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Chapter 18: Problem 54

CP BIO 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 \(\mathrm{m}\) in Lake Michigan (which is fresh water), what will be the volume at the surface of an \(\mathrm{N}_{2}\) bubble that occupied 1.0 \(\mathrm{mm}^{3}\) 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 only to the changing water pressure, not to any temperature difference, an assumption that is reasonable, since we are warm-blooded creatures.)

Short Answer

Expert verified
The bubble's volume expands from 1.0 mm鲁 to 3.42 mm鲁, which could be problematic.

Step by step solution

01

Understand the Problem

We need to determine how the volume of a nitrogen bubble changes as a scuba diver ascends from a depth of 25 m to the surface of Lake Michigan. The initial volume of the bubble at this depth is 1.0 mm鲁. We assume the change in the bubble's volume is due to changes in pressure, according to Boyle's Law, as temperature effects are negligible.
02

Recall Boyle's Law

Boyle's law states that for a given amount of gas at constant temperature, the product of its pressure and volume is a constant. This can be expressed as \(P_1 V_1 = P_2 V_2\), where \(P_1\) and \(P_2\) are the initial and final pressures, and \(V_1\) and \(V_2\) are the initial and final volumes, respectively.
03

Calculate the Initial Pressure at Depth

The pressure at the depth of 25 m includes both the atmospheric pressure and the pressure due to the water column. The pressure underwater can be calculated using the formula \(P = P_0 + \rho g h\), where \(P_0\) is the atmospheric pressure (101,325 Pa), \(\rho\) is the density of water (approximately 1000 kg/m鲁), \(g\) is the acceleration due to gravity (9.81 m/s虏), and \(h\) is the depth (25 m).
04

Calculate Pressure Values

Calculate \(P = 101325 + (1000 \times 9.81 \times 25)\). This results in a total pressure of approximately 346,550 Pa at 25 m depth.
05

Calculate the Final Pressure at Surface

At the water surface, the pressure is equal to the atmospheric pressure, \(P_2 = 101,325\) Pa.
06

Apply Boyle's Law

Using Boyle's Law, \(P_1 V_1 = P_2 V_2\), substitute the values to find \(V_2\). The initial pressure \(P_1\) is 346,550 Pa, and \(V_1\) is 1.0 mm鲁. Solve for \(V_2\): \(346550 \times 1.0 = 101325 \times V_2\). Thus, \(V_2 = \frac{346550}{101325} \times 1.0\).
07

Calculate Final Volume

Calculate \(V_2 = 3.42\, \text{mm}^3\). This is the volume of the bubble at the surface.
08

Analyze the Problem

The volume of the nitrogen bubble has expanded from 1.0 mm鲁 to 3.42 mm鲁. This significant increase in volume can be problematic as it may cause physiological issues if bubbles form in tissues.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gas Laws
Gas laws are fundamental principles in chemistry and physics that explain how gases behave under various conditions. One of the key gas laws is Boyle's Law.
Boyle's Law states that the pressure of a gas is inversely proportional to its volume, when temperature remains constant. This is mathematically represented by the equation: \[ P_1 V_1 = P_2 V_2 \]where \(P_1\) and \(P_2\) are the initial and final pressures, and \(V_1\) and \(V_2\) are the initial and final volumes of the gas.
The law implies that if the pressure increases, the volume decreases, and vice versa.
This principle plays a crucial role in understanding the behavior of gases in different environments, such as under the water during a dive.
Pressure in Fluids
The concept of pressure in fluids is vital for understanding how gases behave underwater.
Pressure is defined as the force exerted per unit area. In fluids, pressure depends on the depth of the fluid, due to the weight of the fluid above.
The pressure in a fluid at a given depth is calculated using the formula:\[ P = P_0 + \rho gh \]where \(P_0\) is the atmospheric pressure at the surface, \(\rho\) is the fluid's density, \(g\) is the acceleration due to gravity, and \(h\) is the depth of the fluid.
As a diver descends deeper underwater, the pressure increases, compressing any gas bubbles present in their body.
When ascending, the pressure decreases, leading to the expansion of these bubbles, which needs careful management to avoid accidents like "the bends."
Decompression Sickness
Decompression sickness, also known as "the bends," occurs when a diver ascends too quickly, causing dissolved gases to form bubbles in their body.
This happens because of the rapid decrease in pressure, which allows nitrogen dissolved in the bloodstream at high pressures to come out of solution and form gas bubbles.
These bubbles can block blood vessels and restrict blood flow, leading to various symptoms ranging from joint pain to severe neurological issues.
To prevent decompression sickness, divers are advised to ascend slowly and follow specific decompression stops. This ensures that the gas has adequate time to escape safely from the bloodstream without forming harmful bubbles.
Nitrogen Bubbles
Nitrogen bubbles are at the heart of decompression sickness.
Underwater, the increased pressure causes more nitrogen to dissolve in the diver's tissues.
If the pressure decreases too rapidly, as when a diver ascends quickly, this dissolved nitrogen expands and forms bubbles in the body.
  • These nitrogen bubbles can cause blockages in the bloodstream or damage to tissues, leading to symptoms commonly associated with the bends.
  • Symptoms might include dizziness, joint pain, or difficulty breathing.
  • Proper diving practices, like controlled ascents and using dive computers, help mitigate this risk by allowing the nitrogen to leave the body gradually, without forming bubbles.
Understanding how these bubbles form and how to prevent them is essential for safe diving practices.

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Most popular questions from this chapter

An empty cylindrical canister 1.50 \(\mathrm{m}\) long and 90.0 \(\mathrm{cm}\) in diameter is to be filled with pure oxygen at \(22.0^{\circ} \mathrm{C}\) to store in a space station. To hold as much gas as possible, the absolute pressure of the oxygen will be 21.0 atm. The molar mass of oxygen is 32.0 \(\mathrm{g} / \mathrm{mol} .\) (a) How many moles of oxygen does this canister hold? (b) For someone lifting this canister, by how many kilograms does this gas increase the mass to be lifted?

The size of an oxygen molecule is about 2.0 \(\times 10^{-10} \mathrm{m}\) Make a rough estimate of the pressure at which the finite volume of the molecules should cause noticeable deviations from ideal-gas behavior at ordinary temperatures \((T=300 \mathrm{K})\) .

(a) Compute the increase in gravitational potential energy for a nitrogen molecule (molar mass 28.0 \(\mathrm{g} / \mathrm{mol} )\) for an increase in elevation of 400 \(\mathrm{m}\) near the earth's surface. (b) At what temperature is this equal to the average kinetic energy of a nitrogen molecule? (c) Is it possible that a nitrogen molecule near sea level where \(T=15.0^{\circ} \mathrm{C}\) could rise to an altitude of 400 \(\mathrm{m} ?\) Is it likely that it could do so without hitting any other molecules along the way? Explain.

For diatomic carbon dioxide gas \(\left(\mathrm{CO}_{2},\) molar mass \right. 44.0 \(\mathrm{g} / \mathrm{mol} )\) at \(T=300 \mathrm{K}\) , calculate (a) the most probable speed \(v_{\mathrm{mp}} ;\) (b) the average speed \(v_{\mathrm{av}} ;(\mathrm{c})\) the root-mean-square speed \(v_{\mathrm{rms}}\)

During a test dive in \(1939,\) prior to being accepted by the U.S. Navy, the submarine Squalus sank at a point where the depth of water was 73.0 m. The temperature at the surface was \(27.0^{\circ} \mathrm{C},\) and at the bottom it was \(7.0^{\circ} \mathrm{C}\) . The density of seawater is 1030 \(\mathrm{kg} / \mathrm{m}^{3} .\) (a) A diving bell was used to rescue 33 trapped crewmen from the Squalus. The diving bell was in the form of a circular cylinder 2.30 \(\mathrm{m}\) high, open at the bottom and closed at the top. When the diving bell was lowered to the bottom of the sea, to what height did water rise within the diving bell? (Hint: You may ignore the relatively small variation in water pressure between the bottom of the bell and the surface of the water within the bell.) (b) At what gauge pressure must compressed air have been supplied to the bell while on the bottom to expel all the water from it?
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