A tank initially holds 100 gallons of a brine solution containing 1 lb. of salt. At t = 0 another brine solution containing 1 pound of salt per gallon is poured into the tank at the rate of 4 gal/min., while the well-stirred mixture leaves the tank at the same rate. Find (a) the amount of salt in the tank at any time t and (b) the time at which the mixture in the tank contains 2 lb of salt.

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

i. Set up an expression for the rate of change of salt concentration.

[tex]\dfrac{\text{d}A}{\text{d}t} = r_{i} - r_{o}\\\\r_{i} = \dfrac{\text{4 gal}}{\text{1 min}} \times \dfrac{\text{1 lb}}{\text{1 gal}} = \text{4 lb/min}\\\\r_{o} = \dfrac{\text{4 gal}}{\text{1 min}} \times \dfrac {A\text{ lb}}{\text{100 gal}} =\dfrac{A}{25}\text{ lb/min}\\\\\dfrac{\text{d}A}{\text{d}t} = 4 - \dfrac{x}{25}[/tex]

ii. Integrate the expression

[tex]\dfrac{\text{d}A}{\text{d}t} = \dfrac{100 - A}{25}\\\\\dfrac{\text{d}A}{100 - A} = \dfrac{\text{d}t}{25}\\\\\int \frac{\text{d}A}{100 - A} = \int \frac{\text{d}t}{25}\\\\-\ln(100 - A) = \dfrac{t}{25} + C[/tex]

iii. Find the constant of integration

[tex]-\ln (100 - A) = \dfrac{t}{25} + C\\\\\text{At $t$ = 0, $A$ = 1, so}\\\\-\ln (100 - 1) = \dfrac{0}{25} + C\\\\C = -\ln 99[/tex]

iv. Solve for A as a function of time.

[tex]\text{The integrated rate expression is}\\\\-\ln (100-A) = \dfrac{t}{25} - \ln 99\\\\\text{Solve for } A\\\\\ln(100 - A) = \ln 99 - \dfrac{t}{25}\\\\100 - A = 99e^{-t/25}\\\\A = \boxed{\mathbf{100 - 99e^{-t/25}}}[/tex]

The diagram shows A as a function of time. The mass of salt in the tank starts at 1 lb and increases asymptotically to 100 lb.

(b) Time to 2 lb salt

[tex]A = 100 - 99e^{-t/25}\\\\2 = 100 - 99e^{-t/25}\\\\99e^{-t/25} = 98\\\\e^{-t/25} = 0.989899\\\\-t/25 = -0.01015\\\\t = 25\times 0.01015 =\text{0.25 min = 15 s}\\\\\text{The tank will contain 2 lb of salt after } \boxed{\textbf{15 s}}[/tex]