The ratio of average translational KE to rotational KE of a linear polyatomic molecule at temperature T is
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
3 : 1
b
5 : 1
c
3 : 2
d
7 : 5
answer is C.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
For poly atomic linear molecule, number of degrees of freedom for translational motion is 3 and that for rotational motion is 2.∴The ratio = 3 x KT22 x KT2=32