Thermodynamics

774 | ThermodynamicsTTs(T,P)∆s = – R u lnPP0Ps°(T,P 0 )(Tabulated)P 0 = 1 atmFIGURE 15–29At a specified temperature, theabsolute entropy of an ideal gas atpressures other than P 0 1 atmcan be determined by subtractingR u ln (P/P 0 ) from the tabulated valueat 1 atm.sThe total entropy generated during a process can be determined by applyingthe entropy balance to an extended system that includes the system itselfand its immediate surroundings where external irreversibilities might beoccurring. When evaluating the entropy transfer between an extended systemand the surroundings, the boundary temperature of the extended systemis simply taken to be the environment temperature, as explained in Chap. 7.The determination of the entropy change associated with a chemical reactionseems to be straightforward, except for one thing: The entropy relationsfor the reactants and the products involve the entropies of the components,not entropy changes, which was the case for nonreacting systems. Thus weare faced with the problem of finding a common base for the entropy of allsubstances, as we did with enthalpy. The search for such a common base ledto the establishment of the third law of thermodynamics in the early partof this century. The third law was expressed in Chap. 7 as follows: Theentropy of a pure crystalline substance at absolute zero temperature is zero.Therefore, the third law of thermodynamics provides an absolute base forthe entropy values for all substances. Entropy values relative to this base arecalled the absolute entropy. The s – ° values listed in Tables A–18 throughA–25 for various gases such as N 2 ,O 2 ,CO,CO 2 ,H 2 ,H 2 O, OH, and O arethe ideal-gas absolute entropy values at the specified temperature and at apressure of 1 atm. The absolute entropy values for various fuels are listed inTable A–26 together with the h – ° f values at the standard reference state of25°C and 1 atm.Equation 15–20 is a general relation for the entropy change of a reactingsystem. It requires the determination of the entropy of each individual componentof the reactants and the products, which in general is not very easyto do. The entropy calculations can be simplified somewhat if the gaseouscomponents of the reactants and the products are approximated as idealgases. However, entropy calculations are never as easy as enthalpy or internalenergy calculations, since entropy is a function of both temperature andpressure even for ideal gases.When evaluating the entropy of a component of an ideal-gas mixture, weshould use the temperature and the partial pressure of the component. Notethat the temperature of a component is the same as the temperature of themixture, and the partial pressure of a component is equal to the mixturepressure multiplied by the mole fraction of the component.Absolute entropy values at pressures other than P 0 1 atm for any temperatureT can be obtained from the ideal-gas entropy change relation writtenfor an imaginary isothermal process between states (T,P 0 ) and (T,P), asillustrated in Fig. 15–29:s 1T,P2 s° 1T,P0 2 R u ln P P 0For the component i of an ideal-gas mixture, this relation can be written assi 1T,P i 2 s° i 1T,P 0 2 R u ln y iP m1kJ>kmol # K2P 0(15–21)(15–22)where P 0 1 atm, P i is the partial pressure, y i is the mole fraction of thecomponent, and P m is the total pressure of the mixture.