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JEE Main Physics Class 11 Thermodynamics Part 2 Questions
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© examsnet.com
Question : 12
Total: 82
A thermodynamic system is taken from an original state
A
to an intermediate state
B
by a linear process as shown in the figure. It's volume is then reduced to the original value from
B
to
C
by an isobaric process. The total work done by the gas from A to B and B to C would be :
[29-Jan-2024 Shift 1]
33800
J
2200
J
800
J
1200
J
Validate
Solution:
👈: Video Solution
Work done
AB
=
1
2
(
8000
+
6000
)
Dyne
∕
cm
2
×
4
m
3
=
(
6000
Dyne
∕
cm
2
)
×
4
m
3
Work done
BC
=
−
(
4000
Dyne
∕
cm
2
)
×
4
m
3
Total work done
=
2000
Dyne
∕
cm
2
×
4
m
3
=
2
×
10
3
×
1
10
5
N
cm
2
×
4
m
3
=
2
×
10
−
2
×
N
10
−
4
m
2
×
4
m
3
=
2
×
10
2
×
4
Nm
=
800
J
© examsnet.com
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