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PHYSICS
Gay-Lussac’s law
Gay-Lussac studied the transformation of gases as explained in the previous section.
The first Gay-Lussac’s law states that:
at constant pressure the volume of a gas increases linearly with temperature.
When a gas is subject to a drop in temperature it transforms from gas to liquid.
This value corresponds to a temperature of absolute zero, the gas will re-expand if the temperature is increased.
Absolute zero is related to the Kelvin scale and corresponds to –273 degrees Celsius.
0
K
= – 273
°C
Above this temperature the volume of a gas re-expands in a linear fashion.
A
coefficient for expansion of gas
defined as
�
, valid for all gases:
�
= 3.663 * 10
-3
°C
-1
approx. equal to 1/273
°C
-1
If we imagine the gas in a closed container, the volume increase of
�
, represents the relative increase in pressure
when its temperature increases by 1
°C
.
The formula indicating this linearity when
p=K
(constant) is:
V
t
=
V
0
* (
1+
α
t)
V
t
is the volume occupied by the gas at temperature
t
°C
V
0
is the volume occupied by the gas at temperature
0
°C
t
is the temperature expressed in
°C
Example 1:
the temperature of a gas of volume 2
dm
3
increases from 273
K (t
i
)
to 373
K (t
f
)
,
∆
t
is 100
K
(which also corresponds to 100
°C
) therefore its volume becomes:
V
t
=
V
0
*
1+ 1
t
1
*
∆
t
V
t
=
V
0
*
1+ 1
273 * 100
V
t
=
2
*
�
1+
�
0,0036
* 100
��
V
t
=
2 * 1,36
V
t
=
2,72
dm
3
the gas volume increases about 36%.
Example 2:
its previous volume experiences a drop in temperature of 100
K.
From373
K
its volume becomes:
V
t
=
V
0
*
1 – 1
373
*
100
V
t
=2,72
*
�
1 –
�
0,0026
* 100
��
V
t
=2,72
* 0,74
V
t
=
2
dm
3
Example 3:
from the final condition of the first case, reducing the temperature with 20
K
(equivalent to 20
°C
)
the volume becomes:
V
t
=
V
0
*
1 – 1
353
*
20
V
t
=
2,72
*
�
1 –
�
0,0028
* 20
��
V
t
=2,72
* 0,94
V
t
=
2,56
dm
3
1
19
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