Question

If a scuba diver fills his lungs to full capacity of 5.5 L when 10 m below the surface, to what volume would his lungs expand if he quickly rose to the surface? Assume, the temperature of the air inside the lungs remains costant. Density of sea water is 1025 kg/m3.

Answer 1

Answer:

10.95 L

Explanation:

Let the pressure at depth 10 m is P1.

V1 = 5.5 L

P1 = Po + h x d x g

P2 is the pressure at the surface

P2 = Po

where, Po be the atmospheric pressure  = 1.01 x 10^5 Pa

Let V2 be the volume at the surface.

Use P1 x V1 = P2 x V2

(Po + 10 x 1025 x 9.8) x 5.5 = Po x V2

(1.01 x 10^5 + 1 x 10^5) x 5.5 = 1.01 x 10^5 x V2

2.01 x 5.5 = 1.01 x V2

V2 = 10.95 L

Answer 2

Answer:

The volume at the surface is 10.97 L.

Explanation:

Given that,

Volume = 5.5 L

Height = 10 m

Density of sea water= 1025 kg/m³

We need to calculate the pressure at that point

Using formula of pressure

$P'=P+\rho gh$

Put the value into the formula

$P'=1.01\times10^{5}+1025\times9.8\times10$

$P'=201450\ Pa$

We need to calculate the volume at the surface

Using equation of ideal gas

$PV= RT$

So, for both condition

$PV=P'V'$

Put the value into the formula

$V=\dfrac{201450\times5.5}{1.01\times10^{5}}$

$V=10.97\ L$

Hence, The volume at the surface is 10.97 L.