Why Volume Matters
A species can change concentration even when its amount does not change. Adding solute-free fluid dilutes a compartment; removing solute-free fluid concentrates it. Conversely, a changing amount does not by itself determine whether concentration rises or falls: the volume may be changing at the same time. A QSP model that uses concentration as its state must account for both changes.
The essential step is to write the balance for the species amount first, then convert that balance to concentration. This keeps the physical meaning of each term visible and avoids treating concentration as if it were an amount. The equations below apply to one well-mixed compartment; the same conversion can be used for each compartment in a larger model.
From an Amount Balance to a Concentration Balance
Assume that amount and volume are differentiable in time and that the compartment is well mixed, so its concentration is spatially uniform. Differentiate \(A=CV\) with respect to time. The product rule gives the general conversion:
The second term in the concentration equation is the volume-change contribution. When volume increases, it lowers concentration at fixed amount; when volume decreases, it raises concentration at fixed amount. The term is not an additional reaction. It arises from expressing an amount balance using concentration as the state.
For a compartment with fluid inflow \(Q_{\mathrm{in}}\) at concentration \(C_{\mathrm{in}}\), fluid outflow \(Q_{\mathrm{out}}\), and net biochemical production rate \(R\), a suitable amount balance is
Here \(R\) is the net amount produced by reactions per unit time, with production positive and consumption negative. The flow terms assume that the incoming fluid carries the species at concentration \(C_{\mathrm{in}}\), and that outgoing fluid carries it at the compartment concentration \(C\). Thus, the compartment is well mixed, and there is no selective removal or addition of the species beyond the listed terms. For example, \(Q_{\mathrm{in}}C_{\mathrm{in}}\) has units of amount per time when \(Q_{\mathrm{in}}\) is volume per time and concentration is amount per volume.
Substituting this amount balance into the conversion gives a concentration equation that remains valid when volume changes for any reason:
If fluid flows alone determine the volume change, with no other source or sink of volume, then \(dV/dt=Q_{\mathrm{in}}-Q_{\mathrm{out}}\). Substitution and collection of terms yields the simpler form
This simplification depends on the stated volume balance. It should not be used automatically when, for example, volume also changes through an unmodeled fluid transfer. In that case, retain the general equation and use the actual \(dV/dt\).
Worked Applications
Worked Example: Solute-Free Expansion With Reaction Production
At an instant, a well-mixed compartment has \(A=54\) nmol and \(V=1.7\) L. Reactions produce the species at a net rate of \(6.0\) nmol/min. Solute-free fluid causes the volume to increase at \(0.050\) L/min, and there are no species-bearing fluid flows. Find the instantaneous concentration rate.
First calculate concentration from the amount and volume:
The amount rate is \(dA/dt=6.0\) nmol/min. Using the general conversion and retaining unrounded values through the arithmetic,
Thus the concentration is increasing at approximately \(2.595\) nmol/(L·min) at this instant. Production alone would raise concentration faster, but expansion dilutes the species. Because production exceeds the dilution contribution here, the net concentration rate remains positive.
Worked Example: Equal Inflow and Outflow
A compartment has \(V=2.0\) L and \(A=50\) nmol, so \(C=25\) nmol/L. Fluid enters at \(Q_{\mathrm{in}}=0.10\) L/min with \(C_{\mathrm{in}}=40\) nmol/L and leaves at \(Q_{\mathrm{out}}=0.10\) L/min. Reactions consume the species at a net rate \(R=-3.0\) nmol/min. Find \(dC/dt\).
The equal flow rates imply \(dV/dt=0\), given the assumption that these flows alone determine volume. The incoming species rate is \((0.10)(40)=4.0\) nmol/min, and the outgoing species rate is \((0.10)(25)=2.5\) nmol/min. Therefore,
The same result follows from the simplified equation:
Although the incoming fluid is more concentrated than the compartment, its contribution does not offset reaction consumption and outflow. The concentration is decreasing at \(0.750\) nmol/(L·min).
Worked Example: Concentration During Expansion With Conserved Amount
Suppose a species amount remains fixed at \(120\) nmol, with no reactions or species-bearing flows. The compartment starts at \(1.5\) L and expands at a constant \(0.020\) L/min. Find the concentration at 10 min and its rate of change at the start.
The volume is \(V(t)=1.5+0.020t\) L when \(t\) is measured in minutes. Since the amount is constant,
At 10 min, \(V=1.5+(0.020)(10)=1.7\) L, so
At the start, \(C(0)=120/1.5=80\) nmol/L. Using \(dA/dt=0\) in the general conversion,
The negative rate is dilution: no species amount is lost, but the same amount occupies increasing volume. The exact expression for \(C(t)\) also shows that the decline is not linear in time.
Assumptions, Diagnostics, and Alternatives
Before using a concentration equation, check that the state represents concentration in a well-mixed compartment and that \(V(t)\) is positive. The balance must account for the processes that change species amount, while the volume equation must account for the processes that change volume. If flows alone determine volume, the equality \(dV/dt=Q_{\mathrm{in}}-Q_{\mathrm{out}}\) must be justified for that model. If volume can change independently, keep its separate rate in the concentration equation.
A useful diagnostic is to reconstruct the implied amount rate from a proposed concentration equation:
This must match the amount balance. If it does not, a volume term may be missing, duplicated, or assigned the wrong sign. Also check units: \(dC/dt\) must have units of concentration per time, while \(dA/dt\) must have units of amount per time. For positive \(C\), increasing volume with fixed amount must produce a negative concentration rate; a positive rate in that case signals an error in the conversion or its implementation.
Two state choices are common. An amount state makes conservation and species-bearing flow terms direct, but observations or rates expressed as concentrations require conversion using the current volume. A concentration state may align more naturally with concentration measurements and concentration-dependent reaction rates, but its equation must include the changing-volume contribution. Neither choice removes the need to model volume consistently. If the compartment is not well mixed, a single concentration may not represent the local concentration carried by outflow, and a spatial or multi-compartment description may be needed.
Common Mistakes
- Using only \(dC/dt=(dA/dt)/V\). This omits the change in concentration caused by changing volume. The correct conversion also subtracts \((C/V)(dV/dt)\).
- Calling the volume term a species loss. Expansion can lower concentration without removing any species amount. It changes the relationship between amount and concentration.
- Using the compartment concentration for inflow. Incoming fluid carries its own concentration \(C_{\mathrm{in}}\); the outgoing fluid carries \(C\) under the well-mixed assumption.
- Assuming volume change equals inflow minus outflow in every model. This holds only when those flows alone determine volume. Other volume sources or sinks must be represented in \(dV/dt\).
- Mixing rounded intermediate values. Use unrounded values in intermediate calculations, or ensure that displayed rounded terms produce the displayed result.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- A compartment contains \(90\) nmol in \(3.0\) L. Its amount is momentarily constant while its volume decreases at \(0.060\) L/min. What is the concentration rate?
- A \(4.0\) L compartment contains \(100\) nmol. Fluid enters at \(0.20\) L/min with concentration \(5.0\) nmol/L and leaves at \(0.20\) L/min. Reactions produce the species at \(2.0\) nmol/min. Find \(dA/dt\) and \(dC/dt\).
- If \(C=12\) nmol/L, \(V=2.0\) L, \(dA/dt=0\), and \(dV/dt=0.10\) L/min, find \(dC/dt\) and state whether concentration rises or falls.
- Explain why the simplified equation with the term \(Q_{\mathrm{in}}(C_{\mathrm{in}}-C)/V\) cannot automatically be used if volume also changes through another process.
Solutions. (1) Here \(C=90/3.0=30\) nmol/L. Thus \(dC/dt=0-(30/3.0)(-0.060)=0.600\) nmol/(L·min): concentration rises as volume decreases. (2) The current concentration is \(100/4.0=25\) nmol/L. The amount rate is \(2.0+(0.20)(5.0)-(0.20)(25)=2.0+1.0-5.0=-2.0\) nmol/min. Since the equal flows give \(dV/dt=0\), \(dC/dt=(-2.0)/4.0=-0.500\) nmol/(L·min). (3) The conversion gives \(dC/dt=0-(12/2.0)(0.10)=-0.600\) nmol/(L·min), so concentration falls. (4) That simplified equation substitutes \(dV/dt=Q_{\mathrm{in}}-Q_{\mathrm{out}}\). If another process changes volume, this substitution is not the actual volume balance; use the general conversion with the full \(dV/dt\) and the appropriate amount balance.