The combustion of benzene $$(l)$$ gives $$C{O_2}\left( g \right)$$  and $${H_2}O\left( l \right)$$ . Given that heat of combustion of benzene at constant volume is $$ - 3263.9\,kJ\,mo{l^{ - 1}}$$   at $${25^ \circ }C$$ ; heat of combustion ( in $$kJ\,mo{l^{ - 1}}$$  ) of benzene at constant pressure will be :
$$\left( {R = 8.314\,J{K^{ - 1}}\,mo{l^{ - 1}}} \right)$$

Solution :
$$\eqalign{ & {C_6}{H_6}\left( {\text{l}} \right) + \frac{{15}}{2}{O_2}\left( g \right) \to 6C{O_2}\left( g \right) + 3{H_2}O\left( {\text{l}} \right) \cr & \Delta {n_g} = 6 - \frac{{15}}{2} = - \frac{3}{2} \cr & \Delta H = \Delta U + \Delta {n_g}RT \cr & = - 3263.9 + \left( { - \frac{3}{2}} \right) \times 8.314 \times 298 \times {10^{ - 3}} \cr & = - 3263.9 + \left( { - 3.71} \right) \cr & = - 3267.6\,kJ\,mo{l^{ - 1}} \cr} $$