Let's solve the given problem step by step: --- ## **Given Data** - \( A_{12} = 1.6 \) - \( A_{21} = 1.2 \) - \( P = 1 \) atm - \( P_1^{sat} = .9 \) atm, \( P_2^{sat} = .4 \) atm - \( x_1 = .4 \) (so \( x_2 = 1 - x_1 = .6 \)) --- ## **1. Activity Coefficient (\(\gamma_1\) and \(\gamma_2\))** ### **Van Laar Model Equations:** \[ \ln \gamma_1 = \frac{A_{12} (A_{21} x_2)^2}{(A_{12} x_1 + A_{21} x_2)^2} \] \[ \ln \gamma_2 = \frac{A_{21} (A_{12} x_1)^2}{(A_{12} x_1 + A_{21} x_2)^2} \] Let's substitute the values: - \( x_1 = .4 \), \( x_2 = .6 \) - \( A_{12} = 1.6 \), \( A_{21} = 1.2 \) #### **Denominator Calculation:** \[ A_{12} x_1 + A_{21} x_2 = 1.6 \times .4 + 1.2 \times .6 = .64 + .72 = 1.36 \] #### **\(\ln \gamma_1\):** \[ \ln \gamma_1 = \frac{1.6 \times (1.2 \times .6)^2}{(1.36)^2} \] \[ 1.2 \times .6 = .72 \] \[ (.72)^2 = .5184 \] \[ 1.6 \times .5184 = .82944 \] \[ (1.36)^2 = 1.8496 \] \[ \ln \gamma_1 = \frac{.82944}{1.8496} = .4486 \] \[ \gamma_1 = \exp(.4486) \approx 1.566 \] #### **\(\ln \gamma_2\):** \[ \ln \gamma_2 = \frac{1.2 \times (1.6 \times .4)^2}{(1.36)^2} \] \[ 1.6 \times .4 = .64 \] \[ (.64)^2 = .4096 \] \[ 1.2 \times .4096 = .49152 \] \[ \ln \gamma_2 = \frac{.49152}{1.8496} = .2658 \] \[ \gamma_2 = \exp(.2658) \approx 1.304 \] --- ## **2. Vapor Phase Composition (\(y_1, y_2\))** **VLE relation:** \[ y_i P = x_i \gamma_i P_i^{sat} \] \[ y_i = \frac{x_i \gamma_i P_i^{sat}}{P} \] ### **\(y_1\):** \[ y_1 = \frac{.4 \times 1.566 \times .9}{1} \] \[ y_1 = .4 \times 1.566 = .6264 \] \[ y_1 = .6264 \times .9 = .5638 \] ### **\(y_2\):** \[ y_2 = \frac{.6 \times 1.304 \times .4}{1} \] \[ y_2 = .6 \times 1.304 = .7824 \] \[ y_2 = .7824 \times .4 = .31296 \] #### **Check Normalization:** \[ y_1 + y_2 = .5638 + .31296 = .8768 \] If you add both, it's less than 1, which indicates the pressure might not be the bubble point pressure. These \(y_i\) values are for the given \(P\), not at the bubble point. --- ## **3. Bubble Point Pressure (\(P_{bubble}\))** \[ P_{bubble} = x_1 \gamma_1 P_1^{sat} + x_2 \gamma_2 P_2^{sat} \] \[ P_{bubble} = .4 \times 1.566 \times .9 + .6 \times 1.304 \times .4 \] \[ = .4 \times 1.4094 + .6 \times .5216 \] \[ = .5638 + .31296 = .8768 \text{ atm} \] --- ## **Final Answers** ### **1. Activity Coefficients** - \(\gamma_1 = 1.566\) - \(\gamma_2 = 1.304\) ### **2. Vapor Phase Composition at \(P = 1 \text{ atm}\)** - \(y_1 = .5638\) - \(y_2 = .31296\) ### **3. Bubble Point Pressure** - \(P_{bubble} = .8768 \text{ atm}\) --- ### **Summary Table** | Quantity | Value | |--------------------|------------| | \(\gamma_1\) | 1.566 | | \(\gamma_2\) | 1.304 | | \(y_1\) | .5638 | | \(y_2\) | .31296 | | \(P_{bubble}\) | .8768 atm | --- Let me know if you need further clarification!