For a three-component system in two phases the degrees of freedom number
Answer: C
Correct answer: three
\[ F = 3 - 2 + 2 = 3. \] With three variables free the system cannot be represented on a plane, which is why ternary systems are drawn as triangular diagrams at fixed temperature and pressure, reducing the free variables to the two composition ratios.
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A one-component system is found to have zero degrees of freedom. The number of phases present must be
Answer: D
Setting F to zero with C equal to one requires P to be three, since 1 minus 3 plus 2 is zero. Four phases would demand a negative degree of freedom, which has no physical meaning, so no quadruple point can exist for one component.
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A two-component condensed system shows one degree of freedom. The number of phases present is
Answer: A
Correct answer: two
\[ F' = C - P + 1 \quad\Rightarrow\quad 1 = 2 - P + 1 \quad\Rightarrow\quad P = 2. \] One degree of freedom in a binary condensed system means two phases - typically a solid and the liquid along a liquidus, where fixing the temperature fixes the composition of each.
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A eutectic of lead and silver contains 2.6 per cent silver. 500 kg of an argentiferous lead melt containing 1 per cent silver, cooled to just above the eutectic temperature, has deposited pure lead amounting to about
Answer: B
Correct answer: 308 kg
The silver stays in the melt, so 500 kg at 1 per cent contains 5 kg of silver, and just above the eutectic that 5 kg is present at 2.6 per cent: \[ \frac{5}{0.026} = 192\ \mathrm{kg\ of\ melt}. \] The rest has crystallised as pure lead: \( 500 - 192 = 308 \) kg. This is the Pattinson process, which concentrates silver by removing lead rather than by extracting silver.
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For that same charge the residual eutectic liquid weighs about
Answer: C
Silver is confined to the liquid, so the liquid mass is the silver in the charge divided by the eutectic silver fraction, 5 over 0.026. Repeating the skim and remelt on successive batches is what concentrates the silver in Pattinson's process.
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A binary melt at 40 per cent B separates into a solid of 10 per cent B and a liquid of 60 per cent B. The fraction that is liquid is
Answer: D
Correct answer: 0.60
By the lever rule the liquid fraction is the distance from the overall composition to the solid, over the whole tie line: \[ \frac{40-10}{60-10} = \frac{30}{50} = 0.60. \] The arm is measured towards the phase whose amount is wanted - taking the other arm gives 0.40, which is the standard error and is offered here as a distractor.
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For that same tie line the fraction that is solid is
Answer: A
It is the distance from the overall composition to the liquid composition over the whole tie line, that is 20 over 50. The two fractions necessarily sum to unity, which is the quickest check on any lever rule calculation.
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Water's triple point occurs at 4.58 mm of mercury. At a pressure of 760 mm the number of phases that can coexist in equilibrium is at most
Answer: B
Correct answer: two
Fixing the pressure at 760 mm removes a variable, so \( F = C - P + 1 = 2 - P \), and \( F \ge 0 \) requires \( P \le 2 \). Three phases coexist only at the triple point’s own pressure of 4.58 mm; at any other pressure the most that can be in equilibrium is two.
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Match the system with what its phase diagram shows.
1. Lead-silver 2. Copper-nickel 3. Phenol-water
P. Continuous solid solution with a lens-shaped field Q. A simple eutectic R. A critical solution temperature
Answer: C
Correct answer: 1-Q, 2-P, 3-R
Three binary systems of three different types. Lead and silver are immiscible as solids and give a simple eutectic. Copper and nickel are miscible in all proportions in both states and give a continuous solid solution with a lens-shaped two-phase field. Phenol and water are two partially miscible liquids and show a critical solution temperature above which they mix completely.
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Match the number of phases with the degrees of freedom for a one-component system.
1. One phase 2. Two phases 3. Three phases
P. One Q. Zero R. Two
Answer: D
Correct answer: 1-R, 2-P, 3-Q
Applying \( F = 3 - P \) in turn: one phase gives two degrees of freedom and appears as an area; two phases give one and appear as a curve; three phases give zero and appear as a point. The dimensionality of each region on the diagram is the phase rule made visible.
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