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Why is entropy not zero for an irreversible adiabatic process though q=0? Ideally entropy is a state function so entropy should be zero.

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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&dq=Modern+Thermodynamics:+From+Heat+Engines+to+Dissipative+Structures&hl=en&sa=X&ei=IveUU8HvH8KVqAbDvILoBA&ved=0CCcQ6AEwAA

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Because there is many way to produce work and there is always irreversibility associated with real production. This is why work and heat variation are not exact differential. The path (the way) used does matter. In your case:

$$ \mathrm{d}U = - \delta W$$

Adiabatic (no heat) irreversible (real) transformation produce entropy. Isentropic means adiabatic and reversible. Therefore you will produce work and irreversibilities (eg.: friction) which will account for internal energy variation.

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If you go between two thermodynamic equilibrium states of a system using an adiabatic reversible process, you won't be able to get between these same two states adiabatically by means of an irreversible process. To get between the same two states using an irreversible process, you will have to remove some heat.

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