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Heat of combustion of ethanol at constant pressure and at temperature TK is found to be $- q J {mol}^{- 1}$ . Hence, heat of combustion (in $J {mol}^{- 1}$ ) of ethanol at the same temperature and at constant volume will be

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RT-q

-(q+RT)

q-RT

q +RT

answer is A.

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Detailed Solution

$\begin{array}{l} C_{2} H_{5} OH (l) + 3 O_{2} (g) ⟶ 2 {CO}_{2} (g) + 3 H_{2} O (l) \\ {Δn}_{g} = - 1 \\ q_{v} = ΔH - {Δn}_{g} RT = - q + RT \end{array}$

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