A chemical reaction turns substance A into substance B. The rate at which the reaction takes place is proportional to the square root of the amount of substance A present.

If ????a is the amount of substance A present (measured in grams) at any time ????t (measured in minutes), set up a differential equation for ????a in terms of ????.t. If there is an arbitrary constant in your equation, call it ????k and make sure that ????>0k>0.

Now, suppose that the chemical reaction is to take place in a large tank of liquid!
The tank contains 25 liters of a solution of substance A dissolved in water. The concentration of the solution is 750 grams per liter. A 200 gram per liter solution of substance A is poured in at 12 liters per minute. At the same time, the solution is drained from the bottom of the tank at 11 liters per minute.

If ????a is the amount of substance A in the tank (measured in grams) at any time ????t (measured in minutes), set up a differential equation for ????a in terms of ????:

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nandhini123 nandhini123 · Answer 1 · 2019-12-28T05:34:33+01:00

Answer:

differential equation fora in terms of t is [tex]\frac{da}{dt} = 12 (0.2) - \frac{11}{v(t)a}[/tex]

Explanation:

Given :

Rate at which the reaction takes place is proportional to the square root of the amount of substance A present.

the amount of substance A present (measured in grams) at any time t = a

orbitary constant = k

Solution

The rate of change of the reaction is

[tex]\frac{d a}{d t} \propto \sqrt{a}[/tex]

[tex]\frac{d a}{d t}= -k \sqrt{a}[/tex]

Thus , if amount present is low then [tex]\frac{da}{dt }[/tex] will be high

Now

Volume = v(t) = 25-(12-11)t

Volume v(t) = 25 –t

[tex]\frac{da}{dt}[/tex] = 12 (200 grams) - 11/(v(t))a

converting 200 gram to kg we get

[tex]\frac{da}{dt}[/tex] = 12 (0.2) - 11/(v(t))a

[tex]\frac{da}{dt} = 12 (0.2) - \frac{11}{v(t)a}[/tex]