Answer to Question #138779 in Differential Equations for Kaii

Question

a substance in a chemical reaction is used up to a rate which is proportional to the amount of the substance present at any time. if 80% of the substance is used up in 4 hours, how much if the substance is left at the end of 120 minutes

Answer 1

Let $A$ be the amount of the substance present at time $t$ .

The rate at which the substance is used up is $-\frac{dA}{dt}$ . The negative sign is due to the fact that the amount of substance is decreasing with time.

According to the problem, the rate $-\frac{dA}{dt}$ is proportional to $A$ .

$\therefore -\frac{dA}{dt}\propto A\\ \Rightarrow -\frac{dA}{dt}=kA$

where $k$ is the proportionality constant.

$\Rightarrow \frac{dA}{A}=-kdt$

Let the initial amount of the substance is $A_0$ .

Integrating both sides,

$\int_{A_0}^A\frac{dA}{A}=-k\int_0^tdt\\ \Rightarrow\left[ \ln A\right]_{A_0}^A=-k[t]_0^t\\ \Rightarrow \ln A-\ln A_0=-kt\\ \Rightarrow \ln \frac{A}{A_0}=-kt\\ \Rightarrow \frac{A}{A_0}=e^{-kt}\qquad ...............(1)$

At $t=4\ hours$ , 80% of the substance is used up and hence 20% of the substance is left.

Therefore, at $t=4\ hours$ , $A=20\%\ of\ A_0=0.20A_0$

Substituting these values in Eq.(1),

$\frac{0.20A_0}{A_0}=e^{-4k}\\ \Rightarrow \ln (0.20)=-4k\\ \Rightarrow k=-\frac{1}{4}\ln (0.20)=0.402\ hour^{-1}$

At the end of 120 minutes, i.e, at $t=2\ hours$ , from Eq.(1),

$\frac{A}{A_0}=e^{-0.402\times 2}\\ \Rightarrow \frac{A}{A_0}=0.478\\ \Rightarrow A=0.478A_0\\ A=47.8\%\ of A_0$

Answer: At the end of 120 minutes 47.8% of the substance is left.