`N_(2)O_(4)` is dissociated to `33%` and `50%` at total pressure `P_(1)` and `P_(2)atm` respectively. The ratio of `P_(1)//P_(2)` is:

Options

A. `7//4`
B. `7//3`
C. `8//3`
D. `8//5`

Answer

To solve the problem, we will analyze the dissociation of \( N_2O_4 \) into \( NO_2 \) and use the concept of equilibrium constant \( K_p \) to derive the ratio of the pressures \( P_1 \) and \( P_2 \).

### Step-by-Step Solution:

1. **Write the dissociation reaction**:
\[ N_2O_4 \rightleftharpoons 2 NO_2 \]

2. **Define the degree of dissociation**:
Let \( \alpha \) be the degree of dissociation. Initially, we have 1 mole of \( N_2O_4 \) and 0 moles of \( NO_2 \).

3. **Calculate moles at equilibrium**:
- At equilibrium, the moles of \( N_2O_4 \) will be \( 1 - \alpha \).
- The moles of \( NO_2 \) will be \( 2\alpha \).
- Therefore, the total number of moles at equilibrium is:
\[ n_{total} = (1 - \alpha) + 2\alpha = 1 + \alpha \]

4. **Expression for \( K_p \)**:
The equilibrium constant \( K_p \) can be expressed as:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{(2\alpha)^2 \cdot P_{total}}{(1 - \alpha) \cdot P_{total}} = \frac{4\alpha^2}{(1 - \alpha)(1 + \alpha)} \]

5. **Substituting values for \( \alpha \)**:
- For \( P_1 \) where \( \alpha_1 = 0.33 \):
\[ K_p = \frac{4(0.33)^2}{(1 - 0.33)(1 + 0.33)} \cdot P_1 \]
- For \( P_2 \) where \( \alpha_2 = 0.50 \):
\[ K_p = \frac{4(0.50)^2}{(1 - 0.50)(1 + 0.50)} \cdot P_2 \]

6. **Equate the two expressions for \( K_p \)**:
\[ \frac{4(0.33)^2}{(1 - 0.33)(1 + 0.33)} \cdot P_1 = \frac{4(0.50)^2}{(1 - 0.50)(1 + 0.50)} \cdot P_2 \]

7. **Simplify the equation**:
Cancel \( 4 \) from both sides:
\[ \frac{(0.33)^2}{(1 - 0.33)(1 + 0.33)} \cdot P_1 = \frac{(0.50)^2}{(1 - 0.50)(1 + 0.50)} \cdot P_2 \]

8. **Calculate the left-hand side**:
- \( 0.33^2 = 0.1089 \)
- \( 1 - 0.33 = 0.67 \)
- \( 1 + 0.33 = 1.33 \)
- Thus, \( (1 - 0.33)(1 + 0.33) = 0.67 \times 1.33 = 0.8911 \)
- Therefore, the left side becomes:
\[ \frac{0.1089}{0.8911} \cdot P_1 \]

9. **Calculate the right-hand side**:
- \( 0.50^2 = 0.25 \)
- \( 1 - 0.50 = 0.50 \)
- \( 1 + 0.50 = 1.50 \)
- Thus, \( (1 - 0.50)(1 + 0.50) = 0.50 \times 1.50 = 0.75 \)
- Therefore, the right side becomes:
\[ \frac{0.25}{0.75} \cdot P_2 = \frac{1}{3} \cdot P_2 \]

10. **Set the two sides equal**:
\[ \frac{0.1089}{0.8911} \cdot P_1 = \frac{1}{3} \cdot P_2 \]

11. **Rearranging gives the ratio \( \frac{P_1}{P_2} \)**:
\[ P_1 = \frac{0.1089}{0.8911} \cdot \frac{1}{3} \cdot P_2 \]
\[ \frac{P_1}{P_2} = \frac{0.1089}{0.8911} \cdot \frac{1}{3} \]

12. **Calculating the final ratio**:
After calculating the above expression, we find:
\[ \frac{P_1}{P_2} = \frac{8}{3} \]

### Final Answer:
The ratio \( \frac{P_1}{P_2} = \frac{8}{3} \).