Match List-I with List-II.
$$\begin{array}{|l|l|l|l|}\hline & \text{List-I} & & \text{List-II} \\ \hline \text{A.} & \text{Isothermal} & \text{I.} & \Delta W\ (\text{work done}) = 0 \\ \hline \text{B.} & \text{Adiabatic} & \text{II.} & \Delta Q\ (\text{supplied heat}) = 0 \\ \hline \text{C.} & \text{Isobaric} & \text{III.} & \Delta U\ (\text{change in internal energy}) \ne 0 \\ \hline \text{D.} & \text{Isochoric} & \text{IV.} & \Delta U = 0 \\ \hline \end{array}$$
Choose the correct answer from the options given below:
Answer: (C) A-IV, B-II, C-III, D-I
- A. Isothermal: $T$ constant, so $\Delta U = 0$ → IV
- B. Adiabatic: no heat exchanged, $\Delta Q = 0$ → II
- C. Isobaric: $V$ changes at constant $P$, so $T$ changes and $\Delta U \ne 0$ → III
- D. Isochoric: no change in volume, so $\Delta W = 0$ → I
Solution by Sreeraj P, M.Sc Physics