{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Thermodynamics filled in class notes_Part_130

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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) ...
View Full Document

{[ snackBarMessage ]}