Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
In non-rigid diatomic molecule with an additional vibrational mode
PhysicsKinetic Theory of GasesAP EAMCETAP EAMCET 2022 (04 Jul Shift 2)
Options:
  • A 81Cv2=49CP2
  • B 49Cv2=25CP2
  • C 49Cv2=81CP2
  • D 25Cv2=49Cp2
Solution:
2145 Upvotes Verified Answer
The correct answer is: 81Cv2=49CP2

In non-rigid diatomic molecule with an additional vibrational mode, total degree of freedom f=3rotational+3translational+1vibrational=7

Therefore, 

CpCv=1+2f=9781Cv2=49CP2

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.