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A study of climbers who reached the summit of Mount Everest without supplemental oxygen showed that the partial pressures of \(\mathrm{O}_{2}\) and \(\mathrm{CO}_{2}\) in their lungs were \(35 \mathrm{mm}\) Hg and \(7.5 \mathrm{mm} \mathrm{Hg}\) respectively. The barometric pressure at the summit was \(253 \mathrm{mm}\) Hg. Assume the lung gases are saturated with moisture at a body temperature of \(37^{\circ} \mathrm{C}\) [which means the partial pressure of water vapor in the lungs is \(\left.P\left(\mathrm{H}_{2} \mathrm{O}\right)=47.1 \mathrm{mm} \mathrm{Hg}\right] .\) If you assume the lung gases consist of only \(\mathrm{O}_{2}, \mathrm{N}_{2}\) \(\mathrm{CO}_{2},\) and \(\mathrm{H}_{2} \mathrm{O},\) what is the partial pressure of \(\mathrm{N}_{2} ?\)

Short Answer

Expert verified
The partial pressure of \(\mathrm{N}_{2}\) is 163.4 mm Hg.

Step by step solution

01

Understand the problem

We need to find the partial pressure of nitrogen (\(\mathrm{N}_{2}\)) in the lungs of climbers at the summit of Mount Everest, given the partial pressures of \(\mathrm{O}_{2}\) (35 mm Hg), \(\mathrm{CO}_{2}\) (7.5 mm Hg), and water vapor (47.1 mm Hg), as well as the total barometric pressure (253 mm Hg).
02

Apply Dalton's Law of Partial Pressures

According to Dalton's Law, the total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual gases. Therefore:\[P_{\text{total}} = P_{\mathrm{O}_{2}} + P_{\mathrm{CO}_{2}} + P_{\mathrm{H}_{2}\mathrm{O}} + P_{\mathrm{N}_{2}}\]Substitute the known values:\[253 = 35 + 7.5 + 47.1 + P_{\mathrm{N}_{2}}\]
03

Solve for the partial pressure of \(\mathrm{N}_{2}\)

Rearrange the equation to solve for \(P_{\mathrm{N}_{2}}\):\[P_{\mathrm{N}_{2}} = 253 - (35 + 7.5 + 47.1)\]Calculate the sum of the known pressures:\[P_{\mathrm{N}_{2}} = 253 - 89.6\]Finally, subtract and find \(P_{\mathrm{N}_{2}}\):\[P_{\mathrm{N}_{2}} = 163.4\, \mathrm{mm} \mathrm{Hg}\]

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Key Concepts

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

Dalton's Law of Partial Pressures
Dalton's Law of Partial Pressures is a fundamental principle in chemistry and physics, particularly when dealing with gas mixtures. It states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of each individual gas in the mixture. This principle is essential when calculating the pressure of gases in a constrained environment, such as the human lungs at high altitudes like Mount Everest.

A partial pressure is the pressure that a gas would have if it alone occupied the entire volume. In the Everest example, climbers deal with a mixture of oxygen ( O_{2} ), carbon dioxide ( CO_{2} ), nitrogen ( N_{2} ), and water vapor ( H_{2}O ). Dalton's Law allows us to deduce the pressure contribution from each gas component. This helps us understand the distribution of pressure among the constituent gases, which is crucial for assessing the overall respiratory efficiency of climbers at high altitudes.

Dalton's Law is mathematically expressed as:
  • P_{ ext{total}} = P_{ O_{2} } + P_{ CO_{2} } + P_{ H_{2}O } + P_{ N_{2} }
Calculating the partial pressures and understanding their individual roles is an integral part of studies focused on high-altitude physiology.
Gas Mixtures
Gas mixtures are omnipresent in our daily life, from the air we breathe to the oxygen mixtures used by mountaineers or divers. A gas mixture is simply a combination of two or more gases. Each gas in the mixture behaves independently, even though they occupy the same space.

In our case, the mixture in the lung air at Everest consists of oxygen, nitrogen, carbon dioxide, and water vapor. Each of these gases contributes differently to the total pressure, in accordance with Dalton's Law. Understanding gas mixtures is crucial because each gas impacts respiratory functions based on its proportion in the mixture.

Here are some fundamental points about gas mixtures:
  • The properties of a gas mixture depend on its components' properties and proportions.
  • Each gas retains its identity and chemical properties within the mixture.
  • Partial pressures help to easily analyze complex mixtures by breaking them down into comprehensible parts.
Mastering gas mixtures is crucial for comprehending phenomena like breathing under variable pressures, significant in high-altitude scenarios.
Oxygen Levels
Oxygen levels, or the partial pressure of oxygen, in our environment play a critical role in our body's ability to sustain life. At sea level, atmospheric oxygen is dense, with a partial pressure that is optimal for human physiology. However, as altitude increases, the atmospheric pressure decreases.

This results in lower oxygen levels available for breathing. On Mount Everest, the available oxygen is dramatically reduced, making it challenging to maintain adequate oxygen levels in the bloodstream without supplemental oxygen.

Here are some important considerations regarding oxygen levels:
  • Low partial pressure of oxygen at high altitudes reduces the oxygen absorption capability of lungs.
  • The body acclimatizes by increasing red blood cell production, but limits still exist.
  • Knowing the exact oxygen partial pressure is crucial for climbers to gauge their physiological limits.
Monitoring oxygen levels is imperative for ensuring safety and performance during high-altitude endeavors.
Barometric Pressure at High Altitude
Barometric pressure, sometimes called atmospheric pressure, is the force per unit area exerted on a surface by the weight of the air above that surface. It decreases as altitude increases, due to the thinner, less dense air. On the peaks of high mountains like Mount Everest, the barometric pressure is significantly lower than at sea level.

For climbers, understanding the implications of lower barometric pressure is important. It influences not just the breathing dynamics but also the climbers' ability to perform physically. At lower pressures, there is less oxygen available per breath, necessitating a consideration of how much exertion their bodies can sustain.

Key aspects of barometric pressure at altitude include:
  • Low barometric pressure affects the boiling point of water, cooking, and other daily activities.
  • Every 1000m ascent leads to a significant drop in atmospheric pressure.
  • Climbers must adapt their techniques and gear to compensate for these changes.
An awareness of barometric pressure's role is indispensable for safe and successful high-altitude climbing.

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Most popular questions from this chapter

Ethane burns in air to give \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CO}_{2}\) $$ 2 \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g})+7 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ (a) Four gases are involved in this reaction. Place them in order of increasing rms speed. (Assume all are at the same temperature.) (b) \(\mathrm{A} 3.26-\mathrm{L}\) flask contains \(\mathrm{C}_{2} \mathrm{H}_{6}\) at a pressure of \(256 \mathrm{mm}\) Hg and a temperature of \(25^{\circ} \mathrm{C}\) Suppose \(\mathrm{O}_{2}\) gas is added to the flask until \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{O}_{2}\) are in the correct stoichiometric ratio for the combustion reaction. At this point, what is the partial pressure of \(\mathbf{O}_{2}\) and what is the total pressure in the flask?

Ethane burns in air to give \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CO}_{2}\) $$ 2 \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g})+7 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ What volume of \(\mathrm{O}_{2}(\mathrm{L})\) is required for complete reaction with \(5.2 \mathrm{L}\) of \(\mathrm{C}_{2} \mathrm{H}_{6}\) ? What volume of \(\mathrm{H}_{2} \mathrm{O}\) vapor (L) is produced? Assume all gases are measured at the same temperature and pressure.

Forty miles above Earth's surface, the temperature is \(250 \mathrm{K},\) and the pressure is only \(0.20 \mathrm{mm} \mathrm{Hg} .\) What is the density of air (in grams per liter) at this altitude? (Assume the molar mass of air is \(28.96 \mathrm{g} / \mathrm{mol.}\)

What volume (in liters) of \(\mathrm{O}_{2}\), measured at standard temperature and pressure, is required to oxidize 0.400 mol of phosphorus \(\left(P_{4}\right) ?\) $$ \mathrm{P}_{4}(\mathrm{s})+5 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{P}_{4} \mathrm{O}_{10}(\mathrm{s}) $$

A halothane-oxygen mixture \(\left(\mathrm{C}_{2} \mathrm{HBrClF}_{3}+\mathrm{O}_{2}\right)\) can be used as an anesthetic. A tank containing such a mixture has the following partial pressures: \(P\) (halothane) \(=170 \mathrm{mm} \mathrm{Hg}\) and \(P\left(\mathrm{O}_{2}\right)=570 \mathrm{mm} \mathrm{Hg}\) (a) What is the ratio of the number of moles of halothane to the number of moles of \(\mathrm{O}_{2} ?\) (b) If the tank contains \(160 \mathrm{g}\) of \(\mathrm{O}_{2}\), what mass of \(\mathrm{C}_{2} \mathrm{HBrClF}_{3}\) is present?

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