Correct option is (B) P → 4; Q → 3; R → 5; S → 2
\(\mathrm{J}\left(\mathrm{P}_0, \mathrm{~V}_0, \mathrm{~T}_0\right)\)
\(\mathrm{K}\left(\mathrm{P}_0, 3 \mathrm{~V}_0, 3 \mathrm{~T}_0\right) \)
\( \mathrm{M}\left(2 \mathrm{P}_0, \frac{\mathrm{V}_0}{2}, \mathrm{~T}_0\right) \)
\( \mathrm{L}\left(2 \mathrm{P}_0, \frac{3 \mathrm{~V}_0}{2}, 3 \mathrm{~T}_0\right)\)
\(\mathrm{P}_0 \mathrm{~V}_0=\mathrm{nRT}_0 \)
\(\mathrm{JK} \rightarrow \text { isobaric } \Rightarrow \mathrm{W}=\mathrm{P}_0\left(2 \mathrm{~V}_0\right)=2 \mathrm{nRT}_0 \)
\(\Delta \mathrm{U}=\frac{3}{2} \mathrm{nR}\left(2 \mathrm{~T}_0\right)=3 n R \mathrm{~T}_0\)
\(\mathrm{KL} \rightarrow \text { isothermal } \rightarrow \mathrm{W}=\mathrm{nR}(3 \mathrm{~T}) \ln \left(\frac{1}{2}\right)=-3 \mathrm{nRT}_0 \ln 2\)
\(\Delta \mathrm{U}=0 \Rightarrow \mathrm{Q}=-3 \mathrm{nRT}_0 \ln 2\)
\( \mathrm{LM} \rightarrow \text { isobaric }=2 \mathrm{P}_0\left(-\mathrm{V}_0\right)=-2 \mathrm{nRT}_0 \)
\(\mathrm{MJ} \rightarrow \text { isothermal } \Rightarrow \mathrm{nRT}_0 \ln 2 ; \Delta \mathrm{U}=0\)
\(\mathrm{WD}_{\text {net }}=-2 \mathrm{nRT}_0\ln2\)
\(\mathrm{P} \rightarrow 4, \mathrm{Q} \rightarrow 3, \mathrm{R} \rightarrow 5, \mathrm{~S} \rightarrow 2\)