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JEE Main 202331 JanuaryEvening Shift
Exam
JEE Main
Year
2023
Date
31 January
Shift
Evening Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2023 · 31 January · Evening Shift · Physics · Thermodynamics

Heat energy of 735J is given to a diatomic gas allowing the gas to...

Heat energy of 735J735J 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:

Options

  1. A525J525J
  2. B441J441J
  3. C572J572J
  4. D735J735J

Correct answer

A. 525J525J

Step-by-step solution

For isobaric process and diatomic gas we can write, \DeltaQ=nCP\DeltaT=735J\DeltaQ = n{C}_{P}\DeltaT = 735J

⇒n7R2\DeltaT=735J⇒nR\DeltaT=210⇒ n\dfrac{7R}{2}\DeltaT = 735J ⇒ nR\DeltaT = 210

Now internal energy will be,

\DeltaU=nCV\DeltaT=n5R2\DeltaT=52×210\DeltaU = n{C}_{V}\DeltaT = n\dfrac{5R}{2}\DeltaT = \dfrac{5}{2} \times 210

=525J= 525J

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