Dissolved molecules of three sizes start in the left chamber. Which ones get through the pores, and how fast does the cell equalise?
Headless runs of the same random-walk engine.
The science
Stokes-Einstein. A sphere of radius r in a fluid of viscosity η has D = kBT / (6π η r). Water's viscosity is taken from a Vogel fit, η(T) = 2.414×10−5 · 10247.8/(T−140) Pa·s (typical; 0.89 mPa·s at 25 °C). With r = 0.21 nm this gives D ≈ 1.17×10−9 m²/s, the textbook order for a small molecule in water. Stokes drag is a continuum formula and over-estimates the drag on molecules this small, so absolute rates are indicative only.
Random walk. Each solute takes independent Gaussian steps Δx = √(2DΔt) ξ in x, y, z (ξ standard normal), giving ⟨r²⟩ = 6Dt in free space. Solutes are dilute and non-interacting (they overlap freely; the picture is of an ideal solution). Walls are mirrors.
Fick's laws. The flux follows the gradient, J = −D ∇c, and concentration obeys ∂c/∂t = D ∇²c. For a step start (left half full) in a reflecting box of half-length L the solution is a cosine series, u(x,t) = 1/2 + Σk odd (2 sin(kπ/2)/(kπ)) cos(kπ(x+L)/2L) e−D(kπ/2L)² t, so the fraction on the right is 1/2 − (4/π²) Σk odd e−D(kπ/2L)² t/k². The dashed curve with the membrane off is this exact result.
Pores. The membrane is an infinitely thin wall at x = 0 with nine circular pores (spacing 4.4 nm). A solute may cross only where its step meets a pore, and only if its centre passes inside radius rpore − rsolute (steric exclusion; a molecule bigger than the pore never passes). A thin-wall pore passes 2 D a of conductance (two access resistances 1/(4Da)); with the finite patch this gives a relaxation rate λ = 4D / (V (1/(n a) + 1/b)), V the volume of each chamber, a = rpore − rsolute, n pores, b the patch radius. The right-side fraction then follows ½ (1 − e−λt). The dashed curve with the membrane on is this estimate; it ignores pore-pore interference and so is only a guide for the nine-pore array (the independent check uses a single pore).
Simplifications. No hydrodynamic hindrance inside the pore, no electrostatics, no solute-solute or solute-wall forces, an infinitely thin membrane drawn with a visual thickness, a nanometre-scale cell (1 BU = 1 nm) so that equalisation takes about a microsecond. Constants are typical values: spot-check them before quoting. Assets built with Blender; scene by Three.js.