Entropy change in a reversible adiabatic process

  • Thread starter Bipolarity
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  • #1
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For a reversible process, I imagine it is correct to say that

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

However, for an adiabatic process, [itex] dq = 0 [/itex].

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

[itex] dS = \frac{dq}{T} = \frac{0}{T} = 0 [/itex]

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

BiP
 

Answers and Replies

  • #2
DrDu
Science Advisor
6,145
818
Entirely correct.
 

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