One mole of a monatomic ideal gas undergoes four thermodynamic processes as shown schematically in the PV-diagram below. Among these four processes, one is isobaric, one is isochoric, one is isothermal and one is adiabatic. Match the processes mentioned in List-I with the corresponding statements in List-II. (P) In process I (1) Work done by the gas is zero. (Q) In process II (2) Temperature of the gas remains unchanged. (R) In process III (3) No heat is exchanged between the gas and its surroundings. (S) In process IV (4) Work done by the gas is $6P_{0} V_{0}$

Step 1: Identify the isobaric process (Process II) and calculate its work done.

Process II is represented by a horizontal line at constant pressure.

$$P = 3P_0 = \text{constant}$$

The work done by the gas during this expansion from $V_0$ to $3V_0$ is:

$$W_{\text{II}} = P \Delta V = 3P_0(3V_0 - V_0) = 6P_0V_0$$

Thus, Process II (Q) matches with statement 4.

Step 2: Identify the isochoric process (Process III) and calculate its work done.

Process III is represented by a vertical line at constant volume.

$$V = 3V_0 = \text{constant}$$

Since there is no change in volume, the work done by the gas is:

$$W_{\text{III}} = \int P \, dV = 0$$

Thus, Process III (R) matches with statement 1.

Step 3: Distinguish between the isothermal and adiabatic processes (Processes I and IV).

Processes I and IV are expansion curves, where one is isothermal and the other is adiabatic.

For a monatomic ideal gas, the adiabatic index is:

$$\gamma = \frac{C_P}{C_V} = \frac{5}{3}$$

Since $\gamma > 1$, the adiabatic curve is steeper than the isothermal curve on a $PV$-diagram.

$$\left| \frac{dP}{dV} \right|_{\text{adiabatic}} > \left| \frac{dP}{dV} \right|_{\text{isothermal}}$$

Process I is the steeper curve, representing the adiabatic process, while Process IV is the less steep curve, representing the isothermal process.

Step 4: Match Processes I and IV with their corresponding statements.

For the adiabatic process (Process I), no heat is exchanged with the surroundings.

$$Q = 0$$

Thus, Process I (P) matches with statement 3.

For the isothermal process (Process IV), the temperature of the gas remains constant.

$$T = \text{constant}$$

Thus, Process IV (S) matches with statement 2.

Step 5: Combine the matching pairs.

We obtain the following matching:

$$\text{P} \to 3, \quad \text{Q} \to 4, \quad \text{R} \to 1, \quad \text{S} \to 2$$

Answer: P $\to$ 3, Q $\to$ 4, R $\to$ 1, S $\to$ 2