SM_4A - Chapter 4 - Section A - Mathcad Solutions 4.1 (a)...

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Q 1.942 10 3 × kJ = 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 - ( 29 B 2 T 0 2 τ 2 1 - ( 29 + C 3 T 0 3 τ 3 1 - ( 29 + = τ Find τ ( 29 = 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 = ICPH 473.15 1373.15 , 5.699 , 0.801 10 3 - , 0.0 , 1.015 - 10 5 , ( 29 5.654 10 3 = ICPH 5.654 10 3 K = H R ICPH = Q n H = Q 470.074kJ = 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 - = ICPH 523.15 1473.15 , 1.213 , 28.785 10 3 - , 8.824 - 10 6 - , 0.0 , ( 29 1.947 10 4 = ICPH 1.947 10 4 K = H R ICPH = Q n H = 61
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τ 2.256 = T τ T 0 = T 1202.8K = Ans. T 1705.4degF = 4.3 Assume air at the given conditions an ideal gas. Basis of calculation is 1 second. P 1 atm = T 0 122 degF = V 250 ft 3 = T 932 K = Convert given values to SI units V 7.079m 3 = T T 32degF - ( ) 273.15K + = T 0 T 0 32degF - ( 29 273.15K + = T 1187.37K = T 0 323.15K = n P V R T 0 = n 266.985mol = For air: A 3.355 = B 0.575 10 3 - = C 0.0 = D 0.016 - 10 5 = ICPH 323.15 773.15 , 3.355 , 0.575 10 3 - , 0.0 , 0.016 - 10 5 , ( 29 1648.702 = τ 3 = (guess) Given Q n R A T 0 τ 1 - ( 29 B 2 T 0 2 τ 2 1 - ( 29 + C 3 T 0 3 τ 3 1 - ( 29 + = τ Find τ ( 29 = τ 2.652 = T τ T 0 = T 1413.8K = Ans. (c) T 0 500 degF = n 40 lbmol = Q 10 6 BTU = Values converted to SI units T 0 533.15K = n 1.814 10 4 × mol = Q 1.055 10 6 × 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 - ( 29 B 2 T 0 2 τ 2 1 - ( 29 + C 3 T 0 3 τ 3 1 - ( 29 + = τ Find τ ( 29 = 62
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4.7 Let step 12 represent the initial reversible adiabatic expansion, and step 23 the final constant-volume heating. T 1 298.15 K = T 3 298.15 K = P 1 121.3 kPa = P 2 101.3 kPa = P 3 104.0 kPa = T 2 T 3 P 2 P 3 = T 2 290.41K = C P 30 joule mol K = (guess) Given T 2 T 1 P 2 P 1 R C P = C P Find C P ( 29 = C P 56.95 joule mol K = Ans. 4.9 M 72.150 86.177 78.114 92.141 84.161 gm mol = T c 469.7 507.6 562.2 591.8 553.6 K = P c 33.70 30.25 48.98 41.06 40.73 bar = T n 309.2 341.9 353.2 383.8 353.9 K
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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_4A - Chapter 4 - Section A - Mathcad Solutions 4.1 (a)...

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