Chapter 7, Problem 13RE

a.

To determine

To find : the concentration at time $t$

Answer to Problem 13RE

The concentration at time $t$ is $C=C_0\cdot e^{-kt}$

Explanation of Solution

Given information : The question is

If $C_0$ is the concentration at time $t=0$

Calculation :

  $\begin{array}{l}C\text{'}\left(t\right)=-kC\\ \frac{dC}{dt}=-kC\\ \frac{dC}{C}=-kdt\\ \ln C=-kt+C_1\end{array}$

Substitute $t=0$ and $C=C_0$

  $\ln C_0=0+C_1$

Substitute back $C_1=\ln C_0$

  $\begin{array}{l}\ln C=-kt+\ln C_0\\ \ln C-\ln C_0=-kt\\ \ln \frac{C}{C_0}=-kt\\ \frac{C}{C_0}=e^{-kt}\\ C=C_0\cdot e^{-kt}\end{array}$

b.

To determine

To calculate : how long does it take to eliminate $90\%$ of the drug

Answer to Problem 13RE

It takes $99.65hours$ to eliminate $90\%$ of the drug

Explanation of Solution

Given information : The question is

If the body eliminates half the drug in $30$ hours

Calculation :

In $t$ hours it takes $C\left(t\right)=C\left(0\right)e^{-kt}$

  $0.5=e^{-30k}$

Take $\ln$ on both sides

  $\begin{array}{l}-30k=\ln 0.5\\ k=2.31\cdot {10}^{-2}\end{array}$

Let the time taken to eliminate $90\%$ of the drug be $T$ hours

Then,

  $\begin{array}{l}0.1=e^{-kt}\\ -kT=\ln 0.1\\ T=\frac{\ln 0.1}{\ln 0.5}=99.65hours\end{array}$