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← JEE Advanced 2022 · Paper 1
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JEE Advanced 2022 · Paper 1, Physics question 17

List I describes thermodynamic processes in four different systems. List II gives the magnitudes (either exactly or as a close approximation) of possible changes in the internal energy of the system due to the process.

Which one of the following options is correct?

List-I

  • I
    10−310^{-3} kg of water at 100∘100^\circC is converted to steam at the same temperature, at a pressure of 10510^5 Pa. The volume of the system changes from 10−6 m310^{-6}\ \mathrm m^3 to 10−3 m310^{-3}\ \mathrm m^3 in the process. Latent heat of water = 2250 kJ/kg.
  • II
    0.2 moles of a rigid diatomic ideal gas with volume VV at temperature 500 K undergoes an isobaric expansion to volume 3V3V. Assume R=8.0 J mol−1K−1R = 8.0\ \mathrm{J\,mol^{-1}K^{-1}}.
  • III
    One mole of a monatomic ideal gas is compressed adiabatically from volume V=13 m3V = \frac13\ \mathrm m^3 and pressure 2 kPa to volume V8\frac V8.
  • IV
    Three moles of a diatomic ideal gas whose molecules can vibrate, is given 9 kJ of heat and undergoes isobaric expansion.

List-II

  • P
    2 kJ
  • Q
    7 kJ
  • R
    4 kJ
  • S
    5 kJ
  • T
    3 kJ
Choose one option
Answer and worked solution

Official answer: C

(I) Q=10−3×2250=2.25Q=10^{-3}\times2250=2.25 kJ and W=105×(10−3−10−6)≈0.1W=10^5\times(10^{-3}-10^{-6})\approx0.1 kJ, so
ΔU≈2.15 kJ≈2 kJ → P.\Delta U\approx2.15\ \text{kJ}\approx2\ \text{kJ}\ \to\ \textbf{P}.

(II) Isobaric expansion to 3V3V raises TT from 500 K to 1500 K:
ΔU=nCvΔT=0.2×52×8×1000=4 kJ → R.\Delta U=nC_v\Delta T=0.2\times\tfrac52\times8\times1000=4\ \text{kJ}\ \to\ \textbf{R}.

(III) Adiabatic compression to V8\frac V8 with γ=53\gamma=\frac53 gives P2=2×85/3=64P_2=2\times8^{5/3}=64 kPa:
ΔU=32(P2V2−P1V1)=32(6424−23)=3 kJ → T.\Delta U=\tfrac32(P_2V_2-P_1V_1)=\tfrac32\left(\tfrac{64}{24}-\tfrac23\right)=3\ \text{kJ}\ \to\ \textbf{T}.

(IV) With vibration, Cv=72RC_v=\frac72R and Cp=92RC_p=\frac92R. At constant pressure:
ΔU=QCvCp=9×79=7 kJ → Q.\Delta U=Q\frac{C_v}{C_p}=9\times\tfrac79=7\ \text{kJ}\ \to\ \textbf{Q}.