Thermodynamics filled in class notes_Part_53

# Thermodynamics filled in class notes_Part_53 - 113 4.3....

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Unformatted text preview: 113 4.3. LEGENDRE TRANSFORMATIONS For the Gibbs equation, (4.1), du = −P dv + T ds, one has y = u, two canonical variables, x1 = v and x2 = s, and two conjugates, ψ1 = −P and ψ2 = T . Thus N = 2, and one can expect 22 − 1 = 3 Legendre transformations. They are τ1,2 τ1 = y − ψ1 x1 = h = h(P, s) = u + P v, enthalpy, (4.97) τ2 = y − ψ2 x2 = a = a(v, T ) = u − T s, Helmholtz free energy, (4.98) = y − ψ1 x1 − ψ2 x2 = g = g (P, T ) = u + P v − T s, Gibbs free energy. (4.99) It has already been shown for the enthalpy that dh = T ds + vdP , so that the canonical variables are s and P . One then also has dh = ∂h ∂s ∂h ∂P ds + P dP, (4.100) s from which one deduces that T= ∂h ∂s , ∂h ∂P v= P . (4.101) s From Eq. (4.101), a second Maxwell relation can be deduced by diﬀerentiation of the ﬁrst with respect to P and the second with respect to s: ∂T ∂P = s ∂v ∂s (4.102) . P The relations for Helmholtz and Gibbs free energies each supply additional useful relations including two new Maxwell relations. First consider the Helmholtz free energy a = u − T s, da = du − T ds − sdT, = (−P dv + T ds) − T ds − sdT, = −P dv − sdT. (4.103) (4.104) (4.105) (4.106) So the canonical variables for a are v and T . The conjugate variables are −P and −s. Thus da = ∂a ∂v dv + T ∂a ∂T dT. (4.107) v So one gets −P = ∂a ∂v , T −s = ∂a ∂T . (4.108) v CC BY-NC-ND. 18 November 2011, J. M. Powers. 114 CHAPTER 4. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS and the consequent Maxwell relation ∂P ∂T = v ∂s ∂v (4.109) . T For the Gibbs free energy g = u + P v −T s, (4.110) =h = h − T s, dg = dh − T ds − sdT, = (T ds + vdP ) −T ds − sdT, (4.111) (4.112) (4.113) =dh = vdP − sdT. (4.114) So for Gibbs free energy, the canonical variables are P and T while the conjugate variables are v and −s. One then has g = g (P, T ), which gives ∂g ∂P dg = dP + T ∂g ∂T dT. (4.115) P So one ﬁnds v= ∂g ∂P −s = , T ∂g ∂T ∂s ∂P . (4.116) . P The resulting Maxwell function is then ∂v ∂T P =− (4.117) T Example 4.6 Canonical Form If h(s, P ) = cP To P Po R/cP exp s cP + (ho − cP To ) , (4.118) and cP , To , R, Po , and ho are all constants, derive both thermal and caloric state equations P (v, T ) and u(v, T ). Now for this material ∂h ∂s ∂h ∂P = To P = s P Po RTo Po CC BY-NC-ND. 18 November 2011, J. M. Powers. R/cP P Po exp s cP , exp s cP R/cP −1 (4.119) . (4.120) ...
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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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