We can write first law of thermodynamics in two forms. d U = T d S − p d Vand d U = d q + d wIt is also true that dw=−pdV therefore TdS=dq for every process irrespective of reversibility. What I am missing here?

Question

We can write first law of thermodynamics in two forms.  d  U  =  T  d  S  −  p  d  Vand  d  U  =  d  q  +  d  wIt is also true that dw=−pdV therefore TdS=dq for every process irrespective of reversibility. What I am missing here?

Demarion Thornton · Accepted Answer

It is true, for all processes whether reversible or not, that:  d  U  =  T  d  S  −  P  d  V  =  d  q  +  d  was long as two of the four variables (T,S,V,P) can be defined for the system.However, it is not true that dq=TdS always; that equality only holds for reversible processes. Likewise, dw=−PdV is only true for reversible processes. If the processes is irreversible, then dq−PdV.Nevertheless, because of the first law, the sum of dq and dw always equals the change in energy, whether the process is reversible or not:  d  U  =  d      q          rev        +  d      w          rev        =  d      q          irrev        +  d      w          irrev      which also is equal to TdS−PdV.

We can write first law of thermodynamics in two forms. dU=TdS−pdV and dU=dq+dw It is also true that dw=−pdV therefore TdS=dq for every process irrespective of reversibility. What I am missing here?

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