HOMOGENEOUS CLOSED SYSTEM

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Transcription:

CHAE II

A closed system is one that does not exchange matter with its surroundings, although it may exchange energy. W n in = 0 HOMOGENEOUS CLOSED SYSEM System n out = 0 Q dn i = 0 (2.1) i = 1, 2, 3,... No internal energy transported accross boundary. All energy exchange between a closed system and its surroundings appears as heat and work. he total energy change of the surroundings equals the net energy transferred to or from it as heat and work.

First and second laws of hermodynamcs: du ds d (2.2) For reversible process: du = ds d (2.3) With ds = dq rev : heat absorbed by the system d = dw rev : work done by the system If the interaction occurs irreversibly: du < ds d (2.4)

he internal energy change can be calculated by integrating eq. (2.2): U S 2 U2 U1 ds d S 1 2 1 (2.5) For process occuring at constant S and : du S, 0 (2.6) At constant S and, U tends toward a minimum in an actual or irreversible process in a closed system, and remains constant in a reversible process. Eq. (2.6) provides a criterion for equilibrium in a closed system.

Definition: H U + (2.7) Differentiating eq. (2.7) yields: dh = du + d + d Combining the above equation with eq. (2.3) leads to: dh = ( ds d) + d + d dh = ds + d (2.8) For a closed system at constant S and : dh,s 0 (2.9)

the Helmholtz free energy (A) is a thermodynamic potential that measures the useful work obtainable from a closed system at a constant temperature and volume. A = the maximum amount of work extractable from a thermodynamic process in which temperature and volume are held constant.. Under these conditions, it is minimized at equilibrium.

Definition: A = U S Differentiating eq. (2.10) yields: (2.10) da = du d(s) = dq + dw ds S d = ds d ds S d da = S d d (2.11) For a closed system at constant dan : da, 0 (2.12)

Definition: G A + (2.13) Gibbs free energy (G) is a thermodynamic potential that measures the "useful" or process-initiating work obtainable from a thermodynamic system at a constant temperature and pressure (isothermal, isobaric). he Gibbs free energy is the maximum amount of nonexpansion work that can be extracted from a closed system; this maximum can be attained only in a completely reversible process.

Differentiating eq. (2.13) yields: dg = da + d() = S d d + d + d dg = S d + d (2.14) For a closed system at constant and : dg, 0 (2.15)

If F = F(x,y), the total differential of F is: dy y F dx x F df x y with y x F M x y F N (2.16) Ndy Mdx F

Further differentiation yields y x F y M 2 x y x F x N 2 y y x x N y M (2.17) Hence from equation: Ndy Mdx F we obtain: y x x N y M (2.17) (2.16)

esume: du = ds d (2.3) dh = ds + d (2.8) da = S d d (2.11) dg = S d + d (2.14) According to eq. (2.17): S S S S S S (2.18) (2.21) (2.20) (2.19)

ENHALY As a function of and, we can express: H H dh, otal differential of eq. (2.22): H d H d (H/) is obtained from the definition of C : H C (2.22) (2.23) (2.24)

dh = ds + d (2.8) Differentiation with respect of at constant yields: S H Combining eq. (2.25) with Maxwell equation (2.21): H (H/) is derived from fundamental equation: (2.25) (2.26) Introducing eqs. (2.24) and (2.26) into eq. (2.23) results in : d d C dh (2.27)

ENOY Entropy as a function of and : S S, otal differential of eq. (2.28): (2.28) ds S d S d (2.29) (S/) is obtained from Maxwell eq. (2.21): S (2.21)

dh = ds + d (2.8) Differentiation with respect of at constant yields: S H Combining eqs. (2.30) with (2.24): (S/) is derived from fundamental equation: (2.30) (2.31) Introducing eqs. (2.21) and (2.31) into eq. (2.29) results in : (2.32) C S d d C ds

INENAL ENEGY As a function of and, we can express: U U, otal differential of eq. (2.33): du U d U d (U/) is obtained from the definition of C : U C (2.33) (2.34) (2.35)

du = ds d (2.3) Differentiation with respect of at constant yields: Combining eqs. (2.36) and Maxwell equation (2.20): (U/) is derived from fundamental equation: (2.36) (2.37) Introducing eqs. (2.35) and (2.37) into eq. (2.34) results in : (2.38) S U U d d C du

Entropy as a function of and : ENOY, S S d S d S ds otal differential of eq. (2.39): (2.39) (2.40) (S/) is obtained from Maxwell eq. (2.20): (2.20) S

du = ds d (2.3) Differentiation with respect of at constant yields: Combining eqs. (2.41) and (2.35): (S/) is derived from fundamental equation: (2.41) (2.42) Introducing eqs. (2.20) and (2.42) into eq. (2.40) results in : (2.43) S U C S d d C ds

Fundamental equation: dg = S d + d (2.14) expressed as functional relation G = G(, ) (2.44) hus the special, or canonical, variables for the Gibbs energy are temperature and pressure. Since these variables can be directly measured and controlled, the Gibbs energy is a thermodynamic property of great potential utility

An alternative form of Eq. (2.14), a fundamental property relation, follows from the mathematical identity: G 1 G d dg 2 d (2.45) Substitution for dg by Eq. (2.14) and for G by Eq. (2.13) gives, after algebraic reduction G d H d 2 d (2.46) he advantage of this equation is that all terms are dimensionless; moreover, in contrast to eq. (2.14), the enthalpy rather than the entropy appears on the right side.

From eq. (2.46): G (2.47) H G (2.48) When G/ is known as a function of and, / and H/ follow by simple differentiation. he remaining properties are given by defining equations. In particular, S H G and U H

No experimental method for the direct measurement of numerical values of G or GI is known, and the equations which follow directly from the Gibbs energy are ofblittle practical use. by definition the residual Gibbs energy is: G = G G ig (2.49) where G and Gig are the actual and the ideal-gas values of the Gibbs energy at the same temperature and pressure.

Other residual properties are defined in an analogous way. he residual volume, for example, is: ig (2.50) Since = Z/, the residual volume and the compressibility factor are related: Z 1 (2.51)

he definition for the generic residual property is: M = M M ig (2.52) where M is the molar value of any extensive thermodynamic property, e.g.,, U, H, S, or G. Note that M and M ig, the actual and ideal-gas properties, are at the same temperature and pressure. Equation (2.46), written for the special case of an ideal gas, becomes ig ig ig G H d d 2 d (2.53)

esidual roperty: G d H d 2 d (2.54) his fundamental property relation for residual properties applies to fluids of constant composition. Useful restricted forms are: G (2.55) H G (2.56)

esidual Gibbs energy: G = H S (2.57) he residual entropy is therefore S H G (2.58) For constant, eq. (2.54) becomes: d G d (contant ) (2.59)

Integration from zero pressure to arbitrary pressure yields: G 0 d (constant ) (2.60) where at the lower limit G / is equal to zero because the zero-pressure state is an ideal-gas state. In view of eq. (2.51): G 0 Z 1 d (constant ) (2.61) Differentiation of eq. (2.61) with respect to temperature gives G Z (2.62)

Combining eqs. (2.62) and (2.54) gives H Z 0 d (constant ) (2.63) he residual entropy is found by combination of eqs. (2.58), (2.61), and (2.63): S Z 0 d Z 0 1 d p (constant ) (2.64)

ESIDUAL OEY FOM IIAL EOS For two-term virial equation Z 1 B From eq. (2.61): G 0 Z 1 d ( konstan) (2.61) We can get: G B (2.65)

Differentiation of eq. (2.56) with respect to at constant gives Introducing eq. (2.62) into eq. (2.56): H H 1 db B 2 d B db d Introducing eqs. (2.65) and (2.67) into eq. (2.58) S G db d B 2 db d (2.66) (2.67) (2.68)

ESIDUAL OEY FOM CUBIC EOS Equations (2.61), (2.62), and (2.64) are incompatible with pressure-explicit equations of state, and must be transformed to make the variable of integration. Z Z d dz d (constant ) 2 d dz Z d 2 (constant ) (2.69) d dz Z d (constant ) (2.70)

Introducing eq. (2.70) into eq. (2.61): G 0 Z 1 dz Z d (constant ) (2.71) he lower limit of the integration is = 0 is the condition of ideal gas: = 0 = and Z = 1 Hence, eq. (2.71) becomes: G Z dz 1 Z Z 1 Z 1 d (constant )

G Z 1 1 1 Z dz Z 1 d (constant ) G Z 1 lnz Z 1 d (constant ) G Z 1 lnz Z 1 d (constant ) (2.72)

Generalized form of cubic eos: b a b b Z 1 b a b b Substituting the above equation into eq. (2.72) yields: G Z 1 lnz 1 b a b b 1 d (constant ) (2.73)

Consider the terms in the bracket of right hand side: 1 b 1 b a 1 b b a 1 b b b 1 1 Integration of the equation: 1 b a b b b 1 1 1 d ln b a ln b b b ln

ln b a ln b b b ln b a ln b b b Introducing the last equation into eq. (2.73): G Z 1 lnz ln b a ln b b b (constant ) (2.74)

he corresponding equation for H follows from Eq. (2.54), which in view of Eq. (2.51) may be written: H 2 d d G Z 1 d Division by d and restriction to constant yields: H 2 (2.75) Z 1 G (constant ) (2.76) Differentiation of eq. (2.69) provides the first derivative on the right, and differentiation of eq. (2.72) provides the second. Substitution leads to: H Z Z 1 d (2.75)

Z is derived from a cubic eos Z b a b b Z a 2 b b 1 Z 2 a b b

b 1 b 1 b a Z 2 he integration part of eq. (2.75): d b 1 b 1 b a d Z b b ln b a

Introducing the last eq. into eq. (2.75): H Z 1 a b ln b b (2.76)

S is calculated using the equation: S H G Z 1 Z 1 lnz a ln b b a ln b ln b b b b S lnz ln b a b ln b b (2.77)

EXAMLE Calculate H and S for n-butane at 500K and 50 bar using the K eos. SOLUION b a b b For the K eos: = 1 = 0 r c 500 425.1 a = 0,42748 b = 0,08664 1.1762 For n-butane: c = 425,1 K c = 37,96 bar

a 2 2 c a c =14066688 0,42748 2 83,14 425,1 37,96 2 b b c c 0,08664 is calculated numerically: 83,14425,1 37,96 80,667 b a b b Initial value of : 83,14 50 500 0 831,4 cm 3 /mole

i b a i1 i 1 b b i1 i i-1 i error 1 831.40 655.25 2.69E-01 2 655.25 602.97 8.67E-02 3 602.97 583.38 3.36E-02 4 583.38 575.44 1.38E-02 5 575.44 572.12 5.80E-03 6 572.12 570.71 2.46E-03 7 570.71 570.11 1.05E-03 8 570.11 569.86 4.48E-04 9 569.86 569.75 1.91E-04 10 569.75 569.70 8.17E-05

ada iterasi ke-10 diperoleh hasil = 569,7 cm 3 /mol Z 50569,7 83,14500 0,6850 H Z 1 a b ln b b H r 0.5 Z 1 0,5 0.5 c a b 1.5 1 r 0.5 0.5 0.5 r 0,5 0.5 r ln b

H Z 1 b 1,5a 0.5 r ln b H 0,6850 1 1,514066688 80,66783,145001.1762 = 1,0833 0.5 ln 569,7 569,7 80,667 H = (8,314 J mol -1 K -1 ) (500 K) ( 1,0833) = 4.503,3 J mol -1

S lnz ln b a b ln b b S lnz ln b 0,5a b 0.5 r ln b = 0,78735 S = (8,314 J mole -1 K -1 ) ( 1,0833) = 6.546 J mole -1 K -1