Q53 / 54 · JEE Advanced 2022 (Paper 1) · Physics

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 kg of water 100 °C is converted to steam at the same temperature, at a pressure of 105Pa. The volume of the system changes from 10-6m3 to 10-3m3 in the process. Latent heat of water =2250 kJ kg-1 (P) 2 kJ
(II) 0.2 moles of a rigid diatomic ideal gas with volume V at temperature 500 K undergoes an isobaric expansion to volume 3 V. Assume 8.0 J mol–1 K–1 (Q) 7 kJ
(III) One mole of a monoatomic ideal gas is compressed adiabatically from volume V=13m3 and pressure 2 kPa to volume V8. (R) 4 kJ
(IV) Three moles of a diatomic ideal gas whose molecules can vibrate, is given 9 kJ of heat and undergoes isobaric expansion. (S) 5 kJ
    (T) 3 kJ

Which one of the following options is correct?

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

Solution

(I)

According to the first law of thermodynamics,

∆Q=∆U+W⇒∆U=∆Q-W

⇒U=ML-PΔV

=10-3×2250-102kP×10-3-10-6m3

=2.25 kJ-0.1 kJ

=2.15 kJ

Therefore, I-P

(II)

For isobaric process,

V1V2=T1T2⇒T2=3×500=1500 K

Now the change in the internal energy will be,

∆U=nCV∆T=0.2×52×8×1000=4 kJ

Therefore, II-R

(III) 

For adiabatic expansion of monoatomic gas γ=53

P1V1γ=P2V2γ⇒2kPa×V053=P2×V0853

⇒P2=64 kPa

Now the change in the internal energy will be,

ΔU=nCvΔT

=3nRΔT2=32P2V2-P1V1=32×64×13×8-2×13

=32×83-23=3 kJ

Therefore, III-T

(IV)

For diatomic gas whose molecules vibrate,

CV=62R and CP=82R

⇒ΔU=nCVΔT=3nRΔT

ΔQ=nCpΔT=4nRΔT

ΔUΔQ=34

ΔU=34×9=6.75≃7 kJ

Therefore, IV-Q

Hence, the correct answer is option: 'C'.

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