The change in internal energy for the combustion of 1.0 mol of octane at a pressure of 1.0 atm is - 5084.1 kj. f the change in enthalpy is - 5074.3 kj, how much work is done during the combustion

Given:

Change in internal energy = ΔU = - 5084.1 kJ

Change in enthalpy = ΔH = - 5074.3 kJ

To determine:

The work done, W

Explanation:

Based on the first law of thermodynamics,

ΔH = ΔU + PΔV

the work done by a gas is given as:

W = - PΔV

Therefore:

ΔH = ΔU - W

W = ΔU-ΔH = - 5084.1 - (-5074.3) = - 9.8 kJ

Ans: Work done is - 9.8 kJ