`{:(,"Column I",,"Column II"),((A),"Isothermal process",(p),q=DeltaU),((B),"Adiabatic process",(p),w=-pPDeltaV),((C ),"Isobaric process",(r),w=DeltaU),((D),"Isochoric process",(s),w=nRT ln(V_(2)//V_(1))):}`
(a) `A -s,B-q,C-r,D-p` (b) `A-s,B-r,C-q,D-p` (c )`A-p, B-r,C-q,D-s` (d) `A-r,B-p,C-s,D-q`

Answer

To solve the problem, we need to match the processes in Column I with their corresponding work expressions in Column II. Let's analyze each process step by step.

### Step 1: Identify the Isothermal Process
- **Isothermal Process (A)**: In an isothermal process, the temperature remains constant (ΔT = 0). The work done (w) during an isothermal process for an ideal gas is given by:
\[ w = nRT \ln\left(\frac{V_2}{V_1}\right) \]
This corresponds to option **(s)**.

### Step 2: Identify the Adiabatic Process
- **Adiabatic Process (B)**: In an adiabatic process, there is no heat exchange with the surroundings (Q = 0). The first law of thermodynamics states:
\[ \Delta U = Q + W \implies \Delta U = W \quad \text{(since Q = 0)} \]
Therefore, the work done (w) is equal to the change in internal energy (ΔU). This corresponds to option **(q)**.

### Step 3: Identify the Isobaric Process
- **Isobaric Process (C)**: In an isobaric process, the pressure remains constant. The work done during an isobaric process is given by:
\[ w = -P \Delta V \]
This corresponds to option **(r)**.

### Step 4: Identify the Isochoric Process
- **Isochoric Process (D)**: In an isochoric process, the volume remains constant (ΔV = 0). Therefore, no work is done:
\[ w = 0 \]
This corresponds to option **(p)**.

### Summary of Matches
- A (Isothermal process) → s (w = nRT ln(V2/V1))
- B (Adiabatic process) → q (w = ΔU)
- C (Isobaric process) → r (w = -PΔV)
- D (Isochoric process) → p (w = 0)

### Final Matching
Thus, the correct option is:
- **(A - s, B - q, C - r, D - p)**