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 3.2 L at STP and the temperature of the water remains the same, what is the
volume 40.32 m below the water’s surface? Answer in units of L
So the new pressure is 101 + (100*40.32/10.2) = ?. Then p1v1 = p2v2
Why do I need to add 101?
Read the problem.
"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;"
So the pressure 10.2m below is 100 + whatever it is at the surface (which is 101 kPa) so the total is 101 + 100 = 201. And at 20.4 it is 301 (that's 101 + 100+100 = 301). Right? The problem has given you the numbers to make a general formula and that's what I did.
0.59
To find the volume of the balloon at 40.32 m below the water's surface, we first need to use the relationship between pressure and volume using Boyle's Law. Boyle's Law states that the pressure and volume of a gas are inversely proportional at constant temperature.
According to the information provided, divers know that the pressure increases about 100 kPa for every 10.2 m of depth. This means that the pressure at 10.2 m is 201 kPa, at 20.4 m it is 301 kPa, and so on.
Now, let's find the pressure at 40.32 m below the water's surface. Since the pressure increases by 100 kPa for every 10.2 m, we can calculate:
Pressure increase per meter = 100 kPa / 10.2 m = 9.8 kPa per meter
Pressure at 40.32 m below the surface = 201 kPa + (9.8 kPa per meter * 40.32 m)
Pressure at 40.32 m below the surface = 201 kPa + 394.176 kPa
Pressure at 40.32 m below the surface = 595.176 kPa
Now that we have the pressure at the given depth, we can use Boyle's Law to find the volume of the balloon. Boyle's Law equation is:
P1 * V1 = P2 * V2
Where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.
At STP (Standard Temperature and Pressure), the pressure is 101.3 kPa and the volume is 3.2 L. We can plug these values into the equation along with the pressure at 40.32 m:
(101.3 kPa) * (3.2 L) = (595.176 kPa) * V2
Solving for V2 will give us the volume at 40.32 m below the water's surface:
V2 = (101.3 kPa * 3.2 L) / (595.176 kPa)
V2 ≈ 0.544 L
Therefore, the volume of the balloon at 40.32 m below the water's surface is approximately 0.544 L.