A binary solution contains `x_1 and x_2` mile fraction of two components having vapour pressure `p_1^@ and p_2^@` in this pure states. The total vapour pressure above the solution is

Options

A. `(p_1^@-P_2^@)x_1+P_2^@`
B. `(p_2^@-P_1^@)x_1+P_2^@`
C. `(p_1^@-P_2^@)x_1+P_1^@`
D. `(p_2^@-P_1^@)x_1+P_1^@`

Answer

To solve the problem regarding the total vapor pressure above a binary solution containing two components with given mole fractions and vapor pressures, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Components**:
- Let \( x_1 \) be the mole fraction of component 1.
- Let \( x_2 \) be the mole fraction of component 2.
- The vapor pressure of component 1 in its pure state is \( P_1^0 \).
- The vapor pressure of component 2 in its pure state is \( P_2^0 \).

2. **Using Raoult's Law**:
- According to Raoult's Law, the partial vapor pressure of each component in the solution can be expressed as:
\[ P_1' = x_1 P_1^0 \]
\[ P_2' = x_2 P_2^0 \]

3. **Total Vapor Pressure**:
- The total vapor pressure \( P \) above the solution is the sum of the partial pressures:
\[ P = P_1' + P_2' = x_1 P_1^0 + x_2 P_2^0 \]

4. **Substituting for \( x_2 \)**:
- Since \( x_1 + x_2 = 1 \), we can express \( x_2 \) as:
\[ x_2 = 1 - x_1 \]
- Substitute \( x_2 \) into the total vapor pressure equation:
\[ P = x_1 P_1^0 + (1 - x_1) P_2^0 \]

5. **Expanding the Equation**:
- Expanding the equation gives:
\[ P = x_1 P_1^0 + P_2^0 - x_1 P_2^0 \]
- Combine like terms:
\[ P = P_2^0 + x_1 (P_1^0 - P_2^0) \]

6. **Final Expression**:
- Thus, the total vapor pressure above the solution can be expressed as:
\[ P = P_2^0 + x_1 (P_1^0 - P_2^0) \]

### Final Answer:
The total vapor pressure above the solution is given by:
\[ P = P_2^0 + x_1 (P_1^0 - P_2^0) \]