CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Pressure-temperature relationship for an ideal gas undergoing adiabatic change is:

γ=Cp/Cv

A
PTγ= constant
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
PT1+γ= constant
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
PT1Tγ= constant
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
P1γTγ= constant
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D P1γTγ= constant
From the first law of thermodynamics, dU=dQ+dW
But for adiabatic process, dQ=0
dU=dW
dU=nCvdT
dW=PdV
nCvdT=PdVnCvdT+PdV=0
For one mole of gas, n=1 CvdT+PdV=0
For ideal gas, PV=RT
P=RTV
CvdT+RTVdV=0
CvdTT+RdVV=0 [CpCv=R]
CvdTT+(CpCv)dVV=0 γ=CpCv
dTT+(CpCv1)dVV=0
dTT+(γ1)dVV=0
1TdT+(γ1)1VdV=0
On integration,
ln T+(γ1)ln V=ln C
ln TVγ1=ln C
TVγ1=C
T(RTP)1γ=C [PV=RT; \ V=RTP]
TγP(1γ)=constant


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Charles' Law
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon