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JEE Main 24 June 2022 Shift 1 Solved Paper
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© examsnet.com
Question : 51
Total: 90
0.056
kg
of Nitrogen is enclosed in a vessel at a temperature of
127
∘
C
. Th amount of heat required to double the speed of its molecules is _____
k
cal.
Take
R
=
2
cal
mole
−
1
K
−
1
)
[24-Jun-2022-Shift-1]
Your Answer:
Validate
Solution:
Because the vessel is closed, it will be an isochoric process.
To double the speed, temperature must be 4 times
(
∨
α
√
T
)
So,
T
f
=
1600
K
,
T
i
=
400
K
number of moles are
56
28
=
2
so
Q
=
n
C
v
∆
T
=
2
×
5
2
×
2
×
1200
=
12000
cal
=
12
K
cal
© examsnet.com
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