In an adiabatic process, the density of a diatomic gas becomes 32 times its initial value. The final pressure of the gas is found to be n times the initial pressure. The value of n 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
32
b
132
c
326
d
128
answer is D.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
If density of certain amount of gas becoming 32 times ⇒V2=V1/32 and it is given that diatomic gas undergoes adiabatic processPVγ=constantPmργ=constantPργ=constant∵ρ=mV, m is constant P∝ργP2P1=ρ2ρ1γnP1P1=32ρ1ρ175n=3275=2575=27=128