Answer
$c_2=0.5625$g/L
Work Step by Step
To find the amount of the chemical in the tank, we obtain:
$$\frac{dA}{dt}=c_1r_1-c_2r_2$$
We are given:
$c_1=0.5$
$r_1=6$
$r_2=4$
$A=20$
and $V(t)=(r_1-r_2)t+V_0=100+2t$
Since $V=200 \rightarrow 200=100+2t \rightarrow t=50$
Putting the values:
$$\frac{dA}{dt}=6\times0.5-4\frac{A}{V}$$
$$\frac{dA}{dt}+\frac{2A}{50+t}=3$$
Solve for $A$:
$$A=e^{-\int \frac{2}{50+t}}dt(c+\int 3e^{\int \frac{2}{50+t }dt}dt)$$
where $c$ is a constant of integration.
$$A=e^{-2\ln (50+t)}dt(c+\int 3e^{\int 2\ln (50+t)}dt)$$
$$A=\frac{1}{(50+t)^2}(c+\int3(50+t)^2dt)$$
$$A=\frac{c}{(50+t)^2}+(50+t)$$
Since $A(0)=100 $
Find $c$:
$100=\frac{c}{50^2} +50\rightarrow c=50^3$
When tank overflows
$$A=\frac{50^3}{(50+t)^2}+(50+t)=\frac{50^3}{100}+100=112.5$$g
When tank is full, the concentration is:
$$c_2=\frac{A}{V}=\frac{112.5}{200}=0.5625$$g/L
