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	List-II
(P) Work done in the complete cyclic process	(1) $R{T}_0-<!--\; -\; -->4R{T}_{0}ln2$
(Q) Change in the internal energy of the gas in the process JK	(2) 0
(R) Heat given to the gas in the process KL	(3) $3R{T}_0$
(S) Change in the internal energy of the gas in the process MJ	(4) $-<!--\; -\; -->2R{T}_{0}ln2$
	(5) $-<!--\; -\; -->3R{T}_{0}ln2$

Correct Option: P → 4; Q → 3; R → 5; S → 2

Let us analyze the cyclic process for $n=1$ mole of a monatomic ideal gas as shown in the pressure-temperature ($P-T$) diagram.
For a monatomic ideal gas:
The molar heat capacity at constant volume is:
$C_V=\frac{3}{2}R$
The molar heat capacity at constant pressure is:
$C_P=\frac{5}{2}R$

1. Analysis of individual processes:

Process J → K: Isobaric heating
From the diagram, the pressure is constant at $P_0$, and the temperature increases from $T_0$ to $3T_0$.
The change in internal energy ($\Delta U_{JK}$) is given by:
$\Delta U_{JK}=n{C}_V\Delta T=1\cdot \left(\frac{3}{2}R\right)(3T_0-T_0)=3R{T}_0$
Thus, (Q) → (3).
The work done during this process is:
$W_{JK}=nR\Delta T=1\cdot R(3T_0-T_0)=2R{T}_0$

Process K → L: Isothermal compression
From the diagram, the temperature remains constant at $3T_0$, and the pressure increases from $P_0$ to $2P_0$.
Since the temperature is constant, the change in internal energy is $\Delta U_{KL}=0$.
The heat given to the gas ($Q_{KL}$) is equal to the work done ($W_{KL}$):
$Q_{KL}=W_{KL}=nRT\ln \left(\frac{P_i}{P_f}\right)=1\cdot R\left(3T_0\right)\ln \left(\frac{P_0}{2P_0}\right)=-3R{T}_0\ln 2$
Thus, (R) → (5).

Process L → M: Isobaric cooling
The pressure is constant at $2P_0$, and the temperature decreases from $3T_0$ to $T_0$.
The work done during this process is:
$W_{LM}=nR\Delta T=1\cdot R(T_0-3T_0)=-2R{T}_0$

Process M → J: Isothermal expansion
The temperature remains constant at $T_0$, and the pressure decreases from $2P_0$ to $P_0$.
Since this is an isothermal process, the change in internal energy is:
$\Delta U_{MJ}=0$
Thus, (S) → (2).
The work done in this process is:
$W_{MJ}=nRT\ln \left(\frac{P_i}{P_f}\right)=1\cdot R{T}_0\ln \left(\frac{2P_0}{P_0}\right)=R{T}_0\ln 2$

2. Work done in the complete cyclic process:
The net work done ($W$) is the sum of the work done in each stage:
$W=W_{JK}+W_{KL}+W_{LM}+W_{MJ}$
Substituting the calculated work values:
$W=2R{T}_0-3R{T}_0\ln 2-2R{T}_0+R{T}_0\ln 2$
Simplifying:
$W=-2R{T}_0\ln 2$
Thus, (P) → (4).

Conclusion:
Matching the items from List-I with List-II:
P → 4
Q → 3
R → 5
S → 2
This corresponds directly to the option: P → 4; Q → 3; R → 5; S → 2.