Two well-mixed compartments exchange a dilute solute through a membrane

Two well-mixed compartments exchange a dilute solute through a membrane. The volumes are $V_1=1\,\mathrm{L}$ and $V_2=3\,\mathrm{L}$, the membrane capacity is $kA=0.4\,\mathrm{L}/\mathrm{min}$, and the initial data are $C_1(0)=12$ and $C_2(0)=0$ (same concentration units). The molar flux from 1 into 2 is $kA(C_1-C_2)$. There is no reaction. 1. Prove that the conserved inventory is $V_1 C_1+V_2 C_2=12$ and that the equilibrium concentration is $C_\infty=3$. Writing $\Delta:=C_1-C_2$, prove \[ \Delta(t)=12\,e^{-8t/15} \] and therefore \[ C_1=3+9\,e^{-8t/15},\qquad C_2=3-3\,e^{-8t/15}. \] 2. At the instant $t=(15/8)\ln 3$, prove $C_1=6$, $C_2=2$, and that the instantaneous membrane flux is $1.6$ (concentration units times $\mathrm{L}/\mathrm{min}$). Confirm that the initial flux is $4.8$. 3. Define the half-time by $\Delta(t_{1/2})=\Delta(0)/2$ and prove $t_{1/2}=(15/8)\ln 2$. 4. Audit both sentences: (i) “the tanks may be treated as equal volume, so the decay rate is $2kA/V$ with $V=2\,\mathrm{L}$ and $C_1,C_2$ are symmetric about $6$”; (ii) “inventory need not be conserved, so $C_\infty=0$ or $C_\infty=12$.” Also refuse a CRE reaction-inventory reading of the two tanks. Do not replace the membrane ODE by a ternary Maxwell–Stefan inversion or by a flux-ledger exact boundary layer.

Inventory $1\cdot 12+3\cdot 0=12$ forces $C_\infty=12/4=3$. The difference decays at $0.4(1+1/3)=8/15$.