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Question: Answered & Verified by Expert
In a constant volume calorimeter, \( 3.5 \mathrm{~g} \) of a gas with molecular weight \( 28 \) was burnt in excess oxygen at \( 298.0 \mathrm{~K} \). The temperature of the calorimeter was found to increase from \( 298.0 \mathrm{~K} \) to \( 298.45 \mathrm{~K} \) 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
ChemistryThermodynamics (C)JEE Main
Solution:
1044 Upvotes Verified Answer
The correct answer is: 9
Energy released by combustion of 3.5 g gas
=2.5×(298.45-298)kJ
Energy released by 1 mole of gas =2.5×0.453.528=9 kJmol-1

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