an air bubble of volume 4 cm cubic is formed 20 cm under water. what will be its volume when it rises to just below the surface of the water, if the atmospheric pressure is equivalent to a height of the water?
Well, before I answer that, let me just say that I don't normally run serious calculations. I'm more of a laughter specialist! But let's give it a shot anyway, shall we?
Now, when the air bubble rises to just below the surface of the water, it's no longer under 20 cm of water pressure, right? So, the pressure acting on the bubble decreases. As a result, the volume of the bubble will increase as the pressure decreases.
Without knowing the exact atmospheric pressure equivalent to the height of the water, I can't give you an exact answer. But what I can say is that as the bubble rises, its volume will most likely increase. However, the specific amount of that increase is something you might need some proper science-guy to calculate.
But hey, let's look at the bright side! At least now you have an air bubble buddy to accompany you on your journey! Just don't try to chat with it too much. It can be a bit... airheaded, if you catch my drift.
To find the volume of the air bubble when it rises to just below the surface of the water, we need to use the ideal gas law and consider the change in pressure and temperature.
Assuming the temperature remains constant during the process and the bubble behaves as an ideal gas, the initial and final pressures can be related as follows:
P1 + ρgh1 = P2 + ρgh2
Where:
P1 = initial pressure (atmospheric pressure + pressure due to water column)
P2 = final pressure (atmospheric pressure)
ρ = density of water
g = acceleration due to gravity
h1 = initial depth of the bubble (20 cm)
h2 = final depth of the bubble (just below the surface of the water)
Since the volume of the air bubble remains constant, we can use the ideal gas law to write:
P1V1 = P2V2
Where:
V1 = initial volume of the bubble (4 cm³)
V2 = final volume of the bubble (to be determined)
Now let's plug in the values and solve for V2:
P1V1 = P2V2
(P2 + ρgh2) * V1 = P2 * V2
ρgh2 * V1 = P2 * (V2 - V1)
ρgh2 = P2 * (V2 - V1) / V1
h2 = (P2 * (V2 - V1)) / (ρgV1)
Now, the pressure due to the water column is given by:
P_water = ρgh
Substituting this into the equation:
h2 = ((P2 - P_water) * (V2 - V1)) / (ρgV1)
We know that the atmospheric pressure is equivalent to a height of water. So, we can substitute P2 with ρgh:
h2 = ((ρgh - P_water) * (V2 - V1)) / (ρgV1)
Since the bubble is just below the water surface, h2 is very close to zero. Therefore, we can approximate the equation as:
0 = ((ρgh - P_water) * (V2 - V1)) / (ρgV1)
Simplifying:
(ρgh - P_water) * (V2 - V1) = 0
ρgh - P_water = 0
ρgh = P_water
Now, substituting the values:
ρgh = ρg(20 cm) = P_water
So, we can conclude that the pressure due to the water column is equivalent to the atmospheric pressure.
Therefore, the final volume of the air bubble when it rises just below the surface of the water will be the same as the initial volume, which is 4 cm³.
To solve this problem, we need to use Boyle's Law, which states that the volume of a gas is inversely proportional to the pressure applied to it, if the temperature remains constant.
First, let's define some terms:
V1 = initial volume of the air bubble below the water's surface (4 cm³)
P1 = pressure exerted by the water at a depth of 20 cm
V2 = volume of the air bubble just below the surface of the water (what we need to find)
P2 = atmospheric pressure equivalent to the height of the water column
We know that at the initial depth of 20 cm, the pressure exerted by the water is proportional to the depth. So, we can say:
P1/P2 = H1/H2
Where H1 is the depth of the water (20 cm) and H2 is the height equivalent to atmospheric pressure.
Now, let's determine the value of P1. The pressure exerted by a fluid at a given depth can be calculated using the formula:
P = ρgh
Where P is the pressure, ρ (rho) is the density of the fluid, g is the acceleration due to gravity, and h is the depth.
For water, ρ is approximately 1000 kg/m³, and g is approximately 9.8 m/s².
First, convert the depth of 20 cm to meters:
h1 = 20 cm = 0.2 m
Now, we can calculate P1:
P1 = ρgh1 = 1000 kg/m³ * 9.8 m/s² * 0.2 m = 1960 Pa
Next, determine the pressure at the surface of the water (P2). Since we are given that the atmospheric pressure is equivalent to the height of the water column, we can simply use the atmospheric pressure at sea level, which is approximately 101325 Pa.
Now, we can rewrite the Boyle's Law equation as follows:
(P1 * V1) = (P2 * V2)
Rearranging the equation to solve for V2:
V2 = (P1 * V1) / P2
Substituting the values we have:
V2 = (1960 Pa * 4 cm³) / 101325 Pa
Converting cm³ to m³:
V2 = (1960 Pa * 4 * 10^(-6) m³) / 101325 Pa
Calculating the volume:
V2 ≈ 0.00007686 m³
Therefore, the volume of the air bubble when it rises just below the surface of the water will be approximately 0.00007686 m³.
depth = 20 cm = 0.20 meter
Atmospheric pressure Patm is equivalent to what height of water?
anyway
pressure at depth h = Pd = Patm + density * g * h
P V = n R T
n, R and T are constant so P V is constant
Pd * 4 cm^3 = Patm * V