One mole of a monatomic ideal gas undergoes the cyclic process $J \rightarrow K \rightarrow L \rightarrow M \rightarrow J$, 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 | ($1$) $R T_0-4 \ R T_0 \ln 2$ |
| ($Q$) 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 $MJ$ | ($4$) $-2 \ R T_0 \ln 2$ |
| ($5$) $-3 \ R T_0 \ln 2$ |
IIT 2024, Advanced
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$J \left( P _0, V _0, T _0\right)$
$K \left( P _0, 3 V _0, 3 T _0\right)$
$M \left(2 P _0, \frac{ V _0}{2}, T _0\right)$
$L \left(2 P _0, \frac{3 V _0}{2}, 3 T _0\right)$
$P _0 V _0= nRT _0$
$JK \rightarrow \text { isobaric } \Rightarrow W = P _0\left(2 V _0\right)=2 nRT \text {. }$
$\Delta U =\frac{3}{2} nR \left(2 T _0\right)=3 nRT _0$
$KL \rightarrow \text { isothermal } \rightarrow W = nR (3 T ) \ln \left(\frac{1}{2}\right)=-3 nRT _0 \ln 2$
$\Delta U =0 \Rightarrow Q =-3 nRT _0 \ln 2$
$LM \rightarrow \text { isobaric }=2 P _0\left(- V _0\right)=-2 nRT _0$
$MJ \rightarrow \text { isothermal } \Rightarrow nRT _0 \ell n 2 ; \Delta U =0$
$WD _{ xat }=-2 nRT _0 \ln 2$
$P \rightarrow 4, Q \rightarrow 3, R \rightarrow 5, S \rightarrow 2$
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