1. The 3 rd Law of Thermodynamics Valentim M. B. Nunes ESTT-IPT May 2015

2. Based on the techniques of classical thermodynamics, or macroscopic thermodynamics, we can calculate the calorimetric entropy : All quantities required for this calculation are normally obtained by calorimetric techniques.

3. Planck’s statement: the entropy of all substances at the absolute zero, in the form of a perfect crystal, is zero: S (0 K) = 0 Nernst statement: For any process,

4. Based on techniques of statistical thermodynamics we can calculate the spectroscopic entropy: When  (E,V,N) = 1, then S = 0. Admitting that at T = 0 exists several complexions equally probable,  0, then:

5. The difference between the spectroscopic and calorimetric entropy is the residual entropy: For example, for solid carbon monoxide, CO, there are two possible molecular orientations, CO and OC. For N molecules of CO, the residual entropy comes:

6. Residual entropy / J.K -1.mol -1 S esp S cal S res CO197.95193.34.65R ln 2 = 5.77 N2ON2O219.99215.14.89R ln 2 = 5.77 NO211.0207.93.10½ R ln2 = 2.88 OCS231.54231.24~ 0~ Rln 1 = 0 H2OH2O188.72185.33.42R ln 3/2 = 3.37

7. The water in the solid state adopts a structure (Ice I) at atmospheric pressure, containing covalent and hydrogen bonds. Each molecule of H 2 O can guide the  bonds in six different directions (tetrahedron). However the probability of a direction be chosen is ½, since each neighbor molecule has only half of the available sites. For the 2 nd H the probability is also ½, so the total probability is ¼. For each molecule exists 6/4 = 3/2 choices. For N molecules we have: