Question

The bacterial growth follows the rate law, $\frac{dN}{dt}=kN$ where $k$ is a constant and '$N$' is the number of bacteria cell at any time. If the population of bacteria (number of cell) is doubled in $5$ min, find the time in which the population will be eight times of the initial one?

Answer 1

Bacterial growth rate is given as:

$\frac{dN}{dt}=kN$ which is a first order reaction

$\Rightarrow \int \frac{dN}{N}=k\int dt$

Integrating both sides, we get

$\Rightarrow ln\frac{N}{N_o}=kt$ where $N_o$ is the initial population

$\Rightarrow kt=ln$ $\frac{N}{N_o}$

After $t=5$ min, $N=2N_o$

$\therefore$ $k\left(5\right)=ln$ $2$ ..........(1)

After time $=t$ min, $N=8N_o$

$\therefore$ $k\left(8\right)=ln$ $8$ ..........(2)

Dividing (2) and (1), we get $\frac{t}{5}=\frac{ln8}{ln2}$, where $ln8=3ln2$

$\frac{t}{5}=\frac{3ln2}{ln2}$ $⟹t=15$ min