One mole of a monatomic ideal gas undergoes the cyclic process J → K → L → M → J, as shown in the $P-T$ diagram. Match the quantities mentioned in List-I with their values in List-II and choose the correct option. [R is the gas constant.] List-I: (P) Work done in the complete cyclic process; (Q) Change in the internal energy of the gas in the process JK; (R) Heat given to the gas in the process KL; (S) Change in the internal energy of the gas in the process MJ. List-II: (1) RT$_{0} − 4$RT$_{0} \ln_{2}; (2) 0; (3) 3$RT$_{0}; (4) −2$RT$_{0} \ln_{2}; (5) −3$RT$_{0} \ln_{2}.$ Options: (1) P→1; Q→3; R→5; S→4. (2) P→4; Q→3; R→5; S→2. (3) P→4; Q→1; R→2; S→2. (4) P→2; Q→5; R→3; S→4.

Step 1: Analyze the process $J \to K$.

This is an isobaric process at constant pressure $P = P_0$, where the temperature increases from $T_0$ to $3T_0$.

For $1\text{ mole}$ of a monatomic ideal gas, the change in internal energy is:

$$\Delta U_{JK} = n C_V \Delta T$$ $$\Delta U_{JK} = 1 \cdot \frac{3}{2}R (3T_0 - T_0)$$ $$\Delta U_{JK} = 3RT_0$$

This matches quantity (Q) with value (3).

The work done in this process is:

$$W_{JK} = P_0(V_K - V_J) = R(3T_0 - T_0)$$ $$W_{JK} = 2RT_0$$

Step 2: Analyze the process $K \to L$.

This is an isothermal process at temperature $T = 3T_0$, where the pressure increases from $P_0$ to $2P_0$.

Since the temperature is constant, the change in internal energy is:

$$\Delta U_{KL} = 0$$

By the First Law of Thermodynamics, the heat given to the gas is:

$$Q_{KL} = W_{KL} = n R (3T_0) \ln\left(\frac{V_L}{V_K}\right)$$ $$Q_{KL} = 3RT_0 \ln\left(\frac{P_K}{P_L}\right)$$ $$Q_{KL} = 3RT_0 \ln\left(\frac{P_0}{2P_0}\right)$$ $$Q_{KL} = -3RT_0 \ln 2$$

This matches quantity (R) with value (5).

Step 3: Analyze the process $L \to M$.

This is an isobaric process at constant pressure $P = 2P_0$, where the temperature decreases from $3T_0$ to $T_0$.

The work done in this process is:

$$W_{LM} = 2P_0(V_M - V_L) = R(T_0 - 3T_0)$$ $$W_{LM} = -2RT_0$$

Step 4: Analyze the process $M \to J$.

This is an isothermal process at temperature $T = T_0$, where the pressure decreases from $2P_0$ to $P_0$.

Since the temperature is constant, the change in internal energy is:

$$\Delta U_{MJ} = 0$$

This matches quantity (S) with value (2).

The work done in this process is:

$$W_{MJ} = n R T_0 \ln\left(\frac{P_M}{P_J}\right)$$ $$W_{MJ} = RT_0 \ln\left(\frac{2P_0}{P_0}\right)$$ $$W_{MJ} = RT_0 \ln 2$$

Step 5: Calculate the total work done in the cyclic process.

Sum the work done in each of the four steps:

$$W_{\text{cycle}} = W_{JK} + W_{KL} + W_{LM} + W_{MJ}$$ $$W_{\text{cycle}} = 2RT_0 - 3RT_0 \ln 2 - 2RT_0 + RT_0 \ln 2$$ $$W_{\text{cycle}} = -2RT_0 \ln 2$$

This matches quantity (P) with value (4).

Answer: The correct matching is (P) $\to$ (4), (Q) $\to$ (3), (R) $\to$ (5), (S) $\to$ (2).