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Test Index
JEE Main Physics Class 11 Kinetic Theory of Gases Part 1 Questions
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Question : 83
Total: 100
A gas molecule of mass
M
at the surface of the Earth has kinetic energy equivalent to
0
°
C
. If it were to go up straight without colliding with any other molecules, how high it would rise? Assume that the height attained is much less than radius of the earth.
(
k
B
.
is Boltzmann constant).
[Online April 19, 2014]
0
273
k
B
2
M
g
546
k
B
3
M
g
819
k
B
2
M
g
Validate
Solution:
Kinetic energy of each molecule,
K
.
E
.
=
3
2
K
B
T
In the given problem,
Temperature,
T
=
0
°
C
=
273
K
Height attained by the gas molecule,
h
=
?
K
.
E
.
=
3
2
K
B
(
273
)
=
819
K
B
2
K
.
E
=
P
E
⇒
819
K
B
2
=
M
g
h
or
h
=
819
K
B
2
M
g
© examsnet.com
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