Question

The growth of Pseudomonas bacteria is modelled as a first-order process with k = 0.035 $\mathit{m}\mathit{i}{\mathit{n}}^{\mathbf{-}\mathbf{1}}$ at 37°C. The initial Pseudomonas population density is $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}$cells/L.

(a) What is the population density after two h?

(b) What is the time required for the population to go from $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathit{t}\mathit{o}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}$cells/ L?

Short answer

(a) Populated density after 2hrs is $7.0\times {10}^{4}cells/L$

(b) The time required is 20min

Step-by-step solution

 Calculation of Population density after 2hr

As the reaction follows first-order kinetics. The rate of the reaction is:

$ln{\left[A\right]}_t=Kt+ln{\left[A\right]}_0$

Given,

K=0.035$mi{n}^{-1}$

The initial concentration or initial density ${\left[A\right]}_0=1.0\times {10}^{-3}cells/L$

Time =2hr

In minutes it becomes 120 min

So,

$$\begin{gathered} ln{\left[A\right]}_t=Kt+ln{\left[A\right]}_0 \\ ln\frac{{\left[A\right]}_t}{{\left[A\right]}_0}=Kt \\ ln\frac{{\left[A\right]}_t}{1.0\times {10}^3}=0.035\left(120\right) \\ {\left[A\right]}_t=7.0\times {10}^{4}cells/L \end{gathered}$$

Hence, the populated density after 2hrs is$7.0\times {10}^{4}cells/L$

Calculation of time required for population

As the reaction follows first-order kinetics. The rate of the reaction is:

$ln{\left[A\right]}_t=Kt+ln{\left[A\right]}_0$

Given,

K=0.035$mi{n}^{-1}$

The initial concentration or initial density ${\left[A\right]}_0=1.0\times {10}^{-3}cells/L$

Final concentration or density ${\left[A\right]}_t=2\times {10}^{3}cells/L$

So,

$$\begin{gathered} ln{\left[A\right]}_t=Kt+ln{\left[A\right]}_0 \\ ln\frac{{\left[A\right]}_t}{{\left[A\right]}_0}=Kt \\ ln\frac{2.0\times {10}^3}{1.0\times {10}^3}=0.035\left(t\right)l \\ ln2.0=0.032t \\ t=20min \end{gathered}$$

Hence, the time required is 20min