`X_(A)` and `X_(B)` are the mole fraction of A and B respectively in liquid phase `y_(A)` and `y_(B)` are the mole fraction of A and B respective in vapour phase. Find out the slope of straight line if a graph is plotted `(1)/(y_(A))` along Y-axis against `(1)/(x_(A))` along X-axis gives straight line `[p_(A)^(@)` and `p_(B)^(@)` are vapour pressure of pure components A and B].

Options

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

Answer

To find the slope of the straight line when plotting \( \frac{1}{y_A} \) against \( \frac{1}{x_A} \), we will follow these steps:

### Step 1: Understand the relationship between mole fractions and partial pressures
The mole fraction of component A in the vapor phase, \( y_A \), can be expressed in terms of the partial pressure of A, \( P_A \), and the total pressure \( P \):
\[ y_A = \frac{P_A}{P} \]
where \( P = P_A + P_B \).

### Step 2: Express the partial pressure \( P_A \)
The partial pressure \( P_A \) can be expressed as:
\[ P_A = x_A \cdot P^0_A \]
where \( P^0_A \) is the vapor pressure of pure component A, and \( x_A \) is the mole fraction of A in the liquid phase.

### Step 3: Substitute \( P_A \) into the equation for \( y_A \)
Substituting \( P_A \) into the equation for \( y_A \):
\[ y_A = \frac{x_A \cdot P^0_A}{x_A \cdot P^0_A + x_B \cdot P^0_B} \]
Since \( x_B = 1 - x_A \), we can rewrite it as:
\[ y_A = \frac{x_A \cdot P^0_A}{x_A \cdot P^0_A + (1 - x_A) \cdot P^0_B} \]

### Step 4: Simplify the expression for \( y_A \)
This simplifies to:
\[ y_A = \frac{x_A \cdot P^0_A}{x_A \cdot P^0_A + P^0_B - x_A \cdot P^0_B} \]
\[ y_A = \frac{x_A \cdot P^0_A}{x_A (P^0_A - P^0_B) + P^0_B} \]

### Step 5: Take the reciprocal of \( y_A \)
Taking the reciprocal gives:
\[ \frac{1}{y_A} = \frac{x_A (P^0_A - P^0_B) + P^0_B}{x_A \cdot P^0_A} \]

### Step 6: Rearrange the equation
Rearranging the equation:
\[ \frac{1}{y_A} = \frac{(P^0_A - P^0_B)}{P^0_A} + \frac{P^0_B}{x_A \cdot P^0_A} \]
This can be expressed in the form of a straight line equation \( y = mx + c \), where:
- \( y = \frac{1}{y_A} \)
- \( x = \frac{1}{x_A} \)
- Slope \( m = \frac{P^0_B}{P^0_A} \)

### Step 7: Conclusion
Thus, the slope of the straight line when plotting \( \frac{1}{y_A} \) against \( \frac{1}{x_A} \) is:
\[ \text{Slope} = \frac{P^0_B}{P^0_A} \]