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

One mole of an ideal monatomic gas undergoes the following four reversible processes:
Step1 - it is first compressed adiabatically from volume 8.0m3 to 1.0m3.
Step2 - then expanded isothermally at temperature T1 to volume 10.0m3.
Step3 - then expanded adiabatically to volume 80.0m3.
Step4 - then compressed isothermally at temperature T2 to volume 8.0m3.
Then T1/T2 is

A
2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
6
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
8
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 4
Given:
V1=8m3
V2=1m3
V3=10m3
V4=80m3

To find T1T2=?
for ourcase T2T4=? (as per our diagram)
Process: {23&41} Isothermal process

T2=T3&T4=T1

Process 12
we know for adiabatic process T1T2=(V2V1)γ1T1T2=(18)γ1 ...(i)
34 ad. expn.
T4T3=(V3V4)γ1=(18)γ1 ...(ii)

from eqn. (i) & (ii)
γ=1.66 T1T2=T4T3T4T2=(18)γ1T4T2=14

( Since T1=T4&[T4T3=(18)γ1])
T2T4=2
That is T1T2=4


777072_739620_ans_08d8e8f955724d75b2f6bac7a40c7181.png

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
The First Law of Thermodynamics
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon