1) What is the reason why dH!=0 for an adiabatic(q=0) reversible process? The mathematical argument is irrefutable and it is clear that it has to do with the process not being isobaric: ΔH=ΔU+PΔV+VΔP , ΔU=work=−PΔV Therefore, ΔH=VΔP and this is not 0. However, I do not understand it conceptually... 2) Consider an adiabatic irreversible process from state A to B. At first I would expect to have dS=0, as there is no heat transfer, However, if we construct a reverse isothermal path that connects state B and A, then S(B→A)=-S(A→B). The entropy calculated for the isothermal path is positive, so the entropy for the irreversible process is negative, which agrees with the clausius inequality. My question is: conceptually, why should a reversible adiabatic process has dS=0 and an irreversible adiabatic process have a dS!=0 if there is no heat transfer? Thanks.