JEE Main 26 Feb 2021 Shift 2 Solved Paper

Given, ideal gas equation, U=3pV+4
Here, U= internal energy,
p= pressure
and V= volume.
By ideal gas equation, pV=nR∆T
=R∆T (∵n=1)...(i)
∆U‌‌=nCv∆T=3pV+4
‌‌=n‌

f

2

R∆T=3pV+4
⇒‌fR∆T‌‌=6pV+8
⇒‌f⋅pV‌‌=6pV+8 [From Eq. (i)]
⇒‌f‌‌=6+‌

8

pV

[From Eq. (i)] As, degree of freedom is more than 6 .
∴ Gas is polyatomic.