Chapter4_A - Chapter 4 Section A Mathcad Solutions 4.1(a...

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Ans. 4.2 (a) T 0 473.15 K := n 10 mol := Q 800 kJ := For ethylene: A 1.424 := B 14.394 10 3 K := C 4.392 10 6 K 2 := τ 2 := (guess) Given Q n R A T 0 τ 1 ( ) B 2 T 0 2 τ 2 1 ( ) + C 3 T 0 3 τ 3 1 ( ) + = τ Find τ ( ) := τ 2.905 = T τ T 0 := T 1374.5K = Ans. (b) T 0 533.15 K := n 15 mol := Q 2500 kJ := For 1-butene: A 1.967 := B 31.630 10 3 K := C 9.873 10 6 K 2 := Chapter 4 - Section A - Mathcad Solutions 4.1 (a) T 0 473.15 K := T 1373.15 K := n 10 mol := For SO2: A 5.699 := B 0.801 10 3 := C 0.0 := D 1.015 10 5 := H R ICPH T 0 T , A , B , C , D , ( ) := H 47.007 kJ mol = Q n H := Q 470.073kJ = Ans. (b) T 0 523.15 K := T 1473.15 K := n 12 mol := For propane: A 1.213 := B 28.785 10 3 := C 8.824 10 6 := D 0 := H R ICPH T 0 T , A , B , C , 0.0 , ( ) := H 161.834 kJ mol = Q n H := Q 1.942 10 3 × kJ = 76
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