# Entropy and Internal Energy in irreversible adiabatic process

I have a question related to title. Please see the following discussion:

In an irreversible process:

1. $$ΔU_\mathrm{irr} = q_\mathrm{irr} + w_\mathrm{irr}$$
2. $$ΔU_\mathrm{rev} = q_\mathrm{rev} + w_\mathrm{rev}$$
3. $$ΔU_\mathrm{irr} = ΔU_\mathrm{rev}$$ ($$∵ U$$ is a state function)
4. $$|w_\mathrm{rev}| \ge |w_\mathrm{irr}|$$; $$w_\mathrm{rev} \le w_\mathrm{irr}$$
5. $$q_\mathrm{irr} = q_\mathrm{rev} = 0$$

And there is a contradiction: $$ΔU_\mathrm{irr} = ΔU_\mathrm{rev}, w_\mathrm{rev} \le w_\mathrm{irr}$$ How is it possible?

Or how is the gap $$|w_\mathrm{rev} - w_\mathrm{irr}|$$ generated? Is this related to entropy?

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• Delta U_rev <> Delta U_irr, as W_rev <> W_irr. // From the same initial state, you cannot reach the same final state by rev. and irr. adiabatic compression. Commented Jun 10 at 13:15
• There is no adiabatic reversible path between the same two end states as far an adiabatic irreversible process. Commented Jun 10 at 13:19

For an ideal gas undergoing an adiabatic free expansion, a simple example of an irreversible adiabatic process, no work is done, so $$q=w=0$$ implying $$\Delta U = w+q = 0$$ and it follows (since the gas is ideal) that T is constant.
For any reversible adiabatic process, on the other hand, work is done, so $$w \ne 0, ~q=0$$ which means that $$\Delta U \ne 0$$ and (since the gas is ideal) T is not constant.
It follows that the final states for reversible and irreversible adiabatic expansions starting from the same state differ, and therefore that $$\Delta U_\mathrm{irr} \ne \Delta U_\mathrm{rev}$$ It follows that point (3) in the question is false (despite $$U$$ being a state function).