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High.
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Let's start the solution.
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In this question, we have given da upon dt is equal to minus ka.
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Now solve this first order differential equation by using variable separable.
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So we get 1 by a da is equal to minus k dt.
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Now by integrating both sides, then we get integration of 1 by a is ln mod a equal to minus k integration of dt is t plus integration constant c.
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Say this one is equation number 1.
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Now ln a naught is equal to minus k into 0 plus c because at time t equal to 0, the undissolved amount is a naught.
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So here we get c is equal to ln mod a naught.
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So put the value of c in equation 1, ln mod a is equal to minus kt plus ln mod a naught.
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So here we get a equal to e power minus kt plus ln mod a naught.
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That is a is equal to a naught e power minus kt by using log property.
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Say this one is number 2.
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Now at time t equal to 1 minute, the undissolved amount is a naught into 100 minus 20 percent.
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That is a naught into 100 minus 20 is 80 upon 100 is equal to a naught e power minus kt.
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So here we get this 00 is cancelled.
02:37
4 upon 5 a naught equal to a naught e power minus kt.
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So we get here 4 by 5 is equal to e power minus kt.
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Now by taking log both sides, so we get ln 4 by 5 is equal to minus k and time t is 1...