Entropy change, simple question

  • Thread starter lesodk
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  • #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,
 

Answers and Replies

  • #2
astrorob
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h ttp://en.wikipedia.org/wiki/Isobaric_process

Try looking there.
 
  • #3
lesodk
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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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entropy change = dS = int( dQ/T )
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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entropy change = dS = int( dQ/T )
where you integrate from state 1, to state 2.
Correct, you can now use this in conjunction with the information which astrorob provided to calculate the entropy change.
The forumula only holds for reversible processes and i don't know if that's the case in my question.
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.

Does that make sense?
 
  • #7
lesodk
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so i shoud insert the value for dQ = nCvdT and by that getting
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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so i shoud insert the value for dQ = nCvdT and by that getting
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?
The only thing that concerns me is the Cv. Since the process is isobaric, you need Cp. Apart from that your expression is correct.
 
  • #9
lesodk
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thanks alot
 

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