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Chemistry 443 Honors
Fall, 2007
Question # 1
11
DIFFERENCES IN GAS LAWS
The idealgas law is generally used for gases “if the pressure is sufficiently low”.
When
this limit is not reached, one has to resort to the socalled realgas laws, one of which is
the gas law of van der Waals.
Let’s compare these two laws under different conditions.
(a) Calculate the pressure of a gas using the idealgas law and the van der Waalsgas law
when one mole of carbon dioxide at temperature is 300K is confined in 1 m
3
.
If you
assume that the van der Waalsgas law gives the “correct” pressure, by what percent is
the calculation using the idealgas law in error?
The pressure is easily calculated by direct substitution for both cases.
Ideal gas
Torr
Pa
m
K
K
mol
J
mol
V
nRT
P
709
.
18
10
49433
.
2
1
300
3144349
.
8
1
3
3
1
1
=
×
=
×
×
=
=
−
−
This is a low pressure, so one might expect that the van der Waals equation would give a
result close to it.
van der Waals gas
Torr
Pa
Pa
Pa
m
mol
m
Pa
mol
mol
m
Pa
mol
m
K
K
mol
J
mol
V
a
n
nb
V
nRT
P
707
.
18
07
.
2494
3658
.
0
10
437
.
2494
)
1
(
3658
.
0
)
1
(
10
8588
.
42
1
1
300
3144349
.
8
1
3
2
3
2
6
2
1
3
6
3
1
1
2
2
=
=
−
×
=
−
×
×
−
×
=
−
−
=
−
−
−
−
−
These are extremely close values, so close that one has to carry the calculation to six
decimal places.
The error in the calculation by using the idealgas equation is
%
011
.
0
100
707
.
18
707
.
18
709
.
18
%
=
×
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
=
Torr
Torr
Torr
error
(b) Repeat the calculations of part (a), except in this case consider that one mole of the
gas is confined in a volume of 1 liter (i.e. it is confined in a volume of
0.001 m
3
).
The calculations proceed by the same method here; the numbers are different.
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View Full DocumentChemistry 443 Honors
Fall, 2007
Question # 1
12
Ideal gas
bar
Torr
Pa
m
K
K
mol
J
mol
V
nRT
P
9433
.
24
18709
10
49433
.
2
001
.
0
300
3144349
.
8
1
6
3
1
1
=
=
×
=
×
×
=
=
−
−
This pressure is significantly higher than that calculated in part a.
The pressure
calculated with the van der Waals’s equation might be expected to be different.
van der Waals gas
bar
Pa
Pa
Pa
m
mol
m
Pa
mol
mol
m
Pa
mol
m
K
K
mol
J
mol
V
a
n
nb
V
nRT
P
402
.
22
10
2402
.
2
10
658
.
3
10
60602
.
2
)
001
.
0
(
3658
.
0
)
1
(
10
8588
.
42
1
001
.
0
300
3144349
.
8
1
6
5
6
2
3
2
6
2
1
3
6
3
1
1
2
2
=
×
=
×
−
×
=
−
×
×
−
×
=
−
−
=
−
−
−
−
−
The error made by using the idealgas law in this situation is
%
3
.
11
100
402
.
22
402
.
22
9433
.
24
%
=
×
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
=
bar
bar
bar
error
This is a much larger difference between the two results.
(c) Usually, the parameters that we can easily control are temperature and pressure, with
the gas volume being the parameter calculated.
Calculate the molar volume of CO
2
at
300 K and 1.000 bar, using the idealgas law and the van der Waalsgas law.
What is the
percentage error made by using the idealgas law under this condition, if you consider
that the van der Waalsgas equation gives the “correct” value?
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 Spring '07
 PhilipJ.Silva
 Chemistry, pH

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