## Calculating change in entropy for a process? (9.3)

Volume: $\Delta S = nR\ln \frac{V_{2}}{V_{1}}$
Temperature: $\Delta S = nC\ln \frac{T_{2}}{T_{1}}$

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### Calculating change in entropy for a process? (9.3)

In 9.3, page 325 it reads:

"To calculate a change in entropy for a process, we find a reversible path between the initial and final states. It is immaterial whether the actual process is irreversible or reversible. Because entropy is a state function, the change for that path will be the same as that for the irreversible path, provided the two paths have the same initial and final states."

Could someone further elaborate/ explain what they are trying to say here?
Thanks! :-)

Ryan Sydney Beyer 2B
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### Re: Calculating change in entropy for a process? (9.3)

I think what the book is trying to say here is that the change in entropy (delta S) is going to be the same for an irreversible path or a reversible path. This is because they said entropy is a state function so all that matter is the final state minus the initial state and nothing in between.

Naveed Zaman 1C
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### Re: Calculating change in entropy for a process? (9.3)

It's just an explanation for how to calculate entropy for certain reactions. We don't worry about whether a reaction can only go in one way or another. This allows us to do things like find the entropy for reactions that are not necessarily common. Again, it has to do with the fact that entropy is a state property.