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JEE Advanced 2016 Paper 2 · Chemistry · Chemical Thermodynamics · hard

One mole of an ideal monatomic gas (CV=3/2R, R=8.314 J mol^-1K^-1) at 300 K and 1 L is allowed to expand to 8 L by two different routes: (I) a…

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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} and 1 L1\ \mathrm{L} is allowed to expand to 8 L8\ \mathrm{L} by two different routes: (I) a reversible adiabatic expansion, and (II) a free expansion into an evacuated, thermally insulated vessel so that the gas finally occupies 8 L8\ \mathrm{L}. Choose the correct statement(s).
  1. A. The final temperature after process (I) is 75 K75\ \mathrm{K}.
  2. B. The work done by the gas in process (I) is about 3.74 kJ3.74\ \mathrm{kJ}.
  3. C. ΔSuniverse\Delta S_{\text{universe}} for process (I) is +17.3 J K−1+17.3\ \mathrm{J\,K^{-1}}.
  4. D. ΔSuniverse\Delta S_{\text{universe}} for process (II) is about +17.3 J K−1+17.3\ \mathrm{J\,K^{-1}}.
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Answer: A, D

(A) Reversible adiabatic: TVγ−1=TV^{\gamma-1}= constant with γ=5/3\gamma=5/3, so T2=300(1/8)2/3=75T_2=300(1/8)^{2/3}=75 K. True. (B) ∣w∣=CVΔT=1.5×8.314×225=2.81|w|=C_V\Delta T=1.5\times8.314\times225=2.81 kJ; the value 3.743.74 kJ results from wrongly using CV×300C_V\times300. False. (C) A reversible adiabatic process has ΔSsys=0\Delta S_{\text{sys}}=0 and ΔSsurr=0\Delta S_{\text{surr}}=0, so ΔSuniverse=0\Delta S_{\text{universe}}=0, not 17.317.3. False. (D) In the free expansion into vacuum with insulation, q=w=0q=w=0, ΔU=0\Delta U=0 and so TT is unchanged for an ideal gas; entropy is a state function, so ΔSsys=Rln⁡(8)=17.29 J K−1\Delta S_{\text{sys}}=R\ln(8)=17.29\ \mathrm{J\,K^{-1}} (computed numerically), while ΔSsurr=0\Delta S_{\text{surr}}=0. Hence ΔSuniverse=+17.3 J K−1\Delta S_{\text{universe}}=+17.3\ \mathrm{J\,K^{-1}}. True. Correct: A and D.
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