91Ó°ÊÓ

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.)

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.