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Section IBasic principles(a) (b)(d)(c)Figure 6.10.Rateconstant for elimination and clearance.(a) This represents a unit of
volume (e.g. one litre); if the rate constant for elimination (k) is 0.4 min−^1 then 0.4 litres will
have drug entirely removed from it every minute (2/5 of the volume, shaded area). (b) After
2/5 of the drug has been removed, the remaining drug is diluted into the full volume, so
reducing its concentration. (c) This represents a drug with an apparent volume of distribution
(Vd) of 9 litres, 2/5 of each of those litres will have drug removed every minute. Clearance is
the product of Vd and k so the clearance in this example is 9×0.4 or 0.36 litres per minute.
(d) In the case of our example the remaining drug disperses into the entire 9 litres and the
concentration falls.