Living By Chemistry: First Edition Textbook
Living By Chemistry: First Edition Textbook
1st Edition
ISBN: 9781559539418
Author: Angelica Stacy
Publisher: MAC HIGHER
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Question
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Chapter U3.13, Problem 3E

(a)

Interpretation Introduction

Interpretation: Whether the volume of the gas increase or decreases needs to be explained.

Concept Introduction: The combined gas law is combination of three gas laws that is Boyle’s law, Charles’ law and Gay-Lussac’s law. According to this law the ratio of temperature to the product of pressure and volume is a constant.

(a)

Expert Solution
Check Mark

Explanation of Solution

The pressure and volume of the gas in the container is 0.97 atm and 0.5 L respectively. The pressure changes to 1.0 atm. The number of moles and temperature remains constant.

The relation between pressure and volume at constant number of moles and temperature is represented by Boyle’s law. According to this law, as volume of the container decreases, pressure tends to increase.

Since, pressure changes from 0.97 atm to 1.0 atm that is pressure increases thus, volume of the gas decreases.

(b)

Interpretation Introduction

Interpretation: The equation used to calculate the new volume of the gas needs to be determined.

Concept Introduction: The combined gas law is combination of three gas laws that is Boyle’s law, Charles’ law and Gay-Lussac’s law. According to this law the ratio of temperature to the product of pressure and volume is a constant.

(b)

Expert Solution
Check Mark

Explanation of Solution

According to Boyle’s law, pressure and volume are inversely proportional to each other.

This is mathematically represented as follows:

  P1V1=P2V2

This can be rearranged as follows:

  V2=P1V1P2

Here, P1 is initial pressure, P2 is final pressure, V1 is initial volume and V2 is final volume.

(c)

Interpretation Introduction

Interpretation: The volume of the gas at pressure 1.0 atm needs to be determined. 4

Concept Introduction: According to Boyle’s law, pressure and volume are inversely proportional to each other.

This is mathematically represented as follows:

  P1V1=P2V2

Here, P1 is initial pressure, P2 is final pressure, V1 is initial volume and V2 is final volume.

(c)

Expert Solution
Check Mark

Explanation of Solution

The volume of the gas at 1.0 atm pressure can be calculated as follows:

  V2=P1V1P2

Putting the values,

  V2=(0.97 atm)(0.5 L)(1 atm)=0.485 L

Thus, final volume of the gas is 0.485 L.

Chapter U3 Solutions

Living By Chemistry: First Edition Textbook

Ch. U3.2 - Prob. 4ECh. U3.2 - Prob. 5ECh. U3.2 - Prob. 6ECh. U3.2 - Prob. 7ECh. U3.3 - Prob. 1TAICh. U3.3 - Prob. 1ECh. U3.3 - Prob. 2ECh. U3.3 - Prob. 3ECh. U3.3 - Prob. 4ECh. U3.3 - Prob. 5ECh. U3.3 - Prob. 6ECh. U3.3 - Prob. 7ECh. U3.3 - Prob. 8ECh. U3.3 - Prob. 9ECh. U3.4 - Prob. 1TAICh. U3.4 - Prob. 1ECh. U3.4 - Prob. 2ECh. U3.4 - Prob. 3ECh. U3.4 - Prob. 4ECh. U3.4 - Prob. 5ECh. U3.4 - Prob. 6ECh. U3.4 - Prob. 7ECh. U3.4 - Prob. 8ECh. U3.4 - Prob. 9ECh. U3.5 - Prob. 1TAICh. U3.5 - Prob. 1ECh. U3.5 - Prob. 2ECh. U3.5 - Prob. 3ECh. U3.5 - Prob. 4ECh. U3.5 - Prob. 5ECh. U3.5 - Prob. 6ECh. U3.5 - Prob. 7ECh. U3.5 - Prob. 8ECh. U3.5 - Prob. 9ECh. U3.5 - Prob. 10ECh. U3.5 - Prob. 11ECh. U3.5 - Prob. 12ECh. U3.6 - Prob. 1TAICh. U3.6 - Prob. 1ECh. U3.6 - Prob. 2ECh. U3.6 - Prob. 3ECh. U3.6 - Prob. 4ECh. U3.6 - Prob. 5ECh. U3.6 - Prob. 6ECh. U3.6 - Prob. 7ECh. U3.6 - Prob. 8ECh. U3.7 - Prob. 1TAICh. U3.7 - Prob. 1ECh. U3.7 - Prob. 2ECh. U3.7 - Prob. 3ECh. U3.7 - Prob. 4ECh. U3.7 - Prob. 5ECh. U3.7 - Prob. 6ECh. U3.8 - Prob. 1TAICh. U3.8 - Prob. 1ECh. U3.8 - Prob. 2ECh. U3.8 - Prob. 3ECh. U3.8 - Prob. 4ECh. U3.8 - Prob. 5ECh. U3.8 - Prob. 6ECh. U3.8 - Prob. 7ECh. U3.8 - Prob. 8ECh. U3.8 - Prob. 9ECh. U3.8 - Prob. 10ECh. U3.9 - Prob. 1TAICh. U3.9 - Prob. 1ECh. U3.9 - Prob. 2ECh. U3.9 - Prob. 3ECh. U3.9 - Prob. 4ECh. U3.9 - Prob. 5ECh. U3.9 - Prob. 6ECh. U3.9 - Prob. 7ECh. U3.9 - Prob. 8ECh. U3.9 - Prob. 10ECh. U3.10 - Prob. 1TAICh. U3.10 - Prob. 1ECh. U3.10 - Prob. 2ECh. U3.10 - Prob. 4ECh. U3.10 - Prob. 5ECh. U3.10 - Prob. 6ECh. U3.10 - Prob. 7ECh. U3.11 - Prob. 1TAICh. U3.11 - Prob. 1ECh. U3.11 - Prob. 2ECh. U3.11 - Prob. 3ECh. U3.11 - Prob. 4ECh. U3.11 - Prob. 5ECh. U3.12 - Prob. 1TAICh. U3.12 - Prob. 1ECh. U3.12 - Prob. 2ECh. U3.12 - Prob. 3ECh. U3.12 - Prob. 4ECh. U3.12 - Prob. 5ECh. U3.12 - Prob. 6ECh. U3.12 - Prob. 7ECh. U3.12 - Prob. 8ECh. U3.13 - Prob. 1TAICh. U3.13 - Prob. 1ECh. U3.13 - Prob. 2ECh. U3.13 - Prob. 3ECh. U3.13 - Prob. 4ECh. U3.13 - Prob. 5ECh. U3.13 - Prob. 6ECh. U3.13 - Prob. 7ECh. U3.14 - Prob. 1TAICh. U3.14 - Prob. 1ECh. U3.14 - Prob. 2ECh. U3.15 - Prob. 1TAICh. U3.15 - Prob. 1ECh. U3.15 - Prob. 2ECh. U3.15 - Prob. 3ECh. U3.15 - Prob. 4ECh. U3.15 - Prob. 5ECh. U3.15 - Prob. 6ECh. U3.15 - Prob. 7ECh. U3.15 - Prob. 8ECh. U3.16 - Prob. 1TAICh. U3.16 - Prob. 1ECh. U3.16 - Prob. 2ECh. U3.16 - Prob. 3ECh. U3.16 - Prob. 4ECh. U3.16 - Prob. 5ECh. U3.16 - Prob. 6ECh. U3.16 - Prob. 7ECh. U3.17 - Prob. 1TAICh. U3.17 - Prob. 1ECh. U3.17 - Prob. 2ECh. U3.17 - Prob. 3ECh. U3.17 - Prob. 4ECh. U3.17 - Prob. 5ECh. U3.17 - Prob. 6ECh. U3.17 - Prob. 7ECh. U3.17 - Prob. 8ECh. U3.18 - Prob. 1TAICh. U3.18 - Prob. 1ECh. U3.18 - Prob. 2ECh. U3.18 - Prob. 3ECh. U3.18 - Prob. 4ECh. U3.18 - Prob. 5ECh. U3.18 - Prob. 6ECh. U3.18 - Prob. 7ECh. U3.18 - Prob. 8ECh. U3.18 - Prob. 9ECh. U3.18 - Prob. 10ECh. U3.18 - Prob. 11ECh. U3.18 - Prob. 12ECh. U3.19 - Prob. 1TAICh. U3.19 - Prob. 1ECh. U3.19 - Prob. 2ECh. U3.19 - Prob. 4ECh. U3 - Prob. SI1RECh. U3 - Prob. SI2RECh. U3 - Prob. SI3RECh. U3 - Prob. SI4RECh. U3 - Prob. SI5RECh. U3 - Prob. SII1RECh. U3 - Prob. SII2RECh. U3 - Prob. SII3RECh. U3 - Prob. SII4RECh. U3 - Prob. SII5RECh. U3 - Prob. SIII1RECh. U3 - Prob. SIII2RECh. U3 - Prob. SIII3RECh. U3 - Prob. SIII4RECh. U3 - Prob. SIII5RECh. U3 - Prob. 1RECh. U3 - Prob. 2RECh. U3 - Prob. 3RECh. U3 - Prob. 4RECh. U3 - Prob. 5RECh. U3 - Prob. 6RECh. U3 - Prob. 7RECh. U3 - Prob. 8RECh. U3 - Prob. 9RECh. U3 - Prob. 10RECh. U3 - Prob. 11RECh. U3 - Prob. 12RECh. U3 - Prob. 13RE
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