Question

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

Answer 1

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