Verified Solution • 2025–26 Session moderate

QJEEM23Pha6eb: JEE Main 2023 Physics - Class 11 JEEM23Pha6eb Solution: Heat energy of is given to a diatomic gas allowing the gas to…

Heat energy of 735 J735 \mathrm{~J} is given to a diatomic gas allowing the gas to expand at constant pressure. Each gas molecule rotates around an internal axis but do not oscillate. The increase in the internal energy of the gas will be :

Multiple Choice Options

  • A 572 J572 \mathrm{~J}
  • B 441 J441 \mathrm{~J}
  • C 525 J525 \mathrm{~J}
  • D 735 J735 \mathrm{~J}

Step-by-Step Verified Solution

ΔQ=nCPΔT=\Delta Q=n C_{P} \Delta T=

n(f2+1)RΔTn\left( {{f \over 2} + 1} \right)R\Delta T

= n(52+1)RΔTn\left( {{5 \over 2} + 1} \right)R\Delta T

= n(72)RΔTn\left( {{7 \over 2}} \right)R\Delta T

Given, ΔQ\Delta Q = 735

⇒\Rightarrow n(72)RΔTn\left( {{7 \over 2}} \right)R\Delta T = 735

⇒\Rightarrow nRΔT=735×27nR\Delta T = 735 \times {2 \over 7}

Also, Δ\Delta U = f2nRΔT{f \over 2}nR\Delta T

= 52nRΔT{5 \over 2}nR\Delta T

= 52×735×27{5 \over 2} \times 735 \times {2 \over 7}

= 525 J

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