<span>Answer:
I guess there is is a typo in your question both change in enthalpy and change in internal energy for the combustion are negative:
â†H = -5074 kJ
â†U = - 5084.2 kJ
Work done to surrounding in a process is given by the integral:
W = â« P dV from initial to final volume
for a constant pressure process the integral simplifies to:
W = P â™ â« dV = Pâ™â†V
Enthalpy is defined as:
H = U + Pâ™V
So the change in enthalpy and the change in internal energy are related as:
â†H = â†U + â†(Pâ™V)
for a constant pressure process:
â†H = â†U + Pâ™â†V
Hence,
â†H = â†U + W
=>
W = â†H - â†U
= -5074 kJ - (-5084.2) kJ
= 10.2 kJ</span>
