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JEE Advanced 2017 Paper 1 · Chemistry · Chemical Thermodynamics · medium

JEE Advanced 2017 Paper 1 Chemistry: Paragraph 2

One mole of an ideal monatomic gas (CV=32RC_V=\tfrac32R, CP=52RC_P=\tfrac52R) at T1=300 KT_1=300\ \mathrm{K} undergoes one of the four reversible processes listed in Column 1. Use R=8.314 J mol−1K−1R=8.314\ \mathrm{J\,mol^{-1}K^{-1}} and ln⁡2=0.693\ln2=0.693.

Column 1 (process): (I) isothermal expansion to twice the initial volume; (II) adiabatic expansion to twice the initial volume; (III) isobaric expansion to twice the initial volume; (IV) heating at constant volume until the temperature has doubled to 600 K600\ \mathrm{K}.

Column 2 (entropy change of the gas): (i) 00; (ii) Rln⁡2R\ln2; (iii) 52Rln⁡2\tfrac52R\ln2; (iv) 32Rln⁡2\tfrac32R\ln2.

Column 3 (heat absorbed by the gas, in J): (P) about 17291729; (Q) 00; (R) about 62366236; (S) about 37413741.

In each question the correct option is the row set (process, its entropy change, its heat absorbed) that satisfies the stated condition; all four options list internally consistent triples of the table.

Question 1

Single correct option

The process in which the heat absorbed by the gas is exactly 53\tfrac53 times the heat absorbed in process (IV) is
  1. A. (I), (ii), (P)
  2. B. (II), (i), (Q)
  3. C. (III), (iii), (R)
  4. D. (IV), (iv), (S)
Show answer and solution

Answer: (C) (III), (iii), (R)

Evaluate the heat in each process. (I) Isothermal: q=RT1ln⁡2=8.314×300×0.693=1729q=RT_1\ln2=8.314\times300\times0.693=1729 J, with ΔS=Rln⁡2\Delta S=R\ln2. (II) Adiabatic reversible: q=0q=0 and ΔS=qrev/T=0\Delta S=q_\mathrm{rev}/T=0. (III) Isobaric: doubling VV at constant PP doubles TT to 600600 K, so q=CPΔT=52(8.314)(300)=6236q=C_P\Delta T=\tfrac52(8.314)(300)=6236 J and ΔS=CPln⁡2=52Rln⁡2\Delta S=C_P\ln2=\tfrac52R\ln2. (IV) Isochoric: q=CVΔT=32(8.314)(300)=3741q=C_V\Delta T=\tfrac32(8.314)(300)=3741 J and ΔS=CVln⁡(600/300)=32Rln⁡2\Delta S=C_V\ln(600/300)=\tfrac32R\ln2. The ratio q(III)/q(IV)=CP/CV=γ=5/3q_{(III)}/q_{(IV)}=C_P/C_V=\gamma=5/3, since both processes have the same temperature rise of 300300 K. The required process is therefore (III), with ΔS=52Rln⁡2\Delta S=\tfrac52R\ln2 and q≈6236q\approx6236 J, so the correct option is C. For the other options the ratios are 0.460.46, 00 and 11.

Question 2

Single correct option

The process in which the work done by the gas is equal to the heat absorbed by it is
  1. A. (I), (ii), (P)
  2. B. (II), (i), (Q)
  3. C. (III), (iii), (R)
  4. D. (IV), (iv), (S)
Show answer and solution

Answer: (A) (I), (ii), (P)

Apply the first law, q=ΔU+wbyq=\Delta U+w_\mathrm{by}, to each process. (I) Isothermal: ΔU=0\Delta U=0 for an ideal gas because UU depends only on TT, so wby=q=RT1ln⁡2=1729w_\mathrm{by}=q=RT_1\ln2=1729 J. This is the required match, with ΔS=Rln⁡2\Delta S=R\ln2. (II) Adiabatic: q=0q=0, but the gas does work −ΔU=32R(T1−T2)-\Delta U=\tfrac32R(T_1-T_2) where T2=300×2−2/3=189T_2=300\times2^{-2/3}=189 K, which is 13841384 J, so work and heat are unequal. (III) Isobaric: w=PΔV=RT1=2494w=P\Delta V=RT_1=2494 J while q=6236q=6236 J, so they differ (the rest raises UU by 37413741 J). (IV) Isochoric: w=0w=0 but q=3741q=3741 J. Hence only the isothermal process has equal heat and work, and the correct option is A. A quick check on the table is that the entropy of an isothermal expansion is qrev/T=Rln⁡2q_\mathrm{rev}/T=R\ln2 as listed in (ii).

Question 3

Single correct option

The process in which the internal energy of the gas decreases is
  1. A. (I), (ii), (P)
  2. B. (II), (i), (Q)
  3. C. (III), (iii), (R)
  4. D. (IV), (iv), (S)
Show answer and solution

Answer: (B) (II), (i), (Q)

For an ideal monatomic gas ΔU=CVΔT=32RΔT\Delta U=C_V\Delta T=\tfrac32R\Delta T, so the sign of ΔU\Delta U is the sign of ΔT\Delta T. (I) Isothermal: ΔT=0\Delta T=0, so ΔU=0\Delta U=0. (II) Reversible adiabatic expansion: q=0q=0 and the gas does work at the expense of its internal energy. From TVγ−1=constantTV^{\gamma-1}=\text{constant} with γ=5/3\gamma=5/3 and V2=2V1V_2=2V_1, T2=300×2−2/3=189T_2=300\times2^{-2/3}=189 K, so ΔT=−111\Delta T=-111 K and ΔU=32(8.314)(−111)=−1384\Delta U=\tfrac32(8.314)(-111)=-1384 J. This process has ΔS=0\Delta S=0 and q=0q=0, which are entries (i) and (Q) of the table. (III) Isobaric expansion: TT doubles to 600600 K, so ΔU=+3741\Delta U=+3741 J. (IV) Isochoric heating: ΔU=q=+3741\Delta U=q=+3741 J. Only process (II) cools the gas, so the correct option is B. Note that (III) and (IV) have the same ΔU\Delta U because UU is a function of temperature only.
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Original question written in the JEE Advanced 2017 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