f.4phy (gases)

Suppose that atm pressure = 105 Pa and g = 10 m/s2

Pressure at water surface = 105 Pa

Pressure at 10 m below water surface = 105 + 1000 x 10 x 10 = 2 x 105 Pa

(Using the static fluid pressure = Density x g x depth)

So by Boyle's law (PV = constant), the volume of his lungs at water surface should be twice that at 10 m below water surface.

2011-02-08 08:31:39 補充：
因為水的 density = 1000 kg/m^3

Let Po be the atmospheric pressure at water surface, and Vo be the volume of the diver's lung there.

Pressure at 10 m under water = (Po + 10 x 1000g) N/m^2
where g is the acceleration due to gravity, taken to be 10 m/s^2

By Boyle's Law,
(Po + 10 x 1000g).Vo = Po.V
where V is the volume of the diver's lung on water surface
i.e. (Po + 10^5)Vo = PoV
V = (Po + 10^5)Vo/Po = (1 + 10^5/Po).Vo

Increase in volume = (V - Vo) = (1 + 10^5/Po).Vo - Vo = 10^5Vo/Po
Percentage increase in volume
= [10^5Vo/Po]/Vo x 100%
= 10^5/Po x 100% = 99%
[if taken atmospheric pressure Po = 1.01 x 10^5 N/m^2]