Skip to 0 minutes and 1 second Welcome to thermodynamics in energy engineering week 4. We are going to focus on solving entropy change problems this week. Before solving the entropy changes in real situations, we need to know one more law of thermodynamics. That is the 3rd law. The 3rd law is the temperature principle. According to the 3rd law, entropy is zero at zero kelvin for pure crystalline substances. Or the entropy changes in chemical reaction at zero kelvin is zero if the reactants and products are pure, crystalline substances. Here, pure means pure elements and compounds with exact stoichiometry. Other statement of the 3rd law is from the Plank. Plank described the 3rd law like this.
Skip to 0 minutes and 55 seconds An entropy of any homogenous substance, which is in complete internal equilibrium, may be taken to be zero at zero kelvin. Complete internal equilibrium indicates the crystalline state free of defects. Now we know the entropy at zero K. Then, entropy at temperatures other than zero K can be calculated from the temperature dependence of entropy. The temperature change is under constant pressure, so it belongs to the entropy change at constant pressure process. Let’s start from the enthalpy H. H is (U+PV). Take the differential on both side. Now, insert the first law, dU equals (delta Q-PdV). Then, dH becomes (delta Q-PdV+PdV+VdP). Cancel out the same thing. Then, (delta Q+VdP) remain. Here, dP is zero at constant pressure. So dH equals to delta Qp.
Skip to 2 minutes and 6 seconds According to the definition of dS, dS is delta Q over T, thus dHp over T. dHp is CpdT. So dS is now CpdT over T. Therefore, delta S is integration of (Cp over T) over temperature.
The 3rd law
The third law is talking about the entropy of pure, perfectly crystalline materials at zero K.
They have zero entropy at zero K, by the third law. Therefore, in classical thermodynamics, thermal motions entirely stop in particles at zero K. Entropy change can be calculated from the knowledge on the heat capacity.