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Standard XII
Chemistry
Real Gases
The compressi...
Question
The compression factor (compressibility factor) for
1
mole of a van der Waals' gas at
0
∘
C
and
100
atmospheric pressure is found to be
0.5
. Assuming that the volume of a gas molecule is negligible, calculate the van der Waals' constant
′
a
′
.
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Solution
We know that,
Z
=
P
V
R
T
0.5
=
100
×
V
0.0821
×
273
V
=
0.112
l
i
t
r
e
According to van der Waals' equation,
(
P
+
a
V
2
)
(
V
−
b
)
=
R
T
for
1
mole
[
100
+
a
(
0.112
)
2
]
[
0.112
−
0
]
−
0.0821
×
273
On solving, we get
a
=
1.253
L
2
m
o
l
−
2
a
t
m
.
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