PrepRung
Sign inSign up free

JEE Advanced 2016 Paper 1 · Chemistry · Chemical Thermodynamics · hard

One mole of an ideal monoatomic gas (Cv=3/2R, R=8.314 J K^-1 mol^-1) at 300 K doubles its volume by three different routes: (i) reversible isothermal…

One or more options correct

One mole of an ideal monoatomic 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} doubles its volume by three different routes: (i) reversible isothermal expansion, (ii) reversible adiabatic expansion, (iii) adiabatic free expansion into vacuum. Choose the correct statement(s). (ln⁡2=0.693\ln2=0.693)
  1. A. In route (i), ΔSsys≈+5.76 J K−1\Delta S_{\mathrm{sys}}\approx+5.76\ \mathrm{J\,K^{-1}} and ΔSsurr≈−5.76 J K−1\Delta S_{\mathrm{surr}}\approx-5.76\ \mathrm{J\,K^{-1}}.
  2. B. In route (iii), ΔSsurr=0\Delta S_{\mathrm{surr}}=0 and ΔSuniverse≈+5.76 J K−1\Delta S_{\mathrm{universe}}\approx+5.76\ \mathrm{J\,K^{-1}}.
  3. C. In route (ii), the final temperature is about 150 K150\ \mathrm{K}.
  4. D. The magnitude of work done by the gas in route (ii) is greater than that in route (i).
Show answer and solution

Answer: A, B

(A) Correct. For the reversible isothermal route ΔSsys=Rln⁡2=8.314×0.693=5.76 J K−1\Delta S_{\mathrm{sys}}=R\ln2=8.314\times0.693=5.76\ \mathrm{J\,K^{-1}}, and since the process is reversible, the surroundings change by −5.76-5.76 so that ΔSuniv=0\Delta S_{\mathrm{univ}}=0. (B) Correct. In free expansion q=w=0q=w=0, so TT stays 300 K300\ \mathrm{K}; SS is a state function, so ΔSsys=Rln⁡2\Delta S_{\mathrm{sys}}=R\ln2 again, while the surroundings do not change, giving ΔSuniv=+5.76\Delta S_{\mathrm{univ}}=+5.76 (irreversible). (C) Incorrect. For reversible adiabatic expansion TVγ−1=TV^{\gamma-1}= constant with γ=5/3\gamma=5/3: T2=300×(1/2)2/3=189 KT_2=300\times(1/2)^{2/3}=189\ \mathrm{K}, not 150 K150\ \mathrm{K}. (D) Incorrect. ∣w(ii)∣=CvΔT=1.5×8.314×111=1384 J|w_{(ii)}|=C_v\Delta T=1.5\times8.314\times111=1384\ \mathrm{J} whereas ∣w(i)∣=RTln⁡2=1729 J|w_{(i)}|=RT\ln2=1729\ \mathrm{J}; the isothermal route does more work because heat flows in to compensate.
Report a mistake

Original question written in the JEE Advanced 2016 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake