Why Unequal Volumes Require an Amount Balance
When material moves between compartments, the transfer removes an amount from one compartment and adds the same amount to the other, assuming there is no reaction or loss during transfer. Concentration is amount per volume, so equal changes in concentration do not generally represent equal amounts when the compartments have different volumes. The safe starting point is therefore to write the balance in amounts and convert to concentrations only afterward.
Consider two well-mixed compartments with constant volumes \(V_1\) and \(V_2\), and concentrations \(C_1(t)\) and \(C_2(t)\) of the same freely exchanging species. Let \(G\) be a positive exchange conductance with units of volume per time. In the linear exchange model, the signed net transfer rate from compartment 1 to compartment 2 is proportional to the concentration difference.
The exchange law is an assumption about the process, not a consequence of conservation alone. It represents linear, passive exchange driven by a concentration difference. Conservation specifies how that transfer changes the two amounts.
Amount Balances and Concentration Equations
Assume each compartment is well mixed, the volumes are fixed, the same species is transferred between them, and there are no reactions, external inputs, or external outputs. With \(J_{12}\) defined as above, the amount lost by compartment 1 is exactly the amount gained by compartment 2:
Since \(A_i=V_iC_i\) and \(V_i\) is constant, \(dA_i/dt=V_i\,dC_i/dt\). Substituting the exchange law gives the concentration equations:
The factors \(1/V_1\) and \(1/V_2\) matter. For the same amount transferred per unit time, a smaller compartment has the larger concentration change per unit time. These equations do not say that \(C_1+C_2\) is constant. The conserved quantity is the total amount \(A_1+A_2=V_1C_1+V_2C_2\).
Worked Example: Instantaneous Transfer With Unequal Volumes
Suppose \(V_1=2\) L, \(V_2=5\) L, \(C_1=8\) mmol/L, \(C_2=2\) mmol/L, and \(G=0.3\) L/h. Find the net transfer rate and the instantaneous concentration rates of change.
The concentration difference is \(C_1-C_2=8-2=6\) mmol/L. Thus the signed net transfer rate is
Because this is positive, the net transfer is from compartment 1 to compartment 2. The amount balances are \(dA_1/dt=-1.8\) mmol/h and \(dA_2/dt=1.8\) mmol/h, which sum to zero. Converting each to a concentration rate gives
The concentration rates are not equal in magnitude, but the amount rates are equal and opposite. Multiplying each concentration rate by its own volume recovers \(-1.8\) mmol/h and \(1.8\) mmol/h.
Conservation, Equilibrium, and the Concentration Difference
Adding the two amount balances shows directly why total amount is conserved:
Write initial concentrations as \(C_1(0)\) and \(C_2(0)\). The conserved total amount is \(V_1C_1(0)+V_2C_2(0)\). If \(G>0\), a stationary state under this exchange-only model requires \(J_{12}=0\), and hence \(C_1=C_2\). The common equilibrium concentration must be the initial total amount divided by the total volume:
This is a volume-weighted average, not an unweighted average of the two initial concentrations. To describe the approach to equilibrium, define \(D=C_1-C_2\). Subtracting the concentration equations yields a linear differential equation:
Set \(\lambda=G(1/V_1+1/V_2)\), which has units of inverse time. If \(D(0)=0\), then \(D(t)=0\) for all \(t\), as is verified by substitution into the differential equation. If \(D(0)\ne0\), separation and integration give \(\ln|D(t)|-\ln|D(0)|=-\lambda t\), so \(D(t)=D(0)e^{-\lambda t}\). Thus the same expression \(D(t)=D(0)e^{-\lambda t}\) also holds in the zero case, without dividing by \(D\).
The two concentrations can be recovered from the conserved total amount and their difference:
These expressions satisfy both \(C_1-C_2=D\) and \(V_1C_1+V_2C_2=(V_1+V_2)C_{\mathrm{eq}}\). For \(G>0\), the difference decays toward zero, while the total amount remains fixed.
Worked Example: Concentrations Approaching a Shared Equilibrium
Use the first example’s volumes, conductance, and initial concentrations: \(V_1=2\) L, \(V_2=5\) L, \(G=0.3\) L/h, \(C_1(0)=8\) mmol/L, and \(C_2(0)=2\) mmol/L. Find both concentrations after 3 h under the model.
The initial total amount is
The equilibrium concentration is \(C_{\mathrm{eq}}=26\ \mathrm{mmol}/(7\ \mathrm{L})=3.714\) mmol/L. The initial difference is \(D(0)=6\) mmol/L, and \(\lambda=(0.3\ \mathrm{L/h})(1/2+1/5\ \mathrm{L}^{-1})=0.21\ \mathrm{h}^{-1}\). Therefore,
Using the reconstruction equations,
The concentrations are closer than their initial values, and compartment 1 still has the higher concentration. The amount check gives approximately \((2)(5.997)+(5)(2.801)=26.00\) mmol, consistent with conservation; the small rounding difference is numerical.
Worked Example: Transfer Reverses When the Gradient Reverses
Let \(V_1=1\) L, \(V_2=3\) L, \(C_1=1.5\) mmol/L, \(C_2=4\) mmol/L, and \(G=0.2\) L/min. Calculate the signed transfer and concentration rates.
Here \(C_1-C_2=1.5-4=-2.5\) mmol/L, so
The negative sign means net transfer is from compartment 2 to compartment 1. The amount rates are \(dA_1/dt=+0.5\) mmol/min and \(dA_2/dt=-0.5\) mmol/min. Consequently,
The sign convention works in either direction: the amount leaving one compartment enters the other, regardless of which concentration is larger.
Choosing the State Representation and Checking the Model
An amount-based formulation is the clearest choice when the primary task is to express transfer and conservation: each flux appears once with opposite signs in the two balances. A concentration-based formulation is often convenient when measurements and kinetic laws are expressed in concentration units, but each amount rate must be divided by the volume of the compartment receiving that concentration state. Either representation is valid when conversions are handled consistently.
| Representation | State | Transfer contribution | Conservation check |
|---|---|---|---|
| Amount-based | \(A_1,A_2\), in amount units | \(-J_{12},+J_{12}\) | \(d(A_1+A_2)/dt=0\) |
| Concentration-based | \(C_1,C_2\), in amount per volume | \(-J_{12}/V_1,+J_{12}/V_2\) | \(d(V_1C_1+V_2C_2)/dt=0\) |
Useful checks are to verify that the transfer rate has units of amount per time, that each concentration derivative has units of concentration per time, and that multiplying each concentration equation by its compartment volume recovers the amount balance. A sign check is also informative: if \(C_1>C_2\) and \(G>0\), compartment 1 must lose amount while compartment 2 gains it.
A common mistake is to write \(dC_1/dt=-dC_2/dt\) because the transfer is conserved. That reasoning treats concentrations as if they were amounts. For unequal volumes, it fails: conservation requires \(V_1\,dC_1/dt+V_2\,dC_2/dt=0\), not that the concentration rates sum to zero. For example, the first worked example has concentration rates \(-0.9\) and \(+0.36\) mmol/(L h), which do not cancel, while the amount rates \(-1.8\) and \(+1.8\) mmol/h do.
This model has limitations. It assumes constant volumes, well-mixed compartments, a single freely exchanging species, and a linear passive exchange law with fixed \(G\). Binding, active transport, nonlinear transport, reactions, external inputs or losses, and changing volumes require additional terms or different assumptions. In particular, if a compartment changes volume, differentiating \(A_i=V_iC_i\) also produces a volume-change contribution; the constant-volume conversion used here would no longer suffice.
Check Your Understanding
Work each problem using the amount balance first. Solutions are provided below the questions.
- Let \(V_1=1\) L, \(V_2=4\) L, \(C_1=10\) mmol/L, \(C_2=1\) mmol/L, and \(G=0.2\) L/h. Find \(J_{12}\) and both instantaneous concentration rates.
- A closed pair of compartments has volumes 3 L and 2 L and initial concentrations 2 mmol/L and 5 mmol/L. Find the total amount and the common concentration at equilibrium.
- If two compartments have equal concentrations but unequal volumes, what is the net transfer rate under this model? Explain why their amounts need not be equal.
- For \(V_1=2\) L, \(V_2=3\) L, \(G=0.12\) L/h, and \(D(0)=4\) mmol/L, calculate \(D(10)\).
Solutions. (1) \(J_{12}=(0.2)(10-1)=1.8\) mmol/h. Thus \(dC_1/dt=-1.8/1=-1.8\) mmol/(L h) and \(dC_2/dt=1.8/4=0.45\) mmol/(L h). (2) The total amount is \((3)(2)+(2)(5)=16\) mmol; the total volume is 5 L, so the equilibrium concentration is \(16/5=3.2\) mmol/L. (3) Equal concentrations make \(J_{12}=G(C_1-C_2)=0\), even if the amounts differ, because amount is concentration multiplied by volume. (4) \(\lambda=0.12(1/2+1/3)=0.1\ \mathrm{h}^{-1}\), so \(D(10)=4e^{-0.1(10)}=4e^{-1}\approx1.472\) mmol/L.