Proof of dU=TdS-PdV: dw=PdV Not dw=PdV+vdP

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Homework Help Overview

The discussion revolves around the thermodynamic equation dU=TdS-PdV, specifically focusing on the differentiation of work done, questioning why dw=PdV is preferred over dw=PdV+vdP.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the definitions and implications of work in thermodynamics, questioning the role of pressure and volume changes in the context of internal energy.

Discussion Status

Some participants are examining the reasoning behind the equations, with one providing a rationale related to the nature of work and internal energy. There is an ongoing exploration of the concepts without a clear consensus.

Contextual Notes

One participant notes the absence of their textbook, which may limit their ability to reference specific details or definitions during the discussion.

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When proving the equation,
dU=TdS-PdV

why is dw= PdV not dw= PdV +vdP?
 
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I think it is as you wrote, but don't quote me. I'm at school and don't have my book infront of me.
 
asdf1 said:
When proving the equation,
dU=TdS-PdV

why is dw= PdV not dw= PdV +vdP?
Work (force x distance) is done only if the pressure (force/area) acts over some change in volume (area x distance moved). If there is no change in volume, the force does not act over any distance - ie. the pressure just builds up but does not move anything (ie. it just increases internal energy). So, VdP is part of the internal energy of the gas, not the work done by the gas.

AM
 
thank you!
 

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