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 pool​ boy, Geoff, tested the water and found the amount of free chlorine to be 2.4 parts per million​ (ppm). ​ Twenty-four hours​ later, Geoff tested the water again and found the amount of free chlorine to be 2.1 ppm. What will be the reading after 2 days​ (that is, 48 ​hours)? 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?

Answer 1

From Uninhibted decay law , expression for decomposition of chlorine in swimming pool is

Where is intitial amount of Chlorine , y is amount of chlorine after t days.

At t = 0 , y = 2.4 ppm

After 24 hours means after one day y = 2.1

So we can write at t = 1 , y = 2.1 ppm

As we write

At t = 0 , y = 2.4 ppm

So we can write expression as:

At t = 1 , y = 2.1

So

So

After t = 2 days

At t = ? , y = 1 ppm

So after 6.5 days approx chlorine level reaches 1 ppm.