`N_2O_4rarr2NO_2`
In the above equation, `alpha` varies with `(D)/(d)` according to:

Options

A.
B.
C.
D.

Explanation

To solve the problem regarding the dissociation of \( N_2O_4 \) into \( 2NO_2 \) and the relationship between the degree of dissociation (\( \alpha \)) and the ratio of vapor densities (\( \frac{D}{d} \)), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction:**
The reaction is given as:
\[ N_2O_4 \rightleftharpoons 2NO_2 \]
Here, one mole of \( N_2O_4 \) dissociates to form two moles of \( NO_2 \).

2. **Define the Degree of Dissociation (\( \alpha \)):**
The degree of dissociation (\( \alpha \)) is defined as the fraction of the original substance that has dissociated. If we start with 1 mole of \( N_2O_4 \), then after dissociation:
- Moles of \( N_2O_4 \) remaining = \( 1 - \alpha \)
- Moles of \( NO_2 \) formed = \( 2\alpha \)

3. **Calculate the Total Moles After Dissociation:**
The total moles after dissociation will be:
\[ \text{Total moles} = (1 - \alpha) + 2\alpha = 1 + \alpha \]

4. **Relate Vapor Density to Moles:**
The vapor density (\( D \)) is related to the number of moles. The vapor density before dissociation is:
\[ D = \text{Vapor density before dissociation} = P_0 \]
After dissociation, the vapor density (\( d \)) is given by:
\[ d = \text{Vapor density after dissociation} = \frac{\text{Total mass}}{\text{Total volume}} = \frac{M(1 + \alpha)}{V} \]
where \( M \) is the molar mass and \( V \) is the volume.

5. **Express \( D \) and \( d \) in Terms of \( P_0 \):**
The vapor density before dissociation is:
\[ D = P_0 \]
The vapor density after dissociation can be expressed as:
\[ d = P_0(1 + \alpha) \]

6. **Formulate the Ratio \( \frac{D}{d} \):**
Now, we can find the ratio of the vapor densities:
\[ \frac{D}{d} = \frac{P_0}{P_0(1 + \alpha)} = \frac{1}{1 + \alpha} \]

7. **Rearranging the Equation:**
Rearranging gives:
\[ D = d(1 + \alpha) \]
This can also be expressed as:
\[ \alpha = \frac{1}{\frac{D}{d}} - 1 \]

8. **Graphical Representation:**
The equation \( \alpha = \frac{1}{\frac{D}{d}} - 1 \) indicates a linear relationship between \( \alpha \) and \( \frac{D}{d} \). As \( \frac{D}{d} \) increases, \( \alpha \) increases, leading to a straight line graph.

### Conclusion:
The graph of \( \alpha \) versus \( \frac{D}{d} \) will be a straight line, confirming that the relationship is linear.