One mole of a monatomic ideal gas is taken along two cyclic processes E → F → G → E and E → F → H → E as shown in the PV diagram. The processes involved are purely isochoric, isobaric, isothermal or adiabatic. Match the paths in List I with the magnitudes of the work done in List II and select the correct answer using the codes given below the lists. List I: P. G → E; Q. G → H; R. F → H; S. F → G. List II$: 1. 160 P0V0 \ln_{2}; 2. 36 P0V0; 3. 24 P0V0; 4. 31 P0V0.$ Options: (1) P→4, Q→3, R→2, S→1; (2) P→4, Q→3, R→1, S→2; (3) P→3, Q→1, R→2, S→4; (4) P→1, Q→3, R→2, S→4.

Step 1: Identify the states and processes from the $PV$ diagram.

For 1 mole of a monatomic ideal gas, the adiabatic index is $\gamma = \frac{5}{3}$. From the given diagram, the coordinates of the states are:

$$F = (V_0, 32P_0)$$ $$E = (V_0, P_0)$$

The states $G$ and $H$ lie on the isobaric line $P = P_0$. The processes are identified as:

• $E \to F$: Isochoric
• $G \to H$ and $H \to E$: Isobaric
• $F \to G$: Isothermal
• $F \to H$: Adiabatic

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Step 2: Calculate the work done for path S ($F \to G$).

Since $F \to G$ is an isothermal process, the temperature remains constant:

$$P_F V_F = P_G V_G$$ $$(32P_0)(V_0) = P_0 V_G$$ $$V_G = 32V_0$$

The work done during this isothermal expansion is:

$$W_{F \to G} = P_F V_F \ln\left(\frac{V_G}{V_F}\right)$$ $$W_{F \to G} = 32P_0 V_0 \ln\left(\frac{32V_0}{V_0}\right)$$ $$W_{F \to G} = 32P_0 V_0 \ln(2^5) = 160P_0 V_0 \ln 2$$

Thus, S matches with 1.

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Step 3: Calculate the work done for path R ($F \to H$).

Since $F \to H$ is an adiabatic process:

$$P_F V_F^{\gamma} = P_H V_H^{\gamma}$$

Substitute $\gamma = \frac{5}{3}$:

$$(32P_0) V_0^{5/3} = P_0 V_H^{5/3}$$ $$V_H = (32)^{3/5} V_0 = 8V_0$$

The work done during this adiabatic expansion is:

$$W_{F \to H} = \frac{P_F V_F - P_H V_H}{\gamma - 1}$$ $$W_{F \to H} = \frac{32P_0 V_0 - P_0 (8V_0)}{\frac{5}{3} - 1}$$ $$W_{F \to H} = \frac{24P_0 V_0}{\frac{2}{3}} = 36P_0 V_0$$

Thus, R matches with 2.

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Step 4: Calculate the magnitude of work done for path Q ($G \to H$).

Since $G \to H$ is an isobaric process at pressure $P_0$:

$$W_{G \to H} = P_0 (V_H - V_G)$$ $$W_{G \to H} = P_0 (8V_0 - 32V_0) = -24P_0 V_0$$

The magnitude of the work done is:

$$|W_{G \to H}| = 24P_0 V_0$$

Thus, Q matches with 3.

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Step 5: Calculate the magnitude of work done for path P ($G \to E$).

Since $G \to E$ is an isobaric process at pressure $P_0$:

$$W_{G \to E} = P_0 (V_E - V_G)$$ $$W_{G \to E} = P_0 (V_0 - 32V_0) = -31P_0 V_0$$

The magnitude of the work done is:

$$|W_{G \to E}| = 31P_0 V_0$$

Thus, P matches with 4.

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Answer: The correct matching is: P $\to$ 4, Q $\to$ 3, R $\to$ 2, S $\to$ 1