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प्रश्न
In a constant volume calorimeter, 3.5 g of gas with molecular weight 28 was burnt in excess oxygen at 298 K. The temperature of the calorimeter was found to increase from 298 K to 298.45 K due to the combustion process. Given that the calorimeter constant is 2.5 kJ K−1. Calculate the enthalpy of combustion of the gas in kJ mol−1.
उत्तर
Given, Ti = 298 K
Tf = 298.45 K
k = 2.5 kJ K−1
m = 3.5 g
Mm = 28
heat evolved = k∆T
∆HC = k (Tf – Ti)
∆HC = 2.5 kJ K−1 (298.45 – 298) K−1
∆HC = 1.125 kJ
∆HC = `1.125/3.5 xx 28` kJ mol−1
∆HC = 9 kJ mol−1
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