/*! 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

Terephthalic acid (TPA), a raw material in the manufacture of polyester fiber, film, and soft drink bottles, is synthesized from \(p\) -xylene (PX) in the process shown below. A fresh feed of pure liquid \(\mathrm{PX}\) combines with a recycle stream containing \(\mathrm{PX}\) and a solution (S) of a catalyst (a cobalt salt) in a solvent (methanol). The combined stream, which contains \(S\) and \(P X\) in a 3: 1 mass ratio, is fed to a reactor in which \(90 \%\) of the \(\mathrm{PX}\) is converted to TPA. A stream of air at \(25^{\circ} \mathrm{C}\) and 6.0 atm absolute is also fed to the reactor. The air bubbles through the liquid and the reaction given above takes place under the influence of the catalyst. A liquid stream containing unreacted \(\mathrm{PX}\), dissolved TPA, and all the S that entered the reactor goes to a separator in which solid TPA crystals are formed and filtered out of the solution. The filtrate, which contains all the \(S\) and \(P X\) leaving the reactor, is the recycle stream. A gas stream containing unreacted oxygen, nitrogen, and the water formed in the reaction leaves the reactor at \(105^{\circ} \mathrm{C}\) and 5.5 atm absolute and goes through a condenser in which essentially all the water is condensed. The uncondensed gas contains 4.0 mole \(\%\) O. (a) Taking \(100 \mathrm{kmol}\) TPA produced/h as a basis of calculation, draw and label a flowchart for the process. (b) What is the required fresh feed rate (kmol PX/h)? (c) What are the volumetric flow rates \(\left(\mathrm{m}^{3} / \mathrm{h}\right)\) of the air fed to the reactor, the gas leaving the reactor, and the liquid water leaving the condenser? Assume ideal-gas behavior for the two gas streams. (d) What is the mass flow rate ( \(\mathrm{kg} / \mathrm{h}\) ) of the recycle stream? (e) Briefly explain in your own words the functions of the oxygen, nitrogen, catalyst, and solvent in the process. (f) In the actual process, the liquid condensate stream contains both water and PX. Speculate on what might be done with the latter stream to improve the economics of the process. [Hint: Note that PX is expensive, and recall what is said about oil (hydrocarbons) and water.]

During your summer vacation, you plan an epic adventure trip to scale Mt. Kilimanjaro in Tanzania. Dehydration is a great danger on such a climb, and it is essential to drink enough water to make up for the amount you lose by breathing. (a) During your pre-trip physical, your physician measured the average flow rate and composition of the gas you exhaled (expired air) while performing light activity. The results were \(11.36 \mathrm{L} / \mathrm{min}\) at body temperature (37^) C) and 1 atm, 17.08 mole\% oxygen, 3.25\% carbon dioxide, 6.12 mole\% \(\mathrm{H}_{2} \mathrm{O},\) and the balance nitrogen. The ambient (inspired) air contained 1.67 mole\% water and a negligible amount of carbon dioxide. Calculate the rate of mass lost through the breathing process (kg/day) and the volume of water in liters you would have to drink per day just to replace the water lost in respiration. Consider your lungs to be a continuous steady-state system, with input streams being inspired air and water and \(\mathrm{CO}_{2}\) transferred from the blood and output streams being expired air and \(\mathrm{O}_{2}\) transferred to the blood. Assume no nitrogen is transferred to or from the blood. (b) You made the trip to Tanzania and completed the climb to Uhuru Peak, the summit of Kilimanjaro, at an altitude of 5895 meters above sea level. The ambient temperature and pressure there averaged \(-9.4^{\circ} \mathrm{C}\) and \(360 \mathrm{mm} \mathrm{Hg},\) and the air contained \(0.46 \mathrm{mole} \%\) water. The molar flow rate of your expired air was roughly the same as it had been at sea level, and the expired air contained \(14.86 \% \mathrm{O}_{2}\) \(3.80 \% \mathrm{CO}_{2},\) and \(13.20 \% \mathrm{H}_{2} \mathrm{O} .\) Calculate the rate of mass lost (g/day) through breathing and water you would have to drink (L/day) just to replace the water lost in respiration. (c) The equality of the molar flow rates of expired air at sea level and at Uhuru Peak is due to a cancellation of effects, one of which would tend to increase the rate at higher altitudes and the other to decrease it. What are those effects? (Hint: Use the ideal-gas equation of state in your solution, and think about how the oxygen concentration at a high altitude would likely affect your breathing rate.)

Chemicals are stored in a laboratory with volume \(V\left(\mathrm{m}^{3}\right) .\) As a consequence of poor laboratory practices, a hazardous species, A, enters the room air (from inside the room) at a constant rate \(\dot{m}_{\mathrm{A}}(\mathrm{g} \mathrm{A} / \mathrm{h})\) The room is ventilated with clean air flowing at a constant rate \(\dot{V}_{\text {air }}\left(\mathrm{m}^{3} / \mathrm{h}\right) .\) The average concentration of A in the room air builds up until it reaches a steady-state value \(C_{\mathrm{A}, \mathrm{r}}\left(\mathrm{g} \mathrm{A} / \mathrm{m}^{3}\right)\) (a) List at least four situations that could lead to A getting into the room air. (b) Assume that the A is perfectly mixed with the room air and derive the formula $$\dot{m}_{\mathrm{A}}=\dot{V}_{\mathrm{air}} C_{\mathrm{A}}$$ (c) The assumption of perfect mixing is never justificd when the enclosed space is a room (as opposed to, say, a stirred reactor). In practice, the concentration of A varies from one point in the room to another: it is relatively high near the point where A enters the room air and relatively low in regions far from that point, including the ventilator outlet duct. If we say that \(C_{\mathrm{A}, \text { duct }}=k C_{\mathrm{A}}\) where \(k < 1\) is a nonideal mixing factor (generally between 0.1 and \(0.5,\) with the lowest value corresponding to the poorest mixing), then the equation of Part (b) becomes $$\dot{m}_{\mathrm{A}}=k \dot{V}_{\mathrm{air}} C_{\mathrm{A}}$$ Use this equation and the ideal-gas equation of state to derive the following expression for the average mole fraction of \(A\) in the room air: $$y_{\mathrm{A}}=\frac{\dot{m}_{\mathrm{A}}}{k \dot{V}_{\mathrm{air}}} \frac{R T}{M_{\mathrm{A}} P}$$ where \(M_{\mathrm{A}}\) is the molecular weight of \(\mathrm{A}\) (d) The permissible exposure level (PEL) for styrene \((M=104.14\) ) defined by the U.S. Occupational Safcty and Health Administration is 50 ppm (molar basis). \(^{21}\) An open storage tank in a polymerization laboratory contains styrene. The evaporation rate from this tank is estimated to be \(9.0 \mathrm{g} / \mathrm{h}\). Room temperature is \(20^{\circ} \mathrm{C}\). Assuming that the laboratory air is reasonably well mixed (so that \(k=0.5\) ), calculate the minimum ventilation rate \(\left(\mathrm{m}^{3} / \mathrm{h}\right)\) required to keep the average styrene concentration at or below the PEL. Then give several reasons why working in the laboratory might still be hazardous if the calculated minimum ventilation rate is used. (e) Would the hazard level in the situation described in Part (d) increase or decrease if the temperature in the room were to increase? (Increase, decrease, no way to tell.) Explain your answer, citing at least two effects of temperature in your explanation.

A small power plant produces \(500 \mathrm{MW}\) of electricity through combustion of coal that has the following composition on a dry basis: 76.2 wt\% carbon, \(5.6 \%\) hydrogen, \(3.5 \%\) sulfur, \(7.5 \%\) oxygen, and the remainder ash. The coal contains 4.0 wt\% water. The feed rate of coal is 183 tons/h, and it is burned with \(15 \%\) excess air at 1 atm, \(80^{\circ} \mathrm{F}\), and \(30.0 \%\) relative humidity. (a) Estimate the volumetric flow rate (ft \(^{3} / \mathrm{min}\) ) of air drawn into the furnace. (b) Effluent gases are discharged from the furnace at \(625^{\circ} \mathrm{F}\) and 1 atm. Estimate the molar (lb-mole/ min) and volumetric (ft \(^{3} / \mathrm{min}\) ) flow rates of gas leaving the furnace. (c) Injection of dry limestone ( \(\mathrm{CaCO}_{3}\) ) into the furnace is being considered as a means of reducing the \(\mathrm{SO}_{2}\) emitted from the plant. The technology calls for \(\mathrm{SO}_{2}\) to react with limestone: $$\mathrm{CaCO}_{3}+\mathrm{SO}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{CaSO}_{4}+\mathrm{CO}_{2}$$ Unfortunately, the process is expected to remove only \(75 \%\) of the \(\mathrm{SO}_{2}\) in the effluent gases, even though the limestone is fed at a rate 2.5 times the stoichiometric amount. What is the required feed rate of limestone? since some of the \(S O_{2}\) is removed from the furnace efflucnt [in contrast to Part (b)], recalculate the molar flow rate and composition of the effluent from the fumace. (d) The gas leaving the furnace passes through an electrostatic precipitator, where particulates from ash and limestone are removed, and then enters a stack (chimney) for release to the atmosphere. What is the gas velocity at a point in the stack where the stack diameter is \(25 \mathrm{ft}\) and the temperature is \(300^{\circ} \mathrm{F}\) ? Does the gas discharged from the stack meet the new Environmental Protection Agency standard that emissions from such power plants contain less than 75 parts of \(\mathrm{SO}_{2}\) per billion?

An innovative engineer has invented a device to replace the hydraulic jacks found at many service stations. A movable piston with a diameter of \(0.15 \mathrm{m}\) is fitted into a cylinder. Cars are raised by opening a small door near the base of the cylinder, inserting a block of dry ice (solid \(\mathrm{CO}_{2}\) ), closing and sealing the door, and vaporizing the dry ice by applying just enough heat to raise the cylinder contents to ambient temperature \(\left(25^{\circ} \mathrm{C}\right)\). The car is subsequently lowered by opening a valve and venting the cylinder gas. The device is tested by raising a car a vertical distance of \(1.5 \mathrm{m}\). The combined mass of the piston and the car is \(5500 \mathrm{kg}\). Before the piston rises, the cylinder contains \(0.030 \mathrm{m}^{3}\) of \(\mathrm{CO}_{2}\) at ambient temperature and pressure ( 1 atm). Neglect the volume of the dry ice. (a) Calculate the pressure in the cylinder when the piston comes to rest at the desired elevation. (b) How much dry ice (kg) must be placed in the cylinder? Use the SRK equation of state for this calculation. (c) Outline how you would calculate the minimum piston diameter required for any elevation of the car to occur if the calculated amount of dry ice is added. (Just give formulas and describe the procedure you would follow- -no numerical calculations are required.)
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