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 2.5kJK- the numerical value for the enthalpy of combustion of the gas in  kJmol-. is:

Since the combustion takes place at constant volume, the energy released ΔU will be given as:ΔU=Z×ΔT×Mm                                  ... (1)Where          Z= heat capacity of the calorimeter =2.5kJK−ΔT= change in temperature =298.45−298.0=0.45KM= molar mass of gas =28g mol −m = mass of the substance =5.3gSubstituting the values in (1), we get :ΔU=2.5kJK−×0.45K×28gmol−5.3g=9.0kJmol−.