# Entropy change in a reversible adiabatic process

1. Sep 28, 2012

### Bipolarity

For a reversible process, I imagine it is correct to say that

$dS = \frac{dq}{T}$ where all quantities refer to system quantities (not the surrounding).

However, for an adiabatic process, $dq = 0$.

Thus, should it be the case that for an adiabatic reversible process,

$dS = \frac{dq}{T} = \frac{0}{T} = 0$

Please correct me if I am wrong, and point out the flaws in my reasoning.

BiP

2. Sep 28, 2012

### DrDu

Entirely correct.