Step 1: Analyzing each case.
(I) Isothermal reversible:
For isothermal reversible expansion, the work done is calculated as:
\[
w = -nRT \ln \frac{V_2}{V_1}
\]
Substituting the values:
\[
w = -1 \times 8.31 \times 300 \times \ln \frac{20}{2} = -8.32 \, \text{kJ}
\]
Thus, the work done is 8.32 kJ. Therefore, the correct match is I-A.
(II) Isothermal irreversible:
For isothermal irreversible expansion, the work done is calculated as:
\[
w = -P \Delta V
\]
Substituting the values:
\[
w = -3 \times 10^3 \times (3 - 1) = -6 \, \text{kJ}
\]
Thus, the work done is 6 kJ. Therefore, the correct match is II-B.
(III) Constant pressure process:
For a constant pressure process, $\Delta H = n C_p \Delta T$. Given $\Delta T = 400$ K and $C_p = 5R/2$, the enthalpy change will be:
\[
\Delta H = 1 \times \frac{5}{2} \times 8.31 \times 400 = 4 \times 10^3 \, \text{J}
\]
Thus, $\Delta H = 4 \, \text{kJ}$. Therefore, the correct match is III-A.
(IV) Ideal gas with Cv:
For an ideal gas, the change in internal energy is given by:
\[
\Delta U = n C_v \Delta T
\]
Substituting the values:
\[
\Delta U = 1 \times \frac{3}{2} \times 8.31 \times 320 = 5.74 \, \text{kJ}
\]
Thus, $\Delta U = 5.74 \, \text{kJ}$. Therefore, the correct match is IV-C.
Step 2: Conclusion.
The correct matches are I-D, II-B, III-A, IV-C, which corresponds to option (2).