Step 1 a) Use stirred tank model with constants \(\displaystyle{V}={4},{R}={0.06},{C}_{{{i}}}={500},{m}_{{{0}}}={0}\)

\(\displaystyle{m}'{\left({t}\right)}=-{\frac{{{0.06}}}{{{4}}}}{m}{\left({t}\right)}+{500}\cdot{0.06}\)

\(\displaystyle{m}{\left({0}\right)}={0}\)

Step 2

b) Solution is given with \(\displaystyle{m}{\left({t}\right)}={\left({m}_{{{0}}}-{C}_{{{i}}}{V}\right)}{e}-{\frac{{{R}}}{{{V}}}}{t}+{C}_{{{i}}}{V}\) Solution is \(\displaystyle{m}{\left({t}\right)}={\left({0}-{500}\cdot{4}\right)}{e}-{\frac{{{0.06}}}{{{4}}}}{t}+{500}\cdot{4}\)

\(\displaystyle{m}{\left({t}\right)}=-{2000}{e}^{{-{0.015}{t}}}+{2000}\) Graph of mass of the drug m(t) in mg over time 19610300291.jpg

Step 3

Graph of concentration of the drug \(\displaystyle{\frac{{{m}{\left({t}\right)}}}{{{V}}}}\) in mg/L over time 19610300292.jpg

Step 4

c) From the first graph it is visible that steady state mass of the drug is 2000 mg.

Step 5 c) 90% of steady state level drug mass is 1800. Solve \(\displaystyle{m}{\left({t}\right)}={1800}\) to obtain t \(\displaystyle{2000}{e}^{{-{0.015}{t}}}+{2000}={1800}\)

\(\displaystyle{e}^{{-{0.015}{t}}}={0.1}\)

\(\displaystyle-{0.015}{t}={\ln{{\left({0.1}\right)}}}t \approx 153.5\)