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JEE Main Physics Class 11 Kinetic Theory of Gases Part 1 Questions
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Question : 22
Total: 100
The volume
V
of an enclosure contains a mixture of three gases,
16
g
of oxygen,
28
g
of nitrogen and
44
g
of carbon dioxide at absolute temperature
T
. Consider
R
as universal gas constant. The pressure of the mixture of gases is
[16 Mar 2021 Shift 1]
88
R
T
V
3
R
T
V
5
2
R
T
V
4
R
T
V
Validate
Solution:
From ideal gas equation,
p
V
=
(
n
1
+
n
2
+
n
3
)
R
T
where,
n
1
=
number of moles of oxygen
=
16
32
n
2
=
number of moles of nitrogen
=
28
28
n
3
=
number of moles of carbon dioxide
=
44
44
⇒
p
V
=
[
16
32
+
28
28
+
44
44
]
R
T
=
[
1
2
+
1
+
1
]
R
T
=
5
2
R
T
⇒
p
=
5
2
R
T
V
This is the required pressure of the mixture of the gases.
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