Q23 / 90 · JEE Main 2019 (08 Apr Shift 2) · Chemistry

5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If CV=28 J K-1, calculate ∆U and ∆pV for the process. (R=8.0 J K-1 mol-1)

  • A
    ∆U=14 J; ∆pV=0.8 J
  • B
    ∆U=14 kJ; ∆pV=18 kJ
  • Correct answer
    C
    ∆U=14 kJ; ∆pV=4 kJ
  • D
    ∆U=2.8 kJ; ∆pV=0.8 kJ

Solution

ΔU=nCVΔT
=5×28×100=14000 J
=14 kJ
Δ(PV)=nRΔT=5×8×100=4000 J=4 kJ

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

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