Two liquids A and B are mixed at temperature T in a certain ratio to form an idela solution. It is found that the partial vapoure pressure of A. i.e, `P_(A)` is equal to `P_(B)`. The vapoure pressure of B for the liquid mixture. Whatis the total vapour pressure of the liquid mixture in terms of `P_(A)^(@) and P_(B)^(@)` ?

Options

A. `(P_(A)^(@)P_(B)^(@))/(P_(A)^(@)+P_(B)^(@))`
B. `(2P_(A)^(@)P_(B)^(@))/(P_(A)^(@)+P_(B)^(@))`
C. `(P_(A)^(@))/(P_(A)^(@)+P_(B)^(@))`
D. `(2P_(B)^(@))/(P_(A)^(@)+P_(B)^(@))`

Answer

To solve the problem, we need to find the total vapor pressure of a mixture of two liquids A and B, given that their partial vapor pressures are equal (i.e., \( P_A = P_B \)). We will express the total vapor pressure in terms of the vapor pressures of the pure components \( P_A^0 \) and \( P_B^0 \).

### Step-by-Step Solution:

1. **Understanding the Ideal Solution**:
- In an ideal solution, the partial vapor pressure of each component is given by Raoult's Law:
\[ P_A = P_A^0 \cdot X_A \]
\[ P_B = P_B^0 \cdot X_B \]
where \( X_A \) and \( X_B \) are the mole fractions of components A and B, respectively.

2. **Given Condition**:
- We know that \( P_A = P_B \). Therefore, we can set the two equations equal to each other:
\[ P_A^0 \cdot X_A = P_B^0 \cdot X_B \]

3. **Expressing Mole Fractions**:
- Since \( X_A + X_B = 1 \), we can express \( X_B \) as:
\[ X_B = 1 - X_A \]
- Substituting this into the equation gives:
\[ P_A^0 \cdot X_A = P_B^0 \cdot (1 - X_A) \]

4. **Rearranging the Equation**:
- Rearranging the above equation, we get:
\[ P_A^0 \cdot X_A + P_B^0 \cdot X_A = P_B^0 \]
- Factoring out \( X_A \):
\[ X_A \cdot (P_A^0 + P_B^0) = P_B^0 \]
- Solving for \( X_A \):
\[ X_A = \frac{P_B^0}{P_A^0 + P_B^0} \]

5. **Finding Total Vapor Pressure**:
- The total vapor pressure \( P_T \) of the mixture is the sum of the partial pressures:
\[ P_T = P_A + P_B \]
- Substituting \( P_A \) and \( P_B \):
\[ P_T = P_A^0 \cdot X_A + P_B^0 \cdot X_B \]
- Since \( X_B = 1 - X_A \):
\[ P_T = P_A^0 \cdot X_A + P_B^0 \cdot (1 - X_A) \]
- Substituting \( X_A \):
\[ P_T = P_A^0 \cdot \frac{P_B^0}{P_A^0 + P_B^0} + P_B^0 \cdot \left(1 - \frac{P_B^0}{P_A^0 + P_B^0}\right) \]

6. **Simplifying the Expression**:
- This simplifies to:
\[ P_T = \frac{P_A^0 \cdot P_B^0}{P_A^0 + P_B^0} + P_B^0 \cdot \frac{P_A^0}{P_A^0 + P_B^0} \]
- Combining the terms gives:
\[ P_T = \frac{P_A^0 \cdot P_B^0 + P_B^0 \cdot P_A^0}{P_A^0 + P_B^0} = \frac{2 P_A^0 \cdot P_B^0}{P_A^0 + P_B^0} \]

7. **Final Answer**:
- Therefore, the total vapor pressure of the liquid mixture is:
\[ P_T = \frac{2 P_A^0 \cdot P_B^0}{P_A^0 + P_B^0} \]