For the dissociation of `PCI_(5)` into `PCI_(3) and Cl_(2)` in gaseous phase reaction, If "d' is the observed vapour density and 'D' theoretical vapour density with `'alpha'` as degree of dissociation. Variation of `(D)/(d)` with `'alpha'` is given by which graph?

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Answer

To solve the problem regarding the dissociation of \( PCl_5 \) into \( PCl_3 \) and \( Cl_2 \) in the gaseous phase, we will analyze the relationship between the observed vapor density \( d \), the theoretical vapor density \( D \), and the degree of dissociation \( \alpha \).

### Step-by-Step Solution:

1. **Write the Reaction:**
The dissociation reaction can be written as:
\[ PCl_5 (g) \rightleftharpoons PCl_3 (g) + Cl_2 (g) \]

2. **Determine the Moles of Gaseous Products:**
From the reaction, we see that 1 mole of \( PCl_5 \) produces 2 moles of gaseous products (\( PCl_3 \) and \( Cl_2 \)). Therefore, if we consider 1 mole of \( PCl_5 \), the total number of moles after dissociation is:
\[ n = 2 \quad (\text{where } n \text{ is the number of moles of gaseous products}) \]

3. **Define the Theoretical and Observed Vapor Densities:**
- Theoretical vapor density \( D \) is calculated based on the initial moles before dissociation.
- The observed vapor density \( d \) changes as \( PCl_5 \) dissociates.

4. **Relate Vapor Densities to Degree of Dissociation:**
The degree of dissociation \( \alpha \) is defined as the fraction of the original substance that has dissociated. The relationship between the vapor densities can be expressed as:
\[ D - d = (n - 1) \cdot d \cdot \alpha \]
Substituting \( n = 2 \):
\[ D - d = (2 - 1) \cdot d \cdot \alpha = d \cdot \alpha \]
Rearranging gives:
\[ D = d + d \cdot \alpha = d(1 + \alpha) \]

5. **Express \( \frac{D}{d} \) in terms of \( \alpha \):**
Dividing both sides by \( d \):
\[ \frac{D}{d} = 1 + \alpha \]

6. **Determine the Relationship:**
This equation shows that \( \frac{D}{d} \) increases linearly with \( \alpha \). Specifically, when \( \alpha = 0 \), \( \frac{D}{d} = 1 \), and as \( \alpha \) approaches 1, \( \frac{D}{d} \) approaches 2.

7. **Graphical Representation:**
- The graph of \( \frac{D}{d} \) (y-axis) versus \( \alpha \) (x-axis) will be a straight line starting from (0, 1) and going up to (1, 2).
- The slope of the line is positive, indicating a direct relationship between \( \frac{D}{d} \) and \( \alpha \).

### Conclusion:
The variation of \( \frac{D}{d} \) with \( \alpha \) is represented by a straight line graph with a positive slope, starting from \( \frac{D}{d} = 1 \) when \( \alpha = 0 \) and reaching \( \frac{D}{d} = 2 \) when \( \alpha = 1 \).