Thermodynamics filled in class notes_Part_76

# Thermodynamics filled in class notes_Part_76 - 5.2...

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Unformatted text preview: 5.2. STOICHIOMETRY 159 Dividing the first row by 3, the second by 2 and the third by − 8 / 3 gives 1 1 3 1 3 1 1 1 2 1 2 1 2 1 1 − 3 4 − 3 8 · ν 1 ν 2 ν 3 ν 4 ν 5 ν 6 = . (5.85) Take bound variables to be ν 1 , ν 2 , and ν 3 and free variables to be ν 4 , ν 5 , and ν 6 . So set ν 4 = ξ 1 , ν 5 = ξ 2 , and ν 6 = ξ 3 , and get 1 0 1 3 0 1 1 0 0 1 · ν 1 ν 2 ν 3 = − ξ 1 3 − ξ 1 2 − ξ 2 2 − ξ 3 2 − ξ 1 + 3 4 ξ 2 + 3 8 ξ 3 . (5.86) Solving, one finds ν 1 ν 2 ν 3 = 1 8 ( − 2 ξ 2 − ξ 3 ) 1 8 (4 ξ 1 − 10 ξ 2 − 7 ξ 3 ) 1 8 ( − 8 ξ 1 + 6 ξ 2 + 3 ξ 3 ) . (5.87) For all the coefficients, one then has ν 1 ν 2 ν 3 ν 4 ν 5 ν 6 = 1 8 ( − 2 ξ 2 − ξ 3 ) 1 8 (4 ξ 1 − 10 ξ 2 − 7 ξ 3 ) 1 8 ( − 8 ξ 1 + 6 ξ 2 + 3 ξ 3 ) ξ 1 ξ 2 ξ 3 = ξ 1 8 4 − 8 8...
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## This note was uploaded on 11/26/2011 for the course EGN 3381 taught by Professor Park-sou during the Fall '11 term at FSU.

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Thermodynamics filled in class notes_Part_76 - 5.2...

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