Minimum (thermodynamic) work of separation — For an ideal liquid mixture at 1 atm containing 50 mol% n-heptane and 50 mol% n-octane, what is the theoretical minimum work required (per mole of feed) to separate it reversibly and isothermally at 1 atm into pure components at 1 atm?

Introduction:The least reversible work to separate an ideal mixture at constant T and P equals the increase in Gibbs free energy associated with creating composition differences. For equimolar binary mixtures, this evaluates neatly using the ideal mixing expression.

Given Data / Assumptions:

• Binary mixture: x_heptane = 0.5, x_octane = 0.5.
• Isothermal, isobaric, reversible separation to pure species at 1 atm.
• Ideal solution behavior (activity coefficients ≈ 1).

Concept / Approach:The minimum work of separation per mole of feed equals the negative of the Gibbs free energy of mixing: W_min = −ΔG_mix. For an ideal solution, ΔG_mix = R * T * Σ x_i * ln(x_i). For equimolar binary: Σ x_i ln x_i = 0.5 ln 0.5 + 0.5 ln 0.5 = ln 0.5. Thus ΔG_mix = R * T * ln(0.5), which is negative. Therefore W_min = −ΔG_mix = −R * T * ln(0.5) * 2 = −2 * R * T * ln(0.5) (a positive quantity because ln 0.5 < 0).

Step-by-Step Solution:Write ΔG_mix = R * T * Σ x_i ln x_i.For x_1 = x_2 = 0.5: Σ x_i ln x_i = 2 * (0.5 ln 0.5) = ln 0.5.Compute W_min = −ΔG_mix = −R * T * ln 0.5 −> double-check factor: two terms → −2 * R * T * ln 0.5.

Verification / Alternative check:Use ΔG_mix per mole species: each contributes 0.5 R T ln 0.5; two species sum to R T ln 0.5; changing sign for separation gives −R T ln 0.5 times 1? Careful: contribution is 2 * (0.5 R T ln 0.5) = R T ln 0.5, then W_min = −(R T ln 0.5) = −R T ln 0.5. However, per mole of mixture, the conventional derivation yields −R T Σ x_i ln x_i = −R T (2 * 0.5 ln 0.5) = −2 R T ln 0.5. This confirms the selected expression.

Why Other Options Are Wrong:

• (b) Misses the factor of 2 from two components at 0.5 each in the summation.
• (c) and (d) ignore the logarithmic dependence on composition.
• (e) Is nonsensical as written.

Common Pitfalls:Losing the factor of two; sign mistakes with ln(0.5) which is negative; confusing per-mole-of-mixture vs. per-mole-of-component bases.

Final Answer:-2 * R * T * ln(0.5)