Question

Anyone who scuba dives is advised not to fly within the next$\mathbf{24}\mathbf{}\mathit{h}$because the air mixture for diving can introduce nitrogen to the bloodstream. Without allowing the nitrogen to come out of solution slowly, any sudden air-pressure reduction (such as during airplane ascent) can result in the nitrogen forming bubbles in the blood, creating the bends, which can be painful and even fatal. Military special operation forces are especially at risk. What is the change in pressure on such a special-op soldier who must scuba dive at a depth of $\mathbf{20}\mathbf{}\mathit{m}$in seawater one day and parachute at an altitude of $\mathbf{7}\mathbf{.}\mathbf{6}\mathbf{}\mathit{k}\mathit{m}$the next day? Assume that the average air density within the altitude range is$\mathbf{0}\mathbf{.}\mathbf{87}\mathbf{}\mathit{k}\mathit{g}\mathbf{}\mathbf{/}{\mathit{m}}^{\mathbf{3}\mathbf{}\mathbf{.}}$

Short answer

The change in pressure on a special–op soldier is$1.4\times {10}^{5}Pa$

Step-by-step solution

The given data

1. The depth at which the special–op soldier must dive in seawater,$d=20m$
2. Height of parachute,$h=7600m$
3. Average density of air,$p=0.87kg/m^3$

Understanding the concept of pressure

We can find the pressure at depth d and height h using the formula for pressure in terms of. The difference between these two pressures gives the change in pressure on a special–op soldier.

Formula:

Pressure at a height hof a substance, $p=pgh$

Calculation of the pressure change

We can find the gauge pressure at the given height and depth.

Calculation for gauge pressure at depthusing equation (i):

$$\begin{gathered} p_{1=}1024\times 9.8\times 20\left(\because Densityofseawater=1024kg/m3\right) \\ =2.01\times {10}^{5}pa \end{gathered}$$

Calculation for gauge pressure at heighth =7600m using equation (i):

$$\begin{gathered} p_{2=}0.87\times 9.8\times 7600 \\ =6.48\times {10}^{4}pa \end{gathered}$$

Thus, the change in the pressure can be given by:

$$\begin{gathered} ∆p=p_{1-}{p}_2 \\ =(2.01\times {10}^5)-(6.48\times {10}^4) \\ =1.362\times {10}^5 \\ ~1.4\times {10}^{5}pa \end{gathered}$$

Therefore, the change in pressure on a special –op soldier is$1.4\times {10}^{5}pa$