What is the relationship between TdS and dU in the thermodynamics identity?

[PeteSampras]

Has some sense write in the thermodynamics identity the terms TdS and dU at the same side of the equation and with the same sign? what would be this sense?For example PdV=TdS+dU

 

[Let'sthink]

TdS and dU

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]

my reason is that in the literature says PdV= TdS-dU

Due to the above affirmation, i want to understand what means that the equation is written as PdV=TdS+dU

 

[Let'sthink]

PdV= TdS-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]

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 ?

 

[Chestermiller]

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 ?

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.

 

[Let'sthink]

PdV is always the work done by the system on the surroundings, which always tends to decrease internal energy,

yes if dV is positive. and also TdS will increase U if dS is positive.