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?
0 --> 1 atm
10 --> 2 atm
20 --> 3 atm
30 --> 4 atm
40 --> 5 atm
20 * 5 = 100 mm^3
because V = nRT/P
Why did the air bubble go to therapy? Because it was feeling under pressure! Now, let's solve this buoyancy puzzle.
For every 10 meters you descend below the surface, the pressure increases by 101 kPa, just like the atmospheric pressure. So, at a depth of 40 meters, the pressure on the air bubble would be 4 times the atmospheric pressure.
Now, let's figure out the volume of the air bubble at the surface. Since the pressure decreases as you ascend, you can use Boyle's Law to solve this. Boyle said that at constant temperature, the product of pressure and volume is always constant.
So, if the pressure at the surface is just the atmospheric pressure (101 kPa) and assuming the temperature remains constant, we can set up the following equation:
Volume(initial) x Pressure(initial) = Volume(final) x Pressure(final)
Substituting the given values:
20 mm^3 x 4 x 101 kPa = Volume(final) x 101 kPa
Now, let's solve for Volume(final):
Volume(final) = (20 mm^3 x 4 x 101 kPa) / 101 kPa
Canceling out the kPa:
Volume(final) = 20 mm^3 x 4
So, the volume of the air bubble just before it reaches the surface would be 80 mm^3. Looks like our bubble got a little bigger during its ascent!
To solve this problem, we need to calculate the change in pressure as the air bubble rises from a depth of 40m to the surface.
1. Determine the pressure at a depth of 40m:
Using the given information, we know that for every 10m depth, the pressure increases by 101 kPa. Therefore, the pressure at a depth of 40m would be:
Pressure at 40m = Atmospheric pressure + (Pressure increase per 10m * Depth in multiples of 10m)
Pressure at 40m = 101 kPa + (101 kPa * 4)
Pressure at 40m = 101 kPa + 404 kPa
Pressure at 40m = 505 kPa
2. Calculate the change in volume:
According to Boyle's Law, the volume of a gas is inversely proportional to its pressure, assuming the temperature remains constant.
Since the pressure decreases as the air bubble rises to the surface, the volume will increase. We can calculate the change in volume using the formula:
P1 * V1 = P2 * V2
Where P1 is the initial pressure, V1 is the initial volume, P2 is the final pressure, and V2 is the final volume.
Substituting the values:
P1 = 505 kPa, V1 = 20 mm^3, P2 = 101 kPa
505 kPa * 20 mm^3 = 101 kPa * V2
V2 = (505 kPa * 20 mm^3) / 101 kPa
V2 ≈ 100 mm^3
Therefore, just before reaching the surface, the air bubble will have a volume of approximately 100 mm^3.
To predict the volume of the air bubble just before it reaches the surface, we can use Boyle's Law, which states that the volume of a gas is inversely proportional to its pressure. Boyle's Law can be expressed as:
P1 x V1 = P2 x V2
Where P1 and P2 are the initial and final pressures respectively, and V1 and V2 are the initial and final volumes respectively.
In this case, the initial pressure is the sum of atmospheric pressure (101 kPa) and the pressure at a depth of 40m (which we can calculate using the given information).
We know that the pressure increases by an amount equal to atmospheric pressure for every 10m descent below the surface. So, for 40m depth, the pressure increase would be 4 times the atmospheric pressure (4 x 101 kPa = 404 kPa).
Let's assume the final pressure just before reaching the surface is atmospheric pressure (101 kPa). We can solve for the final volume using the formula:
P1 x V1 = P2 x V2
(101 kPa + 404 kPa) x 20 mm^3 = 101 kPa x V2
(505 kPa) x 20 mm^3 = 101 kPa x V2
V2 = (505 kPa x 20 mm^3) / 101 kPa
V2 ≈ 100 mm^3
Therefore, the predicted volume of the air bubble just before it reaches the surface is approximately 100 mm^3.