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A person at rest inhales 0.50 L of air with each breath at a pressure of 1.00 atm and a temperature of 20.0\(^\circ\)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 2000 m but the temperature is still 20.0\(^\circ\)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.

Short Answer

Expert verified
(a) Approximately 2.59 x 10^21 oxygen molecules per breath at sea level. (b) Approximately 1.97 x 10^21 oxygen molecules per breath at 2000 m. (c) Shortness of breath occurs as fewer oxygen molecules are inhaled at high altitudes.

Step by step solution

01

Calculate the number of moles of air inhaled at sea level

Use the ideal gas law to calculate the number of moles. The formula is \( PV = nRT \), where \( P = 1.00 \) atm, \( V = 0.50 \times 10^{-3} \) m\(^3\), \( R = 8.314 \) J/(mol·K), and \( T = 293.15 \) K (since 20.0\( ^\circ\)C = 293.15 K).\[n = \frac{PV}{RT} = \frac{(1.00 \times 101325) \times (0.50 \times 10^{-3})}{8.314 \times 293.15} \approx 0.0205 \text{ moles}\]
02

Calculate the number of oxygen molecules at sea level

Since the air is 21.0% oxygen, calculate the number of moles of oxygen. \[n_{O_2} = 0.21 \times 0.0205 \approx 0.004305 \text{ moles}\]Now, convert moles to molecules using Avogadro's number, \(6.022 \times 10^{23}\) molecules/mole.\[N_{O_2} = 0.004305 \times 6.022 \times 10^{23} \approx 2.59 \times 10^{21} \text{ oxygen molecules}\]
03

Determine the pressure at 2000 m elevation

Use the barometric formula approximating that the pressure decreases by about 12% per 1000 meters.\[P_{2000m} = (1.00 - 0.12 \times 2) \times 1.00 \approx 0.76 \text{ atm}\]
04

Calculate the number of moles of air inhaled at 2000 m

Use the ideal gas law with the calculated pressure.\[n_{2000m} = \frac{P_{2000m} \cdot V}{RT} = \frac{(0.76 \times 101325) \times (0.50 \times 10^{-3})}{8.314 \times 293.15} \approx 0.0156 \text{ moles}\]
05

Calculate the number of oxygen molecules at 2000 m

Calculate the moles of oxygen: \[n_{O_2, 2000m} = 0.21 \times 0.0156 \approx 0.003276 \text{ moles}\]Convert this to molecules:\[N_{O_2, 2000m} = 0.003276 \times 6.022 \times 10^{23} \approx 1.97 \times 10^{21} \text{ oxygen molecules}\]
06

Explain why shortness of breath occurs at high altitudes

The body requires a constant number of oxygen molecules per second to function properly. Since fewer oxygen molecules are inhaled per breath at high altitudes, one may experience 'shortness of breath' as the body attempts to maintain oxygen intake by increasing breath rate or depth.

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Key Concepts

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

Understanding Oxygen Molecules
Oxygen molecules are essential for human survival. They consist of two oxygen atoms bonded together, forming the molecule \({O}_2\). This molecule is crucial because it participates in cellular respiration—the process where cells produce energy. When we breathe in air, only about 21% is composed of oxygen. Thus, each breath you take includes a mix of different gases, with oxygen being a vital component.

At the molecular level, to understand how many oxygen molecules enter your system, we need to use Avogadro’s number, which states that one mole of any substance contains approximately \[6.022 \times 10^{23}\text{ molecules/mole}\]. So, by calculating the number of moles in each breath and knowing the oxygen proportion, we can determine the count of \({O}_2\) molecules inhaled. Using the Ideal Gas Law \(PV = nRT\), we can find the moles of air and thus calculate the \({O}_2\) molecules.
The Barometric Formula and Air Pressure
The barometric formula helps us estimate how air pressure changes with altitude. Pressure tends to decrease as you go higher in altitude because there's less air above to exert force downward. Often, it's approximated that air pressure decreases by about 12% for every 1000 meters of elevation.

This effect is significant when calculating how many oxygen molecules one inhales at different altitudes since reduced pressure allows fewer molecules to occupy the same volume. For example, if you move up to 2000 meters, pressure might only be 76% of what it is at sea level. Using this adjusted pressure in the Ideal Gas Law allows us to assess the change in moles of air, and by extension, the \({O}_2\) molecules inhaled.
High Altitude Breathing and Its Effects
Breathing at high altitudes presents unique challenges. When you're at higher elevations, the body receives fewer oxygen molecules with each breath due to the reduced air pressure. This does not mean that the percentage of oxygen changes—it remains around 21%, but fewer molecules are packed into each lungful of air.

This decrease is why some people report feeling 'shortness of breath' on mountains or high altitudes. The body has to work harder to get the necessary oxygen for normal function, sometimes leading to rapid or deep breaths.
The adjustment period varies among individuals, but acclimatization can occur, allowing the body to adapt to lower oxygen levels over time, usually by producing more red blood cells to better transport oxygen throughout the body.

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Most popular questions from this chapter

A hot-air balloon stays aloft because hot air at atmospheric pressure is less dense than cooler air at the same pressure. If the volume of the balloon is 500.0 m\(^3\) and the surrounding air is at 15.0\(^\circ\)C, what must the temperature of the air in the balloon be for it to lift a total load of 290 kg (in addition to the mass of the hot air)? The density of air at 15.0\(^\circ\)C and atmospheric pressure is 1.23 kg/m\(^3\).

(a) What is the total translational kinetic energy of the air in an empty room that has dimensions 8.00 m \(\times\) 12.00 m \(\times\) 4.00 m if the air is treated as an ideal gas at 1.00 atm? (b) What is the speed of a 2000-kg automobile if its kinetic energy equals the translational kinetic energy calculated in part (a)?

Calculate the volume of 1.00 mol of liquid water at 20\(^\circ\)C (at which its density is 998 kg/m\(^3\)), and compare that with the volume occupied by 1.00 mol of water at the critical point, which is 56 \(\times\) 10\({^-}{^6}\) m\(^3\). Water has a molar mass of 18.0 g/mol.

The rate of \(effusion\)-that is, leakage of a gas through tiny cracks-is proportional to \(v_{rms}\) . If tiny cracks exist in the material that's used to seal the space between two glass panes, how many times greater is the rate of \(He\) leakage out of the space between the panes than the rate of \(Xe\) leakage at the same temperature? (a) 370 times; (b) 19 times; (c) 6 times; (d) no greater-the \(He\) leakage rate is the same as for \(Xe\).

Estimate the number of atoms in the body of a 50-kg physics student. Note that the human body is mostly water, which has molar mass 18.0 g/mol, and that each water molecule contains three atoms.

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