Question

The Bends. If deep-sea divers rise to the surface too quickly, nitrogen bubbles in their blood can expand and prove fatal. This phenomenon is known as the bends. If a scuba diver rises quickly from a depth of 25 m in Lake Michigan (which is fresh water), what will be the volume at the surface of an N₂ bubble that occupied 1.0 mm³ in his blood at the lower depth? Does it seem that this difference is large enough to be a problem? (Assume that the pressure difference is due to only the changing water pressure, not to any temperature difference. This assumption is reasonable, since we are warm-blooded creatures.)

Short answer

The volume of the bubble change from 1 mm³ to 3.4 mm^3.

This is a large change in volume and would have a serious effect on the body of the divers.

Step-by-step solution

Step 1

The pressure at the bottom of the lake differs than the pressure at the top surface. So let's consider the pressure at the bottom p₁ and at the top be p₂. We would employ a relation between$p_1,V_1,p_{2}and{V}_2$

${\mathit{p}}_{\mathbf{1}}\mathbf{,}{\mathit{V}}_{\mathbf{1}}\mathbf{,}{\mathit{p}}_{\mathbf{2}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{V}}_{\mathbf{2}}$

We were given the value of V₁ but we want to calculate p₁ and p₂.

For p₂, it is the same atmospheric pressure at the surface of the earth and equals 1.013 x 10⁵ Pa.

For p₁, there is a change in the pressure due to the depth of the lake h and the pressure increased as the drivers diving more depth so this change can be employed by:

$\mathbf{∆}\mathbf{p}\mathbf{=}\mathbf{\rho gh}$

where $\rho$ is the density of water, g is the gravity of acceleration and h is the lake depth.

$\begin{array}{r}{p}_1=p_2+\rho gh\\ =1.013\times {10}^5\mathrm{Pa}+1000\mathrm{kg}/{\mathrm{m}}^3\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 25\mathrm{m}\\ =3.463\times {10}^5\mathrm{Pa}\end{array}$

Step 2

After calculating the pressure at the bottom of the lake p1 and at the surface p2 substitute with all the values of p₁, p₂ and given value of the volume of the bubble at the bottom of the lake v₁ and solve for v₂ to find the volume of the nitrogen bubble at the surface of the lake where

$\begin{array}{r}{V}_2=\frac{p_1{V}_1}{p_2}\\ =\frac{3.463\times {10}^5\mathrm{Pa}\times 1{\mathrm{mm}}^3}{1.013\times {10}^5\mathrm{Pa}}\\ =3.4{\mathrm{mm}}^3\end{array}$

The volume of the bubble change from 1 mm3 to 3.4 mm3 and this is a large change in volume and would have a serious effect on the body of the divers.