wiz-icon
MyQuestionIcon
MyQuestionIcon
5
You visited us 5 times! Enjoying our articles? Unlock Full Access!
Question

An ideal gas of mass m and temperature T1 undergoes a reversible isothermal process from an initial pressure P1 to final pressure P2. The heat loss during the process is Q. The entropy change ΔS of the gas is

A
mRIn(P2P1)QT1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
mRIn(P1P2)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
mRIn(P2P1)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
zero
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B mRIn(P1P2)
We know that
dS=δQT
Intergrating between states 1 and 2 gives
21ΔS=21δQT
S2S1=1T21δQ T=C
=1T×Q12=mRTInV2V2T
mRInV2V1=mRInp1p2
where p1p2=V2V1 for isothermal process

Alternatively:
We know that the entropy change in terms of temperautre and pressure rations.
S2S1=mcpInT2T1S2S1=mRInp2p1
where T1=T2 for isothermal process
S2S1=mRInp2p1=mRInp1p2

flag
Suggest Corrections
thumbs-up
1
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Enthalpy
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon