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Thermodynamics filled in class notes_Part_80

# Thermodynamics filled in class notes_Part_80 - 167 5.3...

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5.3. FIRST LAW ANALYSIS OF REACTING SYSTEMS 167 The first law gives U 2 U 1 = 1 Q 2 1 W 2 , (5.162) U 2 U 1 = 0 , (5.163) n H 2 O u H 2 O n H 2 u H 2 n O 2 u O 2 = 0 , (5.164) n H 2 O ( h H 2 O RT 2 ) n H 2 ( h H 2 RT 1 ) n O 2 ( h O 2 RT 1 ) = 0 , (5.165) 2 h H 2 O 2 h H 2 bracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright =0 h O 2 bracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright =0 + R ( 2 T 2 + 3 T 1 ) = 0 , (5.166) 2 h H 2 O + (8 . 314) (( 2) T 2 + (3) (298)) = 0 , (5.167) h H 2 O 8 . 314 T 2 + 3716 . 4 = 0 , (5.168) h o f,H 2 O + Δ h H 2 O 8 . 314 T 2 + 3716 . 4 = 0 , (5.169) 241826 + Δ h H 2 O 8 . 314 T 2 + 3716 . 4 = 0 , (5.170) 238110 + Δ h H 2 O 8 . 314 T 2 = 0 . (5.171) At this point, one begins an iteration process, guessing a value of T 2 and an associated Δ h H 2 O . When T 2 is guessed at 5600 K , the left side becomes 6507 . 04. When T 2 is guessed at 6000 K , the left side becomes 14301 . 4. Interpolate then to arrive at T 2 = 5725 K. (5.172) This is an extremely high temperature. At such temperatures, in fact, one can expect other species to co-exist in the equilibrium state in large quantities. These may include

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Thermodynamics filled in class notes_Part_80 - 167 5.3...

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