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The triple point \((T_{I}, P_{I})\) is shown in a schematic phase diagram (pressure (P) - temperature (T) plot) for one component system. GS, GL and GV are the free energies of solid, liquid and vapor, respectively. At a constant pressure, \(P_{I}\), the correct free energy (G) versus temperature (T) plot is

Pressure-temperature phase diagram with solid, liquid and vapor regions meeting at the triple point TI, PI

  • A.G versus T at P = PI: three falling lines crossing at TI; GS is the shallowest, GL the middle and GV the steepest
  • B.G versus T at P = PI: three falling lines crossing at TI; GS is the shallowest, GV the middle and GL the steepest
  • C.G versus T at P = PI: three falling lines crossing at TI; GL is the shallowest, GS the middle and GV the steepest
  • D.G versus T at P = PI: three falling lines crossing at TI; GV is the shallowest, GL the middle and GS the steepest

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.

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