Arguing work is an inexact differential

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Homework Help Overview

The discussion revolves around the concept of inexact differentials in thermodynamics, specifically focusing on the term pdV in the context of the fundamental thermodynamic relation TdS=pdV+dU. The original poster seeks to understand why pdV is considered inexact while other terms like dS and dU are exact differentials.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand the nature of pdV as an inexact differential and questions the reasoning behind this classification. Participants explore what additional terms might be needed to convert pdV into an exact differential, specifically considering the term Vdp.

Discussion Status

Participants are actively engaging with the concepts, with some suggesting potential terms to add to pdV to achieve exactness. There is a recognition that pdV is not exact unless certain conditions are met, indicating a productive exploration of the topic.

Contextual Notes

The original poster expresses limited understanding of the topic, indicating a need for clarification on the distinction between exact and inexact differentials in thermodynamics.

Sekonda
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Hey,

Given the fundamental thermodynamic relation (dN=0) ; TdS=pdV+dU, I have to argue that pdV is inexact. I know that dS and dU are exact differentials and that the integral of an exact differential around a closed loop will give a zero result, not too sure where to start, the question is only a couple of marks so the answer need only be brief but my understanding is limited as to why pdV, the work, has to be an inexact differential.

Any help appreciated!

Thanks.
 
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Hey Sekonda! :smile:

What would you add to pdV to make it an exact differential? :wink:
 
Vdp?

I'm still not sure I'm going to have to think about this a bit more!
 
Is this the correct term to add to pdV to make it an exact differential; if so, how does this imply that pdV is inexact?
 
Sekonda said:
Is this the correct term to add to pdV to make it an exact differential

Yes, pdV + Vdp is an exact differential … it's exactly d(pV).

So that means that (unless Vdp = 0, ie p is constant) pdV is not exact. :smile:
 
Ahh yes that does make sense! I'm always bad at 'arguing', 'showing' or 'proofing' questions; thanks man.

Cheers!
 

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