Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 10 - Gases - Exercises: 10.37b

Answer

6464.9 grams of air

Work Step by Step

We have the Volume of the lungs along with the pressure and temperature. First we convert the temperature from 0C to Kelvin by adding 273 so we have 273K. Then we can use ideal gas law: PV = nRT or in this case n = PV/RT = (1atm) * (5.0 x 10^3L) / (0.0821 * 273K) =223.08 moles Then since the molar mass is 28.98 g/mol we have 223.08 moles * (28.98 g/ 1mol) = 6464.9 grams of air
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.