Let's solve the problem **step by step**: --- ## **Given** - Pure liquids A and B: Immiscible - C is soluble in A and B. - 1 mole each of A, B C (total 3 moles- Equilibrium Two liquid phases: - **α-phase**: A + C - **β-phase**: B + C - Excess molar Gibbs free energies: \[ \left(\frac{g^E}{RT}\right)_\alpha = .4\, x_A x_C \] \[ \left(\frac{g^E}{RT}\right)_\beta = .8\, x_B x_C \] --- ## **(a) Activity Coefficients of C** Let: - \( n_C^\alpha \): moles of C in α-phase - \( n_C^\beta \): moles of C in β-phase In each phase: - α-phase: contains A and C (moles: \( n_A^\alpha, n_C^\alpha \)) - β-phase: contains B and C (moles: \( n_B^\beta, n_C^\beta \)) **Mole fractions in α-phase:** \[ x_A^\alpha = \frac{n_A^\alpha}{n_A^\alpha + n_C^\alpha},\quad x_C^\alpha = \frac{n_C^\alpha}{n_A^\alpha + n_C^\alpha} \] **Mole fractions in β-phase:** \[ x_B^\beta = \frac{n_B^\beta}{n_B^\beta + n_C^\beta},\quad x_C^\beta = \frac{n_C^\beta}{n_B^\beta + n_C^\beta} \] **Activity coefficient (Margules model):** \[ g^E = RT\, x_i x_C A \] For α-phase: \( A = .4 \), for β-phase: \( A = .8 \) \[ \ln \gamma_C^\alpha = \frac{\partial (g^E/RT)}{\partial n_C^\alpha} \] But with Margules-type excess free energy: \[ \ln \gamma_C^\alpha = .4\, x_A^\alpha \] \[ \ln \gamma_C^\beta = .8\, x_B^\beta \] **So, the activity coefficients:** \[ \boxed{ \ln \gamma_C^\alpha = .4 \left(\frac{n_A^\alpha}{n_A^\alpha + n_C^\alpha}\right) } \] \[ \boxed{ \ln \gamma_C^\beta = .8 \left(\frac{n_B^\beta}{n_B^\beta + n_C^\beta}\right) } \] --- ## **(b) Phase Equilibrium Condition** At equilibrium, the chemical potential of C must be equal in both phases: \[ \mu_C^\alpha = \mu_C^\beta \] \[ \mu_C^* + RT \ln x_C^\alpha + RT \ln \gamma_C^\alpha = \mu_C^* + RT \ln x_C^\beta + RT \ln \gamma_C^\beta \] \[ \ln x_C^\alpha + \ln \gamma_C^\alpha = \ln x_C^\beta + \ln \gamma_C^\beta \] \[ x_C^\alpha \gamma_C^\alpha = x_C^\beta \gamma_C^\beta \] Substitute the expressions from part (a): \[ x_C^\alpha \exp\left[.4 x_A^\alpha\right] = x_C^\beta \exp\left[.8 x_B^\beta\right] \] Let: - \( n_A^\alpha + n_C^\alpha = n^\alpha \) - \( n_B^\beta + n_C^\beta = n^\beta \) Mass balances: - \( n_A^\alpha + n_A^\beta = 1 \) (but A only in α: \( n_A^\beta = \implies n_A^\alpha = 1 \)) - \( n_B^\alpha + n_B^\beta = 1 \) (B only in β: \( n_B^\alpha = \implies n_B^\beta = 1 \)) - \( n_C^\alpha + n_C^\beta = 1 \) So, - \( n^\alpha = 1 + n_C^\alpha \) - \( n^\beta = 1 + n_C^\beta \) So, \[ x_A^\alpha = \frac{1}{1 + n_C^\alpha}, \quad x_C^\alpha = \frac{n_C^\alpha}{1 + n_C^\alpha} \] \[ x_B^\beta = \frac{1}{1 + n_C^\beta}, \quad x_C^\beta = \frac{n_C^\beta}{1 + n_C^\beta} \] Now substitute into equilibrium condition: \[ \left( \frac{n_C^\alpha}{1 + n_C^\alpha} \right) \exp \left[ .4 \frac{1}{1 + n_C^\alpha} \right] = \left( \frac{n_C^\beta}{1 + n_C^\beta} \right) \exp \left[ .8 \frac{1}{1 + n_C^\beta} \right] \] with \( n_C^\alpha + n_C^\beta = 1 \). This **implicit equation** can be solved numerically for \( n_C^\alpha \), \( n_C^\beta \). --- ## **(c) Minimum Work to Separate Mixture** Minimum work (reversible separation) is equal to the **decrease in Gibbs free energy** (mixing): \[ W_{min} = \Delta G_{mix} \] Where: \[ \Delta G_{mix} = RT \left[ \sum_{i,ph} n_i^{ph} \ln x_i^{ph} + n^{ph} \frac{g^E}{RT} \right] \] Sum over all components and both phases. **For this problem:** - α-phase: \( n_A^\alpha = 1 \), \( n_C^\alpha \) - β-phase: \( n_B^\beta = 1 \), \( n_C^\beta = 1 - n_C^\alpha \) So: \[ \Delta G_{mix}/RT = n_A^\alpha \ln x_A^\alpha + n_C^\alpha \ln x_C^\alpha + (n_A^\alpha + n_C^\alpha) \cdot .4 x_A^\alpha x_C^\alpha \] \[ + n_B^\beta \ln x_B^\beta + n_C^\beta \ln x_C^\beta + (n_B^\beta + n_C^\beta) \cdot .8 x_B^\beta x_C^\beta \] Plug the values from (b) into this formula for the final answer. --- ## **Summary Table** | Part | Formula/Expression | |------|-------------------| | (a) | \( \ln \gamma_C^\alpha = .4 x_A^\alpha \), \( \ln \gamma_C^\beta = .8 x_B^\beta \) | | (b) | \( x_C^\alpha \exp(.4 x_A^\alpha) = x_C^\beta \exp(.8 x_B^\beta) \), with \( n_C^\alpha + n_C^\beta = 1 \) | | (c) | \( \Delta G_{mix}/RT = \) (see above, sum over both phases) | --- ## **Final Answers** 1. **Activity coefficients:** \[ \boxed{ \ln \gamma_C^\alpha = .4 \left(\frac{1}{1 + n_C^\alpha}\right) } \] \[ \boxed{ \ln \gamma_C^\beta = .8 \left(\frac{1}{1 + n_C^\beta}\right) } \] 2. **Mole fractions:** Solve numerically: \[ \left( \frac{n_C^\alpha}{1 + n_C^\alpha} \right) \exp \left[ .4 \frac{1}{1 + n_C^\alpha} \right] = \left( \frac{n_C^\beta}{1 + n_C^\beta} \right) \exp \left[ .8 \frac{1}{1 + n_C^\beta} \right] \] with \( n_C^\alpha + n_C^\beta = 1 \). 3. **Minimum work (in units of \( RT \)):** \[ \Delta G_{mix}/RT = \Big[ n_A^\alpha \ln x_A^\alpha + n_C^\alpha \ln x_C^\alpha + (n_A^\alpha + n_C^\alpha) \cdot .4 x_A^\alpha x_C^\alpha \Big] + \Big[ n_B^\beta \ln x_B^\beta + n_C^\beta \ln x_C^\beta + (n_B^\beta + n_C^\beta) \cdot .8 x_B^\beta x_C^\beta \Big] \] --- **Numerical values** can be found by solving the equation in (b). If you want to see the numerical solution, let me know!