Chapter11_B

# Chapter11_B - Chapter 11 - Section B - Non-Numerical...

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Chapter 11 - Section B - Non-Numerical Solutions 11.6 Apply Eq. (11.7): ¯ T i · ∂( nT ) n i ¸ P , T , n j = T µ n n i T , P , n j = T ¯ P i · nP ) n i ¸ P , T , n j = P µ n n i T , P , n j = P 11.7 ( a )L e t m be the mass of the solution, and define the partial molar mass by: ¯ m i µ m n i T , P , n j Let M k be the molar mass of species k . Then m = 6 k n k M k = n i M i + 6 j n j M j ( j 6= i ) and µ m n i T , P , n j = · n i M i ) n i ¸ T , P , n j = M i Whence, ¯ m i = M i ( b ) Define a partial specific property as: ˜ M i µ M t m i T , P , m j = µ M t n i T , P , m j µ n i m i T , P , m j If M i is the molar mass of species i , n i = m i M i and µ n i m i T , P , m j = 1 M i Because constant m j implies constant n j , the initial equation may be written: ˜ M i = ¯ M i M i 11.8 By Eqs. (10.15) and (10.16), ¯ V 1 = V + x 2 dV dx 1 and ¯ V 2 = V x 1 1 Because V = ρ 1 then 1 = 1 ρ 2 d ρ 1 whence ¯ V 1 = 1 ρ x 2 ρ 2 d ρ 1 = 1 ρ µ 1 x 2 ρ d ρ 1 = 1 ρ 2 µ ρ x 2 d ρ 1 ¯ V 2 = 1 ρ + x 1 ρ 2 d ρ 1 = 1 ρ µ 1 + x 1 ρ d ρ 1 = 1 ρ 2 µ ρ + x 1 d ρ 1 With ρ = a 0 + a 1 x 1 + a 2 x 2 1 and d ρ 1 = a 1 + 2 a 2 x 1 these become: ¯ V 1 = 1 ρ 2 [ a 0 a 1 + 2 ( a 1 a 2 ) x 1 + 3 a 2 x 2 1 ] and ¯ V 2 = 1 ρ 2 ( a 0 + 2 a 1 x 1 + 3 a 2 x 2 1 ) 679

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11.9 For application of Eq. (11.7) all mole fractions must be eliminated from the given equation by the relation x i = n i / n : nM = n 1 M 1 + n 2 M 2 + n 3 M 3 + n 1 n 2 n 3 n 2 C For ¯ M 1 , · ∂( ) n 1 ¸ T , P , n 2 , n 3 = M 1 + n 2 n 3 C " 1 n 2 2 n 1 n 3 µ n n 1 T , P , n 2 , n 3 # Because n = n 1 + n 2 + n 3 , µ n n 1 T , P , n 2 , n 3 = 1 Whence, ¯ M 1 = M 1 + n 2 n 3 n 2 h 1 2 n 1 n i C and ¯ M 1 = M 1 + x 2 x 3 [1 2 x 1 ] C Similarly, ¯ M 2 = M 2 + x 1 x 3 [1 2 x 2 ] C and ¯ M 3 = M 3 + x 1 x 2 [1 2 x 3 ] C One can readily show that application of Eq. (11.11) regenerates the original equation for M .Th e infinite dilution values are given by: ¯ M i = M i + x j x k C ( j , k 6= i ) Here x j and x k are mole fractions on an i -free basis. 11.10 With the given equation and the Dalton’s-law requirement that P = i p i , then: P = RT V 6 i y i Z i For the mixture, P = ZRT / V . These two equations combine to give Z = i y i Z i . 11.11 The general principle is simple enough: Given equations that represent partial properties ¯ M i , ¯ M R i ,or ¯ M E i as functions of com- position, one may combine them by the summability relation to yield a mixture property. Application of the defining (or equivalent) equations for partial properties then regenerates the given equations if and only if the given equations obey the Gibbs/Duhen equation. 11.12 ( a ) Multiply Eq. ( A ) of Ex. 11.4 by n ( = n 1 + n 2 ) and eliminate x 1 by x 1 = n 1 /( n 1 + n 2 ) : nH = 600 ( n 1 + n 2 ) 180 n 1 20 n 3 1 ( n 1 + n 2 ) 2 Form the partial derivative of with respect to n 1 at constant n 2 : ¯ H 1 = 600 180 20 · 3 n 2 1 ( n 1 + n 2 ) 2 2 n 3 1 ( n 1 + n 2 ) 3 ¸ = 420 60 n 2 1 ( n 1 + n 2 ) 2 + 40 n 3 1 ( n 1 + n 2 ) 3 Whence, ¯ H 1 = 420 60 x 2 1 + 40 x 3 1 Form the partial derivative of with respect to n 2 at constant n 1 : ¯ H 2 = 600 + 20 2 n 3 1 ( n 1 + n 2 ) 3 or ¯ H 2 = 600 + 40 x 3 1 680
( b ) In accord with Eq. (11.11),

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## Chapter11_B - Chapter 11 - Section B - Non-Numerical...

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