Thermodynamics

Chapter 13 | 689This is 33 percent lower than the assumed value. Therefore, we shouldrepeat the calculations, using the new value of V m . When the calculationsare repeated we obtain 0.738 m 3 after the second iteration, 0.678 m 3 afterthe third iteration, and 0.648 m 3 after the fourth iteration. This value doesnot change with more iterations. Therefore,V m 0.648 m 3Discussion Notice that the results obtained in parts (b), (c), and (d ) arevery close. But they are very different from the ideal-gas values. Therefore,treating a mixture of gases as an ideal gas may yield unacceptable errors athigh pressures.13–3 PROPERTIES OF GAS MIXTURES:IDEAL AND REAL GASESConsider a gas mixture that consists of 2 kg of N 2 and 3 kg of CO 2 . Thetotal mass (an extensive property) of this mixture is 5 kg. How did we do it?Well, we simply added the mass of each component. This example suggestsa simple way of evaluating the extensive properties of a nonreacting idealorreal-gas mixture: Just add the contributions of each component of themixture (Fig. 13–11). Then the total internal energy, enthalpy, and entropyof a gas mixture can be expressed, respectively, asU m akS m ak(13–13)(13–14)(13–15)By following a similar logic, the changes in internal energy, enthalpy, andentropy of a gas mixture during a process can be expressed, respectively, as¢U m aki1¢H m aki1¢S m aki1i1H m aki1i1U i akS i ak¢U i ak¢H i ak¢S i aki1i1i1i1i1H i aki1m i u i akm i h i akm i s i akm i ¢u i akm i ¢h i akm i ¢s i ak(13–16)(13–17)(13–18)Now reconsider the same mixture, and assume that both N 2 and CO 2 are at25°C. The temperature (an intensive property) of the mixture is, as you wouldexpect, also 25°C. Notice that we did not add the component temperatures todetermine the mixture temperature. Instead, we used some kind of averagingscheme, a characteristic approach for determining the intensive properties ofa mixture. The internal energy, enthalpy, and entropy of a mixture per unitmass or per unit mole of the mixture can be determined by dividing the equationsabove by the mass or the mole number of the mixture (m m or N m ). Weobtain (Fig. 13–12)i1i1i1N i s i1kJ>K2i1i1i1N i u i1kJ2N i h i 1kJ2N i ¢u i1kJ2N i ¢h i1kJ2N i ¢ s i1kJ>K22 kmol A6 kmol BU A =1000 kJU B =1800 kJU m =2800 kJFIGURE 13–11The extensive properties of a mixtureare determined by simply adding theproperties of the components.2 kmol A3 kmol Bū A =500 kJ/kmolū B =600 kJ/kmolū m =560 kJ/kmolFIGURE 13–12The intensive properties of a mixtureare determined by weighted averaging.