Answer
See answers below
Work Step by Step
a) The concentration at any time is:
$$c=\frac{A}{w}$$
To find the amount of the chemical in the tank, we obtain:
$$\frac{dA}{dt}=kr-cr$$
$$\frac{dA}{dt}+cr=kr$$
$$\frac{dA}{dt}+\frac{r}{w}A=kr$$
Find the solution $A$:
$$A=e^{-\int \frac{r}{w}}dt(c_1+\int kre^{\int \frac{r}{w}dt}dt)$$
where $c$ is a constant of integration.
$$A=e^{-\frac{r}{w}}dt(c_1+kwe^{\frac{rt}{w}})$$
Since $A(0)=0 $
Find $c$:
$A(0)=c_1+kw \rightarrow c_1=A_0-kw$
The amount of chemical in the tank at time $t$ is:
$$A=e^{-\frac{rt}{w}}(A-kw+kwe^{\frac{rt}{w}})$$
b) When $t \rightarrow \infty$:
$$\lim A=\lim (A-kw)e^{\frac{-rt}{w}}+kw=kw$$
$$\rightarrow \frac{kw}{w}=k$$
We can say the concentration of chemical in the tank approaches k g/L
