Chemical Thermodynamics Part 3

Concept:Work done in isothermal processes for an ideal gas depends on whether the process is reversible or irreversible and on the external pressure.Explanation:For any process, work done on the system is given by $w = -\int_{V_i}^{V_f} P_{\text{ext}} \, dV$.A. Reversible isothermal expansion: $P_{\text{ext}} = P = \frac{nRT}{V}$. So $w = -\int_{V_i}^{V_f} \frac{nRT}{V} dV = -nRT \ln\frac{V_f}{V_i}$. This matches List‑II II.B. Free expansion: $P_{\text{ext}} = 0$, so $w = 0$. This matches List‑II I.C. Irreversible expansion against constant external pressure: $P_{\text{ext}}$ is constant, so $w = -P_{\text{ext}} (V_f - V_i)$. This matches List‑II III.D. Irreversible compression: again constant $P_{\text{ext}}$, and $V_f < V_i$. Using the same formula $w = -P_{\text{ext}}(V_f - V_i) = P_{\text{ext}}(V_i - V_f)$. This is exactly the expression given in List‑II IV: $w = -P_{\text{ext}}(V_i - V_f)$. So D matches IV.Thus A → II, B → I, C → III, D → IV.Answer:A‑II, B‑I, C‑III, D‑IV. Correct option is C.