Explanation Calculation: Let x ( t ) denote the number of grams of the compound C present at time t . The given conditions are x ( 0 ) = 0 and x ( 5 ) = 10 gm . For x gm of C 2 x 3 gm of A and 1 x 3 gm of B are used. Thus, amount of A and B after a time t is 40 − 2 x 3 and 50 − 1 x 3 respectively. Calculate the rate at which compound C is formed. d x d t ∝ ( 40 − 2 x 3 ) ( 50 − 1 x 3 ) d x d t ∝ 2 9 ( 60 − x ) ( 150 − x ) d x d t = 2 9 k ( 60 − x ) ( 150 − x ) Separate the variables of the above equation. d x ( 60 − x ) ( 150 − x ) = 2 9 k d t 1 90 [ 1 60 − x − 1 ( 150 − x ) ] d x = 2 9 k d t Integrate the above equation. ∫ 1 90 [ 1 60 − x − 1 ( 150 − x ) ] d x = 2 9 k d t − 1 90 ln | 60 − x | + 1 90 ln | 150 − x | = 2 9 k t + c − ln | 60 − x | + ln | 150 − x | = 20 k t + 90 c 150 − x 60 − x = c 1 e 20 k t (1) Substitute x ( 0 ) = 0 in Equation (1). 150 − 0 60 − 0 = c 1 e 20 ( 0 ) c 1 = 5 2 Substitute 5 2 for c 1 in Equation (1)

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 2.8, Problem 11E

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The limiting amount of C after a long time and the amount formed in 20 minutes.

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Chapter 2.8, Problem 10E

Chapter 2.8, Problem 12E

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Chapter 2 Solutions

Advanced Engineering Mathematics

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The limiting amount of C after a long time and the amount formed in 20 minutes.

Solution Summary: The author calculates the limiting value of C after a long time and the amount formed in 20 minutes.

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The limiting amount of C after a long time and the amount formed in 20 minutes.

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Chapter 2 Solutions