Unformatted text preview: 4.10. IRREVERSIBLE ENTROPY PRODUCTION IN A CLOSED SYSTEM 135 Now take a partial molar derivative and analyze to get
T ,P,nj =
= 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
T ,P,nj ∂nk
δik , , (4.369) k =1 , (4.370) T ,P,nj (4.371) k =1 G
n ∂ni T ,P,nj n
gi − .
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)
(4.380) 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),
dU |S,V ≤ 0.
For processes which are isoenergetic and isochoric, the second law expression, Eq. (4.383),
dS |U,V ≥ 0.
Now using Eq. (4.300) to eliminate dS in Eq. (4.385), one can express the second law as 1
dU + dV −
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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