An air bubble released from a diver's breathing apparatus at a depth of 40 m with diameter 2.0. When it reaches the surface water has Diameter of 3.4.
Show volume of air bubble is approx 5.
Done!
Easy...
V=4/3 x 3.14 x 1^3
V=4.18cm^3
V=4/3 x 3.14 x 1.7^3
V=20.56
So 20.56/4.18= 4.9 = approx 5.
Q
Hence Calculate the increase in pressure experienced by the diver when descending to a depth of 40m, assuming that temperature is constant.
Oh and BTW: Atmospheric pressure= 101kPa.
Not my homework!
2007-03-01 04:20:11 · 3 answers · asked by Miss LaStrange 5 in Science & Mathematics ➔ Physics
According to Boyle's Law
PV at constant temperature.
101 x 5 = 505
2007-03-01 05:13:45 · answer #1 · answered by Anonymous · 0⤊ 0⤋
If you assume the air is an ideal gas, then PV = nRT.
The amount of gas in the bubble and the temperature are both constant, so nRT is the same at 40 m down and at the surface. Therefore, let P1, V1 be the pressure and volume at 40 m, and P2, V2 be the pressure and volume at the surface.
P1 * V1 = nRT = P2 * V2
P1 * V1 = P2 * V2
P1/P2 = V2/V1
Since you already calculated V2/V1 (approximately 5), you have P2/P1 is about 5. In other words, the pressure at 40 m is about 5 atmospheres, or 505 kPa.
2007-03-01 12:38:12 · answer #2 · answered by Grizzly B 3 · 0⤊ 0⤋
What's the bubble got to do with it ?
1 m³ has a base area of 100 x 100 = 10,000 cm²
The height of 1 m is 100 cm.
Each cm² therefore has 100 cm above it. 1cm³ (1cc) = 1 gram
A height of 100 cm is therefore exerting a pressure of 100g/cm²
40 m = 4000 cm height acting on a single cm² = 4000g/cm²
Pressure at 40 m = 4kg/cm² (405.2 kPa)
2007-03-01 12:39:13 · answer #3 · answered by Norrie 7 · 0⤊ 0⤋