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Thermodynamics filled in class notes_Part_69

# Thermodynamics filled in class notes_Part_69 - 145 4.11...

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Unformatted text preview: 145 4.11. EQUILIBRIUM IN A TWO-COMPONENT SYSTEM The pressure at equilibrium is P2 = = = (nN2 + nN )RT , V (0.888154 kmole + 0.223693 kmole) 8.314 498.84 111.185 kP a. (4.465) kJ kmole K (6000 K ) , (4.466) (4.467) The pressure has increased because there are more molecules with the volume and temperature being equal. The molar concentrations ρi at equilibrium, are ρN = kmole 0.223693 kmole mole = 4.484 × 10−4 = 4.484 × 10−7 , 3 3 498.84 m m cm3 (4.468) ρN 2 = kmole 0.888154 kmole mole = 1.78044 × 10−3 = 1.78044 × 10−6 . 3 3 498.84 m m cm3 (4.469) Now consider the heat transfer. One knows for the isochoric process that 1Q2 = U2 − U1 . The initial energy is given by U1 = nN2 uN2 , (4.470) = nN2 (hN2 − RT ), (4.471) = (1 kmole) 2.05848 × 105 = 1.555964 × 105 kJ. kJ kJ − 8.314 kmole kmole K (6000 K ) , (4.472) (4.473) The energy at the ﬁnal state is U2 = = = = nN 2 u N 2 + nN u N , nN2 (hN2 − RT ) + nN (hN − RT ), (4.474) (4.475) (0.888154 kmole) 2.05848 × 105 (4.476) kJ kJ (6000 K ) − 8.314 kmole kmole K kJ kJ +(0.223693 kmole) 5.9727 × 105 (6000 K ) , − 8.314 kmole kmole K 2.60966 × 105 kJ. (4.477) (4.478) So 1Q2 = = = U2 − U1 , 2.60966 × 105 kJ − 1.555964 × 105 kJ, 1.05002 × 105 kJ. (4.479) (4.480) (4.481) Heat needed to be added to keep the system at the constant temperature. This is because the nitrogen dissociation process is endothermic. One can check for second law satisfaction in two ways. Fundamentally, one can demand that Eq. (4.377), dS ≥ δQ/T , be satisﬁed for the process, giving 2 S2 − S1 ≥ 1 δQ . T (4.482) CC BY-NC-ND. 18 November 2011, J. M. Powers. 146 CHAPTER 4. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS For this isothermal process, this reduces to S2 − S1 1Q2 ≥ T (nN2 sN2 + nN sN )|2 − (nN2 sN2 + nN sN )|1 yN2 P Po yN2 P − nN2 so 2 − R ln T,N Po PN2 nN2 so 2 − R ln T,N Po PN2 − nN2 so 2 − R ln T,N Po nN2 so 2 − R ln T,N nN2 so 2 − R ln T,N − nN2 so 2 − R ln T,N nN2 RT Po V nN2 RT Po V + nN + nN + nN + nN + nN yN P Po yN P so − R ln T,N Po PN so − R ln T,N Po PN so − R ln T,N Po so − R ln T,N + nN so − R ln T,N nN RT Po V so − R ln T,N nN RT Po V =0 1Q2 ≥ T , (4.483) , (4.484) , (4.485) , (4.486) . (4.487) 2 1 1Q2 ≥ T 2 1 2 1Q2 ≥ ≥ T 1Q2 T 1 Now at the initial state nN = 0 kmole, and RT /Po /V has a constant value of kJ 8.314 kmole K (6000 K ) RT 1 = =1 , Po V (100 kP a)(498.84 m3 ) kmole (4.488) so Eq. (4.487) reduces to nN2 so 2 − R ln T,N nN 2 1 kmole nN 1 kmole nN 2 − R ln 1 kmole + nN so − R ln T,N − n N 2 so 2 T,N 2 1 ≥ 1Q2 T , (4.489) ((0.888143) (292.984 − 8.314 ln (0.88143)) + (0.223714) (216.926 − 8.314 ln (0.223714)))|2 − ((1) (292.984 − 8.314 ln (1)))|1 19.4181 kJ kJ ≥ 17.5004 . K K ≥ 105002 , 6000 (4.490) Indeed, the second law is satisﬁed. Moreover the irreversible entropy production of the chemical reaction is 19.4181 − 17.5004 = +1.91772 kJ/K. For the isochoric, isothermal process, it is also appropriate to use Eq. (4.394), dA|T,V ≤ 0, to check for second law satisfaction. This turns out to give an identical result. Since by Eq. (4.307), A = U − T S , A2 − A1 = (U2 − T2 S2 ) − (U1 − T1 S1 ). Since the process is isothermal, A2 − A1 = U2 − U1 − T (S2 − S1 ). For A2 − A1 ≤ 0, one must demand U2 − U1 − T (S2 − S1 ) ≤ 0, or U2 − U1 ≤ T (S2 − S1 ), or S2 −S1 ≥ (U2 −U1 )/T . Since 1Q2 = U2 −U1 for this isochoric process, one then recovers S2 −S1 ≥ 1Q2 /T. CC BY-NC-ND. 18 November 2011, J. M. Powers. ...
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