Irrev. adiabatic process entropy >0 or =0?

[sparkle123]

My textbook says that
Delta S (sys) > 0 for irrev. ad. proc., closed syst
(or see http://www.britannica.com/EBchecked/topic/5898/adiabatic-process if you don't believe me)

but since Delta S = 0 for reversible adiabatic process and entropy is a state function,
shouldn't Delta S = 0 for irreversible adiabatic process = 0 too?

 

[rock.freak667]

I don't understand your justification for ΔS = 0 for an irreversible process.

By virtue of being an irreversible process ΔS > 0.

 

[sparkle123]

i thought entropy being a state function means it is path dependent so as long as initial and final states are the same, ΔS is the same.
thus ΔS for adiabatic reversible should equal ΔS for adiabatic irreversible (for same initial, and final states) ?
Thanks so much!

 

[Andrew Mason]

i thought entropy being a state function means it is path dependent so as long as initial and final states are the same, ΔS is the same.
thus ΔS for adiabatic reversible should equal ΔS for adiabatic irreversible (for same initial, and final states) ?

Entropy is indeed a state function. But are the final states the same for an adiabatic quasi-static (reversible) expansion and an adiabatic free (irreversible) expansion? Hint: apply the first law to determine the change in internal energy in each case.

AM