wiz-icon
MyQuestionIcon
MyQuestionIcon
8
You visited us 8 times! Enjoying our articles? Unlock Full Access!
Question

When one mole of monoatomic ideal gas at temperature T undergoes adiabatic change reversibly, change in volume is from 1 L to 5L, the final temperature in Kelvin would be:

A
T52/3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
T+23×0.0821
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
T
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
T23×0.0821
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A T52/3
We know the relation for an adiabatic reversible process :
T1T2=(V2V1)γ1
where, T1= initial temperature
T2= final temperature
V1= initial volume
V2= final volume
γ=CpCv=53 (for monoatomic gas)
so,
T1T2=(51)(5/3)1=5(2/3)
T2=T15(2/3)

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon