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CGPET 2015 Solved Paper
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© examsnet.com
Question : 24
Total: 150
Two identical containers
A
and
B
with frictionless pistons contain the same ideal gas at the same temperature and the same volume
V
. The mass of the gas in
A
is
m
A
and that in
B
is
m
B
. The gas in each container is now allowed to expand isothermally to the same final volume
2
V
. The changes in the pressure in
A
and
B
are to be found
∆
p
and 1.5
∆
p
respectively, then relation for masses will be
4
m
A
=
9
m
B
2
m
A
=
3
m
B
3
m
A
=
2
m
B
9
m
a
=
4
m
B
Validate
Solution:
For isothermal expansion of an ideal gas
pV = constant
⇒
p
Δ
V
+
V
Δ
p
=
0
For container A,
p
iA
(
V
)
+
V
(
Δ
p
)
=
0
⇒
p
iA
=
−
V
Δ
p
V
=
−
Δ
p
.....(i)
where, p_i is initial pressure.
For container B
p
iB
(
V
)
+
V
(
1.5
Δ
p
)
=
V
⇒
p
iB
=
−
1.5
Δ
p
....(ii)
From Eqs. (i) and (ii), we have
p
iA
=
p
iB
1.5
=
2
3
p
iB
⇒
m
A
=
2
3
mB
⇒
3
m
A
=
2
m
B
[
∵
p
∞
m
]
© examsnet.com
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