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JEE Advanced 2017 Paper 1 · Physics · Thermodynamics and Kinetic Theory of Gases · hard

JEE Advanced 2017 Paper 1 Physics: Paragraph 1

One mole of a monatomic ideal gas (γ=5/3\gamma=5/3) starts at (P0,V0,T0)(P_0,V_0,T_0) and expands quasi-statically until its volume is 2V02V_0 by one of four processes listed in Column 1. Column 2 gives the molar heat capacity CC of the gas along the process (taking the isothermal process as C→∞C\to\infty), and Column 3 gives the heat absorbed by the gas in the process.

Column 1: (I) isothermal expansion; (II) isobaric expansion; (III) reversible adiabatic expansion; (IV) expansion along a straight line P=kVP=kV passing through the origin.
Column 2: (i) C=0C=0; (ii) C=52RC=\tfrac52R; (iii) C=2RC=2R; (iv) C=∞C=\infty.
Column 3: (P) 00; (Q) P0V0ln⁡2P_0V_0\ln2; (R) 52P0V0\tfrac52P_0V_0; (S) 6P0V06P_0V_0.

Each question asks for the one row-consistent combination, i.e. a process of Column 1 together with its correct entries from Columns 2 and 3.

Question 1

Single correct option

Which combination describes the process in which the final temperature of the gas is 4T04T_0?
  1. A. (IV) (ii) (S)
  2. B. (II) (ii) (R)
  3. C. (IV) (iii) (S)
  4. D. (I) (iv) (S)
Show answer and solution

Answer: (C) (IV) (iii) (S)

The process P=kVP=kV gives PV=kV2PV=kV^2, so T∝V2T\propto V^2 and doubling the volume multiplies TT by 44: this is process (IV). The polytropic exponent is PV−1=PV^{-1}= const, i.e. x=−1x=-1, so C=CV+R1−x=32R+12R=2RC=C_V+\dfrac{R}{1-x}=\tfrac32R+\tfrac12R=2R, which is (iii). Work done is ∫kV dV=32P0V0\int kV\,dV=\tfrac32P_0V_0 and ΔU=32R(4T0−T0)=92P0V0\Delta U=\tfrac32R(4T_0-T_0)=\tfrac92P_0V_0, so Q=6P0V0Q=6P_0V_0, which is (S). In the isobaric case the final temperature is only 2T02T_0; the isothermal one stays at T0T_0. Hence the combination is (IV) (iii) (S). A frequent error is to use CPC_P for every expansion; here the final temperature uniquely fixes the polytropic process.

Question 2

Single correct option

Which combination describes the process in which the final pressure of the gas equals 2−5/3P02^{-5/3}P_0?
  1. A. (III) (i) (Q)
  2. B. (III) (iii) (P)
  3. C. (I) (iv) (P)
  4. D. (III) (i) (P)
Show answer and solution

Answer: (D) (III) (i) (P)

For a reversible adiabatic process PVγ=PV^{\gamma}= const, so Pf=P0(V02V0)5/3=2−5/3P0≈0.315P0P_f=P_0\left(\dfrac{V_0}{2V_0}\right)^{5/3}=2^{-5/3}P_0\approx0.315P_0. The isothermal process gives P0/2P_0/2, the isobaric one P0P_0 and the P=kVP=kV line gives 2P02P_0, so only process (III) fits. In an adiabatic process there is no heat exchange, so C=0C=0, entry (i), and the heat absorbed is zero, entry (P). The other combinations fail because they pair the adiabatic process with a non-zero heat in (Q), with the heat capacity 2R2R of the straight-line process, or pair the isothermal process with the zero heat. The answer is (III) (i) (P).

Question 3

Single correct option

Which combination describes the process in which the heat absorbed by the gas is exactly equal to the work done by it?
  1. A. (II) (ii) (Q)
  2. B. (I) (i) (Q)
  3. C. (I) (iv) (R)
  4. D. (I) (iv) (Q)
Show answer and solution

Answer: (D) (I) (iv) (Q)

By the first law Q=ΔU+WQ=\Delta U+W, so Q=WQ=W requires ΔU=0\Delta U=0, i.e. no change in temperature for an ideal gas: this is the isothermal process (I). Along an isotherm absorbing heat while TT does not change needs an infinite heat capacity, so C=∞C=\infty, entry (iv). The work is W=RT0ln⁡2V0V0=P0V0ln⁡2W=RT_0\ln\dfrac{2V_0}{V_0}=P_0V_0\ln2, so Q=P0V0ln⁡2Q=P_0V_0\ln2, entry (Q). For the other processes the work is P0V0P_0V_0 (isobaric, with Q=52P0V0Q=\tfrac52P_0V_0), 1.5P0V01.5P_0V_0 (straight line, with Q=6P0V0Q=6P_0V_0) and 1.5P0V0(1−2−2/3)1.5P_0V_0(1-2^{-2/3}) (adiabatic, with Q=0Q=0), none equal. The answer is (I) (iv) (Q).
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