Thermodynamics filled in class notes_Part_130

Thermodynamics filled in class notes_Part_130 - 267 7.2....

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Unformatted text preview: 267 7.2. ADIABATIC, ISOCHORIC KINETICS consider A and B to be isomers of an identical molecular species. So we have N = 2 species reacting in J = 1 reactions. The number of elements L here is irrelevant. 7.2.1.1 One-step reversible kinetics Let us insist our reaction process be isochoric and adiabatic, and commence with only A present. The reaction kinetics with β = 0 are −E RT dρA = − a exp dt ρA − 1 ρ, Kc B (7.259) =k =r dρB dt −E RT = a exp ρA − 1 ρ, Kc B (7.260) =k =r ρA (0) = ρA , ρB (0) = 0. (7.261) (7.262) For our alternate compact linear algebra based form, we note that r = a exp −E RT ρA − 1 , ρ Kc B (7.263) and that d dt ρA ρB = −1 (r ). 1 (7.264) Performing the decomposition yields d dt ρA ρA + ρB −1 (r ). 0 = (7.265) Expanded, this is 10 11 d dt ρA ρB = −1 (r ). 0 (7.266) Combining Eqs. (7.259-7.260) and integrating yields d (ρ + ρ B ) = 0 , dt A ρA + ρB = ρA , ρB = ρA − ρA . (7.267) (7.268) (7.269) CC BY-NC-ND. 18 November 2011, J. M. Powers. 268 CHAPTER 7. KINETICS IN SOME MORE DETAIL Thus, Eq. (7.259) reduces to −E RT dρA = −a exp dt 1 ρ − ρA Kc A ρA − . (7.270) Scaling, Eq. (7.270) can be rewritten as d d(at) 7.2.1.2 ρA ρA = − exp − E 1 RTo T /To ρA 1 − ρA Kc 1− ρA ρA . (7.271) First law of thermodynamics Recall the first law of thermodynamics and neglecting potential and kinetic energy changes: dU ˙ ˙ = Q − W. (7.272) dt ˙ Here E is the total internal energy. Because we insist the problem is adiabatic Q = 0. ˙ = 0. Thus we have Because we insist the problem is isochoric, there is no work done, so W dU = 0. (7.273) dt Thus, we find U = Uo . (7.274) Recall the total internal energy for a mixture of two calorically perfect ideal gases is U = nA uA + nB uB , nA nB =V uA + uB , V V = V (ρA uA + ρB uB ) , PA PB = V ρA hA − + ρB (hB − ρA ρB = V ρA hA − RT + ρB (hB − RT , (7.275) (7.276) (7.277) , (7.278) (7.279) o o = V ρA cP (T − To ) + hTo ,A − RT + ρB (cP (T − To ) + hTo ,B − RT o o = V (ρA + ρB )(cP (T − To ) − RT ) + ρA hTo ,A + ρB hTo ,B , o (7.281) o = V (ρA + ρB )((cP − R)T − cP To ) + ρA hTo ,A + ρB hTo ,B , o (7.282) o = V (ρA + ρB )((cP − R)T − (cP − R + R)To ) + ρA hTo ,A + ρB hTo ,B , o o = V (ρA + ρB )(cv T − (cv + R)To ) + ρA hTo ,A + ρB hTo ,B , o o = V (ρA + ρB )cv (T − To ) + ρA (hTo ,A − RTo ) + ρB (hTo ,B − RTo ) , = V (ρA + ρB )cv (T − To ) + ρA uo o ,A + ρB uo o ,B . T T CC BY-NC-ND. 18 November 2011, J. M. Powers. ,(7.280) (7.283) (7.284) (7.285) (7.286) ...
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This note was uploaded on 11/26/2011 for the course EGN 3381 taught by Professor Park-sou during the Fall '11 term at FSU.

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