Phase Diagrams Questions and Answers

Answer: D

The solidification range is 150 °C.

The solidification range is the temperature span between the start of freezing (the liquidus) and the end of freezing (the solidus) for the alloy's composition.

Liquidus at 15 % B. The liquidus is a straight line from 700 °C at 0 % to the eutectic at 400 °C and 30 %, falling 10 °C per 1 % B:

\[ T_{L} = 700 - 10 \times 15 = 550\ ^{\circ}\text{C} \]

Solidus at 15 % B. The alpha solidus ends at 10 % B on the eutectic line. An alloy richer than 10 % B finishes freezing at the eutectic temperature, 400 °C.

So the range is \(550 - 400 = 150\) °C. 250 °C is the superheat, 800 minus 550; 400 °C is the eutectic temperature itself; and 800 °C is the pouring temperature, which plays no part in the freezing range.

Answer: B

The match is P-5, Q-3, R-4, S-2.

• (P) L → α + β: a liquid gives two solids on cooling, the eutectic reaction.
• (Q) L + α → β: a liquid and a solid combine into a new solid, the peritectic reaction.
• (R) γ → α + β: the solid-state analogue of the eutectic, the eutectoid reaction (as in steel, austenite to ferrite and cementite).
• (S) α + β → γ: two solids combine into one, the solid-state analogue of the peritectic, the peritectoid.

A monotectic is L₁ → L₂ + α, a liquid splitting into a second liquid and a solid, which matches none of the four reactions; options that use it for Q are therefore wrong.

Answer: C

There are 3 types.

The phase rule \(F = C - P + 2\) with C = 2 and both temperature and pressure free gives F = 0 for P = 4: an invariant point in a binary system then involves four phases. The ways one group of phases can turn into another are:

• \(\alpha \leftrightarrow \beta + \gamma + \delta\)
• \(\alpha + \beta \leftrightarrow \gamma + \delta\)
• \(\alpha + \beta + \gamma \leftrightarrow \delta\)

That is three types, just as three phases at constant pressure give the two types in the stem. 2 is the constant-pressure count, and 1 and 4 do not match the ways of splitting four phases into two non-empty sides.

Answer: C

Figure (ii) corresponds to T3.

Equilibrium phases are found by common tangents to the free energy curves. In figure (ii) a single straight tangent touches the \(\alpha\), \(L\) and \(\beta\) curves at once, so three phases coexist in equilibrium. In a binary system at fixed pressure that happens only at the invariant eutectic temperature, which is T3 in figure (i).

At T1 the liquid curve would lie below both solid curves over most of the range. At T2 there would be two separate tangents, \(\alpha\)-\(L\) and \(L\)-\(\beta\), with a single-phase liquid range between them. At T4 the liquid curve would lie above everything, and a single tangent would join only \(\alpha\) and \(\beta\).

Answer: B

The correct combination is α - ice; β - water; γ - vapour.

On a P-T diagram the solid occupies the low-temperature, high-pressure region (α), the liquid the higher-temperature, high-pressure region (β), and the vapour the low-pressure region below both (γ). The three boundaries meet at the triple point.

The diagram also shows water's anomaly: the ice-water boundary leans backwards (negative slope), because ice is less dense than water, so raising the pressure lowers the melting point. Every other assignment puts vapour at high pressure or ice at high temperature, which contradicts this layout.

Answer: A

The correct G-X plot is (A).

The diagram is an isomorphous system, with liquidus and solidus rising from the lower-melting component on the left to the higher-melting one on the right. At \(T_0\), alloys near the left are liquid and those near the right are solid, with a two-phase L+S range between.

So the liquid free-energy curve must be lower on the left and the solid curve lower on the right, the two curves crossing so that a common tangent spans the L+S range. Plot (A) shows exactly this: the L curve's minimum on the left, the S curve's minimum on the right, and the curves crossing between them.

In (B) the liquid is lower everywhere, which would make every composition liquid at \(T_0\). In (C) the solid is lower over nearly all compositions. In (D) the roles are reversed, putting the solid on the low-melting side.

Answer: A

The correct plot is (A): at \(T_{I}\) all three lines cross, with \(G_{V}\) steepest, \(G_{L}\) next and \(G_{S}\) shallowest.

At constant pressure \((\partial G/\partial T)_{P} = -S\), so each line falls with a slope equal to the phase's entropy. Vapor has the highest entropy, liquid less and solid the least: \(S_{V} \gt S_{L} \gt S_{S}\). At the triple point pressure all three phases coexist at \(T_{I}\), so their free energies are equal there and the three lines meet at one point.

As a check, plot (A) has the solid lowest below \(T_{I}\) and the vapor lowest above it, which matches the phase diagram along \(P = P_{I}\): solid below the triple point, vapor above. Plots (B), (C) and (D) give the liquid or the solid the steepest line, which would require a larger entropy than the vapor.

Answer: A

The correct G - X plot is (A).

T₀ lies below the eutectic temperature, where the equilibrium phases are α at the A-rich side, β at the B-rich side, and a two-phase α + β mixture between them. No liquid is stable at any composition.

In the free energy diagram this requires the α curve to have its minimum near A, the β curve its minimum near B, and the liquid curve to lie entirely above the common tangent drawn to the α and β curves. Plot (A) shows exactly this.

In (B) and (C) the liquid curve reaches below the solid curves, so liquid would be stable at intermediate compositions, which holds only above the eutectic temperature. (D) places the curves labelled β near A and α near B, the reverse of the phase diagram.