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Chapter 5: Problem 38

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.)

Short Answer

Expert verified
At sea level, you would lose about 1.489 kilograms of water or 1.489 liters of water through respiration per day. At the summit of Kilimanjaro, you would lose about 1551.23 grams or 1.551 liters of water per day. At higher altitudes, the lower ambient pressure increases the volume of the expired gases, while the lower oxygen concentration would likely increases the breathing rate.

Step by step solution

01

Calculate Total Expired Moles At Sea Level

Start by converting the volume of expired gas from liters per minute to liters per day - \( 11.36 \, \mathrm{L/min} = 16340 \, \mathrm{L/day} \). Now, use the ideal gas law equation \( PV = nRT \) to find total moles of expired gases at sea level. Here, P is the pressure (1 atm), V is the volume (16340 L), R is the ideal gas constant (0.0821 L atm/mol K) and T is the temperature in Kelvin (310 K since 37C = 310K). Solving for n provides the total moles of expired gas.
02

Calculate Mass Loss At Sea Level

Multiply the moles of the expired gases by the mole percentage of water in the expired gases to find the moles of water expelled. Then, multiply the moles of water by the molar mass of water (18 g/mol) to get the mass of water in grams lost through respiration per day. Assume this mass is entirely due to loss of water via respiration. Convert this mass to kilograms per day.
03

Convert Mass To Volume At Sea Level

Next, convert the mass of water lost to volume by using the density of water (1 g/cm3). Convert from cubic centimeters to liters.
04

Repeat Steps 1-3 For Uhuru Peak

To find the mass of water lost through respiration at Uhuru Peak, repeat steps 1-3, but use the conditions at Uhuru Peak. The temperature is -9.4C = 263.75K, and the pressure is 360 mmHg = 0.47 atm. Then, the volume of the expired gases is the same as at sea level - 16340 L. Calculate the volume of water lost in liters.
05

Discuss Altitude Effects

At higher altitudes, the effect of decreased ambient pressure would tend to increase the volume of the expired gases, as per Boyle's Law. However, the lower oxygen concentration would likely increase your breathing rate, thereby increasing the rate of expiration.

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a foundational principle in chemical engineering processes and describes how gases behave under different conditions. This law is expressed as \( PV = nRT \), where:
  • \( P \) is the pressure of the gas,
  • \( V \) is the volume,
  • \( n \) is the number of moles,
  • \( R \) is the ideal gas constant \((0.0821 \, \text{L atm/mol K})\), and
  • \( T \) is the temperature in Kelvin.
This equation helps calculate the number of moles of gas under set conditions. When climbing, such as on the Mt. Kilimanjaro journey, it is crucial to consider how pressure and temperature shifts impact the behavior of gases involved in respiration. At sea level, the strong presence of atmospheric pressure contributes to more compact gas volumes compared to noticeably expanded volumes in higher-altitude settings due to reduced pressures.
This characteristic is particularly significant when evaluating the respiratory process, wherein oxygen uptake and carbon dioxide expulsion play critical roles.
Respiration Process
The respiration process is a vital biological mechanism in which living organisms take in oxygen and expel carbon dioxide. It involves the exchange of gases between the body's cells and the environment. During light activity, like when preparing for a trek, the measured flow rate of expired air plays a critical role. The gas composition includes oxygen, carbon dioxide, and water vapor, among others. Breathing out is a continuous process where water, as vapor, is naturally lost. This necessitates replenishment to maintain hydration levels, especially during physically demanding activities such as mountain climbing. As observed in the exercise, enough water should be consumed daily to match the water expelled during respiration.
Interestingly, at higher altitudes, factors such as ambient pressure and temperature alter how gases are exchanged. While the total molar flow rate may remain the same, variations in oxygen concentration at such altitudes can affect the rate at which you breathe. This could mean a faster breathing rate to compensate for lower oxygen levels, resulting in more frequent exhalations and potential water loss.
Mass Transfer
Mass transfer is the process of moving mass from one location to another, crucial in many chemical engineering applications and physiological processes. In respiratory systems, it's essential to study how oxygen enters the lungs and how carbon dioxide exits. Inspired air contains oxygen that is transferred to the bloodstream, while carbon dioxide is simultaneously transferred from the blood to the alveolar spaces to be expelled. This interaction can be studied using mass transfer principles to understand how efficiently the body retains and expels different gases.
At high altitudes, such as Uhuru Peak on Mt. Kilimanjaro, transferring mass becomes challenging due to decreased ambient pressure. The inequality in pressure gradients between the inspired and expired air affects how gases are exchanged, often requiring deeper breaths or increased breathing frequency. Consequently, maintaining optimal hydration becomes critical as more frequent respiration may result in increased water loss.
Therefore, understanding mass transfer during respiration helps in planning for hydration needs during activities involving significant altitude changes.

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Most popular questions from this chapter

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.

Methanol is synthesized from carbon monoxide and hydrogen in the reaction $$\mathrm{CO}+2 \mathrm{H}_{2} \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}$$ The fresh feed to the system, which contains only \(\mathrm{CO}\) and \(\mathrm{H}_{2},\) is blended with a recycle stream containing the same species. The combined stream is heated and compressed to a temperature \(T(\mathrm{K})\) and a pressure \(P(\mathrm{kPa})\) and fed to the reactor. The percentage excess hydrogen in this stream is \(H_{\mathrm{xs}}\). The reactor effluent \(-\) also at \(T\) and \(P-\) goes to a scparation unit where essentially all of the methanol produced in the reactor is condensed and removed as product. The unreacted \(\mathrm{CO}\) and \(\mathrm{H}_{2}\) constitute the recycle stream blended with the fresh feed. Provided that the reaction temperature (and hence the rate of reaction) is high enough and the idealgas equation of state is a reasonable approximation at the reactor outlet conditions (a questionable assumption), the ratio $$K_{p c}=\frac{p_{\mathrm{CH}, \mathrm{OH}}}{p_{\mathrm{CO}} p_{\mathrm{H}}^{2}}$$ approaches the equilibrium value, which is given by the expression $$K_{p}(\mathrm{T})=1.390 \times 10^{-4} \exp \left(21.225+\frac{9143.6}{T}-7.492 \ln T+4.076 \times 10^{-3} T-7.161 \times 10^{-8} T^{2}\right)$$ In these equations, \(p_{i}\) is the partial pressure of species \(i\) in kilopascals \(\left(i=\mathrm{CH}_{3} \mathrm{OH}, \mathrm{CO}, \mathrm{H}_{2}\right)\) and \(T\) is in Kelvin. (a) Suppose \(P=5000 \mathrm{kPa}, T=500 \mathrm{K},\) and the percentage excess of hydrogen in the feed to the reactor \(\left(H_{\mathrm{xz}}\right)=5.0 \% .\) Calculate \(\dot{n}_{4}, \dot{n}_{5},\) and \(\dot{n}_{6},\) the component flow rates \((\mathrm{kmol} / \mathrm{h})\) in the reactor effluent. [Suggestion: Use the known value of \(H_{x s}\), atomic balances around the reactor, and the equilibrium relationship, \(K_{p c}=K_{p}(T),\) to write four equations in the four variables \(\dot{n}_{3}\) to \(\dot{n}_{6} ;\) use algebra to eliminate all but \(\dot{n}_{6} ;\) and use Goal Seck or Solver in Excel to solve the remaining nonlinear equation for \(\dot{n}_{6} .\) J Then calculate component fresh feed rates \(\left(\dot{n}_{1} \text { and } \dot{n}_{2}\right)\) and the flow rate (SCMH) of the recycle stream. (b) Prepare a spreadsheet to perform the calculations of Part (a) for the same basis of calculation (100 kmol CO/h fed to the reactor) and different specified values of \(P(\mathrm{kPa}), T(\mathrm{K}),\) and \(H_{\mathrm{xs}}(\%)\) The spreadsheet should have the following columns: A. \(P(\mathrm{kPa})\) B. \(T(\mathrm{K})\) C. \(H_{x s}(\%)\) D. \(K_{p}(T) \times 10^{8} .\) (The given function of \(T\) multiplied by \(10^{8} .\) When \(T=500 \mathrm{K}\), the value in this column should be 91.113.) E. \(K_{p} P^{2}\) F. \(\dot{n}_{3} .\) The rate (kmol/h) at which \(\mathrm{H}_{2}\) enters the reactor. G. \(\dot{n}_{4}\). The rate ( \(\mathrm{kmol} / \mathrm{h}\) ) at which CO leaves the reactor. H. \(\dot{n}_{5}\). The rate (kmol/h) at which \(\mathrm{H}_{2}\) leaves the reactor. I. \(\dot{n}_{6}\). The rate \((\mathrm{kmol} / \mathrm{h})\) at which methanol leaves the reactor. J. \(\dot{n}_{\text {lot. The total molar flow rate }(\mathrm{kmol} / \mathrm{h}) \text { of the reactor effluent. }}\) K. \(K_{p c} \times 10^{8} .\) The ratio \(y_{\mathrm{M}} /\left(y_{\mathrm{CO}} y_{\mathrm{H}_{2}}^{2}\right)\) multiplied by \(10^{8} .\) When the correct solution has been attained, this value should equal the one in Column E. L. \(K_{p} P^{2}-K_{p r} P^{2} .\) Column E-Column K, which equals zero for the correct solution. M. \(\dot{n}_{1}\). The molar flow rate \((\mathrm{kmol} / \mathrm{h})\) of \(\mathrm{CO}\) in the fresh feed. N. \(\dot{n}_{2}\). The molar flow rate \((\mathrm{kmol} / \mathrm{h})\) of \(\mathrm{H}_{2}\) in the fresh feed. O. \(\dot{V}_{\text {rec }}(\text { SCMH })\). The flow rate of the recycle stream in \(\mathrm{m}^{3}(\mathrm{STP}) / \mathrm{h}\). When the correct formulas have been entered, the value in Column I should be varied until the value in Column L equals 0 . Run the program for the following nine conditions (three of which are the same): \(\cdot\) \(T=500 \mathrm{K}, H_{\mathrm{xs}}=5 \%,\) and \(P=1000 \mathrm{kPa}, 5000 \mathrm{kPa},\) and \(10,000 \mathrm{kPa}\) \(\cdot\) \(P=5000 \mathrm{kPa}, H_{x s}=5 \%,\) and \(T=400 \mathrm{K}, 500 \mathrm{K},\) and \(600 \mathrm{K}\) \(\cdot\) \(T=500 \mathrm{K}, P=5000 \mathrm{kPa},\) and \(H_{\mathrm{xs}}=0 \%, 5 \%,\) and \(10 \%\) Summarize the effects of reactor pressure, reactor temperature, and excess hydrogen on the yield of methanol (kmol M produced per \(100 \mathrm{kmol}\) CO fed to the reactor). (c) You should find that the methanol yield increases with increasing pressure and decreasing temperature. What cost is associated with increasing the pressure? (d) Why might the yield be much lower than the calculated value if the temperature is too low? (e) If you actually ran the reaction at the given conditions and analyzed the reactor effluent, why might the spreadsheet values in Columns \(\mathrm{F}-\mathrm{M}\) be significantly different from the measured values of these quantities? (Give several reasons, including assumptions made in obtaining the spreadsheet values.)

The oxidation of nitric oxide $$\mathrm{NO}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{NO}_{2}$$ takes place in an isothermal batch reactor. The reactor is charged with a mixture containing 20.0 volume percent NO and the balance air at an initial pressure of \(380 \mathrm{kPa}\) (absolute). (a) Assuming ideal-gas behavior, determine the composition of the mixture (component mole fractions) and the final pressure (kPa) if the conversion of NO is 90\%. (b) Suppose the pressure in the reactor eventually equilibrates (levels out) at \(360 \mathrm{kPa}\). What is the equilibrium percent conversion of NO? Calculate the reaction equilibrium constant at the prevailing temperature, \(K_{p}\left[(\mathrm{atm})^{-0.5}\right]\), defined as $$K_{p}=\frac{\left(p_{\mathrm{NO}_{2}}\right)}{\left(p_{\mathrm{NO}}\right)\left(p_{\mathrm{O}_{2}}\right)^{0.5}}$$ where \(p_{i}(\mathrm{atm})\) is the partial pressure of species \(i\left(\mathrm{NO}_{2}, \mathrm{NO}, \mathrm{O}_{2}\right)\) at equilibrium. (c) Assuming that \(K_{\mathrm{p}}\) depends only on temperature, estimate the final pressure and composition in the reactor if the feed ratio of NO to \(\mathrm{O}_{2}\) and the initial pressure are the same as in \(\operatorname{Part}(\) a), but the feed to the reactor is pure \(\mathrm{O}_{2}\) instead of air. (d) Replace the partial pressures in the expression for \(K_{\mathrm{p}}\), and use the result to explain how reactor pressure influences the conversion of NO to \(\mathrm{NO}_{2}\)

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?

An ideal-gas mixture contains \(35 \%\) helium, \(20 \%\) methane, and \(45 \%\) nitrogen by volume at 2.00 atm absolute and \(90^{\circ} \mathrm{C}\). Calculate (a) the partial pressure of each component, (b) the mass fraction of methane, (c) the average molecular weight of the gas, and (d) the density of the gas in \(\mathrm{kg} / \mathrm{m}^{3}\).
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