Explanation Given: Volume of the tire, V tier = 855 mL = 0.855 L [ ∵ 1 L = 1000 mL ] Total pressure, P total = 125 psi Temperature, T = 25 ∘ C = 25 + 273.15 = 298.15 K Average molar mass of air = 28.8 g / m o l Explanation: Convert the total pressure of tire from 125 p s i into atm by using the following relation: 14.7 p s i = 1 a t m ∴ 125 p s i = 1 a t m × 125 p s i 14.7 p s i = 8.5034 a t m Use the ideal gas law to calculate the total moles of gas whether air or helium that can be filled in the tire of volume 855 m L ( o r 0.855 L ) at a pressure of 125 p s i ( o r 8.5034 a t m ) and a temperature of 25 ∘ C ( o r 298.15 K ) . According to the ideal gas law, P V = n R T ⇒ n = P V R T Substitute P = 8.5034 atm , V = 0.855 L , R = 0.08206 L atm/(mol K) , and T = 298.15 K in the above expression to get n = 8.5034 atm × 0.855 L 0.08206 L atm mol K × 298.15 K = 0.29716 mol Thus, the moles of gas whether air or helium that can be filled in the tire are 0.29716 m o l . These moles are same for air as well as helium. Convert these moles of air and helium into their mass by using their molar masses. Mass of air = Moles of air × Average molar mass of air Substitute moles of air = 0

Chemistry: A Molecular Approach, Books a la Carte Edition (4th Edition)

4th Edition

ISBN: 9780134113593

Author: Nivaldo J. Tro

Publisher: PEARSON

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Chapter 5, Problem 107E

Interpretation Introduction

To determine: The mass of air and helium present in the tire, and the mass difference between air- and helium-filled tires.

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Chapter 5, Problem 106E

Chapter 5, Problem 108E

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Chemistry: A Molecular Approach, Books a la Carte Edition (4th Edition)

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