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Thermodynamics filled in class notes_Part_64

Thermodynamics filled in class notes_Part_64 - 4.10...

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Unformatted text preview: 4.10. IRREVERSIBLE ENTROPY PRODUCTION IN A CLOSED SYSTEM 135 Now take a partial molar derivative and analyze to get ∂g ∂ni = T ,P,nj = = = 1 ∂G n ∂ni 1 ∂G n ∂ni 1 ∂G n ∂ni 1 ∂G n ∂ni T ,P,nj − G ∂n n2 ∂ni T ,P,nj − G∂ n2 ∂ni T ,P,nj − G n2 T ,P,nj − G n2 N k =1 N , (4.368) T ,P,nj N nk T ,P,nj ∂nk ∂ni δik , , (4.369) k =1 , (4.370) T ,P,nj (4.371) k =1 G 1 ∂G − 2, n ∂ni T ,P,nj n 1 g gi − . = n n = (4.372) (4.373) Multiplying by n and rearranging, one gets gi = g + n ∂g ∂ni . (4.374) T ,P,nj A similar result holds for other properties. 4.10 Irreversible entropy production in a closed system Consider a multicomponent thermodynamic system closed to mass exchanges with its surroundings coming into equilibrium. Allow the system to be exchanging work and heat with its surroundings. Assume the temperature diﬀerence between the system and its surroundings is so small that both can be considered to be at temperature T . If δQ is introduced into the system, then the surroundings suﬀer a loss of entropy: dSsurr = − δQ . T (4.375) The system’s entropy S can change via this heat transfer, as well as via other internal irreversible processes, such as internal chemical reaction. The second law of thermodynamics requires that the entropy change of the universe be positive semi-deﬁnite: dS + dSsurr ≥ 0. (4.376) Eliminating dSsurr , one requires for the system that dS ≥ δQ . T (4.377) CC BY-NC-ND. 18 November 2011, J. M. Powers. 136 CHAPTER 4. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS Consider temporarily the assumption that the work and heat transfer are both reversible. Thus, the irreversible entropy production must be associated with internal chemical reaction. Now the ﬁrst law for the entire system gives dU = δQ − δW, = δQ − P dV, δQ = dU + P dV. (4.378) (4.379) (4.380) Note because the system is closed, there can be no species entering or exiting, and so there is no change dU attributable to dni . While within the system the dni may not be 0, the net contribution to the change in total internal energy is zero. A non-zero dni within the system simply re-partitions a ﬁxed amount of total energy from one species to another. Substituting Eq. (4.380) into Eq. (4.377) to eliminate δQ, one gets dS ≥ 1 (dU + P dV ), T (4.381) =δQ T dS − dU − P dV dU − T dS + P dV ≥ 0, ≤ 0. (4.382) (4.383) Eq. (4.383) involves properties only and need not require assumptions of reversibility for processes in its derivation. In special cases, it reduces to simpler forms. For processes which are isentropic and isochoric, the second law expression, Eq. (4.383), reduces to dU |S,V ≤ 0. (4.384) For processes which are isoenergetic and isochoric, the second law expression, Eq. (4.383), reduces to dS |U,V ≥ 0. (4.385) Now using Eq. (4.300) to eliminate dS in Eq. (4.385), one can express the second law as 1 P 1 dU + dV − T T T N i=1 =dS − 1 T ≥ 0, µi dni (4.386) U,V N µi dni i=1 ≥ 0. (4.387) irreversible entropy production The irreversible entropy production associated with the internal chemical reaction must be the left side of Eq. (4.387). Often the irreversible entropy production is deﬁned as σ , with CC BY-NC-ND. 18 November 2011, J. M. Powers. ...
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