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,
The discussion revolves around calculating the entropy change for an isobaric process, given specific parameters such as initial and final temperatures, number of moles, and heat capacity.
Participants are actively engaging with the concepts and questioning the assumptions behind the formulas. Some guidance has been offered regarding the use of heat capacities and the nature of the process, but no consensus has been reached on the final approach to the calculation.
There is uncertainty about whether the process in question is reversible, which affects the applicability of the entropy change formula. Additionally, the distinction between using Cv and Cp for the isobaric process is under discussion.
Correct, you can now use this in conjunction with the information which astrorob provided to calculate the entropy change.lesodk said: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.lesodk said:The forumula only holds for reversible processes and i don't know if that's the case in my question.
The only thing that concerns me is the Cv. Since the process is isobaric, you need Cp. Apart from that your expression is correct.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?