Match Isothermal Thermodynamic Processes with Expression

To solve this matching question, let us analyze each process in List - I and find its corresponding mathematical expression in List - II:

1. A. Reversible expansion (Isothermal): For an isothermal process involving an ideal gas, the internal energy change is zero ($\Delta U = 0$). By the First Law of Thermodynamics: $\Delta U = q + w = 0 \implies q = -w$ The work done in a reversible isothermal expansion from volume $V_1$ to $V_2$ is given by: $w = -nRT \ln \left(\frac{V_2}{V_1}\right)$ Therefore, the heat absorbed is: $q = nRT \ln \left(\frac{V_2}{V_1}\right)$ Thus, A matches with II.

2. B. Free expansion (Isothermal): In a free expansion, the gas expands against zero external pressure ($P_{\text{ext}} = 0$), so the work done is $w = 0$. Since the process is also isothermal, $\Delta U = 0$. From the First Law of Thermodynamics ($\Delta U = q + w$): $0 = q + 0 \implies q = 0$ Thus, B matches with I.

3. C. Irreversible Compression: Work done during an irreversible process against a constant external pressure $P_{\text{ext}}$ is expressed as: $w = -P_{\text{ext}} \Delta V = -P_{\text{ext}} (V_2 - V_1) = -P_{\text{ext}} (V_1 - V_2)$ Thus, C matches with III.

4. D. Cyclic reversible: Entropy $S$ is a state function, so for any cyclic process, the total change in entropy is zero: $\Delta S = \oint \frac{dq_{\text{rev}}}{T} = 0 \implies \frac{q_{\text{rev}}}{T} = 0$ Thus, D matches with IV.

Combining these matches gives: $\text{A-II, B-I, C-III, D-IV}$

This corresponds to Option C.