In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen

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 $2.5 \mathrm{~kJ} \mathrm{~K}^{-1}$, the numerical value for the enthalpy of combustion of the gas in $\mathrm{kJ} \mathrm{mol}^{-1}$ is\

Professor PC Thomas Classes · Answer

(9) : Given,\ $C_{V}=2.5 \mathrm{~kJ} \mathrm{~K}^{-1}=2500 \mathrm{~J} \mathrm{~K}^{-1}$\ $\Delta T=T_{2}-T_{1}=298.45-298=0.45 \mathrm{~K}$\ $\Delta H$ due to combustion of 3.5 g gas\ $=C_{V} 	imes \Delta T=2500 	imes 0.45=1125 \mathrm{~J}$\ Given, molecular weight of gas $=28$.\ $	herefore \quad 28 \equiv 1$ mole\ Hence, $\Delta H$ due to combustion of 1 mole of gas $$ =\frac{1125}{3.5} 	imes 28=9000 \mathrm{~J} $$ $	herefore \quad \Delta H$ in $\mathrm{kJ} \mathrm{mol}^{-1}=9 \mathrm{~kJ} \mathrm{~mol}^{-1}$.\