The change in internal energy for the combustion of 1.0 mol of octane at a pressure of 1.0 atm is 5084.5kJ . If the cha...
The change in internal energy for the combustion of 1.0 mol of octane at a pressure of 1.0 atm is 5084.5kJ . If the change in enthalpy is 5074.3kJ , how much work is done during the combustion?
Homework Answers
Enthalpy is defined as H = U + pV, so
dH = dU + pdV + Vdp.
Assuming constant pressure of 1.0 atm, the final term
vanishes.
Work is defined as dW = pdV, so
dH = dU + dW
dW = dH - dU
Thus,
dW = 5074.3 - 5084.5
dW = -10.2 KJ
Add Answer to:
The change in internal energy for the combustion of 1.0 mol of octane at a pressure of 1.0 atm is 5084.5kJ . If the cha...
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