The following temperature/composition data were obtained for a mixture of two liquids A and B at 1.00 atm, where x is the mole fraction in the liquid and y the mole fraction in the vapour at equilibrium. θ/°C 125 130 135 140 145 150 xA 0.91 0.65 0.45 0.30 0.18 0.098 yA 0.99 0.91 0.77 0.61 0.45 0.25 The boiling points are 124 °C for A and 155 °C for B. Plot the temperature–composition diagram for the mixture. What is the composition of the vapour in equilibrium with the liquid of composition (i) xA = 0.50 and (ii) xB = 0.33?

To analyze the given temperature-composition data for the mixture of liquids A and B, we will plot the temperature-composition diagram (also known as a phase diagram) and find the vapor compositions corresponding to specified liquid compositions.

### Step 1: Plot the Temperature-Composition Diagram

**Data Overview:**
- Given temperatures (θ) and mole fractions for components A (xA) and B (xB = 1 - xA):
  
| θ (°C) | xA   | yA   |
|--------|------|------|
| 125    | 0.91 | 0.99 |
| 130    | 0.65 | 0.91 |
| 135    | 0.45 | 0.77 |
| 140    | 0.30 | 0.61 |
| 145    | 0.18 | 0.45 |
| 150    | 0.098| 0.25 |

- Boiling points: A = 124 °C, B = 155 °C.

### Step 2: Generate the Diagram

To create the temperature-composition diagram, we will:
1. Plot \( x_A \) (mole fraction of A) on the x-axis.
2. Plot \( y_A \) (mole fraction of A in vapor) on the y-axis.
3. Mark the boiling points of A and B.

### Step 3: Calculate Vapor Composition for Given Liquid Compositions

1. **For \( x_A = 0.50 \)**:
   - We will interpolate the yA value corresponding to \( x_A = 0.50 \) using the provided data.

2. **For \( x_B = 0.33 \)**:
   - Since \( x_B = 1 - x_A \), we find \( x_A = 0.67 \) and similarly interpolate for yA.

### Step 4: Interpolation

#### For \( x_A = 0.50 \):
- This lies between \( x_A = 0.45 \) (135°C, \( y_A = 0.77 \)) and \( x_A = 0.65 \) (130°C, \( y_A = 0.91 \)).
- Using linear interpolation:
  
\[
y_A = y_A(0.45) + \left( \frac{y_A(0.65) - y_A(0.45)}{x_A(0.65) - x_A(0.45)} \right) \times (0.50 - 0.45)
\]

Calculating:
\[
y_A = 0.77 + \left( \frac{0.91 - 0.77}{0.65 - 0.45} \right) \times (0.50 - 0.45)
\]

#### For \( x_B = 0.33 \) (or \( x_A = 0.67 \)):
- This lies between \( x_A = 0.65 \) (130°C, \( y_A = 0.91 \)) and \( x_A = 0.45 \) (135°C, \( y_A = 0.77 \)).
  
\[
y_A = y_A(0.65) + \left( \frac{y_A(0.45) - y_A(0.65)}{x_A(0.45) - x_A(0.65)} \right) \times (0.67 - 0.65)
\]

Calculating:
\[
y_A = 0.91 + \left( \frac{0.77 - 0.91}{0.45 - 0.65} \right) \times (0.67 - 0.65)
\]

### Summary Calculations

Let's perform these calculations:

1. **For \( x_A = 0.50 \)**:
   - Calculate \( y_A \).

2. **For \( x_A = 0.67 \)** (or \( x_B = 0.33 \)):
   - Calculate \( y_A \).

### Execution of Calculations

Calculating both interpolations:

1. **For \( x_A = 0.50 \)**:
\[
y_A = 0.77 + \left( \frac{0.91 - 0.77}{0.65 - 0.45} \right) \times (0.50 - 0.45)
\]
\[
y_A = 0.77 + \left( \frac{0.14}{0.20} \right) \times 0.05
\]
\[
y_A = 0.77 + 0.07 = 0.84
\]

2. **For \( x_A = 0.67 \)**:
\[
y_A = 0.91 + \left( \frac{0.77 - 0.91}{0.45 - 0.65} \right) \times (0.67 - 0.65)
\]
\[
y_A = 0.91 + \left( \frac{-0.14}{-0.20} \right) \times 0.02
\]
\[
y_A = 0.91 + 0.14 = 0.92
\]

### Final Results

1. **Vapor composition for \( x_A = 0.50 \)**: \( y_A \approx 0.84 \) (so \( y_B = 0.16 \)).
2. **Vapor composition for \( x_B = 0.33 \)**: \( y_A \approx 0.92 \) (so \( y_B = 0.08 \)).

### Summary

- The temperature-composition diagram can be plotted using the data, marking the vapor compositions for the specified liquid compositions.
  
Do you want more details about the plot or any specific part of the calculation? Here are some related questions:

1. How do phase diagrams relate to Raoult's Law?
2. What is the significance of boiling point in distillation?
3. Can you explain the concept of azeotropes?
4. How does pressure affect boiling points and phase behavior?
5. What are the practical applications of temperature-composition diagrams in industry?

**Tip:** Always check for the presence of azeotropes when working with mixtures, as they can complicate distillation processes.

The following temperature/composition data were obtained for a mixture of two liquids A and B at 1.00 atm, where x is the mole fraction in the liquid and y the mole fraction in the vapour at equilibrium. θ/°C 125 130 135 140 145 150 xA 0.91 0.65 0.45 0.30 0.18 0.098 yA 0.99 0.91 0.77 0.61 0.45 0.25 The boiling points are 124 °C for A and 155 °C for B. Plot the temperature–composition diagram for the mixture. What is the composition of the vapour in equilibrium with the liquid of composition (i) xA = 0.50 and (ii) xB = 0.33?

Explore the temperature-composition diagram for a mixture of two liquids, A and B, with given data points at various temperatures. Learn to calculate vapor compositions at specified liquid compositions: xA = 0.50 and xB = 0.33.

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