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JEE Main Chemistry Class 11 Chemical Thermodynamics Part 2 Questions
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© examsnet.com
Question : 83
Total: 88
2.2
g
of nitrous oxide
(
N
2
O
)
gas is cooled at a constant pressure of
1
atm
from
310
K
to
270
K
causing the compression of the gas from
217.1
mL
to
167.75
mL
. The change in internal energy of the process,
∆
U
is '
−
x
′
J
. The value of '
x
' is ______[nearest integer]
(Given : atomic mass of
N
=
14
g
mol
−
1
and of
O
=
16
g
mol
−
1
. Molar heat capacity of
N
2
O
is
100
J
K
−
1
mol
−
1
)
[29-Jun-2022-Shift-2]
Your Answer:
Validate
Solution:
∆
T
=
−
40
K
∆
U
=
q
+
w
=
100
×
2.2
44
(
−
40
)
−
(
−
49.39
)
×
10
−
3
×
101.325
=
−
200
+
5
=
−
195
J
x
=
195
© examsnet.com
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