Verified Solution • 2025–26 Session moderate

QJEEAdv22Ph4b75: JEE Advanced 2022 Physics - Class 11 JEEAdv22Ph4b75 Solution: List I describes thermodynamic processes in four different…

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.

List-I List-II
(I) 10−3 kg10^{-3} \mathrm{~kg} of water at 100∘C100^{\circ} \mathrm{C} is converted to steam at the same temperature, at a pressure of 105 Pa10^{5} \mathrm{~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=2250\, \mathrm{~kJ} / \mathrm{kg} .
 (P) 2 kJ2 \mathrm{~kJ}
(II) 0.20.2 moles of a rigid diatomic ideal gas with volume VV at temperature 500 K500 \mathrm{~K} undergoes an isobaric expansion to volume 3 V3 \mathrm{~V} . Assume R=8.0Jmol−1 K−1R=8.0 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1} .
 (Q) 7kJ7 k J
(III) One mole of a monatomic ideal gas is compressed adiabatically from volume V=13 m3V=\frac{1}{3} \mathrm{~m}^{3} and pressure 2kPa2 \mathrm{kPa} to volume V8\frac{V}{8} .
 (R) 4 kJ4 \mathrm{~kJ}
(IV) Three moles of a diatomic ideal gas whose molecules can vibrate, is given 9 kJ9 \mathrm{~kJ} of heat and undergoes isobaric expansion.
 (S) 5 kJ5 \mathrm{~kJ}
(T) 3 kJ3 \mathrm{~kJ}

Which one of the following options is correct?

Multiple Choice Options

  • A I →\rightarrow T, II →\rightarrow R, III →\rightarrow S, IV →\rightarrow Q
  • B I →\rightarrow S, II →\rightarrow P, III →\rightarrow T, IV →\rightarrow P
  • C I →\rightarrow P, II →\rightarrow R, III →\rightarrow T, IV →\rightarrow Q
  • D I →\rightarrow Q, II →\rightarrow R, III →\rightarrow S, IV →\rightarrow T

Step-by-Step Verified Solution

U=ML−PΔVU = ML - P\Delta V

=10−3×2250−102kP×(10−3−10−6)= {10^{ - 3}} \times 2250 - {10^2}kP \times ({10^{ - 3}} - {10^{ - 6}}) m3

= 2.25 kJ −- 0.1 kJ

= 2.15 kJ

I →\to P

(II) CV=5R2{C_V} = {{5R} \over 2} (rigid diatomic)

For isobaric expansion

V∝TV \propto T

V1V2=T1T2⇒V3V=500T2⇒T2=1500 K{{{V_1}} \over {{V_2}}} = {{{T_1}} \over {{T_2}}} \Rightarrow {V \over {3V}} = {{500} \over {{T_2}}} \Rightarrow {T_2} = 1500\,K

ΔU=nCVΔT=0.2×5×82×(1500−500)\Delta U = n{C_V}\Delta T = 0.2 \times {{5 \times 8} \over 2} \times (1500 - 500) J = 4 kJ

II →\to R

(III) Adiabatic expansion (γ=53)\left( {\gamma = {5 \over 3}} \right)

P1V1γ=P2V2γ⇒{P_1}V_1^\gamma = {P_2}V_2^\gamma \Rightarrow (2 kPa) ×V05/3=P2×(V08)5/3\times V_0^{5/3} = {P_2} \times {\left( {{{{V_0}} \over 8}} \right)^{5/3}}

P2=64{P_2} = 64 kPa

ΔU=nCVΔT\Delta U = n{C_V}\Delta T

=3nRΔT2=32(P2V2−P1V1)=32×(64×13×8−2×13)= {{3nR\Delta T} \over 2} = {3 \over 2}({P_2}{V_2} - {P_1}{V_1}) = {3 \over 2} \times \left( {64 \times {1 \over {3 \times 8}} - 2 \times {1 \over 3}} \right)

=32×(83−23)=3= {3 \over 2} \times \left( {{8 \over 3} - {2 \over 3}} \right) = 3 kJ

III →\to T

(IV) For isobaric expaision,

ΔU=nCVΔT=72 nRΔT\Delta U = n{C_V}\Delta T = {7 \over 2}\,nR\Delta T

ΔQ=nCPΔT=92 nRΔT\Delta Q = n{C_P}\Delta T = {9 \over 2}\,nR\Delta T

ΔUΔQ=79{{\Delta U} \over {\Delta Q}} = {7 \over 9}

ΔU=79ΔQ=7\Delta U = {7 \over 9}\Delta Q = 7 kJ

IV →\to Q

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