Thermodynamics

692 | ThermodynamicsandV O2 a NR uT 1b 10.219 kmol218.314 kPa # m 3 >kmol # K21313 K2 5.70 m 3P 1 O 2100 kPaV N2 a NR uT 1b 10.143 kmol218.314 kPa # m 3 >kmol # K21293 K2 2.32 m 3P 1 N 2150 kPaV m V O2 V N2 5.70 2.32 8.02 m 3Thus,P m N m R u T mV m 10.362 kmol2 18.314 kPa # m 3 >kmol # K21305.2 K28.02 m 3 114.5 kPaDiscussion We could also determine the mixture pressure by using P m V m m m R m T m , where R m is the apparent gas constant of the mixture. Thiswould require a knowledge of mixture composition in terms of mass or molefractions.EXAMPLE 13–4Exergy Destruction during Mixing of Ideal GasesO 225°C200 kPaCO 225°C200 kPaFIGURE 13–15Schematic for Example 13–4.An insulated rigid tank is divided into two compartments by a partition, asshown in Fig. 13–15. One compartment contains 3 kmol of O 2 , and theother compartment contains 5 kmol of CO 2 . Both gases are initially at 25°Cand 200 kPa. Now the partition is removed, and the two gases are allowedto mix. Assuming the surroundings are at 25°C and both gases behave asideal gases, determine the entropy change and exergy destruction associatedwith this process.Solution A rigid tank contains two gases separated by a partition. Theentropy change and exergy destroyed after the partition is removed are to bedetermined.Assumptions Both gases and their mixture are ideal gases.Analysis We take the entire contents of the tank (both compartments) asthe system. This is a closed system since no mass crosses the boundary duringthe process. We note that the volume of a rigid tank is constant, andthere is no energy transfer as heat or work. Also, both gases are initially atthe same temperature and pressure.When two ideal gases initially at the same temperature and pressure aremixed by removing a partition between them, the mixture will also be at thesame temperature and pressure. (Can you prove it? Will this be true for nonidealgases?) Therefore, the temperature and pressure in the tank will still be25°C and 200 kPa, respectively, after the mixing. The entropy change ofeach component gas can be determined from Eqs. 13–18 and 13–25:→0¢S m a ¢S i a N i ¢ s i a N i a c p,i ln T i,2 RT u ln P i,2bi,1 P i,1R u a N i ln y i,2P m,2P i,1R u a N i ln y i,2since P m,2 P i,1 200 kPa. It is obvious that the entropy change is independentof the composition of the mixture in this case and depends on only