LaGrange Multiplier Method for Equilibrium

3. LaGrange Multiplier Method for Equilibrium

a. U, V, m System One can use the LaGrange multiplier method to maximize the entropy. In case an analysis involves several nonreacting subsystems containing several species so that

1 ,N 2 )+S (U ,V ,N 1 ,N 2 ) + ..., (143) the entropy may be maximized subject to the constraint that U=U (1) +U (2) +...= Constant,

S=S (1) (U (1) ,V (1) ,N (1) (1),

(144) V=V (1) +V (2) +...= Constant, and

(145) N 1 =N (1)

(146) Using the LaGrange multiplier method and Eqs. (143) to (146),

1 +N 1 +...= Constant.

1 ,N 2 ,...) + S (U ,V ,N 1 ,N 2 ) + ... .+

λ (2) U (U –(U +U +..))+ λ V (V –(V +V +...)) + λ (N – (N 1 +N 1 +...)). (147)

N1

The maximization process requires that ∂S/∂U (1 )= 0, ∂S/∂U (2) =0, …Differentiating Eq. (147), ∂S/∂U (1) = ∂S (1) /∂U (1) –λ = 0 or ∂S (1) /∂U (1) = 1/T (1) U =λ U .

(148) ∂S/∂U (2) = ∂S (2) /∂U (2) +λ

U = 0, or ∂S (2) /∂U (2) = 1/T (2) =λ U .

Therefore, T (1) =T (2) =....., that represents the thermal equilibrium condition. Since

∂S/∂V (1) = ∂S /∂V –λ

V = 0 or ∂S /∂V =P /T =λ V . (150)

∂S/∂V (2) = ∂S /∂V –λ

V = 0, or ∂S /∂V =P /T =λ V . (151)

Since T (2) =T = …, P =P = …, mechanical equilibrium condition between different phases. Furthermore,

∂S/∂N (1)

1 = ∂S /∂N 1 –λ N1 = 0 or ∂S /∂N 1 =µ 1 /T = –λ(N 1 ). (152)

∂S/∂N (1)

2 = ∂S / ∂N 2 +λ N2 = 0 or ∂S /∂N 2 =µ 2 /T = –λ(N 2 ). (153)

Since the temperatures within the subsystems are identical, µ (1)

1 =µ 2 = …, i.e., phase 1 is in equilibrium and no chemical reaction occurs. Repeating the process for the other subsystems,

1 =µ 1 /T (2) = –λ(N 1 ). (154) (2) = ∂S (2) /∂N (2) +λ = 0 or ∂S (2) /∂N (2)

∂S/∂N (2) 1 = ∂S (2) /∂N (2) 1 –λ N1 = 0 or ∂S (2) /∂N (2)

∂S/∂N (2)

2 2 N2

2 =µ 2 /T = –λ(N 2 ). (155) Furthermore, we assume identical values of λ so that , T (1) =T (2) = … , and µ (1) 1 =µ (2) 1 , ... ,

µ (1) 2 =µ (2) 2 , etc. Therefore, in the nonreacting subsystems, the equilibrium condition requires the temperatures, pressures, and chemical potentials in all of the subsystems to, respectively, equal one another.

A similar procedure can be adopted to determine the equilibrium condition at given T, P and N.

b. T, P, m System viii. One Component

Consider N moles of a pure substance (say, H 2 O) kept at 0ºC and constant pressure P (say, 0.6 kPa, the triple point pressure). The substance attains equilibrium in multiple phases (e.g., the water forms three – π = 3 – phases: solid ,liquid and gas). In general, the number of

moles in each phase is different (say, N (l) , N (g) , N (s) ,) may change, and the Gibbs energy is minimized at equilibrium. For the composite system that includes all the phases

G=G (1) (T,P,N (1) )+G (2) (T,P,N (2) ) + ... = G(T,P, N (1) ,N (2) ...N (π) ), (156) which is to be minimized subject to the constraint N = ΣN (j) = constant. We again use the La- Grange multiplier method and form the function

F = G + λ(ΣN (j) – N), so that (157)

(158) Similarly, for the other phases

∂F/∂N (1) = ∂G/∂N +λ= g + λ = 0.

(159) Therefore,

∂F/∂N (2) = ∂G/∂N (2) +λ= g (2) + λ = 0.

(160) implying that at equilibrium the molal Gibbs function is identical for all species.

g (1) = g (1) = ... = – λ,

ix. Multiple Components The Gibbs energy

G=G (1) (T,P,N (1) (1)

1 ,N 2 ,...,N K )+G (T,P,N 1 ,N 2 ,...,N K )+…+

G (π) (T,P,N 1 (π) ,N 2 (π) , ... ,N K (π) )

1 ,N 2 , ... ,N K ,N 1 ,N 2 , ... ,N K ,N 1 (π) ,N 2 (π) , ... ,N (π) ). (162) We must minimize G subject to the constraints

= G(T,P,N (2)

K =N K +N K + ... N K (π) . Therefore, F=G+

1 (N 1 +N (2) 1 +... + N 1 (π) –N 1 )+

2 (N 1 +N (2) 1 +... + N 1 (π) –N 2 )+ ... +

K (N 1 +N 1 +... + N 1 (π) –N K ); and

∂F/∂N (1)

∂F/∂N (2)

1 = 0 = ˆg 1 + λ 1 ,

so that

The partial molal Gibbs function for each component must be identical in all of the phases at equilibrium.

Chapter 4