Step-by-Step Solution

Step 1: Analyze Column 1 (Work Done Expressions)

The question defines $W$ as the work done on the system. The general formula for work done on the system is $W = -\int P dV$.

• (I) $W_{1 \rightarrow 2} = \frac{1}{\gamma-1}(P_2V_2 - P_1V_1)$ For an adiabatic process, the work done by the gas is $W_{by} = \int P dV = \frac{P_1V_1 - P_2V_2}{\gamma-1}$. The work done on the gas is $W_{on} = -W_{by} = -\frac{P_1V_1 - P_2V_2}{\gamma-1} = \frac{P_2V_2 - P_1V_1}{\gamma-1}$. This expression corresponds to an adiabatic process.

• (II) $W_{1 \rightarrow 2} = -P_2(V_2 - V_1)$ For an isobaric (constant pressure) process, $P$ is constant. Let $P = P_1 = P_2$. The work done on the system is $W_{on} = -\int_{V_1}^{V_2} P dV = -P(V_2 - V_1) = -P_2(V_2 - V_1)$. This expression corresponds to an isobaric process.

• (III) $W_{1 \rightarrow 2} = 0$ Work is done only when there is a change in volume. If $W=0$, it implies $dV=0$, which means the volume is constant. This corresponds to an isochoric process.

• (IV) $W_{1 \rightarrow 2} = -nRT \ln(\frac{V_2}{V_1})$ For an isothermal process, temperature $T$ is constant. From the ideal gas law, $P = \frac{nRT}{V}$. The work done by the gas is $W_{by} = \int_{V_1}^{V_2} P dV = \int_{V_1}^{V_2} \frac{nRT}{V} dV = nRT \ln(\frac{V_2}{V_1})$. The work done on the gas is $W_{on} = -W_{by} = -nRT \ln(\frac{V_2}{V_1})$. This expression corresponds to an isothermal process.

Step 2: Match Column 1 with Column 2 (Process Name)

Based on the analysis in Step 1:

• (I) matches with (iv) Adiabatic
• (II) matches with (iii) Isobaric
• (III) matches with (ii) Isochoric
• (IV) matches with (i) Isothermal

Step 3: Analyze Column 3 (P-V Diagrams)

We analyze the path from state 1 to state 2 in each diagram:

• (P): The path is a horizontal line, which means pressure is constant. This is an isobaric process.
• (Q): The path is a curve. The overall cycle resembles an Otto cycle, where the expansion (1 to 2) and compression strokes are adiabatic. The slope of an adiabatic process ($dP/dV = -\gamma P/V$) is steeper than that of an isothermal process ($dP/dV = -P/V$). The curve 1->2 is steep, characteristic of an adiabatic process.
• (R): The path is a curve, less steep than the one in (Q). This is characteristic of an isothermal process.
• (S): The path is a vertical line, which means volume is constant. This is an isochoric process.

Step 4: Match Column 2 with Column 3

Based on the analysis in Step 3:

• (i) Isothermal matches with (R)
• (ii) Isochoric matches with (S)
• (iii) Isobaric matches with (P)
• (iv) Adiabatic matches with (Q)

Step 5: Identify the Correct Combination

We now combine the matches to find the correct triplets and check against the given options.

• Adiabatic: (I) - (iv) - (Q)
• Isobaric: (II) - (iii) - (P)
• Isochoric: (III) - (ii) - (S)
• Isothermal: (IV) - (i) - (R)

Let's evaluate the given options:

• A: $\left( {{\rm I}V} \right)\left( {ii} \right)\left( S \right)$ - Incorrect. (IV) is Isothermal, while (ii) and (S) are Isochoric.
• B: $\left( {{\rm I}{\rm I}{\rm I}} \right)\left( {ii} \right)\left( S \right)$ - Correct. (III) is the work for an isochoric process, (ii) is the name for an isochoric process, and (S) is the P-V diagram for an isochoric process.
• C: $\left( {{\rm I}{\rm I}} \right)\left( {iv} \right)\left( P \right)$ - Incorrect. (II) and (P) are Isobaric, but (iv) is Adiabatic.
• D: $\left( {{\rm I}{\rm I}} \right)\left( {iv} \right)\left( R \right)$ - Incorrect. (II) is Isobaric, (iv) is Adiabatic, and (R) is Isothermal.

Thus, the only correct combination listed is B.