0.16 g of methane was subjected to combustion at 27°C in a bomb calorimeter. The temperature of the calorimeter system (including water) was found to rise by 0.5°C. Calculate the heat of combustion of methane (i) at constant volume and (ii) at constant pressure. The thermal capacity of the calorimeter system is `17.7 kJ K^(-1)(R = 8.314 JK^(-1) mol^(-1))`.

Explanation

To solve the problem, we need to calculate the heat of combustion of methane at both constant volume and constant pressure. Let's break down the solution step by step.

### Step 1: Calculate the heat absorbed by the calorimeter
The heat absorbed by the calorimeter (q) can be calculated using the formula:
\[ q = C \times \Delta T \]
where:
- \( C \) is the thermal capacity of the calorimeter system (17.7 kJ/K),
- \( \Delta T \) is the change in temperature (0.5°C).
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