Solution :
Given data :
[tex]c_{in}[/tex] = 1 g/L
[tex]r_{in}[/tex] = 5 L/min
[tex]r_{out}[/tex] = 5 L/min
[tex]$v_0$[/tex] = 250 L
[tex]A_0[/tex] = 20 g
∴ [tex]r_{net} = r_{in}- r_{out}[/tex]
= 5 - 5
= 0
[tex]c_{out} = \frac{A}{250} \ g/L[/tex]
Now, [tex]\frac{dA}{dt}=(r_{in} \times c_{in}) - (r_{out} \times c_{out})[/tex]
[tex]$\frac{dA}{dt} = 5-5\left(\frac{A}{250}\right)$[/tex]
[tex]\frac{dA}{dt}+5 \left(\frac{A}{250}\right) = 5[/tex]
[tex]\frac{dA}{dt}+5 \left(\frac{A}{250}\right) = 5 \text{ with} \ A_0 = 20[/tex]
Integrating factor = exp(5 t/250)
Therefore,
[tex]A \times \exp (5t \ /250) = \text{integral of}\ 5 \times \exp (5t / 250) + C[/tex]
Put [tex]A_0=250+C[/tex]
C = -230
[tex]A \times \exp(5t/250) = 250 \exp(5t/250) + (-230)[/tex]
[tex]A(t) = 250-230 \exp(-5t/250)[/tex]
[tex]A(t) = 250-230e^{\left(\frac{-t}{50}\right)} \ g[/tex]
