for every 10 m you descend below the surface of water, the pressure on you will increase by an amount equal to atmospheric pressure 101kpa. an air bubble in a lake has a volume of 20 mm^3 at a depth of 40m. predict what volume it will have just before reaches the surface?
at surface pressure = 1 atm
gains 4 atm so at 5 atm at 40 m
pressure is 1/5 so volume is 5 times at surface
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To calculate the predicted volume of the air bubble just before it reaches the surface, we need to understand how pressure and volume are related in this scenario.
According to Boyle's Law, the pressure and volume of a gas are inversely proportional when temperature remains constant. In this case, as the air bubble rises to the surface of the lake, the pressure decreases, causing the volume to increase accordingly.
Given:
- Atmospheric pressure = 101 kPa
- Depth of the air bubble initially = 40 m
- Volume of the air bubble initially = 20 mm³
To determine the final volume of the air bubble, we can use the following formula:
P1 × V1 = P2 × V2
Where:
- P1 = Initial pressure
- V1 = Initial volume
- P2 = Final pressure (atmospheric pressure)
- V2 = Final volume (what we want to find)
Let's substitute the given values into the formula:
(Pressure at 40m + Atmospheric pressure) × Initial volume = Atmospheric pressure × Final volume
(40 × 10 kPa + 101 kPa) × 20 mm³ = 101 kPa × Final volume
(401 kPa) × 20 mm³ = 101 kPa × Final volume
Now, let's convert the units consistently. Since the pressure is in kilopascals and it's easier to work with cubic centimeters (cc) rather than millimeters (mm³), we can convert the volume to cc:
1 mm³ = 0.001 cc
20 mm³ × 0.001 cc/mm³ = 0.02 cc
401 kPa × 0.02 cc = 101 kPa × Final volume
8.02 cc × 101 kPa = 401 kPa × Final volume
810.02 kPa cc = 401 kPa × Final volume
Now, solve for Final volume:
Final volume = (810.02 kPa cc) / (401 kPa)
Final volume ≈ 2.02 cc
Therefore, the predicted volume of the air bubble just before it reaches the surface is approximately 2.02 cc.