Divers know that the pressure exerted by
the water increases about 100 kPa with every
10.2 m of depth. This means that at 10.2 m
below the surface, the pressure is 201 kPa;
at 20.4 m below the surface, the pressure is
301 kPa; and so forth. If the volume of a
balloon is 2.2 L at STP and the temperature
of the water remains the same, what is the
volume 59.81 m below the water’s surface?
Do you know what this means?
School Subject:
To determine the volume 59.81 m below the water's surface, we need to make use of Boyle's Law, which states that the volume of a gas is inversely proportional to its pressure, assuming constant temperature.
Boyle's Law equation:
P₁V₁ = P₂V₂
Where:
P₁ = initial pressure (at surface level)
V₁ = initial volume (at surface level)
P₂ = final pressure (at a certain depth below the surface)
V₂ = final volume (to be determined)
First, let's calculate the final pressure 59.81 m below the water's surface using the given information. We know that the pressure increases by 100 kPa for every 10.2 m of depth. So, the number of 10.2 m increments from the surface to 59.81 m below is:
(Number of 10.2 m increments) = (59.81 m) / (10.2 m) = 5.857 approx.
Now, we can calculate the final pressure at a depth of 59.81 m below the surface:
Final pressure = (Number of 10.2 m increments) * (Pressure increase per 10.2 m increment) + (Pressure at surface level)
Final pressure = (5.857) * (100 kPa) + (101 kPa) (starting pressure at the surface)
Final pressure = 588.57 kPa + 101 kPa
Final pressure = 689.57 kPa
Now we have the final pressure (P₂ = 689.57 kPa) and the initial volume (V₁ = 2.2 L) at STP. We can use Boyle's Law to find the final volume (V₂) at this depth:
P₁V₁ = P₂V₂
(101 kPa) * (2.2 L) = (689.57 kPa) * (V₂)
To find V₂, rearrange the equation:
V₂ = (P₁V₁) / P₂
V₂ = (101 kPa * 2.2 L) / 689.57 kPa
V₂ ≈ 0.321 L (rounded to three decimal places)
Therefore, the volume 59.81 m below the water's surface is approximately 0.321 L.