The plot of `(1)/(Y_(A))" Vs "(1)/(x_(A))((1)/(Y_(A))" on y - axis")` where A and B form a ideal solution. Y is mole fraction in vapour phase and X is mole fraction in liquid phase, is linear with slope and inercept respectively

Options

A. `(P_(A)^(0))/(P_(B)^(0)) and (P_(A)^(0)-P_(B)^(0))/(P_(B)^(0))`
B. `(P_(A)^(0))/(P_(B)^(0)) and (P_(B)^(0)-P_(A)^(0))/(P_(B)^(0))`
C. `(P_(B)^(0))/(P_(A)^(0)) and (P_(A)^(0)-P_(B)^(0))/(P_(A)^(0))`
D. `(P_(B)^(0))/(P_(A)^(0)) and (P_(B)^(0)-P_(A)^(0))/(P_(B)^(0))`

Answer

To solve the problem, we need to analyze the relationship between the mole fractions in the vapor and liquid phases of an ideal solution. We will derive the linear relationship between \( \frac{1}{Y_A} \) and \( \frac{1}{X_A} \) and identify the slope and intercept.

### Step-by-Step Solution:

1. **Understanding the Variables**:
- Let \( Y_A \) be the mole fraction of component A in the vapor phase.
- Let \( X_A \) be the mole fraction of component A in the liquid phase.
- For an ideal solution, the total pressure \( P \) can be expressed in terms of the vapor pressures of the components and their mole fractions.

2. **Using Raoult's Law**:
- According to Raoult's Law, the partial pressure of component A in the vapor phase can be given as:
\[ P_A = P^0_A \cdot X_A \]
- Similarly, for component B, we have:
\[ P_B = P^0_B \cdot X_B \]
- The total pressure \( P \) is the sum of the partial pressures:
\[ P = P_A + P_B \]

3. **Expressing Mole Fractions**:
- The mole fraction of B in the vapor phase can be expressed as:
\[ Y_B = 1 - Y_A \]
- The mole fraction of B in the liquid phase can be expressed as:
\[ X_B = 1 - X_A \]

4. **Equating Total Pressures**:
- Since both expressions represent the total pressure, we can equate them:
\[ \frac{P^0_A \cdot X_A}{Y_A} = \frac{P^0_B \cdot (1 - X_A)}{(1 - Y_A)} \]

5. **Rearranging the Equation**:
- Rearranging gives us:
\[ P^0_A \cdot Y_B = P^0_B \cdot X_B \]
- Substituting \( Y_B = 1 - Y_A \) and \( X_B = 1 - X_A \):
\[ P^0_A \cdot (1 - Y_A) = P^0_B \cdot (1 - X_A) \]

6. **Finding the Relationship**:
- Rearranging further leads to:
\[ \frac{1}{Y_A} = \frac{P^0_B}{P^0_A} \cdot \frac{1}{X_A} + \frac{P^0_A - P^0_B}{P^0_A} \]

7. **Identifying Slope and Intercept**:
- The equation is now in the form of \( y = mx + c \):
\[ \frac{1}{Y_A} = \left(\frac{P^0_B}{P^0_A}\right) \cdot \frac{1}{X_A} + \left(\frac{P^0_A - P^0_B}{P^0_A}\right) \]
- Here, the slope \( m \) is \( \frac{P^0_B}{P^0_A} \) and the intercept \( c \) is \( \frac{P^0_A - P^0_B}{P^0_A} \).

### Final Result:
- **Slope**: \( \frac{P^0_B}{P^0_A} \)
- **Intercept**: \( \frac{P^0_A - P^0_B}{P^0_A} \)