JEE Advance - Physics (2024 - Paper 1 Online - No. 14)
One mole of a monatomic ideal gas undergoes the cyclic process $\mathrm{J} \rightarrow \mathrm{K} \rightarrow \mathrm{L} \rightarrow \mathrm{M} \rightarrow \mathrm{J}$, 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.
[ $\mathcal{R}$ is the gas constant.]
| List-I | List-II |
|---|---|
| (P) Work done in the complete cyclic process | (1) $RT_0 - 4RT_0 \ln 2$ |
| (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) $3RT_0$ |
| (S) Change in the internal energy of the gas in the process MJ | (4) $-2RT_0 \ln 2$ |
| (5) $-3RT_0 \ln 2$ |
$\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 5 ; \mathrm{S} \rightarrow 4$
$\mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 5 ; \mathrm{S} \rightarrow 2$
$\mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow 1 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 2$
$\mathrm{P} \rightarrow 2 ; \mathrm{Q} \rightarrow 5 ; \mathrm{R} \rightarrow 3 ; \mathrm{S} \rightarrow 4$
Explanation
$$\begin{aligned}
& \mathrm{J}\left(\mathrm{P}_0, \mathrm{~V}_0, \mathrm{~T}_0\right) \\
& K\left(P_0, 3 V_0, 3 T_0\right) \\
& \mathrm{M}\left(2 \mathrm{P}_0, \frac{\mathrm{V}_0}{2}, \mathrm{~T}_0\right) \\
& \mathrm{L}\left(2 \mathrm{P}_{\circ}, \frac{3 \mathrm{~V}_0}{2}, 3 \mathrm{~T}_0\right) \\
& \mathrm{P}_0 \mathrm{~V}_0=\mathrm{nRT}_0 \\
& \mathrm{JK} \rightarrow \text { isobaric } \Rightarrow \mathrm{W}=\mathrm{P}_0\left(2 \mathrm{~V}_0\right)=2 \mathrm{nRT}_0 \\
& \Delta \mathrm{U}=\frac{3}{2} \mathrm{nR}\left(2 \mathrm{~T}_0\right)=3 \mathrm{nRT}_0 \\
& \mathrm{KL} \rightarrow \text { isothermal } \rightarrow \mathrm{W}=\mathrm{nR}(3 \mathrm{~T}) \ell \mathrm{n}\left(\frac{1}{2}\right)=-3 \mathrm{nRT}_0 \ell \mathrm{n} 2 \\
& \Delta \mathrm{U}=0 \Rightarrow \mathrm{Q}=-3 \mathrm{nRT}_0 \ell \mathrm{n} 2 \\
& \mathrm{LM} \rightarrow \text { isobaric }=2 \mathrm{P}_0\left(-\mathrm{V}_0\right)=-2 \mathrm{nRT}_0 \\
& \mathrm{MJ} \rightarrow \text { isothermal } \Rightarrow \mathrm{nRT}_0 \ln 2 ; \Delta \mathrm{U}=0 \\
& \mathrm{WD}_{\text {nes }}=-2 \mathrm{nRT}_0 \ln 2 \\
& \mathrm{P} \rightarrow 4, \mathrm{Q} \rightarrow 3, \mathrm{R} \rightarrow 5, \mathrm{~S} \rightarrow 2 \\
&
\end{aligned}$$
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