Pressure over ideal binary liquid mixture containing `10` moles each of liquid `A` and `B` is gradually decreased isothermally. If `"P"_("A")^(@)=200"mm Hg " "and"" P"_("B")^(@)=100"mm Hg`. Find the pressure at which half of the liquid is converted into vapour.

Options

A. `150` mm Hg
B. 166.5 mmHg
C. 133 mmHg
D. 141.4 mmHg

Answer

To solve the problem of finding the pressure at which half of the liquid is converted into vapor in an ideal binary liquid mixture of liquids A and B, we can follow these steps:

### Step 1: Understand the Given Data
- Moles of liquid A (nA) = 10
- Moles of liquid B (nB) = 10
- Total moles (n_total) = nA + nB = 20
- Vapor pressure of pure A (P_A^0) = 200 mm Hg
- Vapor pressure of pure B (P_B^0) = 100 mm Hg

### Step 2: Define the Situation
When half of the liquid is converted into vapor, we have:
- Moles of liquid A remaining = nA - yA
- Moles of liquid B remaining = nB - yB
- Moles of vapor A = yA
- Moles of vapor B = yB

Given that half of the liquid is converted into vapor, we have:
- yA + yB = 10 (since half of the total moles of liquid, which is 20, is 10)

### Step 3: Write the Relationships
Using Raoult's Law, we can express the partial pressures of A and B:
- Partial pressure of A (P_A) = x_A * P_A^0
- Partial pressure of B (P_B) = x_B * P_B^0

Where:
- x_A = mole fraction of A in the liquid phase
- x_B = mole fraction of B in the liquid phase

### Step 4: Calculate Mole Fractions
The mole fractions can be expressed as:
- x_A = (nA - yA) / (n_total - (yA + yB)) = (10 - yA) / (20 - 10) = (10 - yA) / 10
- x_B = (nB - yB) / (n_total - (yA + yB)) = (10 - yB) / (20 - 10) = (10 - yB) / 10

### Step 5: Substitute into Raoult's Law
Using the mole fractions in Raoult's Law:
- P_A = (10 - yA) / 10 * 200
- P_B = (10 - yB) / 10 * 100

### Step 6: Total Pressure
The total pressure (P_total) is given by:
\[ P_{total} = P_A + P_B \]

Substituting the expressions from Raoult's Law:
\[ P_{total} = \left( \frac{10 - yA}{10} \cdot 200 \right) + \left( \frac{10 - yB}{10} \cdot 100 \right) \]

### Step 7: Substitute yB
Since yB = 10 - yA, we can substitute this into the equation:
\[ P_{total} = \left( \frac{10 - yA}{10} \cdot 200 \right) + \left( \frac{10 - (10 - yA)}{10} \cdot 100 \right) \]
\[ P_{total} = \left( \frac{10 - yA}{10} \cdot 200 \right) + \left( \frac{yA}{10} \cdot 100 \right) \]

### Step 8: Simplify the Equation
Now, simplify the equation:
\[ P_{total} = \frac{2000 - 200yA + 100yA}{10} \]
\[ P_{total} = \frac{2000 - 100yA}{10} \]
\[ P_{total} = 200 - 10yA \]

### Step 9: Solve for yA
To find the pressure at which half of the liquid is converted into vapor, we need to find yA when yA = 5 (since half of 10 moles is 5 moles):
\[ P_{total} = 200 - 10(5) \]
\[ P_{total} = 200 - 50 \]
\[ P_{total} = 150 \text{ mm Hg} \]

### Conclusion
The pressure at which half of the liquid is converted into vapor is **150 mm Hg**.