This preview shows page 1. Sign up to view the full content.
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 BYNCND. 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 BYNCND. 18 November 2011, J. M. Powers. R/cP P
Po exp s
cP , exp s
cP R/cP −1 (4.119)
. (4.120) ...
View
Full
Document
This note was uploaded on 11/26/2011 for the course EGN 3381 taught by Professor Parksou during the Fall '11 term at FSU.
 Fall '11
 ParkSou
 Dynamics, Gate

Click to edit the document details