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A 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 mm Hg and 7.5 mm Hg, respectively. The barometric pressure at the summit was 253 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 \(\mathbf{O}_{2}, \mathbf{N}_{2}, \mathbf{C O}_{2},\) 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

Identify Known Pressures

We know the partial pressures for each component, except for nitrogen. The partial pressures provided are: \( P(O_2) = 35 \, \text{mm Hg} \), \( P(CO_2) = 7.5 \, \text{mm Hg} \), and \( P(H_2O) = 47.1 \, \text{mm Hg} \).
02

Apply Dalton's Law of Partial Pressures

Dalton's Law states that the total pressure is the sum of the partial pressures of all gases present. Therefore, \( P_{\text{total}} = P(O_2) + P(CO_2) + P(N_2) + P(H_2O) \).
03

Substitute Known Values into Dalton's Equation

Substituting the known values into the equation from Step 2 gives: \( 253 = 35 + 7.5 + P(N_2) + 47.1 \).
04

Solve for the Partial Pressure of Nitrogen

To find \( P(N_2) \), rearrange the equation from Step 3: \( P(N_2) = 253 - (35 + 7.5 + 47.1) \).
05

Calculate the Value of \( P(N_2) \)

Substitute and compute: \( P(N_2) = 253 - 89.6 = 163.4 \, \text{mm Hg} \).

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Key Concepts

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

Partial Pressure
Partial pressure is a crucial concept in chemistry and physics, especially when it comes to understanding gas behavior. It refers to the pressure that a specific gas in a mixture of gases would exert if it were the only gas present in a given volume. This concept is useful because it allows us to analyze the contributions of each individual gas in a mixture to the total pressure.

Dalton's Law of Partial Pressures is the principle that describes how to calculate the total pressure of a gas mixture. According to Dalton's Law, the total pressure (\( P_{\text{total}} \) ) is the sum of the partial pressures of all gases involved.
  • \( P_{\text{total}} = P_{O_2} + P_{CO_2} + P_{N_2} + P_{H_2O} \)
By breaking down the total pressure into partial pressures, we can isolate and analyze the properties of individual gases. This concept is essential for students to grasp when studying gas mixtures and understanding how different gases interact within a system.
Barometric Pressure
Barometric pressure, also known as atmospheric pressure, is the pressure exerted by the weight of air in the atmosphere above us. It plays a vital role in weather patterns, altitude, and even the physiological behavior of gases at various altitudes.

At higher altitudes, such as the summit of Mount Everest, barometric pressure decreases significantly due to the thinning atmosphere. This decrease in pressure affects the partial pressures of gases and, consequently, human physiology. It is because of this lower pressure that climbers often require supplemental oxygen to maintain adequate \( O_2 \) levels for survival. Understanding barometric pressure is essential for navigating situations where atmospheric conditions vary, such as in aviation, meteorology, and mountaineering.
Nitrogen Partial Pressure
Nitrogen partial pressure is the specific pressure exerted by nitrogen in a mixture of gases. Nitrogen (\( N_2 \)) is the most abundant gas in Earth’s atmosphere, making its partial pressure a significant component of total atmospheric pressure.

When considering a scenario such as the summit of Mount Everest, where there is only a limited amount of air available, nitrogen still constitutes a major portion of the gas mixture in the lungs. To determine the partial pressure of nitrogen in such a mixture, we use Dalton’s Law.
The calculation involves knowing the total pressure and subtracting the known partial pressures of other gases like \( O_2 \), \( CO_2 \), and \( H_2O \).
Through this process, we found the nitrogen partial pressure to be 163.4 mm Hg. This calculation helps illustrate how each gas contributes to the overall pressure and can influence breathing—and ultimately survival—for climbers at high altitudes.

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Most popular questions from this chapter

You have a sample of \(\mathrm{CO}_{2}\) in flask \(\mathrm{A}\) with a volume of \(25.0 \mathrm{mL} .\) At \(20.5^{\circ} \mathrm{C},\) the pressure of the gas is \(436.5 \mathrm{mm}\) Hg. To find the volume of another flask, B, you move the \(\mathrm{CO}_{2}\) to that flask and find that its pressure is now \(94.3 \mathrm{mm}\) Hg at \(24.5^{\circ} \mathrm{C} .\) What is the volume of flask \(\mathrm{B} ?\)

Ammonia gas is synthesized by combining hydrogen and nitrogen: $$ 3 \mathrm{H}_{2}(\mathrm{g})+\mathrm{N}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{g}) $$ (a) If you want to produce 562 g of \(\mathrm{NH}_{3}\), what volume of \(\mathrm{H}_{2}\) gas, at \(56^{\circ} \mathrm{C}\) and \(745 \mathrm{mm}\) Hg, is required? (b) Nitrogen for this reaction will be obtained from air. What volume of air, measured at \(29^{\circ} \mathrm{C}\) and \(745 \mathrm{mm}\) Hg pressure, will be required to provide the nitrogen needed to produce \(562 \mathrm{g}\) of \(\mathrm{NH}_{3} ?\) Assume the sample of air contains 78.1 mole \(\%\) N \(_{2}\)

Sodium azide, the explosive compound in automobile air bags, decomposes according to the following equation: $$2 \mathrm{NaN}_{3}(\mathrm{s}) \rightarrow 2 \mathrm{Na}(\mathrm{s})+3 \mathrm{N}_{2}(\mathrm{g})$$ What mass of sodium azide is required to provide the nitrogen needed to inflate a \(75.0-\mathrm{L}\) bag to a pressure of 1.3 atm at \(25^{\circ} \mathrm{C} ?\)

A xenon fluoride can be prepared by heating a mixture of \(\mathrm{Xe}\) and \(\mathrm{F}_{2}\) gases to a high temperature in a pressureproof container. Assume that xenon gas was added to a \(0.25-\mathrm{L}\) container until its pressure reached \(0.12 \mathrm{atm}\) at \(0.0^{\circ} \mathrm{C} .\) Fluorine gas was then added until the total pressure reached 0.72 atm at \(0.0^{\circ} \mathrm{C}\). After the reaction was complete, the xenon was consumed completely, and the pressure of the \(\mathrm{F}_{2}\) remaining in the container was 0.36 atm at \(0.0^{\circ} \mathrm{C} .\) What is the empirical formula of the xenon fluoride?

\(\mathrm{Ni}(\mathrm{CO})_{4}\) can be made by reacting finely divided nickel with gaseous CO. If you have CO in a \(1.50-\mathrm{L}\). flask at a pressure of \(418 \mathrm{mm}\) Hg at \(25.0^{\circ} \mathrm{C},\) along with \(0.450 \mathrm{g}\) of Ni powder, what is the theoretical yield of \(\mathrm{Ni}(\mathrm{CO})_{4} ?\)

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