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Chapter 18: Problem 14

A diver observes a bubble of air rising from the bottom of a lake (where the absolute pressure is 3.50 atm) to the surface (where the pressure is 1.00 atm). The temperature at the bottom is \(4.0^{\circ} \mathrm{C},\) and the temperature at the surface is \(23.0^{\circ} \mathrm{C}\) , (a) What is the ratio of the volume of the bubble as it reaches the surface to its volume at the bottom? (b) Would it be safe for the diver to hold his breath while ascending from the bottom of the lake to the surface? Why or why not?

Short Answer

Expert verified
(a) The volume ratio is approximately 3.74. (b) It is unsafe to hold breath while ascending due to expanding lung air.

Step by step solution

01

Understand the Given Parameters

We are given the absolute pressures at the bottom and surface of the lake, which are 3.50 atm and 1.00 atm, respectively. The temperatures given are 4.0掳C at the bottom and 23.0掳C at the surface. We need to find the ratio of the volume of the bubble at the surface to its volume at the bottom.
02

Convert Temperatures to Kelvin

To apply the ideal gas law, we need to convert the temperatures from Celsius to Kelvin. - Temperature at the bottom: 4掳C = 4 + 273.15 = 277.15 K. - Temperature at the surface: 23掳C = 23 + 273.15 = 296.15 K.
03

Apply the Combined Gas Law

The pressure-volume-temperature relationship for a gas is given by the combined gas law: \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \).- Let's denote: - \( P_1 = 3.50 \) atm, \( T_1 = 277.15 \) K. - \( P_2 = 1.00 \) atm, \( T_2 = 296.15 \) K.- We need to find the ratio \( \frac{V_2}{V_1} \).
04

Rearrange for the Volume Ratio

Rearrange the equation to find the volume ratio:\[ \frac{V_2}{V_1} = \frac{P_1 \times T_2}{P_2 \times T_1} \]. Substitute the known values:\[ \frac{V_2}{V_1} = \frac{3.50 \times 296.15}{1.00 \times 277.15} \].
05

Calculate the Ratio

Calculate the values:\[ \frac{V_2}{V_1} = \frac{3.50 \times 296.15}{277.15} \approx 3.74 \]. This is the ratio of the bubble's volume at the surface to its volume at the bottom.
06

Safety Considerations for Holding Breath

It would not be safe for the diver to hold his breath while ascending. As the diver rises, the pressure decreases, causing the air in the lungs to expand. This expansion could potentially lead to lung over-expansion injuries if the diver holds his breath.

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Key Concepts

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

Pressure and Volume
When a gas, like the air in a bubble from a diver, moves from one place to another under different pressure conditions, its volume changes. This is all about the relationship between pressure and volume in gases.
When the bubble is at the bottom of the lake, it experiences a higher pressure of 3.50 atm. As it rises to the surface, the pressure decreases to 1.00 atm.
Because of this, the volume of the bubble will increase, as there is less pressure compressing it.
You'll find this pressure-volume relationship in Boyle's Law, which we'll explore further in another section.
  • As pressure decreases, volume increases.
  • As pressure increases, volume decreases.
Temperature Conversion
Understanding temperatures in the context of gas laws requires converting from Celsius to Kelvin. This is crucial since Kelvin is the temperature scale used in gas law calculations.
In our problem, at the bottom of the lake, the temperature is 4掳C, which converts to 277.15 K by adding 273.15. At the surface, the temperature is 23掳C, which is 296.15 K.
  • Always add 273.15 to convert Celsius to Kelvin.
  • Using Kelvin helps in deriving accurate volume or pressure values in the combined gas law.
Converting temperatures ensures accuracy and consistency when applying gas laws like the Combined Gas Law.
Boyle's Law
Boyle's Law is a crucial principle in understanding how gases behave under pressure changes. It states that the volume of a gas is inversely proportional to its pressure, as long as the temperature remains constant.
This means if you increase the pressure on a gas, its volume decreases. Conversely, if you reduce the pressure, the volume increases.
  • This principle helps predict how the bubble's volume changes as it ascends.
  • While the temperature also plays a role, Boyle's Law provides clear insight into the basic pressure-volume aspect.
Using Boyle's Law, we can easily understand the expansion of the air bubble. However, remember that in real situations, temperature changes are also significant, thus, the combined gas law is often used.
Safety in Diving
Diving safety is crucial and understanding gas behavior underwater can prevent potential hazards. When a diver ascends, the pressure decreases, causing the air in their lungs or equipment to expand.
It is essential for divers not to hold their breath while surfacing. If a diver holds their breath, the expanding air could cause damage to the lungs, also known as over-expansion injuries or even result in burst lungs.
  • Never hold your breath while ascending in water.
  • Keep breathing normally to allow expanding air to escape safely.
Paying attention to these safety aspects could mean the difference between a safe dive and a hazardous situation.

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Most popular questions from this chapter

A person at rest inhales 0.50 \(\mathrm{L}\) of air with each breath at a pressure of 1.00 atm and a temperature of \(20.0^{\circ} \mathrm{C}\) . The inhaled air is 21.0\(\%\) oxygen. (a) How many oxygen molecules does this person inhale with each breath? (b) Suppose this person is now resting at an elevation of \(2,000 \mathrm{m}\) but the temperature is still \(20.0^{\circ} \mathrm{C}\). Assuming that the oxygen percentage and volume per inhalation are the same as stated above, how many oxygen molecules does this person now inhale with each breath? (c) Given that the body still requires the same number of oxygen molecules per second as at sea level to maintain its functions, explain why some people report "shortness of breath" at high elevations.

For carbon dioxide gas \(\left(\mathrm{CO}_{2}\right),\) the constants in the van der Waals equation are \(a=0.364 \mathrm{J} \cdot \mathrm{m}^{3} / \mathrm{mol}^{2}\) and \(b=4.27 \times 10^{-5} \mathrm{m}^{3} / \mathrm{mol} .\) (a) If 1.00 \(\mathrm{mol}\) of \(\mathrm{CO}_{2} \mathrm{gas}\) at 350 \(\mathrm{K}\) is confined to a volume of 400 \(\mathrm{cm}^{3}\) , find the pressure of the gas using the ideal-gas equation and the van der Waals equation. (b) Which equation gives a lower pressure? Why? What is the percentage difference of the van der Waals equation result from the ideat-gas equation result? (c) The gas is kept at the same temperature as it expands to a volume of 4000 \(\mathrm{cm}^{3} .\) Repeat the calculations of parts (a) and (b). (d) Explain how your calculations show that the van der Waals equation is equivalent to the ideat-gas equation if \(n / V\) is small.

(a) Oxygen \(\left(\mathrm{O}_{2}\right)\) has a molar mass of 32.0 \(\mathrm{g} / \mathrm{mol}\) . What is the average translational kinetic energy of an oxygen molecule at a temperature of 300 \(\mathrm{K} ?\) (b) What is the average value of the square at a of its speed? (c) What is the root-mean-square speed? (d) What is the momentum of an oxygen molecule traveling at this speed? (e) Suppose an oxygen molecule traveling at this speed bounces back and forth between opposite sides of a cubical vessel 0.10 \(\mathrm{m}\) on a side. What is the average force the molecule exerts on one of the walls of the container? (Assume that the molecule's velocity is perpendicular to the two sides that it strikes.) (f) What is the aver- age force per unit area? (g) How many oxygen molecules traveling at this speed are necessary to produce an average pressure of 1 \(\mathrm{atm} ?\) (h) Compute the number of oxygen molecules that are actually contained in a vessel of this size at 300 \(\mathrm{K}\) and atmospheric pressure. (i) Your answer for part (h) should be three times as large as the answer for part (g). Where does this discrepancy arise?

(a) A deuteron, \(^{2}_{1} \mathrm{H},\) is the nucleus of a hydrogen isotope and consists of one proton and one neutron. The plasma of deuterons in a nuclear fusion reactor must be heated to about 300 million \(\mathrm{K}\). What is the ms speed of the deuterons? Is this a significant fraction of the speed of light \(\left(c=3.0 \times 10^{8} \mathrm{m} / \mathrm{s}\right) ?\) (b) What would the temperature of the plasma be if the deuterons had an ms speed equal to 0.10\(c ?\)

A flask with a volume of 1.50 \(\mathrm{L}\) , provided with a stopock, contains ethane gas \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) at 300 \(\mathrm{K}\) and atmospheric pressure \(\left(1.013 \times 10^{5} \mathrm{Pa}\right) .\) The molar mass of ethane is 30.1 \(\mathrm{g} / \mathrm{mol}\) . The system is warmed to a temperature of \(380 \mathrm{K},\) with the stopcock open to the atmosphere. The stopcock is then closed, and the flask is cooled to its original temperature. (a) What is the final pressure of the ethane in the flask? (b) How many grams of ethane remain in the flask?
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