Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Physics
Work done per mol in an isothermal change is
Q. Work done per mol in an isothermal change is
2587
271
Thermodynamics
Report Error
A
RT
lo
g
10
V
1
V
2
22%
B
RT
lo
g
10
V
2
V
1
6%
C
RT
lo
g
e
V
1
V
2
61%
D
RT
lo
g
e
V
2
V
1
12%
Solution:
For isothermal process
p
V
=
RT
⇒
p
=
V
RT
=
W
=
p
d
V
=
∫
v
1
v
2
V
RT
d
V
=
RT
lo
g
e
V
1
V
2