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JEE Main Physics Class 11 Thermodynamics Part 2 Questions
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© examsnet.com
Question : 43
Total: 82
Heat energy of
735
J
is given to a diatomic gas allowing the gas to expand at constant pressure. Each gas molecule rotates around an internal axis but do not oscillate. The increase in the internal energy of the gas will be:
[31-Jan-2023 Shift 2]
525
J
441
J
572
J
735
J
Validate
Solution:
∆
Q
=
n
C
P
∆
T
=
735
J
⇒
5
n
R
∆
T
2
=
735
J
∆
U
=
n
C
V
∆
T
=
3
2
(
n
R
∆
T
)
=
3
2
×
2
5
×
735
=
441
J
© examsnet.com
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