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JEE Main Physics Class 11 Thermodynamics Part 1 Questions
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© examsnet.com
Question : 52
Total: 100
One mole of an ideal monoatomic gas is taken along the path ABCA as shown in the PV diagram. The maximum temperature attained by the gas along the path BC is given by
[Online April 16, 2018]
25
8
P
0
V
0
R
25
4
P
0
V
0
R
25
16
P
0
V
0
R
5
8
P
0
V
0
R
Validate
Solution:
Equation of the BC
P
=
P
0
−
2
P
0
V
0
(
V
−
2
V
0
)
using
P
V
=
n
R
T
Temperature,
T
=
P
0
V
−
2
P
0
V
2
V
0
+
4
P
0
~
V
1
×
R
(
∵
n
=
1
mole given
)
T
=
P
0
F
[
5
V
−
2
V
2
V
0
]
d
T
d
V
=
0
⇒
5
−
4
V
V
0
=
0
⇒
V
=
5
4
V
0
T
=
P
0
R
[
5
×
5
V
0
4
−
2
V
0
×
25
16
V
0
2
]
=
25
8
P
0
V
0
R
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