   Chapter 13, Problem 108AP ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
1 views

# helium tank contains 25 . 2 L of helium m 8 . 4 0 atm pressure. Determine how many 1 . 5 0 − L balloons at 755 mm Hg can be inflated with the gas in the tank, assuming that the tank will also have to contain He at 755 mm Hg after the balloons are filled (that is, it is not possible to empty the tank completely). The temperature is 25  ° C in all cases.

Interpretation Introduction

Interpretation:

A helium tank contains 25.2 L of helium at 8.40 atm pressure. The number of 1.50 L balloons at 755 mm Hg that can be inflated with the gas in the tank should be calculating by assuming that the tan will also have to contain Helium at 755 mm Hg after the balloons are filled.

Concept Introduction:

The ideal gas equation is as follows:

PV=nRT

Where, P = Pressure of the ideal gas (atm)

V = Volume of the ideal gas (L)

n = number of moles of ideal gas present (mol)

R = Universal gas constant (L atm/K)

T = Temperature in Kelvin (K)

According to Boyle’s Law, at constant temperature the pressure of an ideal gas is inversely proportional to its volume.

P1 V1 = P2 V2.

Explanation

1 atm = 760 mm Hg

755 mm Hg = 755 mm Hg760 mm Hg = 0.9934 atm

When the pressure inside the tank reduced to 0.9934 atm, the volume occupied by helium gas;

P1V1= P2V2 8.40 atm × 25.2 L = 0.9934 atm × V2

V2=8.40 atm × 25

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