# physics

Assume all temperatures to be exact, and neglect significant figures for small changes in dimension.
A diver releases an air bubble of volume 4.0cm3cm^3 from a depth of 16mm below the surface of a lake, where the temperature is 8.0∘C.

What is the volume of the bubble when it reaches just below the surface of the lake, where the temperature is 22∘C^\circ C?

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1. neglect significant figures for small changes in dimension. >> I cannot believer an instructor actually gave those words to you. For instance, how PRECISE is 4.0 cm^3. What you should have been taught, is the 4 is known with precision, and the 0 is estimated, so the actual measurement can vary from 3.60 to 4.49. So what exactly does one mean on neglect small changes? The world does not work that way.

So given your instructors missue of significant figures, I would take all this to mean
temperatures are infinitely precise (in the real world, they are not, they are measurements also)
other dimensions are to be taken as infinitely precise also.

Pressure1*volume1/Pressure1=P2*V2/T2

you are given V1 as 4.0cm^3
you are looking for V2
pressure 1 is hinted at with depth
you are given Temp2, and temp1

now pressure of the lake depth.
Now I wonder about the depth, do you really mean 16 millimeters below the surface? I doubt it, as your temp of 8C cant be that cold 16mm from the surface temp of 22C.

So assume 16 meters.

pressure at 16m:
pressure due to water depth: density water* depth=1000kg/m^3, and we will again assume that is <infinitely > precise, and does not change with temperature.

pressureduetoDepth=16KPa
total pressure at depth=16kPa+101.3kPa
P1=117.3kPa

finally, the math...

V2=T2P1V1/T1P2= (273+22)(117.3)(4.0)/(273*101.3)

estimating in my mind, about a 20 percent increase over 4.0, let me check..Hmmm, not bad estimating.

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