A mixture of a mol of `C_(3) H_(8)` and `b` mol of `C_(2) H_(4)` was kept is a container of `V L` exerts a pressure of 4.93 atm at temperature `T`. Mixture was burnt in presence of `O_(2)` to convert `C_(3) H_(8)` and `C_(2) H_(4)` into `CO_(2)` in the container at the same temperature. The pressure of gases after the reaction and attaining the thermal equilirium with atomsphere at temperature `T` was found to be 11.08 atm.
The moles of `O_(2)` needed for combustion at temperature `T` is equal to

Options

A. `14 a`
B. `14 b`
C. `15 a`
D. `12 b`

Explanation

To solve the problem, we need to determine the moles of \( O_2 \) required for the complete combustion of a mixture of propane (\( C_3H_8 \)) and ethene (\( C_2H_4 \)).

### Step-by-step Solution:

1. **Understanding the Initial Conditions:**
- We have \( A \) moles of \( C_3H_8 \) and \( B \) moles of \( C_2H_4 \) in a container of volume \( V \) liters.
- The initial pressure of the mixture is given as \( P_1 = 4.93 \, \text{atm} \).

2. **Using the Ideal Gas Law:**
- According to the ideal gas law, \( PV = nRT \).
- For the initial state, we can write:
\[ 4.93V = (A + B)RT \quad \text{(Equation 1)} \]

3. **Combustion Reaction:**
- The combustion reactions for the gases are:
- For propane:
\[ C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O \]
- For ethene:
\[ C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O \]
- From these reactions, we can determine the moles of \( O_2 \) required:
- \( 5A \) moles of \( O_2 \) for \( A \) moles of \( C_3H_8 \)
- \( 3B \) moles of \( O_2 \) for \( B \) moles of \( C_2H_4 \)
- Therefore, the total moles of \( O_2 \) required is:
\[ O_2 = 5A + 3B \]

4. **Final Conditions After Combustion:**
- After combustion, the pressure of the gases is \( P_2 = 11.08 \, \text{atm} \).
- The only gas present after combustion is \( CO_2 \), and the total moles of \( CO_2 \) produced is:
\[ 3A + 2B \]
- Using the ideal gas law again, we can write:
\[ 11.08V = (3A + 2B)RT \quad \text{(Equation 2)} \]

5. **Setting Up the Ratio:**
- Dividing Equation 2 by Equation 1 gives:
\[ \frac{11.08}{4.93} = \frac{3A + 2B}{A + B} \]
- Simplifying this, we find:
\[ 11.08(A + B) = 4.93(3A + 2B) \]
- Expanding and rearranging:
\[ 11.08A + 11.08B = 14.79A + 9.86B \]
\[ 11.08A - 14.79A + 11.08B - 9.86B = 0 \]
\[ -3.71A + 1.22B = 0 \]
\[ B = 3.04A \quad \text{(approximately)} \]

6. **Substituting Back to Find Moles of \( O_2 \):**
- Now substitute \( B \) back into the equation for \( O_2 \):
\[ O_2 = 5A + 3(3.04A) = 5A + 9.12A = 14.12A \]

7. **Final Result:**
- The moles of \( O_2 \) needed for combustion at temperature \( T \) is approximately:
\[ O_2 \approx 14A \]

### Conclusion:
The moles of \( O_2 \) needed for combustion is \( 14A \).