Advanced Uses Of Boyle’s Law
We know from our previous examples that according to Boyle’s Law, an inverse relationship exists between the volume and the pressure of a gas. Thus, pressure (P) times volume (V) will equal a constant (K).
That means that at any depth, whether at 20 feet or even 100 feet the pressure times the volume must be equal. We can put this into an equation where P1 and V1 are the pressure and volume at the first depth and P2 and V2 are the pressure and volume at the second depth.
Let’s try an example that uses this relationship. Let’s say an Aquarius scientist has filled a balloon to one cubic foot at a depth of 66 feet. If our scientist wants to send it to the surface with a sample they collected, what will be the volume of the balloon at the surface? Let’s plug our numbers in and see.
Recall from our previous lessons that seawater exerts a weight of .445 pounds per foot of depth. So, our pressure at the initial depth (P1) is equal to 29.37 psi. This is gauge pressure, so we need to add an additional 14.7 psi to account for atmospheric pressure and get our absolute pressure, which is 44.07 psi.
At the surface we know the pressure is 14.7 psi. This is our P2. Our initial volume is 1 cubic foot, our V1. So let’s plug all these numbers into our equation.
Now all we have to do is solve for V2. We do this by dividing 44.07 by 14.7 to get our unknown V2 which is equal to 3 cubic feet. So, our balloon has expanded to three times its original size. As you can see, we can use the relationship, P1 x V1 = P2 x V2, to calculate pressure or volume at any depth.
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Lesson design by Lucas Gillispie.