asdf1
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When proving the equation,
dU=TdS-PdV
why is dw= PdV not dw= PdV +vdP?
dU=TdS-PdV
why is dw= PdV not dw= PdV +vdP?
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.
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.
One participant notes the absence of their textbook, which may limit their ability to reference specific details or definitions during the discussion.
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.asdf1 said:When proving the equation,
dU=TdS-PdV
why is dw= PdV not dw= PdV +vdP?