Chapter 10.8, Problem 2.3ACP

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# To stay aloft, a blimp must achieve neutral buoyancy; that is, its density must equal that of the surrounding air. The density of the blimp is its total weight (blimp, helium and air, passengers, and ballast) divided by its volume. Assume that the gross weight of the blimp includes the blimp’s structure and the helium, but does not include the air in the ballonets or the weights of the passengers and ballast. If the ballonets are filled with 12,000 ft3 (340 m3) of air at 1.00 atm and 25 °C, what additional weight (of passengers and ballast) is required for neutral buoyancy?

Interpretation Introduction

Interpretation:

The additional weight of passengers and ballast required for neutral buoyancy has to be given if the ballonets are filled with 12000ft3 of air at 1atmand25°C

Concept Introduction:

Density of a gas can be find out with the help of ideal gas equation which is

PV=nRT

n in the given equation can be written as mM Since amount of any substance is given by dividing mass with molar mass.

PV=mMRTmV=PMRTdensity=PMRT

The common equation for density is:

density=massvolume

By knowing the density we can calculate the mass.

Explanation

Given:

The volume of the inner envelope of Goodyear blimp is 5740â€‰m3

Volume of the ballonets is 340â€‰m3

â€‚Â pâ€‰â€‰=â€‰1â€‰atmTâ€‰â€‰â€‰=â€‰â€‰25Â°Câ€‰â€‰=â€‰298â€‰K

â€‚Â volumeâ€‰ofâ€‰theâ€‰blimpâ€‰=â€‰â€‰volumeâ€‰ofâ€‰innerâ€‰envelope-volumeâ€‰ofâ€‰ballonetsâ€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰=â€‰â€‰5740â€‰m3â€‰â€‰â€‰â€‰-â€‰â€‰â€‰340â€‰m3â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰=â€‰â€‰5400â€‰m3

Density of dry air is calculated and found to be 1.184â€‰g/Lâ€‰â€‰â€‰=â€‰1

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