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Physics · Thermodynamics

JEE Advanced 2024 — Paper 1 — Question 31

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. Match the quantities mentioned in List-I with their values in List-II and choose the correct option. [ RR is the gas constant.]

List-IList-II
(P) Work done in the complete cyclic process(1) RT0−4RT0ln⁡2R 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) 3RT03 R T_{0}
(S) Change in the internal energy of the gas in the process MJ(4) −2RT0ln⁡2\quad-2 R T_{0} \ln 2
(5) −3RT0ln⁡2\quad-3 R T_{0} \ln 2
Question figure
  1. Option A:

    P→1;Q→3;R→5;S→4\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 5 ; \mathrm{S} \rightarrow 4

  2. Option 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

    Correct
  3. Option C:

    P→4;Q→1;R→2;S→2\mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow 1 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 2

  4. Option D:

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

Answer: B

Step-by-step solution

(P)JK⇒ΔWJK=PΔV=nRΔT=1.R(3T0−T0)=2RT0\quad(P) J K \Rightarrow \Delta W_{J K}=P \Delta V=n R \Delta T=1 . R\left(3 T_{0}-T_{0}\right)=2 R T_{0}

LM⇒ΔWLM=PΔV=nRΔT=1.R(T0−3 T0)=−2R T0\mathrm{LM} \Rightarrow \Delta \mathrm{W}_{\mathrm{LM}}=\mathrm{P} \Delta \mathrm{V}=\mathrm{nR} \Delta \mathrm{T}=1 . \mathrm{R}\left(\mathrm{T}_{0}-3 \mathrm{~T}_{0}\right)=-2 R \mathrm{~T}_{0}

KL⇒ΔWKL=nRTℓ(PiPf)=R3T0ℓn(P02P0)=−3RT0ℓn2\mathrm{KL} \Rightarrow \Delta \mathrm{W}_{\mathrm{KL}}=n R T \ell\left(\frac{P_{i}}{P_{f}}\right)=R 3 T_{0} \ell n\left(\frac{P_{0}}{2 P_{0}}\right)=-3 R T_{0} \ell n 2

MJ⇒Δ WMJ=nRTℓn(PiPf)=RT0ℓn(2P0P0)=RT0ℓn2\mathrm{MJ} \Rightarrow \Delta \mathrm{~W}_{\mathrm{MJ}}=\mathrm{nRT} \ell \mathrm{n}\left(\frac{\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{\mathrm{f}}}\right)=\mathrm{RT}_{0} \ell \mathrm{n}\left(\frac{2 \mathrm{P}_{0}}{\mathrm{P}_{0}}\right)=\mathrm{RT}_{0} \ell \mathrm{n} 2

Total work done ΔW=−2RT0ℓn2\Delta \mathrm{W}=-2 \mathrm{R} \mathrm{T}_{0} \ell \mathrm{n} 2

(Q) ΔUJK=nCVΔT=1⋅32R(3 T0−T0)=3R T0\Delta \mathrm{U}_{\mathrm{JK}}=\mathrm{nC}_{\mathrm{V}} \Delta \mathrm{T}=1 \cdot \frac{3}{2} \mathrm{R}\left(3 \mathrm{~T}_{0}-\mathrm{T}_{0}\right)=3 R \mathrm{~T}_{0}

(R) ΔQKL=ΔWKL+ΔUKL=ΔWKL+0=−3RT0ℓn2\Delta \mathrm{Q}_{\mathrm{KL}}=\Delta \mathrm{W}_{\mathrm{KL}}+\Delta \mathrm{U}_{\mathrm{KL}}=\Delta \mathrm{W}_{\mathrm{KL}}+0=-3 \mathrm{R} \mathrm{T}_{0} \ell \mathrm{n} 2

(S) ΔUMJ=nCVΔT=0\Delta \mathrm{U}_{\mathrm{MJ}}=\mathrm{nC}_{\mathrm{V}} \Delta \mathrm{T}=0

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2024
Paper
Paper 1
Subject
Physics
Chapter
Thermodynamics
Topic
Calculation of Work, Heat and Internal Energy

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