Question

Consider the phase diagram of NH₃.

Triple point: 195.9 K and 0.069 atm

Normal boiling point: -33.34 ^oC

Normal freezing point: -77.73 ^oC

Density of liquid: 0.618g/mL

Density of solid: 0.817 g/mL

(a) Determine the heat of fusion, the heat of vaporization, and the heat of sublimation at the triple point.

(b) Determine the entropy change of fusion, vaporization, and sublimation at the triple point.

(c) Plot the phase diagram in the temperature range T: [-90 ^oC, 130 ^oC]. For pressure (in unit of atm), use ln (p) instead of p.

Answer 1

(a) Heat of fusion = (17 g/(0.817 g.mL^-1)}*L/1000 mL * 0.069 atm * 101.325 J/L.atm/(273-77.73) K = 7.45*10^-4 J/K * 195.9 K = 0.146 J

Heat of vaporization = (17 g/(0.817 g.mL^-1)}*L/1000 mL * 0.069 atm * 101.325 J/L.atm/(273-33.34) K = 6*10^-6 J/K * 195.9 K = 0.0012 J

Now, the heat of sublimation = heat of fusion + heat of vaporization = 0.146 J + 0.0012 J = 0.1472 J