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JEE Main Physics Class 11 Thermodynamics Part 1 Questions
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© examsnet.com
Question : 88
Total: 100
An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume
V
1
and contains ideal gas at pressure
P
1
and temperature
T
1
. The other chamber has volume
V
2
and contains ideal gas at pressure
P
2
and temperature
T
2
. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be
[2008]
T
1
T
2
(
P
1
V
1
+
P
2
V
2
)
P
1
V
1
T
2
+
P
2
V
2
T
1
P
1
V
1
T
1
+
P
2
V
2
T
2
P
1
V
1
+
P
2
V
2
P
1
V
1
T
2
+
P
2
V
2
T
1
P
1
V
1
+
P
2
V
2
T
1
T
2
(
P
1
V
1
+
P
2
V
2
)
P
1
V
1
T
1
+
P
2
V
2
T
2
Validate
Solution:
Here
Q
=
0
and
W
=
0
. Therefore from first law of thermodynamics
Δ
U
=
Q
+
W
=
0
Internal energy of first vessle + Internal energy of second vessel = Internal energy of combined vessel
n
1
C
v
T
1
+
n
2
C
v
T
2
=
(
n
1
+
n
2
)
C
v
T
∴
T
=
n
1
T
1
+
n
2
T
2
n
1
+
n
2
For first vessel
n
1
=
P
1
V
1
R
T
1
and for second vessle
n
2
=
P
2
V
2
R
T
2
∴
T
=
P
1
V
1
R
T
1
×
T
1
+
P
2
V
2
R
T
2
×
T
2
P
1
V
1
R
T
1
+
P
2
V
2
R
T
2
=
T
1
T
2
(
P
1
V
1
+
P
2
V
2
)
P
1
V
1
T
2
+
P
2
V
2
T
1
© examsnet.com
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