Heat of combustion of ethanol at constant pressure and at temperature T K (= 298 K) is found to be -q J mol-1. Hence, heat of combustion in J mol-1 of ethanol at the same temperature at constant volume will be

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RT-q

-(RT+q)

q-RT

q+RT

answer is A.

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Detailed Solution

C2H5OH l+3O2g → 2CO2g + 3H2Ol ∆ng=2-3=-1 so ∆U=∆H-∆ngRT ( since ∆H=-q) =-q+RT

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