CO2} is linear triatomic molecule. The average K.E. at temperature T will be: D) 7KT / 2 C) 6kT / 2 B) 5kT / 2 A) 3kT / 2 what will be the ans.?? is A is the ans??? pls justify..
New: Direct link to apply for JEE Main 2025 registration for session 1
Also Check: Crack JEE Main 2025 - Join Our Free Crash Course Now!
JEE Main 2025: Sample Papers | Syllabus | Mock Tests | PYQs | Video Lectures
JEE Main 2025: Preparation Guide | High Scoring Topics | Study Plan 100 Days
Hello,
It is very important to understand the concept of Degree of Freedom before you approach the question.
------- DOF or Degree of freedom in layman terms is basically no. Of ways in which a molecule can attain energy.
-------- So energy can be acquired by Translational motion, rotational motion and vibratory motion ( only at high temperatures).
-------- So, for a diatomic molecule and triatomic linear molecule like CO2 following DOF are associated with each form of motion:
• Translational motion - 3
• Rotational motion - 2
• Vibratory Motion - 2 (Only at High Temperature)
-------- Also energy associated with 1 DOF as per Maxwell's Law of equipartition of energy = 1/2KT ( K = Boltzmann Constant)
--------- Hence total Energy for CO2 molecule at normal temperature is 5/2KT.
I hope you find the answer helpful.
Thank you.
Related Questions
Know More about
Joint Entrance Examination (Main)
Eligibility | Application | Preparation Tips | Question Paper | Admit Card | Answer Key | Result | Accepting Colleges
Get Updates Brochure
Your Joint Entrance Examination (Main) brochure has been successfully mailed to your registered email id “”.
