wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A non conducting piston divides an adiabatic container into two equal parts. Such that piston is in equilibrium and temperature on both sides is also same as T0 . Now the 1st chamber is heated such that piston moves very slowly until volume of 2nd chamber is reduced to 14th of the initial. In both chambers, the same monatomic gas is filled such that the number of moles are same. (24/3=2.5)

Open in App
Solution

W.D on piston = 0
WL+WR=0
for ||nd chamber Adiabatic process
T0V0531=Tf.(V04)531
Tf=T0423=T0(2)23=2.5T0
WR=ΔU=n32R(2.5T0T0)=94nRT0
So. WL=94nRT0P0V530=P(V04)53
Pf=P0(4)53=P0(4)53
P04537V04Tf=P0V0T0Tf=7.(4)23T0
TL=352T0
Heat supllied
Q=94nRT0+n32R(3521)T0=(94+994)nRT
=1084nRT0=27nRT0

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
First Law of Thermodynamics
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon