Let's solve each part step by step: --- ## **Given Data** - **Three liquids:** \(A\), \(B\) (immiscible), and \(C\) (soluble in both). - **Mix 1 mole each** of \(A\),B\), and \(C\). - **Two phases form**: - \(\alpha\)-phase: contains \(A\) and \(C\) - \(\beta\)-phase: contains \(B\) and \(C\) - **Excess 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 Species \(C\) in Both Phases** ### **Step 1: Molar Gibbs Free Energy of Mixing** For a mixture, the molar Gibbs free energy (per mole) is: \[ \frac{g}{RT} = \sum_i x_i \ln x_i + \frac{g^E}{RT} \] where \(\frac{g^E}{RT}\) is the excess term. ### **Step 2: Activity Coefficient** The activity coefficient \(\gamma_C\) is given by: \[ \ln\gamma_C = \frac{1}{RT} \left( \frac{\partial n\,g^E}{\partial n_C} \right)_{T,P,n_{i\neq C}} \] For a binary or ternary system, with \(g^E/RT = f(x_A, x_B, x_C)\): #### **For the \(\alpha\)-phase:** \[ \frac{g^E}{RT} = .4 x_A x_C \] \[ \ln\gamma_C^\alpha = \frac{\partial}{\partial n_C} \left( n \cdot .4 x_A x_C \right) \] Let \(n = n_A^\alpha + n_C^\alpha\), \(x_A = n_A^\alpha / n\), \(x_C = n_C^\alpha / n\). Let’s compute: \[ n \cdot x_A x_C = n_A^\alpha n_C^\alpha / n \] So, \[ n \cdot .4 x_A x_C = .4 \frac{n_A^\alpha n_C^\alpha}{n} \] Taking the derivative with respect to \(n_C^\alpha\): Let’s use the quotient rule: \[ \frac{\partial}{\partial n_C^\alpha} \left( \frac{n_A^\alpha n_C^\alpha}{n} \right) = \frac{n_A^\alpha \cdot n - n_A^\alpha n_C^\alpha}{n^2} = \frac{n_A^\alpha (n - n_C^\alpha)}{n^2} = \frac{n_A^\alpha (n_A^\alpha)}{n^2} = \frac{(n_A^\alpha)^2}{n^2} = (x_A^\alpha)^2 \] Thus, \[ \ln\gamma_C^\alpha = .4 (x_A^\alpha)^2 \] #### **For the \(\beta\)-phase:** Similarly, \[ \frac{g^E}{RT} = .8 x_B x_C \] \[ \ln\gamma_C^\beta = .8 (x_B^\beta)^2 \] --- **Final Answers for (a):** \[ \boxed{ \ln\gamma_C^\alpha = .4\, (x_A^\alpha)^2 \qquad \ln\gamma_C^\beta = .8\, (x_B^\beta)^2 } \] --- ## **(b) Mole Fractions of \(C\) in \(\alpha\) and \(\beta\) phases (Phase equilibrium condition)** At equilibrium, the chemical potential of \(C\) in both phases is equal: \[ \mu_C^\alpha = \mu_C^\beta \] For each phase: \[ \mu_C = \mu_C^\circ + RT \ln x_C + RT \ln \gamma_C \] So at equilibrium: \[ \ln x_C^\alpha + \ln \gamma_C^\alpha = \ln x_C^\beta + \ln \gamma_C^\beta \] \[ \ln \left( \frac{x_C^\alpha \gamma_C^\alpha}{x_C^\beta \gamma_C^\beta} \right) = \] \[ x_C^\alpha \gamma_C^\alpha = x_C^\beta \gamma_C^\beta \] Substitute the expressions for \(\gamma_C^\alpha\) and \(\gamma_C^\beta\): \[ x_C^\alpha \exp[.4 (x_A^\alpha)^2] = x_C^\beta \exp[.8 (x_B^\beta)^2] \] --- ## **(c) Minimum Work Required (in \(RT\)) to Separate Mixture into Pure Species** The minimum work is given by the entropy of mixing (neglecting excess entropy, only ideal solution): For 1 mole each (total 3 moles mixed): \[ \Delta_{mix} S = -R \sum_{i=A,B,C} x_i \ln x_i \] For 1 mole each: \(x_i = 1/3\) for each: \[ \Delta_{mix} S = -R \left[ 3 \times \frac{1}{3} \ln \left( \frac{1}{3} \right) \right] = -R \ln \left( \frac{1}{3} \right) = R \ln 3 \] The minimum work is \( -T \Delta_{mix} S = RT \ln 3 \) But since we have non-ideal mixing, **the total minimum work is the change in Gibbs free energy**: \[ \Delta_{mix} G = nRT \sum x_i \ln x_i + nRT \cdot (\text{average } g^E/RT) \] But to separate, we need to reverse the mixing, so the work is: \[ W_{min} = -\Delta_{mix} G = - nRT \sum x_i \ln x_i - nRT \cdot (\text{average } g^E/RT) \] But since the mixture separates into two phases, and then into pure components, the minimum work is the sum of: - **Work to separate the two phases into their pure components** (reverse mixing). - **Work to split the two phases into the three pure species**. But typically, for the minimum work to split into pure components, it is: \[ W_{min} = nRT \sum_{i=1}^{3} x_i \ln x_i + nRT \sum_{phases} y_j \left( \frac{g^E}{RT} \right)_j \] Where \(y_j\) is the fraction of total moles in phase \(j\). So, the answer is: \[ \boxed{ W_{min} = 3RT \ln 3 + RT \left[ y_\alpha \left( \frac{g^E}{RT} \right)_\alpha + y_\beta \left( \frac{g^E}{RT} \right)_\beta \right] } \] Where \(y_\alpha\), \(y_\beta\) are the mole fractions of total moles in \(\alpha\) and \(\beta\) phase, and \(\left( \frac{g^E}{RT} \right)_\alpha, \left( \frac{g^E}{RT} \right)_\beta\) are the average excess Gibbs energies of the phases. --- ## **Summary Table** | Part | Final Answer | |------|--------------| | (a) | \(\ln\gamma_C^\alpha = .4\, (x_A^\alpha)^2\) <br> \(\ln\gamma_C^\beta = .8\, (x_B^\beta)^2\) | | (b) | \(x_C^\alpha \exp[.4 (x_A^\alpha)^2] = x_C^\beta \exp[.8 (x_B^\beta)^2]\) | | (c) | \(W_{min} = 3RT \ln 3 + RT \left[ y_\alpha \left( \frac{g^E}{RT} \right)_\alpha + y_\beta \left( \frac{g^E}{RT} \right)_\beta \right]\) | If you want the explicit values for (b) or (c), more information about the distribution of species between the phases is needed, which typically requires solving simultaneous equations based on total moles and equilibrium. Let me know if you want to proceed further!