Consider a tank used in certain hydrodynamic experiments. After one experiment the tankcontains 200 L of a dye solution with a concentration of 1 g/L. To prepare for the next experi-ment, the tank is to be rinsed with fresh water flowing in at a rate of 2 L/min, the well-stirredsolution flowing out at the same rate. Find the time that will elapse before the concentrationof dye in the tank reaches 1% of its original value.

[tex]\text{Let A = mass of dye after t min}\\\text{and }r_{i} = \text{rate of dye coming into tank}\\\text{and }r_{o} =\text{rate of dye going out of tank}[/tex]

1. Set up an expression for the rate of change of dye concentration.

[tex]\dfrac{\text{d}A}{\text{d}t} = r_{i} - r_{o}\\\\\text{The fresh water is entering with with no dye, so}\\ r_{i} = 0\\r_{o} = \dfrac{\text{2 L}}{\text{1 min}} \times \dfrac {A\text{ g}}{\text{200 L}} =\dfrac{A}{100}\text{ g/min}\\\\\dfrac{\text{d}A}{\text{d}t} = -\dfrac{A}{100}[/tex]

2. Integrate the expression

[tex]\dfrac{\text{d}A}{\text{d}t} = \dfrac{-A}{100}\\\\\dfrac{\text{d}A}{-A} = \dfrac{\text{d}t}{100}\\\\-\int \frac{\text{d}A}{A} = \int \frac{\text{d}t}{100}\\\\\int \frac{\text{d}A}{A} = -\int \frac{\text{d}t}{100}\\\\\ln A = -\dfrac{t}{100} + C[/tex]

3. Find the constant of integration

[tex]\ln A = -\dfrac{t}{100} + C\\\\\text{At t = 0, A = 1 g/L}\\\\-\ln 1 = -\dfrac{0}{100} + C\\\\C = 0[/tex]

4. Solve for A as a function of time.

[tex]\text{The integrated rate expression is}\\\\\ln A = -\dfrac{t}{100}\\\\\text{Solve for }A}\\A = e^{-t / 100}[/tex]

5. Calculate time to reach 1 % of original concentration

The initial concentration was 1 g/L, so the new concentration is 0.01 g/L

[tex]0.01 = e^{-t/100}\\\\\ln(0.01) = -\dfrac{t}{100} \\\\-4.605 = -\dfrac{t}{100}\\\\t = 100\times 4.60 =\textbf{460 min = 7 h 40 min}\\\text{The concentration of dye in the tank will be reduced}\\\text{to 1 \% after } \boxed{\textbf{7 h 40 min}}[/tex]

The diagram shows the concentration starting at 1 g//mL and decreasing asymptotically to 0.01 g/mL after 460 min.