Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 100

Air entering the lungs ends up in tiny sacs called alveoli. It is from the alveoli that oxygen diffuses into the blood. The average radius of the alveoli is \(0.0050 \mathrm{~cm}\) and the air inside contains 14 percent oxygen. Assuming that the pressure in the alveoli is 1.0 atm and the temperature is \(37^{\circ} \mathrm{C},\) calculate the number of oxygen molecules in one of the alveoli.

Short Answer

Expert verified
Provided that all calculations are done, the obtained number of oxygen molecules in one alveolus should be found by concluding step 4.

Step by step solution

01

Convert temperature to Kelvin

For calculations involving the ideal gas law, it is necessary to use temperatures in Kelvin. The Celsius-to-Kelvin conversion formula is \(K = C + 273.15\). Applying this conversion to our temperature, we find that the temperature in Kelvin is \(37^{\circ} \mathrm{C} + 273.15 = 310.15 \mathrm{K}\)
02

Calculate the volume of an alveoli

Alveoli have a spherical shape, so their volume can be calculated using the formula \( V = \frac{4}{3} \pi r^3 \), where r is the radius. Substituting the given radius \( r = 0.0050 \mathrm{~cm} = 0.000050 \mathrm{~m} \) , the volume of one alveolus becomes \( V = \frac{4}{3} \pi (0.000050^3) = 5.24 \times 10^{-10} \mathrm{~m^3}\)
03

Calculate the number of moles of oxygen in the alveolus

We know from the ideal gas law that \( n = \frac{PV}{RT} \), where P is the pressure (in this case 1.0 atm converted to N/m^2 is 101325 N/m^2), V the volume calculated in Step 2, T the temperature in K from step 1, and R the ideal gas constant (\( 8.314 N \cdot m / mol \cdot K\) ). Remember that the mole we find represents all the air in the alveolus, but we are only interested in the oxygen, which makes up 14% of this air. Thus, we need to multiply this result by 0.14 to find the moles of oxygen in an alveolus.
04

Calculate the number of molecules in a mole

We can find the number of molecules from the number of moles because 1 mole of any substance contains Avogadro's number of molecules ( approximately \(6.022 \times 10^{23}\) molecules). Therefore, multiplying the number of moles by Avogadro's number, we obtain the number of oxygen molecules in an alveolus.

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.

Start your free trial

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

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

A gas-filled balloon having a volume of \(2.50 \mathrm{~L}\) at \(1.2 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) is allowed to rise to the stratosphere (about \(30 \mathrm{~km}\) above the surface of Earth), where the temperature and pressure are \(-23^{\circ} \mathrm{C}\) and \(3.00 \times 10^{-3}\) atm, respectively. Calculate the final volume of the balloon.

A sample of nitrogen gas kept in a container of volume \(2.3 \mathrm{~L}\) and at a temperature of \(32^{\circ} \mathrm{C}\) exerts a pressure of \(4.7 \mathrm{~atm} .\) Calculate the number of moles of gas present.

At \(27^{\circ} \mathrm{C}, 10.0\) moles of a gas in a \(1.50-\mathrm{L}\) container exert a pressure of 130 atm. Is this an ideal gas?

A gas evolved during the fermentation of glucose (wine making) has a volume of \(0.78 \mathrm{~L}\) when measured at \(20.1^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm} .\) What was the volume of this gas at the fermentation temperature of \(36.5^{\circ} \mathrm{C}\) and 1.00 atm pressure?

List the physical characteristics of gases.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation
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