Entropy in an isobar transformation

AI Thread Summary
The discussion focuses on the variation of entropy during an isobaric transformation, specifically for an ideal gas. It references the equation TdS = dU + PdV, indicating that with constant pressure, the change in entropy can be expressed as ΔS = ∫(dU/T) + P∫(dV/T). For an ideal gas, the relationships PV = nRT and dU = nCvdT are utilized to derive the entropy change. Participants confirm the calculations and express appreciation for the clarity of the explanation. The conversation effectively clarifies the entropy variation in isobaric processes for ideal gases.
alialice
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What is the variation of entropy in an isobar transformation?
 
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alialice said:
What is the variation of entropy in an isobar transformation?
Are we dealing with an ideal gas? If so, we can work it out from: TdS = dU + PdV

If P is constant then:

\Delta S = \int \frac{dU}{T} + P\int \frac{dV}{T}

If this is an ideal gas where PV = nRT, then we can substitute nRT/V for P (or nRdT/P for dV) and nCvdT for dU. Can you work that out?

AM
 
Yes, it's just so!
Thank you very much! :)
 

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