For an isothermal reversible expansion, the change in internal energy ΔU=0. From the first law of thermodynamics, q=−w. The work done is w=−nRTln(V1V2), which gives q=nRTln(V1V2). Thus, A matches II.
For free expansion, the external pressure is zero (pext=0), so the work done w=0. Since the process is isothermal, ΔU=0. Therefore, the heat exchanged q=0. Thus, B matches I.
For an irreversible compression, the work done against a constant external pressure is given by w=−pextΔV. This corresponds to the expression in III. Thus, C matches III.
For a cyclic reversible process, the initial and final states are the same, so the change in entropy ΔS=0. Since ΔS=Tqrev, we get Tqrev=0. Thus, D matches IV.
The correct matching is A-II, B-I, C-III, D-IV.
Answer: A-II, B-I, C-III, D-IV