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JEE Advanced 2024Paper 1
Exam
JEE Advanced
Year
2024
Paper
Paper 1
Subject
Physics
Chapter
Thermodynamics

JEE Advanced 2024 · Paper 1 · Physics · Thermodynamics

One mole of a monatomic ideal gas undergoes the cyclic process J...

One mole of a monatomic ideal gas undergoes the cyclic process J→K→L→M→J\mathrm{J} \rightarrow \mathrm{K} \rightarrow \mathrm{L} \rightarrow \mathrm{M} \rightarrow \mathrm{J}, as shown in the P-T diagram. One mole of a monatomic ideal gas undergoes the cyclic process J arrow K arrow L arrow M arrow J , as shown in the P-T Match the quantities mentioned in List-I with their values in List-II and choose the correct option. [R\mathcal{R} is the gas constant.] (
List-IList-II
(P) Work done in the complete cyclic process(1) \mathcal{R} T_0-4 \mathcal{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 \mathcal{R} T_0
(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
)

Options

  1. AP→1;Q→3;R→5;S→4\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 5 ; \mathrm{S} \rightarrow 4
  2. BP→4;Q→3;R→5;S→2\mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 5 ; \mathrm{S} \rightarrow 2
  3. CP→4\mathrm{P} \rightarrow 4; Q→1;R→2;S→2\mathrm{Q} \rightarrow 1 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 2
  4. DP→2;Q→5;R→3;S→4\mathrm{P} \rightarrow 2 ; \mathrm{Q} \rightarrow 5 ; \mathrm{R} \rightarrow 3 ; \mathrm{S} \rightarrow 4

Correct answer

B. P→4;Q→3;R→5;S→2\mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 5 ; \mathrm{S} \rightarrow 2

Step-by-step solution

J(P0,V0,T0)K(P0,3V0,3T0)M(2P0,V02,T0)L(2P0,3V02,3T0)P0V0=nRT0JK→ isobaric ⇒W=P0(2V0)=2nRT0ΔU=32nR(2T0)=3nRT0KL→ isothermal →W=nR(3T)ln⁡(12)=−3nRT0ℓn2ΔU=0⇒Q=−3nRT0ln⁡2LM→ isobaric =2P0(−V0)=−2nRTe0MJ→ isothermal ⇒nRTn0ln⁡2;ΔU=0WD net =−2nRTe0ln⁡2P→4,Q→3,R→5,S→2\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}
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