Professor Claude Lupis July 1, 2002 3.025 Thermodynamics and Kinetics of Materials Example of calculations: H, S and G versus T Magnesium melts at 922oK with a heat of fusion equal to 8.95 kJ/mol and boils at 1363oK with a heat of vaporization equal to 145.25 kJ/mol. The heat capacity Cp of the liquid and of the gas can be assumed to be constant at 32.6 J/oK.mol and 20.8 J/oK.mol, respectively, but is temperature dependent for the solid. For the latter, Cp is measured to be 24.9 J/oK.mol at room temperature and 30.8 J/oK.mol at 800oK; assume that a linear variation of Cp with T is justified. a/ Derive analytical representations of the enthalpy, entropy and Gibbs free energy for all three phases of Mg. b/ Plot carefully and accurately the enthalpy, entropy and Gibbs free energy versus temperature in the interval 298 - 1800oK. Indicate on your plots all relevant features (e.g., entropies of fusion and vaporization). (The pressure is fixed at 1 atm) 2 a/ Calculation of the Enthalpy In the temperature interval: 298 < T < 922 where Mg is solid; in J/oK.mol, Cp is equal to 24.9 at 298oK and 30.8 at 800oK, a change of 5.9 for 502o. A linear interpolation yields: Csp = 24.9 + 5.9 (T - 298)/502 = 24.9 + 1.18x10-2(T - 298) or: (J/oK.mol) Csp = 21.4 + 1.18x10-2T Hence: HsT H o298 = [21.4 + 1.18x10-2T] dT = 21.4(T-298) + 0.59x10-2 (T2 - 2982) o HsT -H298 = -6900 + 21.4 T + 0.59 10-2 T2 (J/mol) At 922oK, Hs922 H o298 = 17,835 J and there is a discontinuity in the enthalpy curve corresponding to the fusion of the solid: Hf = 8950 J/mol. Hence, at 922oK: H922 H o298 = 17835 + 8950 = 26785 J. In the temperature interval: 922 < T < 1363 where the liquid is stable, Cp is equal to 32.6 J/oK.mol and: H T H o298 = 26785 + Cp dT = 26785 + 32.6 ( T - 922) or: o H T -H298 = -3270 + 32.6 T (J/mol) At 1363oK, H1363 H o298 = 41160 J/mol and there is a discontinuity which corresponds to the vaporization of the Mg: Hvap = 145,250 J/mol. g H o298 = 41160 + 145250 = 186410 J/mol. At T = 1363, H1363 At T > 1363 , where the gas is the stable phase, Cgp = 20.8 J/oK.mol and: H gT H o298 = 186410 + Cgp dT = 186410 + 20.8 (T - 1363) or: o HgT -H298 = 158,060 + 20.8 T (J/mol) 3 Calculation of the Entropy For the analytical representation of the entropy S, we note that: SsT So298 = ( Csp /T) dT and for 298 < T < 922 , replacing Csp by its analytical representation: SsT So298 = [(21.4/ T) + 1.18 10-2] dT or: SsT So298 = 21.4 ln (T/298) + 1.18 10-2 (T - 298) For So298 = 32.7 J/oK.mol, it can be rewritten: S sT = -92.7 + 21.4 ln T + 1.18 10-2 T (J/oK.mol) At T = 922 oK, Ss922 = 64.27 J/oK.mol. At that temperature there is a discontinuity arising from the S of fusion. It is calculated by: Sf = Hf / Tf = 8950 / 922 = 9.71 J/oK.mol Hence: S922 = 64.27 + 9.71 = 73.98 J/oK.mol. Proceeding as before, in the temperature interval 922 < T < 1363 , we have: ST S922 = 32.6 ln (T/922) and: S T = -148.6 + 32.6 ln T (J/oK.mol) At the boiling point, T = 1363, S1363 = 86.69 J/oK.mol. The Svap is equal to 145,250/1363 or 106.57 J/oK.mol. Hence: g S1363 = 86.69 + 106.57 = 193.26 J/oK.mol. Above the boiling point, T > 1363 , for Cgp equal to 20.8 J/oK.mol: g SgT S1363 = SgT - 193.26 = 20.8 ln (T/1363) or: S gT = 43.1 + 20.8 ln T (J/oK.mol) 4 Calculation of the Gibbs Free Energy To calculate the Gibbs free energy, we note that: G = H - TS, and that: G - H o298 = (H H o298 ) - TS In each temperature interval, we have the functional dependence of (H H o298 ) and S on T. Hence, we immediately deduce the analytical representations of G. For 298 < T < 922: Gs - H o298 = - 6900 + 11.41 T - 0.59 10-2 T2 -21.4 T ln T (J/mol) For 922 < T < 1363: G - H o298 = -3270 + 181.2 T -32.6 T ln T (J/mol) For T > 1363: Gg - H o298 = 158060 - 22.3 T - 20.8 T ln T (J/mol) One should verify that at 922oK, Gs - H o298 is equal to G - H o298 , and similarly that at T = 1363oK, G - H o298 = Gg - H o298 . Of course, it is possible to extrapolate the expressions of G beyond the temperature intervals where they have been derived (for example for a supercooled liquid). b/ See attached figures.