Box 9.3

The Effect of a Vacuum on Entrained Gas

Bubble-Volume Increase

Bubble-volume increase is expressed by a simplified form of the
general gas law, when ignoring the effect of temperature:

P1V1 =P2V2

where
. P = pressure in absolute terms, psia
. V = volume of the bubble

Using a 2-mm-diameter entrained gas bubble, with 1-mm radius,
and 14.7 as absolute pressure, from (P1V1 = P2V2 ) it can be seen
that reducing the atmospheric pressure by one-half doubles the
volume of the bubble.

volume of a sphere = 4/3Πr³ or 4:189r³

P1 × V1 = P2 × V2

So the radius of the bubble with twice the volume becomes 2.099 mm.

Bubble Surface Area Increase

To go a bit further and relate volume to radius to final surface area:
surface area of the bubble; S = 4r² or 12.566r². 

But the surface area increase will be the ratio of the radii squared or
4.406/1.

So, decreasing the pressure from 1 atm (14.7 psia or 29.9 in. Hg)
to ½ atm (7.35 psia or 14.95 in. Hg) doubles the volume of the
bubble and increases the surface area of the original 2-mm-diameter
bubble by 4.4 times.

Two things are happening simultaneously: The bubble volume
is getting larger, so it will rise to the surface faster, where there is
further decreased pressure; and the skin, or surface area, of the
bubble is becoming larger and weaker.