Thermodynamics Seminar Presentation PDF

SEMINAR PRESENTATION THERMODYNAMICS Apparent Exceptions To Third Law Exceptions to So equal to zero are explained in terms of "frozen-in" disorder. For example, a linear molecule such as carbon monoxide can take two possible orientations on a lattice site, CO or OC. Orientations on adjace...

SEMINAR PRESENTATION THERMODYNAMICS Apparent Exceptions To Third Law Exceptions to So equal to zero are explained in terms of "frozen-in" disorder. For example, a linear molecule such as carbon monoxide can take two possible orientations on a lattice site, CO or OC. Orientations on adjacent sites such as COOC or OCCO represent a slightly higher energy state than ordered orientations such as COCO and are therefore favored at higher temperatures.  While there is a tendency for the crystal to move toward the low-energy, ordered state on cooling, the rate at which molecular orientations proceed slows to a standstill and a state of "frozen-in" disorder results at zero absolute temperature. 1. Crystals of CO, N2O,NO,H20 etc do not have perfect order even at 0K Thus their entropy is not equal to zero 𝐶𝑂 − 𝐶 0 − 𝐶𝑂 2. Unattainability of absolute zero temperature By any ideal process, it is impossible to bring any system to absolute zero temperature performing a finite no of operations. This is called principle of Unattainability of absolute zero temperature APPLICATION OF THIRD LAW It help us to calculate the entropies of substances at any temperature at constant pressure DETERMINATION OF ABSOLUTE ENTROPIES – SOLIDS An infinitesimal entropy change (dS) of the system in a process is given by We know that molar heat capacity C of a system is given by  (2) (3) At constant pressure (4) Substituting equation (4) in (1) (5) According to third law , the absolute entropy of a perfectly crystalline substance is zero.integrate (5) between the appropriate limits  𝑆𝑇 𝑇 𝑇 𝐶𝑝 ∫ ∫ 𝑆 𝑇 =∫ 𝑐 𝑃 ⅆ ( ln 𝑇 ) ⅆ 𝑆= ⅆ𝑇 0 0 0 𝑇  LIQUIDS AND GASES The total absolute entropy of a substance in liquid or gaseous state at TK at atmospheric pressure may be taken as the sum of all the entropy changes that occur during its reversible transformation from the crystalline state at 0K to the current state 𝑇 1 ¿ Δ 𝑆1 =∫ 𝐶 𝑝 ⅆ ln 𝑇 0 2) 3) Hence absolute entropy of liquid is given by 4) 5) 6)  Hence absolute entropy of gases at TK + RESIDUAL ENTROPY It is the entropy of the crystal which it posses at 0K S residual is the difference between experimental and calculated value It can be calculated by BOLTZMAN ENTROPY EQUATION Where k = boltzman constant W = thermodynamic probability and gives the number of the possible orientations of the molecule. For example consider CO molecule W= = Avogadro number