For the dissociation of `PCl_(5)` into `PCl_(3)` and `Cl_(2)` in gaseous phase reaction , if d is the observed vapour density and D the theoretical vapour density with `'alpha'` as degree of dissociation ,variaton of D/d with `'alpha'` is given by ?

Options

A. (a)
B. (b)
C. (c)
D. (d) none of these

Answer

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

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The dissociation reaction can be represented as:
\[ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \]
Here, one mole of \( PCl_5 \) produces one mole of \( PCl_3 \) and one mole of \( Cl_2 \). Thus, the total number of moles after dissociation is \( n = 2 \).

2. **Defining Degree of Dissociation**:
The degree of dissociation (\( \alpha \)) is defined as the fraction of the original substance that has dissociated. If we start with 1 mole of \( PCl_5 \), after dissociation:
- Moles of \( PCl_5 \) left = \( 1 - \alpha \)
- Moles of \( PCl_3 \) produced = \( \alpha \)
- Moles of \( Cl_2 \) produced = \( \alpha \)

Therefore, the total moles at equilibrium:
\[ \text{Total moles} = (1 - \alpha) + \alpha + \alpha = 1 + \alpha \]

3. **Calculating Theoretical Vapor Density**:
The theoretical vapor density (\( D \)) is calculated using the formula:
\[ D = \frac{M}{2} \quad \text{(where \( M \) is the molar mass)} \]
For \( PCl_5 \), the molar mass is 208.24 g/mol, hence:
\[ D = \frac{208.24}{2} = 104.12 \text{ g/L} \]

4. **Calculating Observed Vapor Density**:
The observed vapor density (\( d \)) can be related to the degree of dissociation:
\[ d = \frac{M}{2(1 + \alpha)} \]
Where \( M \) is the molar mass of the products. Since we have \( PCl_3 \) and \( Cl_2 \), we can calculate the effective molar mass at equilibrium.

5. **Relating \( D \) and \( d \)**:
We can express the relationship between \( D \) and \( d \) in terms of \( \alpha \):
\[ \alpha = \frac{D - d}{(n - 1)d} \]
Substituting \( n = 2 \):
\[ \alpha = \frac{D - d}{d} \]
Rearranging gives:
\[ \alpha = \frac{D}{d} - 1 \]

6. **Final Relationship**:
The relationship can be expressed as:
\[ \frac{D}{d} = \alpha + 1 \]
This shows that as \( \alpha \) increases, \( \frac{D}{d} \) also increases linearly.

### Conclusion:
The variation of \( \frac{D}{d} \) with \( \alpha \) is linear, with a slope of 1 and an intercept of 1 on the graph of \( \frac{D}{d} \) versus \( \alpha \).