Question

A semi-permeable, immovable membrane separates two compartments of equal volumes V. Initially Nₐ molecules of component A are in the left-hand compartment and Nᵦ molecules of component B are in the right-hand compartment; both are ideal gases. The membrane is impermeable to B but permeable to A. Find the fraction of A molecules on the LHS and the osmotic pressure Π i.e., the pressure difference) across the membrane at equilibrium as functions of T, V, Nₐ, and Nᵦ

Answer 1

At equilibrium, the fraction of A molecules will be equally distributed (1/2) on both sides of the membrane since it is semi-permeable to only A.

Step-by-step explanation:

The student's question pertains to osmosis through a semi-permeable membrane which allows only one type of molecule to pass through. A compartment contains Nṣ molecules of component A and a separate compartment with Nβ molecules of component B. At equilibrium, with only A able to pass through, the fraction of A molecules on the left-hand side (LHS) can be determined from the equilibrium condition where the chemical potentials on both sides are equal.

For ideal gases, the chemical potential is given by μ = μ° + RTln(p/p°), where R is the gas constant, T is the temperature, and p is the partial pressure of the gas. Balancing the chemical potential for A on both sides and solving for pressure, we get that the pressures will be equal. Because the volume is the same on both sides and A can move freely, the fraction of A will be 1/2 on each side established by the number of molecules.

The osmotic pressure (Π) can then be found using the van't Hoff equation Π = iCRT, where i is the van't Hoff factor (which is 1 for an ideal solution), C is the concentration of the solute (which is Nβ/V for component B), and T is the temperature. Substituting these values, we obtain the osmotic pressure as a function of T, V, Nṣ, and Nβ.