Entropy change in adiabatic process
The 2nd law is 

$d S \ge \displaystyle \frac{d q}{T}$

so the best you can do is zero for a reversible process.   Or if you don't like inequalities, 

$d S = \displaystyle \frac{d q}{T} + d S_{irr}$

where $d S_{irr}$ is the entropy production due to the irreversible processes within the system.

You can find a intuitive in depth discussion in Prigogine's book:

Modern Thermodynamics: From Heat Engines to Dissipative Structures

http://books.google.com/books?id=6qFxngEACAAJ&amp;dq=Modern+Thermodynamics:+From+Heat+Engines+to+Dissipative+Structures&amp;hl=en&amp;sa=X&amp;ei=IveUU8HvH8KVqAbDvILoBA&amp;ved=0CCcQ6AEwAA