compositions, so the compositions lie on a horizontal line

one straight line is tangent to both curves, so μ_A and μ_B are each the same in α and in β

each curve has zero slope at its contact point, so each phase sits at its own minimum of G

the tangents to the two curves are parallel, although they cut the composition axes at different values

A regular solution has ΔH_mix = Ωx(1 − x) with Ω > 0 and the ideal entropy of mixing. At a temperature below T_c = Ω/(2R), which statement about its molar Gibbs energy curve is correct?

The curve has a local maximum at x = 0.5 between two minima, so equilibrium at x = 0.5 is a two-phase mixture

The curve has a single minimum at x = 0.5, so the solution is a single phase at every composition of the system

The curve has two minima, yet an alloy at x = 0.5 stays stable as one homogeneous phase because of the mixing entropy

d²G/dx² is positive at x = 0.5 and negative close to the two pure components

In the lead–tin system the eutectic is at 61.9 wt% Sn and 183 °C, and the maximum solubility of Sn in lead-rich α at that temperature is 18.3 wt% Sn. An alloy of 30 wt% Sn is cooled slowly from the liquid to just below 183 °C. Its microstructure is

primary α grains surrounded by an (α + β) eutectic mixture

single-phase α, since the alloy contains less tin than the eutectic

primary β grains surrounded by an (α + β) eutectic mixture

an entirely lamellar (α + β) eutectic with no primary phase

primary α grains surrounded by an (α + β) eutectic mixture

In the Fe–Fe₃C system the eutectoid is at 0.76 wt% C and 727 °C. A 1.0 wt% C steel is cooled very slowly from the single-phase austenite field. Between the temperature at which the first new phase appears and 727 °C, the new phase that forms is

proeutectoid cementite, forming mainly along the austenite grain boundaries

proeutectoid ferrite, forming at the austenite grain boundaries as the steel cools

martensite, because the carbon content of the steel is above the eutectoid value

proeutectoid cementite, forming mainly along the austenite grain boundaries

pearlite colonies only, with no separate proeutectoid phase

In a eutectic system of A and B, the phases L, α and β are in equilibrium at the eutectic temperature T_E. On the molar Gibbs energy–composition plot drawn at T_E,

one straight line is tangent to all three curves, at the three compositions of the invariant reaction

the liquid curve lies above both solid curves at every composition, so no liquid can appear

one straight line is tangent to all three curves, at the three compositions of the invariant reaction

three separate common tangents exist, one for each pair of phases, lying at three different heights

the α–β common tangent passes above the liquid curve, so the liquid is the more stable phase at its own composition

Which of the following statements about thermodynamic functions and equilibrium are correct?

For a closed system doing only pressure–volume work at constant T and P, equilibrium corresponds to a minimum of the Gibbs energy

Gibbs energy G is extensive, whereas the chemical potential μ_i is intensive

For an ideal solution ΔH_mix = 0, so mixing is driven entirely by the increase in configurational entropy

The third law fixes the entropy of every substance, including a glass, at exactly zero at 0 K

For a closed system doing only pressure–volume work at constant T and P, equilibrium corresponds to a minimum of the Gibbs energy

Gibbs energy G is extensive, whereas the chemical potential μ_i is intensive

For an ideal solution ΔH_mix = 0, so mixing is driven entirely by the increase in configurational entropy

Which of the following statements about the Gibbs phase rule are correct? (Use F = C − P + 1 for binary systems at constant pressure and F = C − P + 2 when T and P can both vary.)

In a binary two-phase field at constant pressure, fixing the temperature fixes the compositions of both phases

In a single-phase field of a binary at constant pressure, the temperature and the composition can both be varied independently

A pure substance has F = 0 at its triple point, where solid, liquid and vapour coexist

In a binary two-phase field at constant pressure, fixing the temperature fixes the compositions of both phases

At the eutectic point of a binary at constant pressure, F = 1, so the temperature can vary slightly while L, α and β coexist

In a single-phase field of a binary at constant pressure, the temperature and the composition can both be varied independently

A pure substance has F = 0 at its triple point, where solid, liquid and vapour coexist

Which of the following statements about tie lines, the lever rule and microconstituents are correct?

At a fixed temperature inside a two-phase field, the compositions of the two phases do not depend on the overall composition of the alloy

If compositions are in at%, the lever rule gives mole fractions, not mass fractions

In a hypoeutectic alloy just below the eutectic temperature, the mass fraction of the eutectic microconstituent equals the mass fraction of liquid just above it

At a fixed temperature inside a two-phase field, the compositions of the two phases do not depend on the overall composition of the alloy

If compositions are in at%, the lever rule gives mole fractions, not mass fractions

In a hypoeutectic alloy just below the eutectic temperature, the mass fraction of the eutectic microconstituent equals the mass fraction of liquid just above it

The fraction of a phase is the tie-line segment on the same side as that phase divided by the whole tie line

Thermodynamics and Phase Equilibria: Laws, Phase Rule, Binary Diagrams and the Free-Energy Construction

This chapter answers the first paragraph of Section XE2.2, “Thermodynamics, Kinetics and Phase Transformations”: “Extensive and intensive thermodynamic properties, laws of thermodynamics, phase equilibria, phase rule, phase diagrams - unary pressure-temperature diagrams, binary temperature-composition diagrams, construction of temperature composition diagrams from free energies, common tangent construction in free energy-composition diagrams, invariant reactions.” It is part of Section XE2.2 of XE2 Materials Science, one of the nine optional sections of the Engineering Sciences (XE) paper, any two of which a candidate chooses in the exam. The chapter teaches the laws in the form materials scientists use them (Gibbs energy, chemical potential, mixing), the Gibbs phase rule, unary pressure-temperature diagrams, binary diagrams with the lever rule and every invariant reaction, and how a diagram is built from free energy-composition curves with the common tangent. The numericals are degrees of freedom, lever-rule phase and microconstituent fractions, mixing entropy and Gibbs energy, and the critical temperature of a regular solution.

1. Properties and the laws of thermodynamics as used in materials

An extensive property scales with the amount of material (mass, volume, U, H, S, G, the heat capacity of a whole sample); an intensive property does not (T, P, density, chemical potential μ, every molar or specific quantity). The ratio of two extensive properties is intensive. A state function depends only on the present state and not on the path: U, H, S and G are state functions, while heat q and work w are not.

First law (convention here: w is work done by the system): dU = δq − δw; at constant pressure q = ΔH with H = U + PV. Second law: dS ≥ δq/T, with equality for a reversible path, so the entropy of an isolated system never decreases. Third law: S → 0 as T → 0 for a perfect crystal in equilibrium, so absolute entropies exist (a glass keeps a residual entropy). The laws combine in the Gibbs energy G = H − TS, with dG = V dP − S dT. At constant T and P a closed system changes spontaneously if ΔG < 0 and is in equilibrium when G is a minimum. G is the working potential of materials science because T and P are the variables we control.

For a change α → β at T_tr, ΔG = ΔH − TΔS = 0 gives ΔS = ΔH/T_tr. Below its melting point the solid has the lower G, above it the liquid has; the two G(T) curves cross at T_m.
Ideal solution: ΔH_mix = 0 and ΔS_mix = −R(x_A ln x_A + x_B ln x_B) per mole, so ΔG_mix = RT(x_A ln x_A + x_B ln x_B) < 0 at every composition. Regular solution: the same ΔS_mix but ΔH_mix = Ω x_A x_B. Ω > 0 means like atoms are preferred (clustering, miscibility gap); Ω < 0 means unlike atoms are preferred (ordering).
Chemical potential μ_i = (∂G/∂n_i) at fixed T, P and the other n_j. Phases α and β of a mixture are in equilibrium when μ_i^α = μ_i^β for every component i. This single condition produces the common tangent below.

ℹ️ Worked: mixing an equiatomic ideal solution

For x_A = x_B = 0.5: ΔS_mix = −R(0.5 ln 0.5 + 0.5 ln 0.5) = R ln 2 = 8.314 × 0.6931 = 5.76 J mol⁻¹ K⁻¹. At 1000 K, ΔG_mix = −TΔS_mix = −5.76 kJ mol⁻¹. The configurational entropy is the same at every temperature, but its weight in G grows with T, which is why high temperature favours mixing.

2. Phases, components, the Gibbs phase rule and unary P–T diagrams

A phase is a physically distinct, homogeneous and mechanically separable part of a system with uniform structure and composition: ice, liquid water and vapour are three phases, whereas many grains of one crystal structure and composition are one phase. A component is an independently variable chemical species; Fe–C is a binary system even though Fe₃C exists, because Fe₃C is built from Fe and C. Equilibrium at fixed T and P means minimum G.

Gibbs phase rule: F = C − P + 2. F is the number of intensive variables (T, P, compositions) that can be changed independently without changing the number of phases. Derivation by counting: each phase has C − 1 independent compositions, so the P phases have P(C − 1) of them, plus T and P, a total of P(C − 1) + 2 variables. Equilibrium imposes C(P − 1) equalities of chemical potential. Hence F = P(C − 1) + 2 − C(P − 1) = C − P + 2. For condensed systems studied at a fixed pressure of 1 atm one variable is already used, so F = C − P + 1.

Degrees of freedom F in typical fields
System and field	Rule used	F	Meaning
Unary P–T, single phase	C − P + 2 = 1 − 1 + 2	2	T and P both free
Unary, on a boundary line	1 − 2 + 2	1	P fixed once T is chosen
Unary, triple point	1 − 3 + 2	0	fixed T and P
Binary, constant P, one phase	C − P + 1 = 2 − 1 + 1	2	T and composition free
Binary, constant P, two phases	2 − 2 + 1	1	compositions fixed by T
Binary, constant P, three phases	2 − 3 + 1	0	invariant point

A unary P–T diagram has areas for single phases (F = 2), lines where two phases coexist (F = 1) and a triple point (F = 0). Every line obeys the Clapeyron equation dP/dT = ΔH/(TΔV). The solid–liquid line is steep and, because ΔV on melting is positive for most materials, slopes up; for water, Ga, Si and Bi the liquid is denser (ΔV < 0) and the line slopes backwards. For vapour equilibrium ΔV ≈ RT/P gives the Clausius–Clapeyron relation d ln P/dT = ΔH_vap/(RT²). A solid with allotropes has several solid areas: at 1 atm iron is bcc up to 912 °C, fcc from 912 to 1394 °C and bcc again up to its melting point, 1538 °C.

⚠️ Count components, not compounds

The two usual slips are counting Fe₃C as a third component of Fe–C (C stays 2) and forgetting which form of the rule the problem needs. If pressure is fixed at 1 atm, or the diagram is a T–C section, use F = C − P + 1. If both T and P are free, use F = C − P + 2.

3. Binary T–C diagrams, tie lines and the lever rule

A binary diagram, drawn at constant pressure, shows the phases present at each temperature and overall composition C₀. In a single-phase field the phase has composition C₀. In a two-phase field, draw the horizontal tie line through (C₀, T) up to the two phase boundaries: its ends give the compositions C_α and C_β of the two phases. They depend on T alone (F = 1) and not on C₀. A mass balance gives the lever rule:

W_α = (C_β − C₀)/(C_β − C_α), W_β = (C₀ − C_α)/(C_β − C_α), W_α + W_β = 1.

Read it as: the fraction of a phase is the opposite lever arm over the whole tie line; the nearer C₀ is to a phase, the more of it there is. Compositions in wt% give mass fractions and compositions in at% give mole fractions. In an isomorphous system (Cu–Ni, complete solid solubility) there is a liquidus above, a solidus below and one α field. On equilibrium cooling from the liquid, the first solid forms at the liquidus temperature and is richer in the higher-melting component than the alloy; liquid and solid compositions then slide along the liquidus and solidus and the last liquid freezes at the solidus temperature, when the solid has composition C₀ again.

⚠️ Lever-rule traps

Use the tie line at the temperature asked, not at another one. Never use the rule in a single-phase field. Put the whole tie-line length (C_β − C_α) in the denominator, and take the far arm for the numerator: it is the commonest slip to divide by the wrong segment and get the fraction of the other phase.

4. Invariant reactions and microconstituents

Three phases in a binary at constant pressure give F = 0: the reaction occurs at one temperature with three fixed compositions, drawn as a horizontal line. The shape of the diagram near the line identifies the reaction: one phase above the line and two below is a splitting reaction (eutectic if the single phase is liquid, eutectoid if solid); two phases above and one below is a combining reaction (peritectic if one of them is liquid, peritectoid if both are solid).

Invariant reactions on cooling
Name	Reaction	Example
Eutectic	L → α + β	Pb–Sn; Fe–C at 4.3 wt% C, 1147 °C: L → γ + Fe₃C
Eutectoid	γ → α + β	Fe–C at 0.76 wt% C, 727 °C: γ → α + Fe₃C (pearlite)
Peritectic	L + α → β	Fe–C, iron-rich end: L + δ → γ
Peritectoid	α + β → γ	rarer; two solids combine
Monotectic	L₁ → L₂ + α	Cu–Pb (liquid miscibility gap)
Syntectic	L₁ + L₂ → β	rare; two liquids combine

A microconstituent is a distinguishable feature of the microstructure, not a phase. An alloy left of the eutectic composition (hypoeutectic) cools to primary (proeutectic) α plus the eutectic mixture (α + β), formed from the liquid that remains at the eutectic temperature. So the mass fraction of the eutectic microconstituent equals the fraction of liquid just above the eutectic temperature, found with the lever rule using the eutectic composition as the liquid end. In Fe–Fe₃C the eutectoid product is lamellar pearlite; a hypoeutectoid steel gives proeutectoid ferrite plus pearlite, and a hypereutectoid steel gives proeutectoid cementite on the austenite grain boundaries plus pearlite.

ℹ️ Worked: slow-cooled 0.40 wt% C steel just below 727 °C

Take ferrite 0.022 wt% C, eutectoid austenite 0.76 wt% C and Fe₃C 6.67 wt% C. Proeutectoid ferrite = (0.76 − 0.40)/(0.76 − 0.022) = 0.488; pearlite = 0.512. Total ferrite (proeutectoid plus the ferrite lamellae) = (6.67 − 0.40)/(6.67 − 0.022) = 0.943 and Fe₃C = 0.057. Pearlite (51 %) is therefore not the same as cementite (5.7 %): pearlite is a mixture of ferrite and cementite.

5. Free energy–composition curves and the common tangent

At fixed T and P draw the molar Gibbs energy G of each phase against x_B. For a solution G = x_A μ_A + x_B μ_B, so the tangent to the curve at any composition cuts the x_B = 0 axis at μ_A and the x_B = 1 axis at μ_B. Two phases α and β are in equilibrium when μ_A^α = μ_A^β and μ_B^α = μ_B^β, that is, when their tangents have the same intercepts on both axes. The two tangents are then one line, the common tangent, and its points of contact give the equilibrium compositions x^α and x^β.

Between the two contact points the straight tangent segment is the Gibbs energy of the α + β mixture (the lever rule on the G scale). It lies below both curves, so the mixture is more stable than any single phase there.
Outside the contact points the single-phase curve lies below the tangent, so that single phase is stable.
The horizontal tie line of the phase diagram at temperature T is the projection of one common tangent; changing T moves the tangent points and so traces the phase boundaries.

🎯 Why every phase dissolves a little of every component

The mixing term RT x ln x has slope RT(ln x + 1), which tends to −∞ as x → 0. The G curve of any solution therefore plunges vertically at both edges, the tangent intercepts there go to −∞, and a common tangent always touches at a small non-zero x. Solubilities can be tiny, but they are never exactly zero at T > 0.

6. Building diagrams from free energy curves

Draw G_L(x) and G_S(x) for many temperatures and mark the common-tangent contacts; plotting those contacts against T gives the diagram. Isomorphous system: above the melting point of both components G_L lies below G_S everywhere, so the alloy is liquid. As T falls the solid curve first dips below the liquid near the higher-melting component; a common tangent now gives a liquid and a solid composition (the liquidus and solidus points); at lower T the two curves cross over the whole range and the alloy is fully solid. Eutectic system: three curves, G_L, G_α and G_β, exist. At the eutectic temperature one straight line touches all three, giving the three compositions of the invariant reaction; above it the liquid curve dips between the two solids and below it the common tangent joins α and β.

Miscibility gap. For a regular solution G(x) = Ωx(1 − x) + RT[x ln x + (1 − x) ln(1 − x)] and d²G/dx² = −2Ω + RT/[x(1 − x)]. At x = 0.5 this is −2Ω + 4RT, which vanishes at the critical temperature T_c = Ω/(2R). Above T_c the curve has one minimum and the solution is single-phase at every composition. Below T_c, with Ω > 0, there are two minima and the symmetric common tangent is horizontal, touching at the two minima. Inside the spinodal (d²G/dx² < 0) any composition fluctuation lowers G, so it decomposes spontaneously; between the contact points and the spinodal a composition is metastable and needs nucleation.

ℹ️ Worked: critical temperature

For Ω = 20 kJ mol⁻¹: T_c = 20 000/(2 × 8.314) = 1203 K. At 900 K the solution is below T_c and separates into two solid solutions; at 1300 K it is a single phase at every x. A larger Ω (stronger like-atom preference) raises T_c.

Key takeaways

Extensive properties scale with amount, intensive ones do not; G = H − TS is the potential at constant T and P, and equilibrium means minimum G with μ_i equal in every phase.
Gibbs phase rule: F = C − P + 2, or C − P + 1 at fixed pressure. A binary at constant pressure is invariant (F = 0) with three phases; count components, not compounds.
Unary P–T lines obey dP/dT = ΔH/(TΔV); the vapour line gives d ln P/dT = ΔH/(RT²), and the melting line slopes backwards when the liquid is denser (water).
The lever rule gives the fraction of a phase as the opposite arm over the whole tie line, in the same units (wt% or at%) as the compositions, and only in two-phase fields.
Invariant reactions: one phase splitting into two on cooling is eutectic (from liquid) or eutectoid (from solid); two combining into one is peritectic or peritectoid; the eutectic microconstituent fraction equals the liquid fraction just above the eutectic line.
A common tangent to two G(x) curves means equal chemical potentials of both components; its contact points are the phase boundaries, and a regular solution with Ω > 0 has a miscibility gap below T_c = Ω/(2R).

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Which one of the following is an intensive property?
1. The enthalpy H of a 1 kg sample of the metal at 300 K
2. The entropy S of a given single crystal of the metal
3. The chemical potential of a component in a solution
4. The heat capacity of a whole casting of the alloy
Show answer
Answer: C — The chemical potential of a component in a solution
Chemical potential, the partial molar Gibbs energy, does not change when the amount of material is changed at fixed T, P and composition, so it is intensive. H and S of a given sample, and the heat capacity of a whole casting, all scale with the amount of material and are extensive. The tempting error is to call enthalpy intensive because it is quoted per kilogram; the per-kilogram value is intensive but the enthalpy of the sample is not.
On the pressure–temperature diagram of pure water the solid–liquid boundary slopes backwards (dP/dT < 0), unlike that of most substances. Using the Clapeyron equation dP/dT = ΔH/(TΔV) for melting, the reason is that
1. the enthalpy of fusion of ice is negative because heat is released when ice melts
2. the volume decreases on melting, since ice is less dense than liquid water
3. the entropy of liquid water is lower than that of ice at the melting point
4. the vapour pressure of ice is higher than that of liquid water at 0 °C
Show answer
Answer: B — the volume decreases on melting, since ice is less dense than liquid water
Melting always absorbs heat (ΔH > 0) and raises entropy, so the sign of the slope is the sign of ΔV. Ice is less dense than water, so ΔV < 0 on melting and dP/dT is negative. A negative enthalpy of fusion or lower entropy of the liquid would be physically wrong: melting needs latent heat and the liquid is the more disordered phase.
At temperature T the tie line through an alloy in a binary α + β field has ends at 20 wt% B (α) and 80 wt% B (β). The alloy contains 35 wt% B. The mass fractions of α and β are
1. α = 0.25 and β = 0.75
2. α = 0.35 and β = 0.65
3. α = 0.56 and β = 0.44
4. α = 0.75 and β = 0.25
Show answer
Answer: D — α = 0.75 and β = 0.25
Lever rule: W_α = (80 − 35)/(80 − 20) = 45/60 = 0.75 and W_β = (35 − 20)/60 = 0.25. The alloy lies close to α, so α dominates. The swapped answer comes from dividing the near arm (15) instead of the far arm (45) by the tie line; using 35 and 65 directly confuses overall composition with phase fraction; and 0.56 results from dividing by 80 instead of the tie-line length.
On cooling, a liquid and a primary solid phase react at one fixed temperature to form a single different solid phase. This invariant reaction is called
1. eutectic
2. eutectoid
3. monotectic
4. peritectic
Show answer
Answer: D — peritectic
Liquid + solid → a different solid is the peritectic reaction (L + α → β). Eutectic splits one liquid into two solids, eutectoid splits one solid into two solids, and monotectic splits one liquid into a second liquid plus a solid. The pattern to remember is two phases above the line and one below for the combining reactions.
Two phases α and β of a binary A–B system are in equilibrium at fixed T and P. On a plot of molar Gibbs energy against composition, the equilibrium compositions of α and β are fixed by the condition that
1. the two curves have equal Gibbs energies at the two compositions, so the compositions lie on a horizontal line
2. one straight line is tangent to both curves, so μ_A and μ_B are each the same in α and in β
3. each curve has zero slope at its contact point, so each phase sits at its own minimum of G
4. the tangents to the two curves are parallel, although they cut the composition axes at different values
Show answer
Answer: B — one straight line is tangent to both curves, so μ_A and μ_B are each the same in α and in β
Equilibrium needs μ_A^α = μ_A^β and μ_B^α = μ_B^β. A tangent's intercepts on the x_B = 0 and x_B = 1 axes are μ_A and μ_B, so both intercepts must match, which makes the two tangents one common line. Equal G at the two compositions would be a horizontal line, which is generally not the tangent. Parallel tangents satisfy only equal slope, that is μ_B − μ_A, and not each potential separately.
A regular solution has ΔH_mix = Ωx(1 − x) with Ω > 0 and the ideal entropy of mixing. At a temperature below T_c = Ω/(2R), which statement about its molar Gibbs energy curve is correct?
1. The curve has a local maximum at x = 0.5 between two minima, so equilibrium at x = 0.5 is a two-phase mixture
2. The curve has a single minimum at x = 0.5, so the solution is a single phase at every composition of the system
3. The curve has two minima, yet an alloy at x = 0.5 stays stable as one homogeneous phase because of the mixing entropy
4. d²G/dx² is positive at x = 0.5 and negative close to the two pure components
Show answer
Answer: A — The curve has a local maximum at x = 0.5 between two minima, so equilibrium at x = 0.5 is a two-phase mixture
d²G/dx² = −2Ω + RT/[x(1 − x)]. At x = 0.5 it is −2Ω + 4RT, which is negative when T < Ω/(2R), so x = 0.5 is a local maximum between two minima; the common tangent is horizontal and the equilibrium state is the two-phase mixture. A single minimum is the case above T_c. Near the pure components the term RT/[x(1 − x)] becomes large and positive, so the curvature is positive there, the opposite of the claim of negative curvature near the pure components.
In the lead–tin system the eutectic is at 61.9 wt% Sn and 183 °C, and the maximum solubility of Sn in lead-rich α at that temperature is 18.3 wt% Sn. An alloy of 30 wt% Sn is cooled slowly from the liquid to just below 183 °C. Its microstructure is
1. single-phase α, since the alloy contains less tin than the eutectic
2. primary β grains surrounded by an (α + β) eutectic mixture
3. an entirely lamellar (α + β) eutectic with no primary phase
4. primary α grains surrounded by an (α + β) eutectic mixture
Show answer
Answer: D — primary α grains surrounded by an (α + β) eutectic mixture
30 wt% Sn is below the eutectic composition (hypoeutectic) but above the 18.3 wt% solubility limit, so the alloy first deposits primary α and the remaining liquid, which reaches 61.9 wt% Sn at 183 °C, freezes as (α + β) eutectic. Single-phase α needs a composition below 18.3 wt% Sn; being below the eutectic composition is not enough. An entirely eutectic structure occurs only at 61.9 wt% Sn, and primary β belongs to alloys on the tin-rich side.
In the Fe–Fe₃C system the eutectoid is at 0.76 wt% C and 727 °C. A 1.0 wt% C steel is cooled very slowly from the single-phase austenite field. Between the temperature at which the first new phase appears and 727 °C, the new phase that forms is
1. proeutectoid ferrite, forming at the austenite grain boundaries as the steel cools
2. martensite, because the carbon content of the steel is above the eutectoid value
3. proeutectoid cementite, forming mainly along the austenite grain boundaries
4. pearlite colonies only, with no separate proeutectoid phase
Show answer
Answer: C — proeutectoid cementite, forming mainly along the austenite grain boundaries
At 1.0 wt% C the steel is hypereutectoid: on cooling it enters the γ + Fe₃C field, where excess carbon leaves austenite as proeutectoid cementite, nucleating at grain boundaries. The austenite left at 727 °C has 0.76 wt% C and becomes pearlite. Proeutectoid ferrite belongs to hypoeutectoid steels, martensite needs rapid quenching and is not an equilibrium product, and pearlite alone is the 0.76 wt% C case.
In a eutectic system of A and B, the phases L, α and β are in equilibrium at the eutectic temperature T_E. On the molar Gibbs energy–composition plot drawn at T_E,
1. the liquid curve lies above both solid curves at every composition, so no liquid can appear
2. one straight line is tangent to all three curves, at the three compositions of the invariant reaction
3. three separate common tangents exist, one for each pair of phases, lying at three different heights
4. the α–β common tangent passes above the liquid curve, so the liquid is the more stable phase at its own composition
Show answer
Answer: B — one straight line is tangent to all three curves, at the three compositions of the invariant reaction
Three-phase equilibrium requires the same μ_A and μ_B in all three phases, so a single line must touch G_α, G_L and G_β; this is the F = 0 condition made visible. Slightly above T_E the α–β tangent would pass above the liquid curve (liquid stable), and slightly below it the liquid curve would rise above that tangent. The statement that the α–β tangent passes above the liquid curve describes a temperature above T_E, and the statement of three separate tangents describes different temperatures, not the eutectic.
Which of the following statements about thermodynamic functions and equilibrium are correct?
1. The third law fixes the entropy of every substance, including a glass, at exactly zero at 0 K
2. For a closed system doing only pressure–volume work at constant T and P, equilibrium corresponds to a minimum of the Gibbs energy
3. Gibbs energy G is extensive, whereas the chemical potential μ_i is intensive
4. For an ideal solution ΔH_mix = 0, so mixing is driven entirely by the increase in configurational entropy
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Answer: B — For a closed system doing only pressure–volume work at constant T and P, equilibrium corresponds to a minimum of the Gibbs energy; C — Gibbs energy G is extensive, whereas the chemical potential μ_i is intensive; D — For an ideal solution ΔH_mix = 0, so mixing is driven entirely by the increase in configurational entropy
Minimum G is the equilibrium criterion at constant T and P. G scales with the amount of material while μ_i does not. An ideal solution has no heat of mixing, so ΔG_mix = −TΔS_mix. The third law applies to a perfect crystal in equilibrium; a glass, or any disordered solid, retains a residual (configurational) entropy at 0 K, so the claim of exactly zero entropy for a glass is wrong.
Which of the following statements about the Gibbs phase rule are correct? (Use F = C − P + 1 for binary systems at constant pressure and F = C − P + 2 when T and P can both vary.)
1. In a binary two-phase field at constant pressure, fixing the temperature fixes the compositions of both phases
2. At the eutectic point of a binary at constant pressure, F = 1, so the temperature can vary slightly while L, α and β coexist
3. In a single-phase field of a binary at constant pressure, the temperature and the composition can both be varied independently
4. A pure substance has F = 0 at its triple point, where solid, liquid and vapour coexist
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Answer: A — In a binary two-phase field at constant pressure, fixing the temperature fixes the compositions of both phases; C — In a single-phase field of a binary at constant pressure, the temperature and the composition can both be varied independently; D — A pure substance has F = 0 at its triple point, where solid, liquid and vapour coexist
Binary, two phases: F = 2 − 2 + 1 = 1, so one variable (T) fixes the rest. Single phase: F = 2, so T and composition are free. Unary triple point: F = 1 − 3 + 2 = 0. With three phases in a binary at constant pressure F = 2 − 3 + 1 = 0, so the eutectic is invariant and its temperature cannot drift; the statement that F = 1 there is wrong.
Which of the following statements about tie lines, the lever rule and microconstituents are correct?
1. At a fixed temperature inside a two-phase field, the compositions of the two phases do not depend on the overall composition of the alloy
2. If compositions are in at%, the lever rule gives mole fractions, not mass fractions
3. In a hypoeutectic alloy just below the eutectic temperature, the mass fraction of the eutectic microconstituent equals the mass fraction of liquid just above it
4. The fraction of a phase is the tie-line segment on the same side as that phase divided by the whole tie line
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Answer: A — At a fixed temperature inside a two-phase field, the compositions of the two phases do not depend on the overall composition of the alloy; B — If compositions are in at%, the lever rule gives mole fractions, not mass fractions; C — In a hypoeutectic alloy just below the eutectic temperature, the mass fraction of the eutectic microconstituent equals the mass fraction of liquid just above it
The tie-line ends depend only on temperature (F = 1 in a two-phase field). A lever-rule ratio carries the units of the compositions, so at% gives mole fractions. The liquid present just above the eutectic temperature all becomes eutectic, so the two fractions are equal. The same-side segment is the wrong arm: the fraction of a phase uses the opposite segment, because the nearer the alloy is to a phase, the more of it there is.
In the Fe–Fe₃C system take the solubility of carbon in ferrite as 0.022 wt% and the eutectoid composition as 0.76 wt% C. A 0.40 wt% C steel is cooled very slowly from the austenite field to just below 727 °C. The mass percentage of proeutectoid ferrite is ____ (give the answer to one decimal place).
Numerical answer — type the value.
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Answer: 48.8
The proeutectoid ferrite is the ferrite that forms before the eutectoid reaction, so the lever rule uses the eutectoid austenite composition at the far end: W = (0.76 − 0.40)/(0.76 − 0.022) = 0.36/0.738 = 0.4878, that is 48.8 %. Using 6.67 wt% C as the far end would instead give the total ferrite (94.3 %), which includes the ferrite lamellae of pearlite.
A ternary system (three components) is studied at a constant pressure of 1 atm. Three phases coexist in equilibrium in a certain composition range. The number of degrees of freedom is ____.
Numerical answer — type the value.
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Answer: 1
At fixed pressure F = C − P + 1 = 3 − 3 + 1 = 1. One variable, for example the temperature, may be chosen freely and then fixes the rest. Using F = C − P + 2 would wrongly give 2, because pressure is already fixed.
A regular solution has ΔH_mix = Ωx(1 − x) per mole with Ω = 24 kJ mol⁻¹ and the ideal entropy of mixing. Take R = 8.314 J mol⁻¹ K⁻¹. Using T_c = Ω/(2R), the critical temperature of the miscibility gap is ____ K (nearest integer).
Numerical answer — type the value.
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Answer: 1443
The curvature of G at x = 0.5 is −2Ω + 4RT, which vanishes at T_c = Ω/(2R) = 24 000/(2 × 8.314) = 24 000/16.628 = 1443.3 K, that is 1443 K. Forgetting the factor 2 in the denominator gives 2887 K.
In the lead–tin system take the maximum solubility of Sn in α at the eutectic temperature as 18.3 wt% and the eutectic composition as 61.9 wt% Sn. A 40 wt% Sn alloy is cooled slowly to just below the eutectic temperature. The mass percentage of the eutectic microconstituent (the α + β mixture formed from liquid) is ____ (give the answer to one decimal place).
Numerical answer — type the value.
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Answer: 49.8
The eutectic microconstituent forms from the liquid left at the eutectic temperature, so its fraction is the liquid fraction just above it: W_L = (40 − 18.3)/(61.9 − 18.3) = 21.7/43.6 = 0.4977, that is 49.8 %. The remaining 50.2 % is primary α. Taking the other arm, (61.9 − 40)/43.6, gives 50.2 %, which is the fraction of primary α, not of the eutectic.
An ideal solution contains the three components A, B and C in equal mole fractions (1/3 each). Take R = 8.314 J mol⁻¹ K⁻¹. At 800 K the magnitude of ΔG_mix per mole of solution is ____ kJ mol⁻¹ (give the answer to two decimal places).
Numerical answer — type the value.
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Answer: 7.31
For an ideal solution ΔH_mix = 0 and ΔS_mix = −R Σ x_i ln x_i = R ln 3 = 8.314 × 1.0986 = 9.134 J mol⁻¹ K⁻¹. Then |ΔG_mix| = TΔS_mix = 800 × 9.134 = 7307 J mol⁻¹ = 7.31 kJ mol⁻¹ (ΔG_mix itself is negative). Using ln 2 for a binary mix would give 4.61 kJ mol⁻¹.