Work done during the combustion of one mole of `CH_(4)` in bomb calorimeter is

Options

A. zero
B. `-101 J`
C. `-24.2J`
D. `-1J`

Answer

To solve the question regarding the work done during the combustion of one mole of methane (CH₄) in a bomb calorimeter, we can follow these steps:

### Step 1: Write the combustion reaction
The combustion of methane can be represented by the following balanced chemical equation:
\[ \text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \]

### Step 2: Identify the change in the number of moles of gas
In the balanced equation, we have:
- Reactants: 1 mole of CH₄ and 2 moles of O₂, which gives a total of 3 moles of gas.
- Products: 1 mole of CO₂ and 0 moles of gaseous water (since water is in liquid form).

Thus, the total number of moles of gas before the reaction is 3, and after the reaction, it is 1.

### Step 3: Calculate the change in the number of moles of gas (Δn)
\[ \Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 1 - 3 = -2 \]

### Step 4: Use the formula for work done in a constant volume process
In a bomb calorimeter, the combustion occurs at constant volume. The work done (W) in a constant volume process can be expressed as:
\[ W = -P \Delta V \]
However, since the volume is constant, the change in volume (ΔV) is equal to zero.

### Step 5: Conclusion on work done
Since ΔV = 0, the work done during the combustion of one mole of methane in a bomb calorimeter is:
\[ W = 0 \]

### Final Answer
The work done during the combustion of one mole of CH₄ in a bomb calorimeter is **0**.

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