At `25^(@)C` 1 mole of butane is heated then `CO_(2) and H_(2)O` liquid is formed work done is

Options

A. 75.6 L atm
B. 85.6 L atm
C. 50.3 L atm
D. None of these

Explanation

To solve the problem, we need to determine the work done when 1 mole of butane (C₄H₁₀) is combusted to form carbon dioxide (CO₂) and water (H₂O) at 25°C.

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The combustion of butane can be represented as:
\[ C_4H_{10} + O_2 \rightarrow CO_2 + H_2O \]
Balancing the equation, we get:
\[ C_4H_{10} + \frac{13}{2} O_2 \rightarrow 4 CO_2 + 5 H_2O \]

2. **Identify the Moles of Gases:**
- **Reactants:**
- Butane (C₄H₁₀): 1 mole (gas)
- Oxygen (O₂): \(\frac{13}{2}\) moles (gas)
- Total moles of reactants = \(1 + \frac{13}{2} = \frac{15}{2}\) moles
- **Products:**
- Carbon Dioxide (CO₂): 4 moles (gas)
- Water (H₂O): 5 moles (liquid, not counted in gaseous moles)
- Total moles of products = 4 moles (only CO₂ is gas)

3. **Calculate the Change in Moles of Gas (ΔNG):**
\[ \Delta N_G = \text{Moles of gaseous products} - \text{Moles of gaseous reactants} \]
\[ \Delta N_G = 4 - \frac{15}{2} = 4 - 7.5 = -3.5 \]

4. **Use the Work Done Formula:**
The work done (W) in a constant pressure process can be calculated using the formula:
\[ W = -\Delta N_G \cdot R \cdot T \]
where:
- \(R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}\) (universal gas constant)
- \(T = 25^\circ C = 298 \, \text{K}\)

5. **Substituting the Values:**
\[ W = -(-3.5) \cdot 0.0821 \cdot 298 \]
\[ W = 3.5 \cdot 0.0821 \cdot 298 \]

6. **Calculate the Work Done:**
\[ W = 3.5 \cdot 0.0821 \cdot 298 \approx 85.6 \, \text{L atm} \]

### Conclusion:
The work done when 1 mole of butane is combusted is approximately **85.6 L atm**. Thus, the correct answer is option 2.