For an adiabatic transformation between state A and B, where Δq=0, and consequently from the first law of thermodynamics dU=Δw, since U is a state function, its variation should be the same whether the process is reversible or irreversible. The possibility to go via an irreversible or irreversible path between the two states seems feasible by what I'm reading on 'Chemical Thermodynamics: Classical, Statistical and Irreversible' by J. Rajaram page 66. For the adiabatic expansion of an ideal gas, the work done by the gas is equal to the decrease in internal energy, -ΔU = –w. However, if an ideal gas is taken from state A to state B by a reversible path as well as an irreversible path, while the change in internal energy is the same because the initial and final states are the same, the work done against external pressure will not be the same. The work done in the irreversible process must be less than that done in the reversible process. The decrease in internal energy in either case will be ΔU = nC_v(T_2 - T_1). dU=Δw seems to indicate that also work to be the same for the reversible and irreversible path. But how can work for the irreversible and reversible process be the same? We all know that the maximum work can be extracted by the reversible path. Hence even if ΔU is equal for the two processes, w should not. So does the equal sign in dU=Δw hold only for reversible processes? If yes, Why? If no, how should I read dU=Δw?