The discussion revolves around the thermodynamic identity involving the terms TdS and dU, specifically examining their placement in the equation and the implications of their signs. Participants explore the relationship between these terms in the context of thermodynamic processes, including reversible work and heat transfer.
Participants express differing views on the correct formulation of the thermodynamic identity and the implications of the signs of TdS and dU. No consensus is reached, and multiple competing interpretations remain present.
Participants rely on different assumptions regarding the definitions of work and heat transfer, which may affect their interpretations of the thermodynamic identity. The discussion does not resolve these assumptions or their implications.
Why do you want them to be on opposite side or have different sign if they are on the same side? What is you reason?PeteSampras said:TdS and dU
It all depends on what you presuppose. Tds - Pdv = dU, presupposes Pdv is the Reversible work done by the system which should decrease its internal energy if heat is not allowed to enter the system. Naturally if Tds is defined as the reversible heat transfer to the system which tends to increase the internal energy of the system, then the equation clearly gives you the net change of internal energy between two infinitesimally close thermodynamic states.PeteSampras said:PdV= TdS-dU
PdV is always the work done by the system on the surroundings, which always tends to decrease internal energy, and TdS is always the heat transferred from the surroundings to the system, which always tends to increase internal energy. So there is only one correct way of writing the equation: $$dU=TdS-PdV$$This equation describes more than just reversible heat and work. It provides the interrelationship between the changes in U, S, and V between any two closely neighboring thermodynamic equilibrium states of a material.PeteSampras said:Then if I write PdV-TdS=dU
The assumption is that PdV is the work done by the system which tends to increase the internal energy, and TdS is the reversible heat from the system to exterior that tends to decreases the internal energy ?
yes if dV is positive. and also TdS will increase U if dS is positive.Chestermiller said:PdV is always the work done by the system on the surroundings, which always tends to decrease internal energy,