the liquid-solid-vapor triple point of a pure component is T triple = 10 C and P
ID: 966133 • Letter: T
Question
the liquid-solid-vapor triple point of a pure component is T triple = 10 C and P triple = 0.05 bar. The molar densities of the liquid and solid phases are pl = 22 kmol/m^3 and ps = 26 kmol/m^3, respectively. The enthalpy change of sublimation of the solid at the triple point is Hsv = 6.22 kJ/mol and the enthalpy change of vaporization of the liquid at the triple point is Hlv = 5.65 kJ/mol. You are asked to calculate the melting temperature of the solid at a pressure of P = 10 bar. Please state explicitly all assumptions on which your calculation is based
Explanation / Answer
Apply claperyon equation
Enthalpy change of sublimation = 6.22 Kj/mol
Enthalpy change of vaporization = 5.65 Kj/mol
= 6.22-5.65
= 0.57 Kj/mol
Molar density = pl= 22 k mol/m3
Molar density = ps= 26 k mol/m3
Vl = 1/ = 1/22 = 0.045 k mol/m3
Vs = 1/ = 1/26 = 0.038 k mol/m3
T = 10 + 273 = 283 K
= 9.95 X 283 (0.045-0.038)/0.57
= 35 K or -238oC
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