Why Begin With Amount?
A compartment model describes how drug moves into, out of, and between modeled spaces. The most direct balance is written for amount: the quantity of drug present. Concentration is often what is measured, but it is amount divided by volume. Keeping those quantities distinct makes it easier to check whether a model’s terms represent the same physical quantity and to interpret what a predicted change means.
Here the model has one well-mixed central compartment. Drug enters that compartment at rate \(R_{\mathrm{in}}(t)\), and elimination removes drug at a rate proportional to the amount currently present. The focus is the balance itself, not a particular solution for the full concentration-time curve.
The units make the distinctions concrete. If amount is in mg and time in h, then \(R_{\mathrm{in}}\) is in mg/h. The elimination rate constant \(k_{\mathrm{el}}\) has units h\(^{-1}\), so \(k_{\mathrm{el}}A_c\) is also in mg/h. If volume is in L, concentration is in mg/L. A rate of change of amount has units mg/h, whereas a rate of change of concentration has units mg/L/h.
Deriving the Amount Balance
Over a short interval from \(t\) to \(t+\Delta t\), the amount at the end equals the amount at the start, plus drug entering, minus drug eliminated. For a sufficiently short interval, the entering amount is approximately \(R_{\mathrm{in}}(t)\Delta t\), and the eliminated amount is approximately \(k_{\mathrm{el}}A_c(t)\Delta t\). Thus,
Divide by \(\Delta t\) and take the limit as the interval shrinks to zero. The resulting differential equation is the one-compartment amount balance:
The left side is the net rate of change in the amount present. The right side is the input rate minus the elimination rate. The equation does not say that elimination stops when input is present: both processes can occur at the same time. It says that the amount changes according to their difference.
This balance assumes a single, well-mixed compartment; an input rate defined for drug entering that compartment; and linear elimination, with the elimination rate proportional to the amount present. It also treats the model parameters as fixed over the time interval considered. A constant volume is additionally required when converting the amount equation into a concentration equation. If drug transfers between multiple modeled compartments, those transfers must also appear in the relevant amount balances.
Worked Example: Finding the Instantaneous Net Amount Change
Suppose a hypothetical patient has \(A_c=240\) mg in the central compartment. At the time of interest, drug is entering at \(R_{\mathrm{in}}=6\) mg/h, and \(k_{\mathrm{el}}=0.15\) h\(^{-1}\). Find the instantaneous rate of change in amount.
First calculate the elimination rate:
Then substitute into the balance:
At that instant, the amount is decreasing at 30 mg/h because elimination exceeds input by 30 mg/h. This is an instantaneous rate, not a claim that the amount will decrease by exactly 30 mg during every later hour: as amount or input changes, the balance can change too.
From Amount to Concentration
If \(V_c\) is constant, then \(A_c(t)=V_cC_c(t)\). Differentiating both sides with respect to time gives \(dA_c/dt=V_c\,dC_c/dt\). Substituting this identity and \(A_c=V_cC_c\) into the amount balance yields the concentration balance:
The division by volume matters: input is specified as amount per time, but concentration changes in concentration per time. The elimination term can be written \(k_{\mathrm{el}}C_c\) in the concentration equation because \(k_{\mathrm{el}}A_c/V_c=k_{\mathrm{el}}C_c\). It is not an amount rate until multiplied by volume.
The conversion assumes \(V_c\) does not change with time. If volume varies, differentiating \(A_c=V_c(t)C_c(t)\) gives \(dA_c/dt=V_c(t)\,dC_c/dt+C_c(t)\,dV_c/dt\). The additional term means the constant-volume concentration equation cannot be used unchanged.
Worked Example: Converting the Balance to Concentration Units
Continue with the hypothetical values \(A_c=240\) mg, \(R_{\mathrm{in}}=6\) mg/h, and \(k_{\mathrm{el}}=0.15\) h\(^{-1}\). Suppose the constant central volume is \(V_c=12\) L. Calculate concentration and its instantaneous rate of change.
The concentration is amount divided by volume:
Use the concentration balance:
This agrees with the amount balance: \((-30\ \mathrm{mg/h})/(12\ \mathrm{L})=-2.5\ \mathrm{mg/L/h}\). The amount is decreasing at 30 mg/h and concentration at 2.5 mg/L/h; these are different rates because they refer to different quantities and units.
Volume Separates Amount From Concentration
Concentration alone does not determine amount unless volume is also known. Two compartments can have the same concentration but contain different amounts if their volumes differ. This distinction is important when interpreting model predictions: a concentration is not a count of how much drug is present.
Worked Example: Equal Concentrations, Different Amounts
Consider two hypothetical central compartments. Compartment A has \(V_c=10\) L and \(C_c=15\) mg/L. Compartment B has \(V_c=20\) L and the same concentration, \(15\) mg/L. For both, suppose \(k_{\mathrm{el}}=0.10\) h\(^{-1}\) and there is no input at the time considered. Find each amount and elimination rate, then compare concentration rates of change.
For Compartment A, \(A_c=V_cC_c=(10\ \mathrm{L})(15\ \mathrm{mg/L})=150\) mg. Its elimination rate is \((0.10\ \mathrm{h}^{-1})(150\ \mathrm{mg})=15\) mg/h. With no input, the amount rate is \(0-15=-15\) mg/h, so the concentration rate is \((-15\ \mathrm{mg/h})/(10\ \mathrm{L})=-1.5\) mg/L/h.
For Compartment B, \(A_c=(20\ \mathrm{L})(15\ \mathrm{mg/L})=300\) mg. Its elimination rate is \((0.10\ \mathrm{h}^{-1})(300\ \mathrm{mg})=30\) mg/h. Its amount rate is \(-30\) mg/h, and its concentration rate is \((-30\ \mathrm{mg/h})/(20\ \mathrm{L})=-1.5\) mg/L/h.
Both compartments have the same concentration and the same concentration rate of change, but Compartment B contains twice the amount and has twice the elimination rate in amount per hour. Comparing concentrations without considering volume would miss that difference.
What the Balance Can and Cannot Say
The equation is useful because it gives an interpretable accounting rule at every time: net change equals input minus elimination. It also provides a unit check before any more detailed model is used. In a one-compartment setting with no other transfers, a positive right-hand side means amount is increasing at that instant, a negative right-hand side means it is decreasing, and a zero right-hand side means the amount is momentarily unchanged.
A zero net rate does not mean that input and elimination are absent. They may be nonzero and equal. Similarly, a negative rate does not mean that the amount is negative; it means only that the amount is declining at that time. The balance describes the local direction and rate of change, not by itself the entire future trajectory.
The amount and concentration forms also show why input should not be subtracted directly from concentration. \(R_{\mathrm{in}}\) has units of amount per time, while \(C_c\) has units of amount per volume. In the concentration equation, input must first be divided by \(V_c\). Likewise, \(k_{\mathrm{el}}A_c\) is an amount-per-time elimination rate, whereas \(k_{\mathrm{el}}C_c\) is a concentration-per-time elimination term.
Common Mistakes and Model Checks
- Mixing amount and concentration. Subtracting \(k_{\mathrm{el}}A_c\) from a concentration or subtracting \(R_{\mathrm{in}}\) from a concentration rate is dimensionally invalid. Check that every term in an equation has the same units.
- Omitting volume in the concentration equation. The input contribution is \(R_{\mathrm{in}}/V_c\), not \(R_{\mathrm{in}}\), when concentration is expressed in amount per volume.
- Treating input as elimination’s opposite. Input and elimination can both be active. Their difference determines the net change; equal rates produce zero net change at that instant.
- Assuming the balance covers unmodeled transfers. If drug can leave the central compartment by transfer to another modeled compartment, that outgoing transfer must be included as a loss in its balance. The equation above includes only the stated input and elimination.
- Using a constant-volume conversion when volume varies. If \(V_c\) changes, the derivative of \(A_c=V_cC_c\) includes a volume-change term. Recheck the concentration equation rather than silently applying the constant-volume form.
A useful diagnostic is to inspect the equation term by term: identify the quantity being balanced, write the units beside each term, and verify that the assumptions match the intended compartment. The amount-balance formulation is a direct accounting approach; the concentration formulation is a rescaled version that is convenient when observations are concentrations. Neither representation repairs a model with missing processes or inappropriate assumptions.
Check Your Understanding
Try each question, then check the solutions below.
- A hypothetical central compartment contains 90 mg. At that instant, input is 5 mg/h and \(k_{\mathrm{el}}=0.20\) h\(^{-1}\). What is the amount rate of change?
- A compartment contains 120 mg in 8 L. What is its concentration? If input is zero and \(k_{\mathrm{el}}=0.10\) h\(^{-1}\), what is the instantaneous concentration rate of change?
- With constant volume, why must an input rate in amount per time be divided by volume when writing its contribution to concentration change?
- At a particular time, input and elimination rates are both 18 mg/h. What does the amount balance imply at that instant, and what does it not imply?
Solutions. (1) Elimination is \((0.20\ \mathrm{h}^{-1})(90\ \mathrm{mg})=18\) mg/h. Therefore \(dA_c/dt=5-18=-13\) mg/h. (2) \(C_c=120\ \mathrm{mg}/8\ \mathrm{L}=15\) mg/L. With no input, \(dC_c/dt=-k_{\mathrm{el}}C_c=-(0.10)(15)=-1.5\) mg/L/h. (3) Dividing \(R_{\mathrm{in}}\), in amount per time, by \(V_c\), in volume, gives amount per volume per time, the units of concentration change. (4) The net amount rate is \(18-18=0\) mg/h, so amount is momentarily unchanged. This does not imply that input and elimination are zero or that the amount must remain unchanged at later times.