A gaseous sample is generally allowed to do only expansion/compression type of work against its surroundings. The work done in case of an irreversible expansion ( in the intermediate stages of expansion/ compression the states of gases are not defined). The work done can be calculated using dw=-P_("ext")dV while in case of reversible process the work done can be calculated using dw=-PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process, since P=(nRT)/(V),so, w=intdw=-int_(V_(i))^(V_(f))(nRT)/(V).dV=-nRT ""In((V_(f))/(V_(i))) Since,dw=PdV, so magnitude of work done can also be calculated by calculating the area under the PV curve of the reversible process in PV diagram. If four identical samples of an ideal gas initially at similar state (P_(0),V_(0),T_(0)) are allowed to expand to double their volumes by four different processes. (P)By isothermal irreversible process (Q) By reversible process having equation P^(2)V= constant (R) By reversible adiabatic process (S)By irreversible adiabatic expansion against constant external pressure. Then in the graph shown the final state is represented by four different points then, the correct match can be :