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Question

The bacterial growth follows the rate law, dNdt=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?

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Solution

Bacterial growth rate is given as:
dNdt=kN which is a first order reaction
dNN=kdt
Integrating both sides, we get
lnNNo=kt where No is the initial population
kt=ln NNo

After t=5 min, N=2No
k(5)=ln 2 ..........(1)
After time =t min, N=8No
k(8)=ln 8 ..........(2)

Dividing (2) and (1), we get t5=ln8ln2, where ln8=3ln2
t5=3ln2ln2 t=15 min

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