How would one calculate the entropy change for an adiabatic irreversible process?
I agree with you over here.  


  $\underbrace{\Delta U}_{\text{state function}} = \underbrace{\Delta W}_{\text{path function}}$
  
  And $W_{\text{reversible}}\neq W_{\text{irreversible}}$   


But this part seems to be wrong.   


  If two states $A$ and $ B$ are connected by a reversible path, they can never be connected by an irreversible path during an adiabatic   


Just because the process is adiabatic irreversible it doesn't mean that the reversible path needs to be adiabatic as well.     

The only requisite for calculating the entropy for an irreversible process is simply that the path taken must be reversible.

So until now we have established that the heat transfer does not have to be 0. For an adiabatic process.     

Even though $W_{\text{irrev}} \neq W_{\text{rev}}$  the extra parts will be balanced by the heat transfer in the reversible process.   

For example let's take adiabatic expansion.  

$$W_{\text{irrev}} &gt; W_{\text{rev}}$$  

Numerically speaking in an absolutely hypothetical situation.   

\begin{align} 
\ce{ ∆U &amp;-&gt; 10~J}\\   
\ce{ W_{irrev} &amp;-&gt; 10 ~J}\\  
\ce{ W_{rev} &amp;-&gt; 5~J}\\  
\end{align}  

So the possible proposed reversible path will automatically have a transfer of heat into the gas . Whose value will be 5J.
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