Entropy in an isolated system

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  • Thread starter Joe Cool
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Hello,

In my textbook I read this example:
A gas in an isolated system expands after pulling out a separating plate, so its volume increases and there is no work or heat exchange.

the entropy of if the ideal gas is
$$\Delta S = n R \ln \frac {V_1} {V_2}$$

and the second law of thermodynamics is
$$\Delta S = \int \frac {\delta Q} T = 0$$

But I didn't understand which of the two formula is the true one?
 

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  • #2
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The second equation only applies if you evaluate the change in entropy between the initial and final equilibrium states along a reversible path between these two states. So to get the entropy change between the initial and final states (resulting from an irreversible path between the same two states), you need to devise a reversible path, and then evaluate the integral for that path. What did you think the subscript "rev" meant in the correct equation: $$\Delta S=\int{\frac{dQ_{rev}}{T}}$$

For a cookbook recipe on how to do this, see my recent Physics Forums Insights article: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/
 
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Mapes
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And I'll weigh in with a note I wrote some time ago on determining entropy (because entropy can be enormously confusing). Your equations appear in sections 5 and 4, respectively. Note that section 4 is restricted to reversible processes. Free expansion isn't reversible.
 
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Thanks a lot, especially for your articles:smile:!
 

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