Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The average lung capacity of a human is 6.0 L. How many moles of air are in your lungs when you are in the following situations? a. At sea level \((T=298 \mathrm{K}, P=1.00 \mathrm{atm})\) b. \(10 . \mathrm{m}\) below water \((T=298 \mathrm{K}, P=1.97 \mathrm{atm})\) c. At the top of Mount Everest \((T=200 . \mathrm{K}, P=0.296 \mathrm{atm})\)

Short Answer

Expert verified
In the three given situations, the number of moles of air in the lungs can be calculated as follows: a. At sea level (T=298 K, P=1.00 atm) - \(n = 0.246 \mathrm{mol}\) b. 10 m below water (T=298 K, P=1.97 atm) - \(n = 0.483 \mathrm{mol}\) c. At the top of Mount Everest (T=200 K, P=0.296 atm) - \(n = 0.108 \mathrm{mol}\)

Step by step solution

01

Write down the given variables

For this situation, we have: - P = 1.00 atm - V = 6.0 L - T = 298 K - R = 0.0821 Latm / Kmol (Ideal Gas Constant)
02

Calculate the number of moles (n) using the Ideal Gas Law

Rearrange the Ideal Gas Law formula to solve for n: \(n = \frac{PV}{RT}\) Plug in the given values and calculate n: \(n = \frac{(1.00 \mathrm{atm})(6.0 \mathrm{L})}{(0.0821 \frac{\mathrm{Latm}}{\mathrm{Kmol}}) (298 \mathrm{K})} = 0.246 \mathrm{mol}\) So, at sea level, there are 0.246 moles of air in the lungs. #b. 10 m below water (T=298 K, P=1.97 atm)#
03

Write down the given variables

For this situation, we have: - P = 1.97 atm - V = 6.0 L - T = 298 K - R = 0.0821 Latm / Kmol (Ideal Gas Constant)
04

Calculate the number of moles (n) using the Ideal Gas Law

Use the rearranged Ideal Gas Law formula to solve for n: \(n = \frac{PV}{RT}\) Plug in the given values and calculate n: \(n = \frac{(1.97 \mathrm{atm})(6.0 \mathrm{L})}{(0.0821 \frac{\mathrm{Latm}}{\mathrm{Kmol}}) (298 \mathrm{K})} = 0.483 \mathrm{mol}\) So, 10 meters below water, there are 0.483 moles of air in the lungs. #c. At the top of Mount Everest (T=200 K, P=0.296 atm)#
05

Write down the given variables

For this situation, we have: - P = 0.296 atm - V = 6.0 L - T = 200 K - R = 0.0821 Latm / Kmol (Ideal Gas Constant)
06

Calculate the number of moles (n) using the Ideal Gas Law

Use the rearranged Ideal Gas Law formula to solve for n: \(n = \frac{PV}{RT}\) Plug in the given values and calculate n: \(n = \frac{(0.296 \mathrm{atm})(6.0 \mathrm{L})}{(0.0821 \frac{\mathrm{Latm}}{\mathrm{Kmol}}) (200 \mathrm{K})} = 0.108 \mathrm{mol}\) So, at the top of Mount Everest, there are 0.108 moles of air in the lungs.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Moles Calculation
When trying to find how many moles of gas are in your lungs in varying situations, you use the Ideal Gas Law. This law is represented by the equation \[ PV = nRT \]where
  • \( P \) is the pressure in atmospheres (atm),
  • \( V \) is the volume in liters (L),
  • \( n \) is the number of moles,
  • \( R \) is the Ideal Gas Constant, and
  • \( T \) is the temperature in Kelvin (K).
To find the number of moles \( n \), you rearrange the formula to\[ n = \frac{PV}{RT} \]Plugging in the given values allows calculation of \( n \) for different conditions. The process involves substituting the pressure, volume, temperature, and using the constant value to solve for the number of moles. Each scenario simply requires plugging in the distinct conditions of pressure and temperature.
Gas Constants
Understanding gas constants is crucial for applying the Ideal Gas Law effectively. The gas constant \( R \) is a fixed value used to correlate the physical properties of gases. For ideal gas calculations in standard conditions, the value of \( R \) is usually \[ R = 0.0821 \frac{\mathrm{L} \cdot \mathrm{atm}}{\mathrm{K} \cdot \mathrm{mol}} \]This constant allows you to bridge the units of pressure, volume, and temperature to find the number of moles.Remember, the Ideal Gas Law assumes perfect gas behavior which is mostly accurate under standard conditions of temperature and pressure, unless the gas undergoes extreme temperatures or pressures where it might deviate from ideal behavior.
Pressure and Temperature Change
Gas behavior significantly depends on changes in pressure and temperature. When the pressure increases, as it would underwater, the number of moles of gas in a confined space increases if volume and temperature are consistent. For instance, in the underwater scenario at 1.97 atm versus the sea level situation at 1.00 atm, while the volume and temperature remain the same, more moles of air are present in the lungs due to higher pressure. Similarly, temperature variations also affect physical properties. Higher temperatures tend to increase the kinetic energy of the gas molecules, stimulating expansion and potentially increasing volume if pressure is constant. At the top of Mount Everest with a lower temperature of 200 K and pressure of 0.296 atm, fewer moles of air exist in the lungs due to the reduced kinetic energy and expansion capacity of the gas. This highlights how pressure and temperature interplay affects the number of moles calculated for different environments.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The partial pressure of \(\mathrm{CH}_{4}(g)\) is 0.175 atm and that of \(\mathrm{O}_{2}(g)\) is 0.250 \(\mathrm{atm}\) in a mixture of the two gases. a. What is the mole fraction of each gas in the mixture? b. If the mixture occupies a volume of 10.5 \(\mathrm{L}\) at \(65^{\circ} \mathrm{C}\) , calculate the total number of moles of gas in the mixture. c. Calculate the number of grams of each gas in the mixture.

Air bags are activated when a severe impact causes a steel ball to compress a spring and electrically ignite a detonator cap. This causes sodium azide \(\left(\mathrm{NaN}_{3}\right)\) to decompose explosively according to the following reaction: $$2 \mathrm{NaN}_{3}(s) \longrightarrow 2 \mathrm{Na}(s)+3 \mathrm{N}_{2}(g)$$ What mass of \(\mathrm{NaN}_{3}(s)\) must be reacted to inflate an air bag to 70.0 \(\mathrm{L}\) at STP?

\(\mathrm{N}_{2} \mathrm{O}\) is a gas commonly used to help sedate patients in medicine and dentistry due to its mild anesthetic and analgesic properties; it is also nonflammable. If a cylinder of \(\mathrm{N}_{2} \mathrm{O}\) is at 10.5 atm and has a volume of 5.00 \(\mathrm{L}\) at \(298 \mathrm{K},\) how many moles of \(\mathrm{N}_{2} \mathrm{O}\) gas are present? The gas from the cylinder is emptied into a large balloon at 745 torr. What is the volume of the balloon at 298 \(\mathrm{K}\) ?

A compressed gas cylinder contains \(1.00 \times 10^{3} \mathrm{g}\) argon gas. The pressure inside the cylinder is \(2050 .\) psi (pounds per square inch) at a temperature of \(18^{\circ} \mathrm{C}\) . How much gas remains in the cylinder if the pressure is decreased to 650 . psi at a temperature of \(26^{\circ} \mathrm{C} ?\)

In Example 5.11 of the text, the molar volume of \(\mathrm{N}_{2}(g)\) at \(\mathrm{STP}\) is given as 22.42 \(\mathrm{Lmol} \mathrm{N}_{2} .\) How is this number calculated? How does the molar volume of He(g) at \(\mathrm{STP}\) compare to the molar volume of \(\mathrm{N}_{2}(g)\) at \(\mathrm{STP}\) (assuming ideal gas behavior)? Is the molar volume of \(\mathrm{N}_{2}(g)\) at 1.000 atm and \(25.0^{\circ} \mathrm{C}\) equal to, less than, or greater than 22.42 \(\mathrm{L} / \mathrm{mol} ?\) Explain. Is the molar volume of \(\mathrm{N}_{2}(g)\) collected over water at a total pressure of 1.000 \(\mathrm{atm}\) and \(0.0^{\circ} \mathrm{C}\) equal to, less than, or greater than 22.42 \(\mathrm{L} / \mathrm{mol}\) ? Explain.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free