   Chapter 13, Problem 134AP ### 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
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# 22 − g sample of neon gas exerts a pressure of 2 .0 atm at a certain temperature and volume. What pressure does a 44 − g sample of argon gas exert at these conditions of temperature and volume?

Interpretation Introduction

Interpretation:

To calculate the pressure of Argon gas at constant conditions of temperature and volume.

Concept Introduction:

The ideal gas equation is;

PV = nRT

• P = pressure.
• T = temperature.
• V = volume.
• R = gas constant = 0.0821 L. atm / K. mole.
• n = number of moles.
• Hence at constant volume and temperature, the pressure is directly proportional to moles of gas. The mathematical expression can be written as;

P1n1=P2n2

Explanation

Given information:

Pressure of Neon gas = P1 = 2.0 atm

Molar mass of Neon gas = 20.1 g/mol

Molar mass of Argon gas = 39.9 g/mol

According to ideal gas equation, pressure of ideal gas has direct relation with number of moles of gas at constant volume and temperature. Hence, under same conditions of volume and temperature, the pressure and moles can be used to calculate the pressure of Argon gas with given mass.

Number of moles of Neon gas = n1 = 22 g20.1 g/mol=1.09 moles

Number of moles of Argon gas = n2 = 44 g39.9 g/mol=1

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