- #1

lesodk

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How do i calculate the entropy change for an isobaric process, if i know

T1, T2, n and Cv ?

Thanks,

T1, T2, n and Cv ?

Thanks,

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- Thread starter lesodk
- Start date

- #1

lesodk

- 16

- 0

How do i calculate the entropy change for an isobaric process, if i know

T1, T2, n and Cv ?

Thanks,

T1, T2, n and Cv ?

Thanks,

- #2

astrorob

- 140

- 0

h ttp://en.wikipedia.org/wiki/Isobaric_process

Try looking there.

Try looking there.

- #3

lesodk

- 16

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There is nothing there about entropy change

- #4

Hootenanny

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What is the definition of entropy?

- #5

lesodk

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where you integrate from state 1, to state 2.

The forumula only holds for reversible processes and i don't know if that's the case in my question.

- #6

Hootenanny

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Correct, you can now use this in conjunction with the information which astrorob provided to calculate the entropy change.entropy change = dS = int( dQ/T )

where you integrate from state 1, to state 2.

Indeed, the formula does only hold for a reversible process, but it may be useful to note that since entropy is a state function it is independent of the path taken and depends only on the initial and final states. Therefore, for any process, even an irreversible one, we can chose a reversible path between the initial and final states that allows us to calculate the change in entropy.The forumula only holds for reversible processes and i don't know if that's the case in my question.

Does that make sense?

- #7

lesodk

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dS = int( (n Cv dT)/T) = n Cv ( (1/T) dT ) = nCv ln(T2/T1) ?

Is that correct, even though the process is not reversible?

- #8

Hootenanny

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The only thing that concerns me is the C

dS = int( (n Cv dT)/T) = n Cv ( (1/T) dT ) = nCv ln(T2/T1) ?

Is that correct, even though the process is not reversible?

- #9

lesodk

- 16

- 0

thanks alot

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