The One-Compartment IV Bolus Model

From an IV Bolus to a Concentration-Time Curve

A single intravenous bolus is an idealized input: the dose enters the systemic circulation at once. The one-compartment model then treats the body as a single, instantaneously mixed space. Under these assumptions, the concentration is highest immediately after administration and declines as drug is eliminated. The resulting equation is simple, but its parameters have distinct meanings: dose sets the initial amount, volume links amount to concentration, and clearance governs the rate of elimination.

The equation describes a structural prediction for an individual under specified parameter values. It is not, by itself, a statistical model for measurement error or between-person variability. Its usefulness depends on whether instantaneous mixing and first-order elimination are reasonable approximations over the times and conditions being considered.

Definition: Let \(D\) be the intravenous bolus dose, \(A(t)\) the amount of drug in the modeled compartment at time \(t\), \(C(t)\) the drug concentration, \(V\) the apparent volume of distribution, and \(CL\) the clearance. In the one-compartment IV bolus model, \(C(t)=A(t)/V\), the initial amount is \(A(0^+)=D\), and elimination is first order with constant \(CL\). The elimination rate constant is \(k=CL/V\).

Here \(D\) and \(A(t)\) have units of amount, such as mg; \(C(t)\) has units of amount per volume, such as mg/L; \(V\) has units of volume, such as L; \(CL\) has units of volume per time, such as L/h; and \(k\) has units of inverse time, such as h\(^{-1}\). The notation \(0^+\) means immediately after the bolus has entered the modeled space. For an ideal bolus, the concentration just before dosing is not the same as the concentration at \(0^+\).

Deriving the Equation

The elimination rate is the amount removed per unit time. Under linear, first-order elimination, that rate is proportional to concentration, with clearance as the proportionality factor. Since the amount in the compartment is \(A(t)=V C(t)\), the amount balance after the bolus is:

\frac{dA(t)}{dt}=-CL\,C(t)=-\frac{CL}{V}A(t)=-kA(t), \qquad A(0^+)=D.

This assumes \(V\) and \(CL\) are constant over the interval, the compartment is well mixed, and there is no further drug input after the bolus. Separating variables and integrating from \(0^+\) to \(t\) gives:

\frac{dA}{A}=-k\,dt \quad\Longrightarrow\quad \int_D^{A(t)}\frac{dA}{A}=-k\int_0^t dt \quad\Longrightarrow\quad \ln\!\left(\frac{A(t)}{D}\right)=-kt.

Exponentiating and using \(C(t)=A(t)/V\) gives the concentration-time equation:

A(t)=D e^{-kt}, \qquad C(t)=\frac{D}{V}e^{-kt} =\frac{D}{V}e^{-(CL/V)t}, \qquad t\geq 0.

Key result: The model predicts \(C(0^+)=D/V\) and an exponential decline with rate constant \(k=CL/V\). Dose and volume determine the initial concentration; the ratio of clearance to volume determines how quickly concentration declines.

The half-life is the time required for concentration to fall to half its value at a given time. Setting \(C(t+t_{1/2})/C(t)=1/2\) in the exponential equation gives \(e^{-kt_{1/2}}=1/2\), and therefore \(t_{1/2}=\ln(2)/k=\ln(2)V/CL\). This is a consequence of the model’s constant first-order rate: each equal interval reduces the concentration by the same proportion, not by the same absolute amount.

Another useful implication is the area under the concentration-time curve from the bolus to complete elimination. Integrating the equation gives:

AUC_{0-\infty} =\int_0^\infty \frac{D}{V}e^{-(CL/V)t}\,dt =\frac{D}{V}\frac{V}{CL} =\frac{D}{CL}.

This result assumes the model applies over the full interval and elimination continues to completion. It shows that, for a fixed dose under linear elimination, clearance determines total exposure in this model. Volume affects the initial concentration and the time course, but it cancels from the total area.

Worked Example: Predicting Concentrations After a Bolus

Suppose a hypothetical patient receives a single \(250\) mg IV bolus. Assume instantaneous mixing in one compartment, \(V=25\) L, and constant first-order clearance \(CL=5\) L/h. Calculate \(C(0^+)\), \(C(2\text{ h})\), \(C(6\text{ h})\), and the model half-life.

First, the rate constant is \(k=CL/V=(5\text{ L/h})/(25\text{ L})=0.200\text{ h}^{-1}\). The units reduce to inverse time, as required. The initial concentration is \(C(0^+)=D/V=(250\text{ mg})/(25\text{ L})=10.0\text{ mg/L}\).

At 2 h:

C(2)=10.0e^{-(0.200)(2)} =10.0e^{-0.400} =6.703\text{ mg/L}.

At 6 h:

C(6)=10.0e^{-(0.200)(6)} =10.0e^{-1.200} =3.012\text{ mg/L}.

The half-life is \(t_{1/2}=\ln(2)/0.200=3.466\) h. The model therefore predicts a concentration of \(5.00\) mg/L after one half-life. These are model predictions, not claims about measured concentrations in an actual patient.

What the Parameters Do

Dose and volume appear together in the initial concentration \(D/V\). Holding volume fixed, doubling the dose doubles every predicted concentration. Holding dose fixed, doubling volume halves the initial concentration. Volume is an apparent proportionality between the amount in the modeled space and the measured concentration; it should not be interpreted as a literal anatomical space that the drug uniformly occupies.

Clearance has a different role. At any concentration \(C(t)\), the model’s elimination rate is \(CL\,C(t)\). A higher clearance removes more amount per unit time at that concentration. In the exponential curve, however, the decline rate is governed by \(k=CL/V\), so clearance does not act alone: for a fixed clearance, a larger volume gives a smaller rate constant and a more gradual proportional decline.

Worked Example: Recovering Parameters From an Idealized Curve

Consider hypothetical, rounded concentrations after a single \(300\) mg IV bolus: \(8.19\) mg/L at \(t=1\) h and \(3.679\) mg/L at \(t=5\) h. Assume the one-compartment model is exact for this illustration, with constant parameters and no additional input. Estimate \(k\), extrapolate \(C(0^+)\), and calculate \(V\) and \(CL\).

Taking logarithms of \(C(t)=C(0^+)e^{-kt}\) gives \(\ln C(t)=\ln C(0^+)-kt\). Thus the log concentration difference between two times identifies the rate constant:

k=\frac{\ln(C(1)/C(5))}{5-1} =\frac{\ln(8.19/3.679)}{4\text{ h}} \approx 0.2001\text{ h}^{-1}.

Use the concentration at 1 h to extrapolate back to the bolus time:

C(0^+)=C(1)e^{k(1\text{ h})} =(8.19\text{ mg/L})e^{(0.2001\text{ h}^{-1})(1\text{ h})} \approx 10.00\text{ mg/L}.

Under instantaneous mixing, \(C(0^+)=D/V\), so \(V=D/C(0^+)=(300\text{ mg})/(10.00\text{ mg/L})\approx30.0\) L. Finally, \(CL=kV=(0.2001\text{ h}^{-1})(30.0\text{ L})\approx6.00\) L/h. Because the two concentrations were rounded, the recovered values are approximate. In real data, measurement error and model misspecification would also affect these calculations.

Worked Example: Same Clearance, Different Volumes

Compare two hypothetical model profiles after the same \(200\) mg bolus. Both have \(CL=4\) L/h, but the first has \(V=20\) L and the second has \(V=40\) L. Assume the same one-compartment and first-order conditions for both. Calculate the initial concentration, rate constant, concentration at 4 h, and total AUC.

For \(V=20\) L, \(C(0^+)=200/20=10.0\) mg/L and \(k=4/20=0.200\) h\(^{-1}\). At 4 h, \(C(4)=10.0e^{-(0.200)(4)}=4.493\) mg/L.

For \(V=40\) L, \(C(0^+)=200/40=5.00\) mg/L and \(k=4/40=0.100\) h\(^{-1}\). At 4 h, \(C(4)=5.00e^{-(0.100)(4)}=3.352\) mg/L.

For each profile, \(AUC_{0-\infty}=D/CL=(200\text{ mg})/(4\text{ L/h})=50.0\text{ mg}\cdot\text{h/L}\). Thus the larger volume produces a lower initial concentration and slower fractional decline, while the two profiles have the same total AUC because dose and clearance are the same.

Checking the Model and Knowing Its Limits

The equation is a useful starting point, not a guarantee that every observed concentration-time profile is mono-exponential. Its assumptions should be considered in the context of the administration and sampling:

Instantaneous mixing: the dose is represented as entering the compartment at \(t=0\), and concentration is \(D/V\) immediately afterward. A real bolus takes time to administer and mix, so very early measurements may not match this idealization.
One well-mixed compartment: the model assigns one concentration to the entire modeled space. A systematic early steep decline followed by a slower decline can indicate that a single exponential does not capture the observed time course.
First-order elimination: the amount eliminated per unit time is proportional to concentration, with constant clearance. A changing apparent slope across concentrations or time can be evidence against this assumption, although data noise and sampling design can also create such patterns.
Constant parameters and no further input: the equation applies to a single bolus with fixed \(CL\) and \(V\) and no subsequent drug entering the modeled space. If the input history differs, this equation alone is not the appropriate description.

A practical diagnostic is to plot measured concentrations against time and, for positive concentrations, inspect log concentration against time. The one-compartment equation implies \(\ln C(t)=\ln(D/V)-kt\), a straight line with slope \(-k\) under the ideal model. Curvature or systematic departures may indicate an inadequate structural model. But a nearly straight line does not prove that the assumptions are true: sparse sampling, limited assay precision, or a restricted time window may conceal departures. Early samples are especially important for assessing the predicted initial level, while later samples help characterize the decline.

The main alternatives correspond to different features of the process. An infusion model represents drug entering over a period rather than instantaneously. A two-compartment model allows concentrations to change as drug distributes between modeled spaces. A model with nonlinear elimination relaxes the assumption that elimination rate is proportional to concentration. These alternatives have additional structure and should be considered when the dose history or observed pattern provides a reason to do so; they should not be inferred solely from a visually imperfect curve.

Common pitfall: It is incorrect to say that clearance alone determines how quickly concentration falls. The concentration decline rate is \(k=CL/V\). Clearance determines elimination rate at a given concentration, while volume also affects the proportional decline and the initial concentration. Conversely, volume alone does not determine total AUC under this linear model: \(AUC_{0-\infty}=D/CL\).

The model’s compact form is valuable because it separates an initial level from a decline rate. Its limitations matter when the data show systematic departures, when the bolus is not effectively instantaneous, or when the parameter assumptions do not hold over the observation period. Parameter estimates from a simple model summarize the data under that model; they do not establish that the body literally consists of one uniform space.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

A hypothetical \(180\) mg bolus is modeled with \(V=30\) L and \(CL=6\) L/h. What are \(C(0^+)\), \(k\), and \(C(3\text{ h})\)?
If \(CL\) stays fixed while \(V\) doubles, what happens to \(C(0^+)\) for a fixed dose, and what happens to \(k\)?
Under this model, what is \(AUC_{0-\infty}\) for a \(120\) mg dose and \(CL=3\) L/h? State the units.
Why might a straight line on a log-concentration-versus-time plot support, but not prove, the one-compartment model?

Solutions. (1) \(C(0^+)=180/30=6.00\) mg/L and \(k=6/30=0.200\) h\(^{-1}\). Therefore \(C(3)=6.00e^{-(0.200)(3)}=3.293\) mg/L. (2) For fixed dose, \(C(0^+)=D/V\) is halved, and \(k=CL/V\) is also halved. (3) \(AUC_{0-\infty}=D/CL=(120\text{ mg})/(3\text{ L/h})=40.0\text{ mg}\cdot\text{h/L}\). (4) The model predicts a straight line because \(\ln C(t)=\ln(D/V)-kt\), but limited or noisy observations may appear linear even when the assumptions do not fully describe the underlying process.

The One-Compartment IV Bolus Model

What You'll Learn