One mole of a monatomic ideal gas undergoes the cyclic process $\mathrm{J} \rightarrow \mathrm{K}…

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.] $\begin{array}{l|l} \text{List-I} & \text{List-II} \\ \hline \text{(P) Work done in the complete cyclic process} & (1) \mathcal{R} T_{0}-4 \mathcal{R} T_{0} \ln 2 \\ \text{(Q) Change in the internal energy of the gas in the process JK} & (2) 0 \\ \text{(R) Heat given to the gas in the process KL} & (3) 3 \mathcal{R} T_{0} \\ \text{(S) Change in the internal energy of the gas in the process MJ} & (4) -2 \mathcal{R} T_{0} \ln 2 \\ & (5) -3 \mathcal{R} T_{0} \ln 2 \end{array}$
  1. $\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 5 ; \mathrm{S} \rightarrow 4$
  2. $\mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 5 ; \mathrm{S} \rightarrow 2$
  3. $\mathrm{P} \rightarrow 4$; $\mathrm{Q} \rightarrow 1 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 2$
  4. $\mathrm{P} \rightarrow 2 ; \mathrm{Q} \rightarrow 5 ; \mathrm{R} \rightarrow 3 ; \mathrm{S} \rightarrow 4$

Solution

$\begin{aligned} & \mathrm{J}\left(\mathrm{P}_0, \mathrm{~V}_0, \mathrm{~T}_0\right) \\ & \mathrm{K}\left(\mathrm{P}_0, 3 \mathrm{~V}_0, 3 \mathrm{~T}_0\right) \\ & \mathrm{M}\left(2 \mathrm{P}_0, \frac{\mathrm{V}_0}{2}, \mathrm{~T}_0\right) \\ & \mathrm{L}\left(2 \mathrm{P}_0, \frac{3 \mathrm{~V}_0}{2}, 3 \mathrm{~T}_0\right) \\ & \mathrm{P}_0 \mathrm{~V}_0=\mathrm{nRT}_0 \\ & \mathrm{JK} \rightarrow \text { isobaric } \Rightarrow \mathrm{W}=\mathrm{P}_0\left(2 \mathrm{~V}_0\right)=2 \mathrm{nRT}_0 \\ & \Delta \mathrm{U}=\frac{3}{2} \mathrm{nR}\left(2 \mathrm{~T}_0\right)=3 \mathrm{nRT}_0 \\ & \mathrm{KL} \rightarrow \text { isothermal } \rightarrow \mathrm{W}=\mathrm{nR}(3 \mathrm{~T}) \ln \left(\frac{1}{2}\right)=-3 \mathrm{nRT}_0 \ell \mathrm{n} 2 \\ & \Delta \mathrm{U}=0 \Rightarrow \mathrm{Q}=-3 \mathrm{nRT}{ }_0 \ln 2 \\ & \mathrm{LM} \rightarrow \text { isobaric }=2 \mathrm{P}_0\left(-\mathrm{V}_0\right)=-2 \mathrm{nRT} \mathrm{e}_0 \\ & \mathrm{MJ} \rightarrow \text { isothermal } \Rightarrow \mathrm{nRT} \mathrm{n}_0 \ln 2 ; \Delta \mathrm{U}=0 \\ & \mathrm{WD} \text { net }=-2 \mathrm{nRT} \mathrm{e}_0 \ln 2 \\ & \mathrm{P} \rightarrow 4, \mathrm{Q} \rightarrow 3, \mathrm{R} \rightarrow 5, \mathrm{~S} \rightarrow 2\end{aligned}$ `

Asked in: JEE Advanced 2024 (Paper 1)

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