Physics Applied to Diving 3.- Charles's Law
Charles's law shows the
relationship between the volume of a gas and its temperature in the Ideal Gas
Model. In SCUBA diving this law gives reason to why your dry suit changes
volume when the ambient temperature does.
This law enunciates
that: "The volume of a fixed mass of gas (at constant pressure) is
directly proportional to the thermodynamic temperature" [1]
The mathematical way to
show this law is:
V=T*k
Where V is the volume of
the gas, T is the ABSOLUTE temperature of the gas and k is a constant value. Analyzing
this law from a molecular point of view, if a gas has a certain temperature the
gas molecules have a certain kinetic energy and in consequence occupy a certain
amount of space. When the temperature increases the molecules move faster, so
in order to maintain the pressure, they need more space. On the other hand, if
temperature decreases molecules move slower so they need less space to maintain
the same amount of pressure.
Calculations
In order to calculate the volume of a gas at a certain temperature only using this law, you need a reference state. Imagine you have a reference state
"1" where the volume is V1
and the temperature is T1,
then you change the temperature to a known value we call T2. This new state where the
temperature is T2 will have a
different unknown volume value that we call V2.
To calculate the value of V2, we must do the procedure shown next.
V1/T1
= k
V2/T2
= k
Knowing that k is a
constant, meaning it doesn't change value, it has the same value in both cases
which gives us:
V1/T1 = k = V2/T2
V1/T1 = V2/T2
The equation shown above
is useful to calculate V1, T1,
V2 or T2 with the proper clearance whenever we have 3 of
the 4 variables. In this case because we want to calculate V2 we do the next clearance.
V2/T2 = V1/T1
(V2*T2)/T2 = (T2*V1)/T1
V2 = (T2*V1)/T1
Example 1.-
If a volume of gas of 5
liters has a 10 °C temperature, what will be the volume of the gas if you heat
it up to 40 °C?
What we know:
V1 = 5 L
T1 = 10 °C
V2 =?
T2 = 40 °C
The first thing we must
do is to change the temperature units to absolute ones.
T1
= 10 + 273.15 = 283.15 K
T2 = 40 +
273.15 = 313.15 K
Then to find the volume
we are looking for we must do some clearance, so:
V1/T1
= V2/T2
(T2*V1)/T1 = (T2*V2)/T2
(T2*V1)/T1 = V2
V2 = (313.15 K * 5 L) / 283.15 K
= 5.53 L
So, the volume of the
gas increments 0.53 L by heating it from 10 °C to 40 °C.
[1] Clusgston Michael
and Flemming Rosalind, Advanced Chemistry, Oxford, 2000, page 115
March 27, 2020
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