Question

While doing an experiment in chemistry lab, Komal found that a substance loses its moisture at a rate proportional to the moisture content. If the same substance loses half of its moisture during the first hour, when will it lose 99% ?

• A. $\frac{ln100}{ln2}$
• B. In 50
• C. $\frac{ln50}{ln2}$
• D. In 25

Answer 1

The correct option is A $\frac{ln100}{ln2}$

We want to help Komal in finding when the substance will lose 99% moisture.
We are given that the rate at which the moisture is escaping/evaporates is proportional to the moisture present in the substance.
Let’s represent amount of moisture by m.
Rate at which the moisture is escaping/evaporates is proportional to the moisture
$\Rightarrow \frac{dm}{dt}=km$, where k is a constant
We will solve this differential equation to get a relation between m and time t.
$\Rightarrow \frac{dm}{m}=kdt$
$\Rightarrow lnm=kt+c$
Let us assume that the amount of moisture at t = 0 is M.
$\Rightarrow lnM=0+c$
Or c = ln M
$\Rightarrow lnm=kt+lnM$
$\Rightarrow ln\frac{m}{M}=kt-----------------\left(1\right)$
We are given one more information that the substance loses half of its moisture in one hour. We will use this to find the value of k.
$t=1hour\Rightarrow m=\frac{M}{2}$
$\Rightarrow ln\left(\frac{\frac{M}{2}}{M}\right)=k\times 1$
$\Rightarrow k=-In2$
We want to find the time at which 99% of moisture is lost.
If 99% of moisture is lost, then 1% of moisture will be remaining
$\Rightarrow m=\frac{M}{100}$
Substituting in (1) , we get
$ln\left(\frac{\frac{M}{100}}{M}\right)=kt=(-ln2)t$
$\Rightarrow -ln100=(-ln2)t$
$\Rightarrow t=\frac{ln100}{ln2}$