Question

Under certain water conditions, the free chlorine(hypochlorous acid, hocl) in a swimming pool decomposes according to the law of uninhibited decay. After shocking a pool, the poolboy, geoff, tested the water and found the amount of free chlorine to be 2.7 parts per million (ppm). Twenty-four hours later, geoff tested the water again and found the amount of free chlorine to be 2.5 ppm. What will be the reading after 4 days (that is, 96hours)? When the chlorine level reaches 1.0 ppm, geoff must shock the pool again. How long can geoff go before he must shock the pool again? After 4 days, or 96 hours, the amount of free chlorine in the pool will be ___ ppm.

Answer 1

Answer:

The function $A(t)=A_{0} e^{kt}$ represents the amount A of a material present at a time t where k<0 represents the rate of decay.

substitute 2.5ppm from $A_{0}$ and 2.2ppm for A(t) and 24 hours or 1 day for t in the equation and find k

$2.2=2.5*e^{k}$

Divide both sides of the equation $2.2=2.5*e^{k}$

$e^{k} =0.88$

Take the natural logarithm of both sides of the equation.

ln(0.88)=ln$(e^{k} )$

ln0.88=k

-0.1278=k

so the value of k is -0.1278

substitute 2.5ppm for $A_{0}$ and t=3 and k=-0.1278

$A(t)=2.5.e^{-0.1272(3)}$

            =$2.5.e^{-0.384}$

            =2.5(0.6811)

            =1.7

so the amount of chlorine after 3 days will be 1.7ppm

$1.0=2.5.e^{-0.1278t}$

divide both sides of the equation by 2.5

$0.4=e^-0.1278t}$

Take the natural logarithm of both sides of the equation

ln0.4=$ln(e^{-0.1278t} )$

ln0.4=-0.1278t

divide both sides of the equation by -0.1278

t=7.17

so the pool must be shocked after 7.17days or 172hours