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(1.3) (1.8) (1.11) (1.14) Fundamental equations for homogeneous closed system consisting of 1 mole:

Fundamental equations for homogeneous closed system consisting of n moles: (3.1) (3.2) (3.3) (3.4)

For a single-phase fluid in a closed system wherein no chemical reactions occur, the composition is necessarily constant, and therefore: (3.5) (3.6)

HOMOGENEOUS OPEN SYSTEM An open system can exchange matter as well as energy with its surroundings. For a closed homogeneous system, we consider G to be a function only of T and P: G = g(T, P) (3.7) In an open system, there are additional independent variables, i.e., the mole numbers of the various components present. nG = g(T, P, n 1, n 2,....., n m ) (3.8) where m is the number of components.

The total differential of eq. (3.8) is (3.9) Where subscript n i refers to all mole numbers and subscript n j to all mole numbers other than the i th. Chemical potential is defined as: (3.10)

For a three-component system:

We may rewrite eq. (3.9) as (3.11) For a system comprising of 1 mole, n = 1 and n i = x i (3.12) Eqs. (3.11) and (3.12) are the fundamental equations for an open system corresponding to eq. (3.1) for a closed system.

Using similar derivations, we can get the following relations: (3.13) (3.14) (3.13) It follows that: (3.16)

For a closed system undergoing a reversible process, the criterion for equilibrium is defined in: Within this closed system, each phase is an open system which is free to transfer mass to each other. Eq. (3.11) may be written for each phase: (3.18) (3.19) (3.17)

Total change of internal energy is the sum of internal energy of each phase in the system: (3.20)

The individual variation d(nS) (1), d(nS) (2), etc. are subject to the constraints of constant total entropy, constant total volume, and constant total moles of each species. These may be written as: (3.21) (3.22) (3.23)

Equations (3.21 – 3.23) can be written as (3.24) (3.25) (3.26)

Eq. (3.20) for a two-phase 3-component system gives: (3.27)

Substituting eqs. (3.24 – 3.26) into eq. (3.27) (3.28)

All variations d(nS) (2), d(nV) (2), dn 1 (2), dn 2 (2), etc., are truly independent. Therefore, at equilibrium in the closed system where d(nU) = 0, it follows that

(3.29) (3.30) (3.31) (3.32) (3.33) Thus, at equilibrium

(3.34)

The definition of a partial molar property, Eq. (3.34), provides the means for calculation of partial properties from solution-property data. Solution properties can be calculated from knowledge of the partial properties. The derivation of this equation starts with the observation that the thermodynamic properties of a homogeneous phase are functions of T, P, and the numbers of moles of the individual species which comprise the phase.

Thus for thermodynamic property M: The total differential of nM is (3.35) (3.36)

Because the first two partial derivatives on the right are evaluated at constant n and because the partial derivative of the last term is given by Eq. (3.34), this equation has the simpler form: (3.37)

Since n i = x i n it follows that: When dn i is replaced by this expression, and d(nM) is replaced by the identity: Equation (3.37) becomes:

The terms containing n are collected and separated from those containing dn to yield: In application, one is free to choose a system of any size (n), and to choose any variation in its size (dn). Thus n and dn are independent and arbitrary. The only way that the left side of this equation can be zero is for each term in brackets to be zero. (3.38)

Therefore: (3.39) (3.40) Eq. (3.34) is an important relations (summability relation) for partial molar properties are.

Since Eq. (3.40) is a general expression for M, differentiation yields a general expression for dM: (3.41) Combining eqs. (3.39) and (3.41) yields Gibbs-Duhem equation : (3.42) For the important special case of changes at constant T and P, it simplifies to: (3.43)

(B) (A) Eq. (3.40) for a binary solution: whence When M is known as a function of xl at constant T and P, the appropriate form of the Gibbs-Duhem equation is Eq. (3.43), expressed here as: (C)

Since x 1 + x 2 = 1, it follows that dx 1 = – dx 2. Eliminating dx 2 in favor of dx 1 in Eq. (B) and combining the result with Eq. (C) gives: (D) or:

Elimination of Eq. (A) yields: (3.44)

Elimination of Eq. (A) yields: (3.45) Thus for binary systems, the partial properties are readily calculated directly from an expression for the solution property as a function of composition at constant T and P.

Example 3.1 Describe a graphical interpretation of eqs. (3.44) and(3.45). Solution

Values of dM/dx 1 are given by the slope of tangent lines. One such tangent line is shown. at x 1 = 1  intercept = I 1 at x 1 = 0  intercept = I 2 As is evident from the figure that two equivalent expressions can be written for the slope: and The first eq. is solved for I 2 ; it combines with the second to give I 1 : and

Comparison of these expressions with eqs. (3.44) and (3.55) shows that: The tangent intercepts give directly the values of two partial properties. at x 1 = 0  and at x 1 = 1  and and

3.2

SOLUTION 3.2 Molar volume of the solution is Since the required volume is V t = 2000 cm 3, the total number of moles required is: Of this, 30% is methanol and 70% is water, n1 = moln2 = mol

The volume of each species is:

(3.46) (3.47)

(3.48)

GIBB’S THEOREM (3.49)

(3.50) (3.51) (3.40):

Equation (1.30) of Chapter 1: (1.30) For ideal gas: (3.52)

For a constant T process (constant T) According to eq. (3.49):

whence By the summability relation, eq. (3.49): Or: (3.53)

This equation is rearranged as the left side is the entropy change of mixing for ideal gases. Since 1/yi >1, this quantity is always positive, in agree- ment with the second law. The mixing process is inherently irreversible, and for ideal gases mixing at constant T and P is not accompanied by heat transfer. (3.54)

Gibbs energy for an ideal gas mixture: Partial Gibbs energy : In combination with eqs. (3.50) and (3.54) this becomes or: (3.55)

An alternative expression for the chemical potential can be derived from eq. (1.14): At constant temperature: (1.14) (constant T) Integration gives: (3.56) Combining eqs. (3.55) and (3.56) results in: (3.57)

The origin of the fugacity concept resides in eq. (3.56), valid only for pure species i in the ideal-gas state. For a real fluid, we write an analogous equation: (3.58) where f i is fugacity of pure species i. Subtraction of eq. (3.56) from Eq. (3.58), both written for the same T and P, gives: (3.59)

Combining eqs. (3.53) with (3.59) gives: The dimensionless ratio f i /P is another new property, the fugacity coefficient, given the symbol  i : (3.60) (3.61) Equation (3.50) can be written as (3.62) The definition of fugacity is completed by setting the ideal-gas-state fugacity of pure species i equal to its pressure: (3.63)

Equation (1.50): (constant T)(1.50) Combining eqs. (3.62) and (1.50) results in: (constant T)(3.63) Fugacity coefficients (and therefore fugacities) for pure gases are evaluated by this equation from PVT data or from a volume-explicit equation of state.

An example of volume-explicit equation of state is the 2- term virial equation: (constant T) Because the second virial coefficient B i is a function of temperature only for a pure species, (constant T)(3.64)

FUGACITY COEFFICIENT DERIVED FROM PRESSURE-EXPLICIT EQUATION OF STATE Use equation (1.62): Combining eqs. (3.63) and (50) gives: (3.65)

Tugas II: Soal no. 3.38(a)dari buku Smith dkk (menghitung koefisien fugasitas fase uap dan cair)

VAPOR/LIQUID EQULIBRIUM FOR PURE SPECIES Eq. (3.58) for species i as a saturated vapor: (3.66) For saturated liquid: (3.67) By difference:

Phase transition from vapor to liquid phase occurs at constant T dan P (P i sat ). According to eq. (4): d(nG) = 0 Since the number of moles n is constant, dG = 0, therefore : Therefore: (3.68) (3.69) For a pure species, coexisting liquid and vapor phases are in equilibrium when they have the same temperature, pressure, and fugacity

An alternative formulation is based on the corresponding fugacity coefficients whence: (3.70) (3.71)

FUGACITY OF PURE LIQUID The fugacity of pure species i as a compressed liquid is calculated in two steps: 1.The fugacity coefficient of saturated vapor is determined from Eq. (3.65), evaluated at P = P i sat and V i = V i sat. The fugacity is calculated using eq. (3.61). (3.65) (3.61)

2.the calculation of the fugacity change resulting from the pressure increase, P i sat to P, that changes the state from saturated liquid to compressed liquid. An isothermal change of pressure, eq. (1.49) is integrated to give: (3.72) According to eq. (46): ( – ) (3.73)

Eq. (3.72) = Eq. (3.73): Since V i, the liquid-phase molar volume, is a very weak function of P at T << T c, an excellent approximation is often obtained when V i is assumed constant at the value for saturated liquid, V i L : (3.74)

Remembering that: The fugacity of a pure liquid is: (3.75)