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JEE Advanced 2025 Paper 2 · Chemistry · Chemical Thermodynamics · medium

One mole of an ideal monatomic gas (CV=3/2R, R=8.314 J mol^-1K^-1) at 300 K is expanded reversibly to double its volume in two ways: (I) isothermally…

One or more options correct

One mole of an ideal monatomic gas (CV=32RC_V=\tfrac32R, R=8.314 J mol−1K−1R=8.314\ \mathrm{J\,mol^{-1}K^{-1}}) at 300 K300\ \mathrm{K} is expanded reversibly to double its volume in two ways: (I) isothermally and (II) adiabatically. Which of the following is/are correct?
  1. A. For process I, ΔSsystem=Rln⁡2≈5.76 J K−1\Delta S_{\text{system}}=R\ln2\approx5.76\ \mathrm{J\,K^{-1}}
  2. B. The work done by the gas in process II is smaller than in process I
  3. C. The final temperature in process II is 150 K150\ \mathrm{K}
  4. D. For process II, ΔSsurroundings=−5.76 J K−1\Delta S_{\text{surroundings}}=-5.76\ \mathrm{J\,K^{-1}}
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Answer: A, B

(A) Correct: ΔS=nRln⁡(V2/V1)=8.314ln⁡2=5.76 J K−1\Delta S=nR\ln(V_2/V_1)=8.314\ln2=5.76\ \mathrm{J\,K^{-1}}. (B) Correct: wI=RTln⁡2=1729 Jw_{I}=RT\ln2=1729\ \mathrm{J}. In II, TVγ−1=TV^{\gamma-1}= const with γ=5/3\gamma=5/3, so T2=300 (1/2)2/3=189.0 KT_2=300\,(1/2)^{2/3}=189.0\ \mathrm{K}, and wII=CV(T1−T2)=1.5×8.314×111.0=1384 Jw_{II}=C_V(T_1-T_2)=1.5\times8.314\times111.0=1384\ \mathrm{J}, which is less than 1729 J1729\ \mathrm{J} because the gas cools and its pressure falls faster. (C) Incorrect: T2=189 KT_2=189\ \mathrm{K}, not 150 K150\ \mathrm{K} (150 K150\ \mathrm{K} would follow from halving the temperature). (D) Incorrect: a reversible adiabatic process has q=0q=0 and ΔSsystem=0\Delta S_{\text{system}}=0, and the surroundings exchange no heat, so ΔSsurroundings=0\Delta S_{\text{surroundings}}=0.
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