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

Three different processes that can occur in an ideal monoatomic gas are shown in the P vs V diagram. The paths are labelled as AB, AC and AD. The change in internal energies during these process are taken as EAB, EAC and EAD and the workdone as WAB,WAC and WAD. The correct relation between these parameters are


A
EAB=EAC<EAD,WAB>0,WAC=0,WAD<0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
EAB=EAC=EAD,WAB>0,WAC=0,WAD<0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
EAB<EAC<EAD,WAB>0,WAC>WAD
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
EAB>EAC>EAD,WAB<WAC<WAD
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B EAB=EAC=EAD,WAB>0,WAC=0,WAD<0
Temperature change ΔT is same for all three processes AB,AC and AD
ΔU=nCvΔT= same
EAB=EAC=EAD
Work done, W=P×ΔV
For process AB volume is increasing WAB>0
For process AD volume is decreasing WAD<0
For process AC volume is constant WAC=0

flag
Suggest Corrections
thumbs-up
3
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