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JEE Advanced 2017 Paper 2 · Chemistry · Chemical Thermodynamics · medium

One mole of a monatomic ideal gas (CV=3/2R, R=8.314 J K^-1 mol^-1) at 300 K and volume 8.0 L expands reversibly and adiabatically to 64.0 L.

One or more options correct

One mole of a monatomic ideal gas (CV=32RC_V=\tfrac32R, R=8.314 J K−1 mol−1R=8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}) at 300 K300\ \mathrm{K} and volume 8.0 L8.0\ \mathrm{L} expands reversibly and adiabatically to 64.0 L64.0\ \mathrm{L}. The correct statement(s) is(are)
  1. A. The final temperature is 75 K75\ \mathrm{K}
  2. B. The entropy change of the gas is zero
  3. C. The work done by the gas is about 2.81 kJ2.81\ \mathrm{kJ}
  4. D. The enthalpy change of the gas is about −2.81 kJ-2.81\ \mathrm{kJ}
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Answer: A, B, C

For a reversible adiabatic process TVγ−1=constantTV^{\gamma-1}=\text{constant} with γ=5/3\gamma=5/3, so T2=T1(V1/V2)2/3=300×(1/8)2/3=300/4=75 KT_2=T_1(V_1/V_2)^{2/3}=300\times(1/8)^{2/3}=300/4=75\ \mathrm{K} (A is correct). A reversible adiabatic process has qrev=0q_{rev}=0, so ΔS=0\Delta S=0 (B is correct). With q=0q=0, ΔU=w\Delta U=w, and ΔU=nCVΔT=1.5×8.314×(−225)=−2806 J\Delta U=nC_V\Delta T=1.5\times8.314\times(-225)=-2806\ \mathrm{J}, so the gas does +2.81 kJ+2.81\ \mathrm{kJ} of work on the surroundings (C is correct). The enthalpy change uses CP=52RC_P=\tfrac52R: ΔH=2.5×8.314×(−225)=−4677 J=−4.68 kJ\Delta H=2.5\times8.314\times(-225)=-4677\ \mathrm{J}=-4.68\ \mathrm{kJ}, not −2.81 kJ-2.81\ \mathrm{kJ}, which is ΔU\Delta U; D is wrong. Note that an isothermal reversible expansion over the same volumes would give ΔS=Rln⁡8=17.3 J K−1\Delta S=R\ln8=17.3\ \mathrm{J\,K^{-1}} instead.
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Original question written in the JEE Advanced 2017 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake