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IIT JEE Advanced 2009 Paper 2
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Question : 31
Total: 57
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
k
J
K
−
1
, the numerical value for the enthalpy of combustion of the gas in
k
J
mol
−
1
is ______.
Your Answer:
Validate
Solution:
The rise in temperature is 298.45 K - 298.0 K = 0.45 K.
Energy released at constant volume due to combustion of 3.5 g of a gas = 2.5 × 0.45.
Hence energy released due to combustion of 28 gram (i.e., 1 mole) of a gas = 2.5 × 0.45 × 28/3.5 = 9 KJ/mol.
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