Thermodynamics

where P i,2 y i,2 P m,2 and P i,1 y i,1 P m,1 . Notice that the partial pressure P i ofeach component is used in the evaluation of the entropy change, not themixture pressure P m (Fig. 13–13).Chapter 13 | 691Partial pressureof component iat state 2EXAMPLE 13–3Mixing Two Ideal Gases in a TankAn insulated rigid tank is divided into two compartments by a partition, asshown in Fig. 13–14. One compartment contains 7 kg of oxygen gas at 40°Cand 100 kPa, and the other compartment contains 4 kg of nitrogen gas at20°C and 150 kPa. Now the partition is removed, and the two gases areallowed to mix. Determine (a) the mixture temperature and (b) the mixturepressure after equilibrium has been established.Solution A rigid tank contains two gases separated by a partition. The pressureand temperature of the mixture are to be determined after the partitionis removed.Assumptions 1 We assume both gases to be ideal gases, and their mixtureto be an ideal-gas mixture. This assumption is reasonable since both theoxygen and nitrogen are well above their critical temperatures and well belowtheir critical pressures. 2 The tank is insulated and thus there is no heattransfer. 3 There are no other forms of work involved.OProperties The constant-volume specific heats of N 2 and O 2 at room temperatureare c v,N2 0.743 kJ/kg · K and c v,O2 0.658 kJ/kg · K (Table A–2a).2Analysis 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 and thusthere is no boundary work done.(a) Noting that there is no energy transfer to or from the tank, the energybalance for the system can be expressed asE in E out ¢E systemP i,∆s i° = s i° ,2 – s i° ,1 – R i lnPartial pressureof component iat state 1—––i,2P i,1FIGURE 13–13Partial pressures (not the mixturepressure) are used in the evaluation ofentropy changes of ideal-gas mixtures.7 kg40°C100 kPaN 24 kg20°C150 kPaPartitionFIGURE 13–14Schematic for Example 13–3.0 ¢U ¢U N2 ¢U O23mc v 1T m T 1 24 N2 3mc v 1T m T 1 24 O2 0By using c v values at room temperature, the final temperature of the mixtureis determined to be14 kg2 10.743 kJ>kg # K21Tm 20°C2 17 kg2 10.658 kJ>kg # K21Tm 40°C2 0T m 32.2°C(b) The final pressure of the mixture is determined from the ideal-gas relationwhereN O2 m O 2M O2N N2 m N 2M N2P m V m N m R u T m7 kg 0.219 kmol32 kg>kmol4 kg 0.143 kmol28 kg>kmolN m N O2 N N2 0.219 0.143 0.362 kmol