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(A) For path $F \rightarrow G$ (isothermal): Work done $W_{FG} = nRT \ln(V_f/V_i)$. Given $P_F = 32P_0$,$V_F = V_0$,$P_G = P_0$,$V_G = 32V_0$. Since $

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$4 \quad 3 \quad 2 \quad 1$

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(A) For path $F \rightarrow G$ (isothermal): Work done $W_{FG} = nRT \ln(V_f/V_i)$. Given $P_F = 32P_0$,$V_F = V_0$,$P_G = P_0$,$V_G = 32V_0$. Since $P_F V_F = P_G V_G$,the temperature is constant. $W_{FG} = P_F V_F \ln(V_G/V_F) = (32P_0 V_0) \ln(32V_0/V_0) = 32 P_0 V_0 \ln(2^5) = 160 P_0 V_0 \ln 2$.For path $G \rightarrow E$ (isobaric): Work done $W_{GE} = P_0 (V_E - V_G) = P_0 (V_0 - 32V_0) = -31 P_0 V_0$. Magnitude is $31 P_0 V_0$.For path $F \rightarrow H$ (adiabatic): $W_{FH} = \frac{nR(T_F - T_H)}{\gamma - 1} = \frac{P_F V_F - P_H V_H}{\gamma - 1}$. For monatomic gas,$\gamma = 5/3$. $P_F V_F = 32 P_0 V_0$. For adiabatic process $P_F V_F^\gamma = P_H V_H^\gamma \Rightarrow (32P_0) V_0^{5/3} = P_0 V_H^{5/3} \Rightarrow V_H = (32)^{3/5} V_0 = 8 V_0$. $W_{FH} = \frac{32 P_0 V_0 - P_0 (8 V_0)}{5/3 - 1} = \frac{24 P_0 V_0}{2/3} = 36 P_0 V_0$.For path $G \rightarrow H$ (isobaric): $W_{GH} = P_0 (V_H - V_G) = P_0 (8V_0 - 32V_0) = -24 P_0 V_0$. Magnitude is $24 P_0 V_0$.Matching: $P-4, Q-3, R-2, S-1$.

List-$I$	List-$II$
$P. \quad G \rightarrow E$	$1. \quad 160 P_0 V_0 \ln 2$
$Q. \quad G \rightarrow H$	$2. \quad 36 P_0 V_0$
$R. \quad F \rightarrow H$	$3. \quad 24 P_0 V_0$
$S. \quad F \rightarrow G$	$4. \quad 31 P_0 V_0$

$List-I$	$List-II$
$(P)$ Work done in the complete cyclic process	$(1)$ $R T_0 - 4 R T_0 \ln 2$
$(Q)$ Change in the internal energy of the gas in the process $JK$	$(2)$ $0$
$(R)$ Heat given to the gas in the process $KL$	$(3)$ $3 R T_0$
$(S)$ Change in the internal energy of the gas in the process $MJ$	$(4)$ $-2 R T_0 \ln 2$
	$(5)$ $-3 R T_0 \ln 2$

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