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. The goal is to plot \( \frac{1}{Y_A} \) against \( \frac{1}{X_A} \) and determine the slope and intercept of this linear relationship.

### 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 partial pressures of components A and B.

2. **Using Raoult's Law**:
- According to Raoult's Law, the partial pressure of component A is given by:
\[ P_A = P_{A0} \cdot X_A \]
- Similarly, for component B:
\[ P_B = P_{B0} \cdot X_B \]
- The total pressure \( P \) is:
\[ P = P_A + P_B = P_{A0} \cdot X_A + P_{B0} \cdot X_B \]

3. **Expressing Mole Fractions**:
- Since \( X_B = 1 - X_A \), we can substitute this into the equation for total pressure:
\[ P = P_{A0} \cdot X_A + P_{B0} \cdot (1 - X_A) \]

4. **Relating Vapor and Liquid Phase**:
- The mole fraction of A in the vapor phase can be expressed as:
\[ Y_A = \frac{P_A}{P} = \frac{P_{A0} \cdot X_A}{P_{A0} \cdot X_A + P_{B0} \cdot (1 - X_A)} \]

5. **Deriving the Equation**:
- Rearranging the equation gives us:
\[ Y_A = \frac{P_{A0} \cdot X_A}{P_{A0} \cdot X_A + P_{B0} - P_{B0} \cdot X_A} \]
- This can be simplified further, leading to a relationship between \( Y_A \) and \( X_A \).

6. **Taking the Reciprocal**:
- We take the reciprocal of \( Y_A \):
\[ \frac{1}{Y_A} = \frac{P_{A0} \cdot X_A + P_{B0} - P_{B0} \cdot X_A}{P_{A0} \cdot X_A} \]
- This can be rearranged to show a linear relationship between \( \frac{1}{Y_A} \) and \( \frac{1}{X_A} \).

7. **Identifying Slope and Intercept**:
- From the linear equation derived, we can identify the slope \( m \) and intercept \( c \):
- Slope \( m = \frac{P_{B0}}{P_{A0}} \)
- Intercept \( c = \frac{P_{A0} - P_{B0}}{P_{A0}} \)

### Conclusion:
The plot of \( \frac{1}{Y_A} \) versus \( \frac{1}{X_A} \) is linear, where:
- **Slope** = \( \frac{P_{B0}}{P_{A0}} \)
- **Intercept** = \( \frac{P_{A0} - P_{B0}}{P_{A0}} \)