SM_14B - Chapter 14 - Section B - Non-Numerical Solutions...

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Chapter 14 - Section B - Non-Numerical Solutions 14.2 Start with the equation immediately following Eq. (14.49), which can be modified slightly to read: ln ˆ φ i = ∂( nG R / RT ) n i nZ ) n i + n ln Z n i + 1 where the partial derivatives written here and in the following development without subscripts are understood to be at constant T , n (or ρ/ n ), and n j . Equation (6.59) after multiplication by n can be written: R = 2 n ( nB ) ³ ρ n ´ + 3 2 n 2 ( nC ) ³ ρ n ´ 2 n ln Z Differentiate: R / ) n i = 2 ³ ρ n ´ ( + n ¯ B i ) + 3 2 ³ ρ n ´ 2 ( 2 n 2 C + n 2 ¯ C i ) n ln Z n i ln Z or R / ) n i = 2 ρ( B + ¯ B i ) + 3 2 ρ 2 ( 2 C + ¯ C i ) n ln Z n i ln Z By definition, ¯ B i · ) n i ¸ T , n j and ¯ C i · ) n i ¸ T , n j The equation of state, Eq. (3.39), can be written: Z = 1 + B ρ + C ρ 2 or = n + n ( ) ³ ρ n ´ + n 2 ( ) ³ ρ n ´ 2 Differentiate: ) n i = 1 + ³ ρ n ´ ( + n ¯ B i ) + ³ ρ n ´ 2 ( 2 n 2 C + n 2 ¯ C i ) or ) n i = 1 + B + ¯ B i ) + ρ 2 ( 2 C + ¯ C i ) When combined with the two underlined equations, the initial equation reduces to: ln ˆ φ i = 1 + B + ¯ B i ) + 1 2 ρ 2 ( 2 C + ¯ C i ) The two mixing rules are: B = y 2 1 B 11 + 2 y 1 y 2 B 12 + y 2 2 B 22 C = y 3 1 C 111 + 3 y 2 1 y 2 C 112 + 3 y 1 y 2 2 C 122 + y 3 2 C 222 Application of the definitions of ¯ B i and ¯ C i to these mixing rules yields: ¯ B 1 = y 1 ( 2 y 1 ) B 11 + 2 y 2 2 B 12 y 2 2 B 22 ¯ C 1 = y 2 1 ( 3 2 y 1 ) C 111 + 6 y 1 y 2 2 C 112 + 3 y 2 2 ( 1 2 y 1 ) C 122 2 y 3 2 C 222 ¯ B 2 =− y 2 1 B 11 + 2 y 2 1 B 12 + y 2 ( 2 y 2 ) B 22 ¯ C 2 2 y 3 1 C 111 + 3 y 2 1 ( 1 2 y 2 ) C 112 + 6 y 1 y 2 2 C 122 + 2 y 2 2 ( 3 2 y 2 ) C 222 606
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In combination with the mixing rules, these give: B + ¯ B 1 = 2 ( y 1 B 11 + y 2 B 12 ) 2 C + ¯ C 1 = 3 ( y 2 1 C 111 + 2 y 1 y 2 C 112 + y 2 2 C 122 ) B + ¯ B 2 = 2 ( y 2 B 22 + y 1 B 12 ) 2 C + ¯ C 2 = 3 ( y 2 2 C 222 + 2 y 1 y 2 C 122 + y 2 1 C 112 ) In combination with the boxed equation these expressions along with Eq. (3.39) allow calculation of ln ˆ φ 1 and ln ˆ φ 2 . 14.11 For the case described, Eqs. (14.1) and (14.2) combine to give: y i P = x i P sat i φ sat i ˆ φ i If the vapor phase is assumed an ideal solution , ˆ φ i = φ i , and y i P = x i P sat i φ sat i φ i When Eq. (3.37) is valid, the fugacity coefficient of pure species i is given by Eq. (11.35): ln φ i = B ii P RT and φ sat i = B P sat i Therefore, ln φ sat i φ i = ln φ sat i ln φ i = B P sat i B P = B ( P sat i P ) For small values of the final term, this becomes approximately: φ sat i φ i = 1 + B ( P sat i P ) Whence, y i P = x i P sat i · 1 + B ( P sat i P ) ¸ or y i P x i P sat i = x i P sat i B ( P sat i P ) Write this equation for species 1 and 2 of a binary mixture, and sum. This yields on the left the difference between the actual pressure and the pressure given by Raoult’slaw: P P ( RL ) = x 1 B 11 P sat 1 ( P sat 1 P ) + x 2 B 22 P sat 2 ( P sat 2 P ) Because deviations from Raoult’s law are presumably small, P on the right side may be replaced by its Raoult’s-law value. For the two terms, P sat 1 P = P sat 1 x 1 P sat 1 x 2 P sat 2 = P sat 1 ( 1 x 2 ) P sat 1 x 2 P sat 2 = x 2 ( P sat 1 P sat 2 ) P sat 2 P = P sat 2 x 1 P sat 1 x 2 P sat 2 =
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This note was uploaded on 01/29/2011 for the course CHEM 101 taught by Professor Brown during the Spring '10 term at The University of Akron.

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SM_14B - Chapter 14 - Section B - Non-Numerical Solutions...

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