Is the statement d F = ( δ Q − T d S ) − p d V − S d T more general than d F = − S d T − p d V

Question

Is the statement   d  F  =  (  δ  Q  −  T  d  S  )  −  p  d  V  −  S  d  T more general than   d  F  =  −  S  d  T  −  p  d  V

Hamnetmj · Accepted Answer

The equation dF=-SdT-pdV describes the mutual variations in these parameters between two closely neighboring (differentially separated) thermodynamic equilibrium states. It doesn&#039;t matter whether the process that took the system between these two thermodynamic equilibrium states was reversible or whether it was a highly tortuous irreversible process path which, in the end, put the final state very close to the initial state. But, for the irreversible path, Q was definitely not equal to TdS.Consider any irreversible process in which the initial and final thermodynamic equilibrium states are not close together, but the system is in contact with a constant temperature reservoir at temperature T, which happens to be the same as the initial temperature of the system. Under these circumstances, the final equilibrium temperature of the system will also be T. For this situation, it follows from the first and 2nd laws of thermodynamics that  Δ  U  =  Q  −  ∫            p              e        x        t              d    V  and  Q  =  T  Δ  S  −  T  σwhere   σ is the entropy generated as a result of irreversibility (always positive). So, if we combine these two equations, we obtain:  Δ  U  =  T  Δ  S  −  T  σ  −  ∫            p              e        x        t              d    V  or, equivalently,  W  =  ∫            p              e        x        t              d    V    =  −  Δ  F  −  T  σFrom this it follows that the maximum work that can be obtained by all processes between the specified initial and final thermodynamics equilibrium states of the system is   −  Δ  F

Is the statement dF=(δQ−TdS)−pdV−SdT more general than dF=−SdT−pdV?

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