   Chapter 13, Problem 70QAP ### 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
21 views

# What mass of new gas would but required to fill a 3 .00 − L flask to a pressure of 925 mm Hg at 26  ° C ? What mass of argon gas would be required to fill a similar flask to the same pressure at the same temperature?

Interpretation Introduction

Interpretation:

The mass of neon gas and argon gas should be calculated.

Concept Introduction:

Dalton’s law of partial pressure: In a container for mixture of gases, the total pressure is equal to the sum of partial pressures of all the gases present in the container. The partial pressure is the pressure exerted by a gas if it is the only gas present in the container.

Let a mixture of three gases with partial pressures P1, P2 and P3, the total pressure of the gas will be sum of these partial pressures as follows:

PT=P1+P2+P3

This is the Dalton’s law of partial pressure.

The behaviour of gases is assumed to be ideal thus, partial pressure of gases can be calculated from an ideal gas equation as follows:

PV=nRT

Here, P is pressure, V is volume, n is number of moles, R is Universal gas constant and T is temperature of the gas.

The pressure exerted by an ideal gas depends on the number of gas particles, this does not depend on the nature of particles of gas. The two important things concluded from it will be:

1. The volume of gases is important.
2. The forces in between the particles of gas is not important.
Explanation

Given Information:

Neon and argon gas is filled in 3.00 L flask with pressure 925 mm Hg and temperature 26 C.

Calculation:

Number of moles of Ne and Ar gas can be calculated using the ideal gas equation as follows:

n=PVRT

First convert the temperature from C to K as follows:

0 C=273.15 K

Thus,

26 C=(26+273.15) K=299.15 K

Putting the values,

n=(925 mm Hg)(3.00 L)(0.082 L atm K1 mol1)(299

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