Question

In a constant volume calorimeter, $3.5$ g of a gas with molecular weight $28$ was burnt in excess oxygen at $298.0$ K. The temperature of the calorimeter was found to increase from $298.0$ K to $298.45$ K due to the combustion process. Given that the heat capacity of the calorimeter is $2.5$ kJ K${}^{-1}$, the numerical value for the enthalpy of combustion of the gas in kJmol${}^{-1}$ is:

Answer 1

Energy release at constant volume due to combustion of 3.5 gm of a gas $=C\times \Delta T=2.5\times 0.45kJ$

Hence, energy released due to the combustion of 28 gm (i.e., 1 mole) of a gas $=2.5\times 0.45\times \frac{28}{3.5}=9kJmo{1}^{-1}$