The graph shows logarithmic readings of pressure and volume for two ideal gases A and B undergoing adiabatic process. It can be concluded that (one of the gases A and B is monoatomic and one is diatomic)
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
A is diatomic
b
B is diatomic
c
B is monatomic
d
Both (1) and (3)
answer is D.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
PVγ=C ln(P)+γln(V)=ln(C) ln(P)=−γln(V)+ln(C)→ straight line γ for B is higher.