/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 An adult takes about 12 breaths ... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 36

An adult takes about 12 breaths per minute, inhaling roughly \(500 \mathrm{mL}\) of air with each breath. The molar compositions of the inspired and expired gases are as follows: $$\begin{array}{lcc} \hline \text { Species } & \text { Inspired Gas (\%) } & \text { Expired Gas (\%) } \\ \hline \mathrm{O}_{2} & 20.6 & 15.1 \\ \mathrm{CO}_{2} & 0.0 & 3.7 \\ \mathrm{N}_{2} & 77.4 & 75.0 \\ \mathrm{H}_{2} \mathrm{O} & 2.0 & 6.2 \\ \hline \end{array}$$ The inspired gas is at \(24^{\circ} \mathrm{C}\) and 1 atm, and the expired gas is at body temperature and pressure \(\left(37^{\circ} \mathrm{C}\right.\) and 1 atm). Nitrogen is not transported into or out of the blood in the lungs, so that \(\left(\mathrm{N}_{2}\right)_{\text {in }}=\left(\mathrm{N}_{2}\right)_{\text {out }}\) (a) Calculate the masses of \(\mathrm{O}_{2}, \mathrm{CO}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\) transferred from the pulmonary gases to the blood or vice versa (specify which) per minute. (b) Calculate the volume of air exhaled per milliliter inhaled. (c) At what rate (g/min) is this individual losing weight by merely breathing? (d) The rate at which oxygen is transferred from the air in the lungs to the blood is roughly proportional to \(\left[\left(p_{\mathrm{O}_{2}}\right)_{\mathrm{air}}-\left(p_{\mathrm{O}_{2}}\right)_{\mathrm{blood}}\right],\) where \(\left(p_{\mathrm{O}_{2}}\right)_{\mathrm{blood}}\) is a quantity related to the concentration of oxygen in the blood. Compared to regions where atmospheric pressure is 14.7 psia, what effect does the atmospheric pressure in Denver, which is approximately 12.1 psi, have on the transport rate and breathing rate? How does the body adjust to address this condition?

Short Answer

Expert verified
(a) The masses transferred per minute are approximately -11.63 g of O2 (into the blood), 11.56 g of CO2 (out of the blood), and 1.98 g of H2O (out of the blood). (b) The volume of air exhaled per milliliter inhaled is approximately 1.045 mL exhaled/mL inhaled. (c) This individual is losing weight by merely breathing at a rate of approximately -1.09 g/min. (d) At higher altitudes, such as Denver, the transport rate of oxygen decreases while the breathing rate increases; the body produces more red blood cells to compensate for the reduced oxygen transport.

Step by step solution

01

Calculation of Mass of O2, CO2, and H2O transferred

First calculate the volume of air breathed per minute, which is \(12 \, \text{breaths/min} \times 500 \, \text{mL/breath} = 6000 \, \text{mL/min}\). Then convert this volume to liters by dividing by 1000, to get \(6.0 \, \text{L/min}\). From the ideal gas law \(PV = nRT\), where \(n\) is the number of moles, you can calculate the number of moles in this volume by rearranging it to \(n = \frac{PV}{RT} = \frac{6.0 \, \text{L} \times 1 \, \text{atm}}{0.0821 \, \text{L atm/(K mol)} \times (273 + 24) \, \text{K}} = 0.242 \, \text{mols/min}\). Next, find the moles of each gas in the inspired and expired air by multiplying the total moles by the percent composition (in decimal form). Then find the difference between the inspired and expired gas to find the change in moles per minute for O2, CO2 and H2O. Finally, convert these values to grams by multiplying by the molar mass of each substance.
02

Calculation of Volume of Air Exhaled Per mL Inhaled

For the volume of air exhaled, we need to keep in mind that the expired gas is at 37掳C and 1 atm. First convert the temperature to Kelvin by adding 273 to 37. Apply the ideal gas law again to find the volume of the expired air, \(V = nRT/P = 0.242 \, \text{mols/min} \times 0.0821 \, \text{L atm/(K mol)} \times (273 + 37) \, \text{K} / 1 \, \text{atm} = 6.27 \, \text{L/min}\). Convert this value to mL by multiplying by 1000, to get \(6270 \, \text{mL/min}\). Finally, calculate the volume of air exhaled per mL of air inhaled by dividing the volume of expired air by the volume of inspired air, \(6270 \, \text{mL/min} / 6000 \, \text{mL/min}\).
03

Calculation of Weight Loss Rate

To calculate the rate of weight loss, add up the amounts of O2, CO2, and H2O transferred to the blood per minute (the differences calculated in step 1), keeping in mind that a loss is represented by a negative value and a gain is represented by a positive value. The sum of these values represents the amount of mass that the person loses per minute due to breathing.
04

Discussion of Effects of Atmospheric Pressure on Respiratory Rate

The transport rate of oxygen is affected by the difference in the partial pressures of oxygen in the air and blood. From the given equation, a decrease in atmospheric pressure would decrease the partial pressure in the air (P_O2 air), making the difference smaller and thus the transfer rate of oxygen from the air to the blood cells is reduced. This is experienced by people living in high-altitude areas, such as Denver, where the atmospheric pressure is about 12.1 psi compared to 14.7 psi at sea level. As a result, residents in such areas have to breathe faster to compensate for the lower oxygen transport rate in each breath, and their bodies adapt by producing more red blood cells to improve oxygen transport in the blood.

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Key Concepts

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

Molar Composition of Gases
Understanding the molar composition of gases is crucial when analyzing respiratory processes. In the context of respiration, each breath of air contains different gaseous components, such as oxygen (O鈧�), carbon dioxide (CO鈧�), nitrogen (N鈧�), and water vapor (H鈧侽). Each of these gases contributes a specific percentage to the total volume of air we inhale and exhale.
For instance, inspired air typically has around 20.6% oxygen, 0% carbon dioxide, 77.4% nitrogen, and 2% water vapor. These percentages change slightly in expired air, reflecting the body's uptake and release of these gases during respiration.
The molar composition helps to determine how much of each gas is retained or expelled with every breath. By calculating the changes in these percentages between inspired and expired air, we can better understand the exchange of gases that occurs in the lungs, where oxygen is absorbed into the blood, and carbon dioxide is expelled into the air during exhalation. These subtle changes are pivotal for maintaining life and are an excellent example of the intricate balance our respiratory system achieves.
Ideal Gas Law Calculations
The ideal gas law is a fundamental equation used to connect various physical properties of gases such as pressure, volume, and temperature through the formula: \(PV = nRT\), where:
  • \(P\) stands for pressure,
  • \(V\) is volume,
  • \(n\) is the number of moles,
  • \(R\) is the ideal gas constant, and
  • \(T\) is the temperature in Kelvin.
In respiratory processes, this law helps us calculate the quantity of each gas component involved. It allows us to convert between the number of moles of gas present and the volume they occupy under certain conditions of temperature and pressure.
For example, when performing calculations of the gas volumes inhaled and exhaled, we first convert the air volumes into moles using the ideal gas law. By understanding the number of moles of gases in inspired and expired air, we can then deduce the exact transfer of gases such as \(O_{2}\), \(CO_{2}\), and \(H_{2}O\) during breathing. This kind of calculation is essential in understanding how our bodies manage to efficiently utilize and dispose of gases, even with changes in environmental conditions.
Effects of Altitude on Respiration
The altitude of a location significantly impacts the pressures experienced in respiratory processes. At higher altitudes, such as Denver, the atmospheric pressure is lower (approximately 12.1 psi) compared to sea level (approximately 14.7 psi). This difference in pressure alters the partial pressure of oxygen in the air.
Partial pressure is crucial for the diffusion of oxygen from the alveoli in the lungs into the bloodstream. If the atmospheric pressure drops, as it does at high altitudes, the partial pressure of oxygen in the inspired air also decreases, reducing the efficiency with which oxygen is transferred into the blood.
To cope with this, individuals acclimatized to high altitudes may experience an increase in breathing rate to take in more air for the same amount of oxygen. Additionally, the body adapts over time by producing more red blood cells, enhancing its capacity to carry oxygen throughout the bloodstream.
This adaptation is an amazing showcase of human physiology, illustrating how our bodies strive to maintain balance and effectively function even when faced with environmental challenges such as reduced oxygen availability.

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Most popular questions from this chapter

The concentration of oxygen in a 5000 -liter tank containing air at 1 atm is to be reduced by pressure purging prior to charging a fuel into the tank. The tank is charged with nitrogen up to a high pressure and then vented back down to atmospheric pressure. The process is repeated as many times as required to bring the oxygen concentration below 10 ppm (i.c., to bring the mole fraction of \(\mathrm{O}_{2}\) below \(10.0 \times 10^{-6}\) ). Assume that the temperature is \(25^{\circ} \mathrm{C}\) at the beginning and end of each charging cycle. When doing \(P V T\) calculations in Parts (b) and (c), use the generalized compressibility chart if possible for the fully charged tank and assume that the tank contains pure nitrogen. (a) Speculate on why the tank is being purged. (b) Estimate the gauge pressure (atm) to which the tank must be charged if the purge is to be done in one charge-vent cycle. Then estimate the mass of nitrogen (kg) used in the process. (For this part, if you can't find the tank condition on the compressibility chart, assume ideal-gas behavior and state whether the resulting estimate of the pressure is too high or too low.) (c) Suppose nitrogen at 700 kPa gauge is used for the charging. Calculate the number of charge-vent cycles required and the total mass of nitrogen used. (d) Use your results to explain why multiple cycles at a lower gas pressure are preferable to a single cycle. What is a probable disadvantage of multiple cycles?

The lower flammability limit (LFL) and the upper flammability limit (UFL) of propane in air at 1 atm are, respectively, 2.3 mole \(\% \mathrm{C}_{3} \mathrm{H}_{8}\) and 9.5 mole \(\% \mathrm{C}_{3} \mathrm{H}_{8} .^{17}\) If the mole percent of propane in a propane-air mixture is between \(2.3 \%\) and \(9.5 \%,\) the gas mixture will burn explosively if exposed to a flame or spark; if the percentage is outside these limits, the mixture is safe-a match may burn in it but the flame will not spread. If the percentage of propane is below the LFL, the mixture is said to be too lean to ignite; if it is above the UFL, the mixture is too rich to ignite. (a) Which would be safer to release into the atmosphere- -a fuel-air mixture that is too lean or too rich to ignite? Explain. (b) A mixture of propane in air containing 4.03 mole \(\% \mathrm{C}_{3} \mathrm{H}_{8}\) is fed to a combustion furnace. If there is a problem in the furnace, the mixture is diluted with a stream of pure air to make sure that it cannot accidentally ignite. If propane enters the furnace at a rate of \(150 \mathrm{mol} \mathrm{C}_{3} \mathrm{H}_{8} / \mathrm{s}\) in the original fuel- air mixture, what is the minimum molar flow rate of the diluting air? (c) The actual diluting air molar flow rate is specified to be \(130 \%\) of the minimum value. Assuming the fuel mixture (4.03 mole\% \(\mathrm{C}_{3} \mathrm{H}_{8}\) ) enters the furnace at the same rate as in Part (b) at \(125^{\circ} \mathrm{C}\) and 131 kPa and the diluting air enters at \(25^{\circ} \mathrm{C}\) and \(110 \mathrm{kPa}\), calculate the ratio \(\left(\mathrm{m}^{3} \text { diluting air) } /\right.\) (m \(^{3}\) fuel gas) and the mole percent of propane in the diluted mixture. (d) Give several possible reasons for feeding air at a value greater than the calculated minimum rate.

A gas cylinder filled with nitrogen at standard temperature and pressure has a mass of \(37.289 \mathrm{g}\). The same container filled with carbon dioxide at STP has a mass of 37.440 g. When filled with an unknown gas at STP, the container mass is \(37.062 \mathrm{g}\). Calculate the molecular weight of the unknown gas, and then state its probable identity.

The quantity of sulfuric acid used globally places it among the most plentiful of all commodity chemicals. In the modern chemical industry, synthesis of most sulfuric acid utilizes elemental sulfur as a feedstock. However, an alternative and historically important source of sulfuric acid was the conversion of an ore containing iron pyrites (FeS_) to sulfur oxides by roasting (burning) the ore with air. The following reactions occurred in an oven: $$\begin{array}{c} 2 \mathrm{FeS}_{2}(\mathrm{s})+\frac{11}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+4 \mathrm{SO}_{2}(\mathrm{g}) \\ \mathrm{SO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{3}(\mathrm{g}) \end{array}$$ The gases leaving the oven were fed to a catalytic converter in which most of the remaining \(\mathrm{SO}_{2}\) produced was oxidized to \(\mathrm{SO}_{3}\). Finally, the gas leaving the converter was sent to an absorption column where the \(S O_{3}\) was taken up by water to produce sulfuric acid \(\left(H_{2} S O_{4}\right)\) (a) The ore fed to the oven was 90.0 wt\% \(\mathrm{FeS}_{2}\), and the remaining material may be considered inert. Dry air was fed to the oven in \(30.0 \%\) excess of the amount required to oxidize all of the sulfur in the ore to \(S O_{3}\). Eighty-five percent of the \(\mathrm{FeS}_{2}\) was oxidized, and \(60 \%\) of the \(\mathrm{SO}_{2}\) produced was oxidized to \(S O_{3}\). Leaving the roaster were (i) a gas stream containing \(S O_{2}, S O_{3}, O_{2},\) and \(N_{2}\) and (ii) a solid stream containing unconverted pyrites, ferric oxide \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right),\) and the inert material. Calculate the required feed rate of air in standard cubic meters per \(100 \mathrm{kg}\) of ore fed to the process. Also determine the molar composition and volume (SCM/100 kg ore) of the gas leaving the oven. (b) The gas leaving the oven entered the catalytic converter, which operated at 1.0 atm. Reaction (2) proceeded to equilibrium, at which point the component partial pressures are related by the expression $$K_{\mathrm{P}}(T)=\frac{p_{\mathrm{SO}_{3}}}{p_{\mathrm{SO}_{3}} p_{\mathrm{O}_{2}}^{0.5}}$$ The gases were first heated to \(600^{\circ} \mathrm{C}\) to accelerate the rate of reaction, and then cooled to \(400^{\circ} \mathrm{C}\) to enhance \(S O_{2}\) conversion. The equilibrium constant \(K_{\mathrm{P}}\) at these two temperatures is 9.53 atm \(^{0.5}\) and 397 atm \(^{0.5}\), respectively. Calculate the equilibrium fractional conversions of \(S O_{2}\) at these two temperatures. (c) Estimate the production rate of sulfuric acid in \(\mathrm{kg} / \mathrm{kg}\) ore if all of the \(\mathrm{SO}_{3}\) leaving the converter was transformed to sulfuric acid. What would this value be if all the sulfur in the ore had been converted?

A gas consists of 20.0 mole \(\% \mathrm{CH}_{4}, 30.0 \% \mathrm{C}_{2} \mathrm{H}_{6},\) and \(50.0 \% \mathrm{C}_{2} \mathrm{H}_{4} .\) Ten kilograms of this gas is to be compressed to a pressure of 200 bar at \(90^{\circ} \mathrm{C}\). Using Kay's rule, estimate the final volume of the gas.
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