`N_2O_4` is dissociated to `33%` and `40%` at total pressure `P_1 and P_2` atm respectively. Then the ratio `P_(1)//P_(2)` is

Options

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

Answer

To solve the problem, we need to analyze the dissociation of \( N_2O_4 \) into \( NO_2 \) and use the given percentages of dissociation to find the ratio \( \frac{P_1}{P_2} \).

### Step-by-Step Solution:

1. **Understanding the Dissociation Reaction**:
The dissociation of \( N_2O_4 \) can be represented as:
\[ N_2O_4 \rightleftharpoons 2 NO_2 \]

2. **Calculating Moles at 33% Dissociation**:
- Initial moles of \( N_2O_4 \) = 1 mole.
- At 33% dissociation, \( 0.33 \) moles of \( N_2O_4 \) dissociate to form \( 0.66 \) moles of \( NO_2 \).
- Remaining \( N_2O_4 \) = \( 1 - 0.33 = 0.67 \) moles.
- Total moles at this point = \( 0.67 + 0.66 = 1.33 \) moles.

3. **Calculating Mole Fractions**:
- Mole fraction of \( NO_2 \) at 33% dissociation:
\[ \chi_{NO_2} = \frac{0.66}{1.33} = \frac{66}{133} = \frac{1}{2} \]
- Mole fraction of \( N_2O_4 \):
\[ \chi_{N_2O_4} = \frac{0.67}{1.33} = \frac{67}{133} \]

4. **Calculating \( K_p \) for 33% Dissociation**:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{(\chi_{NO_2} \cdot P_1)^2}{\chi_{N_2O_4} \cdot P_1} \]
Substituting the values:
\[ K_p = \frac{\left(\frac{1}{2} P_1\right)^2}{\frac{67}{133} P_1} = \frac{\frac{1}{4} P_1^2}{\frac{67}{133} P_1} = \frac{133}{268} P_1 \]

5. **Calculating Moles at 40% Dissociation**:
- At 40% dissociation, \( 0.4 \) moles of \( N_2O_4 \) dissociate to form \( 0.8 \) moles of \( NO_2 \).
- Remaining \( N_2O_4 \) = \( 1 - 0.4 = 0.6 \) moles.
- Total moles = \( 0.6 + 0.8 = 1.4 \) moles.

6. **Calculating Mole Fractions for 40% Dissociation**:
- Mole fraction of \( NO_2 \):
\[ \chi_{NO_2} = \frac{0.8}{1.4} = \frac{8}{14} = \frac{4}{7} \]
- Mole fraction of \( N_2O_4 \):
\[ \chi_{N_2O_4} = \frac{0.6}{1.4} = \frac{6}{14} = \frac{3}{7} \]

7. **Calculating \( K_p \) for 40% Dissociation**:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{\left(\frac{4}{7} P_2\right)^2}{\frac{3}{7} P_2} \]
Substituting the values:
\[ K_p = \frac{\frac{16}{49} P_2^2}{\frac{3}{7} P_2} = \frac{16}{49} \cdot \frac{7}{3} P_2 = \frac{16 \cdot 7}{49 \cdot 3} P_2 = \frac{16}{21} P_2 \]

8. **Equating \( K_p \)**:
Since \( K_p \) is constant for a given temperature:
\[ \frac{133}{268} P_1 = \frac{16}{21} P_2 \]

9. **Finding the Ratio \( \frac{P_1}{P_2} \)**:
Rearranging gives:
\[ \frac{P_1}{P_2} = \frac{16 \cdot 268}{21 \cdot 133} \]
Simplifying:
\[ \frac{P_1}{P_2} = \frac{32}{21} \]

### Final Answer:
\[ \frac{P_1}{P_2} = \frac{32}{21} \]