Homework Help: Entropy change, simple question

lesodk · Apr 21, 2008

How do i calculate the entropy change for an isobaric process, if i know
T1, T2, n and Cv ?

Thanks,

astrorob · Apr 21, 2008

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

Try looking there.

lesodk · Apr 21, 2008

There is nothing there about entropy change

Hootenanny · Apr 21, 2008

What is the definition of entropy?

lesodk · Apr 21, 2008

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 thats the case in my question.

Hootenanny · Apr 21, 2008

lesodk said: ↑

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.

lesodk said: ↑

The forumula only holds for reversible processes and i don't know if thats 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?

lesodk · Apr 21, 2008

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?

Hootenanny · Apr 21, 2008

lesodk said: ↑

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 C_v. Since the process is isobaric, you need C_p. Apart from that your expression is correct.

lesodk · Apr 21, 2008

thanks alot