ATKINS' PHYSICAL CHEMISTRY
ATKINS' PHYSICAL CHEMISTRY
11th Edition
ISBN: 9780190053956
Author: ATKINS
Publisher: Oxford University Press
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Chapter 5, Problem 5A.3P

(a)

Interpretation Introduction

Interpretation:

The entropy of mixing of a gaseous mixture with mass percentage composition 75.5(N2), 23.2(O2), and 1.3(Ar), has to be calculated.

Concept Introduction:

In the mixing of two or more perfect gases the entropy of mixing is calculated by the formula below.

    ΔSmix=nRxilnxi

The value of ΔVmix and ΔHmix is always zero for mixing of perfect gases.  As the number of gases being mixed increases, the entropy also increases.

(a)

Expert Solution
Check Mark

Answer to Problem 5A.3P

The entropy of mixing of a gaseous mixture with mass percentage composition 75.5(N2), 23.2(O2), and, 1.3(Ar) is 5.052J K-1_.

Explanation of Solution

The entropy of mixing of perfect gases (ΔSmix) is calculated by the formula given below.

    ΔSmix=nR(xN2lnxN2+xO2lnxO2+xArlnxAr)(1)

Where,

n is the number of moles.

R is the gas constant. (8.314JK1mol1).

xN2, xO2, xAr are the mole fractions of N2, O2, and Ar.

The mass percentage of N2 is 75.5.

Therefore, the mole fraction of N2 (xN2) is calculated below.

  Molefraction=Masspercentage100=75.5100=0.755

The mass percentage of O2 is 23.2.

Therefore, the mole fraction of O2 (xO2) is calculated below.

  Molefraction=Masspercentage100=23.2100=0.232

The mass percentage of Ar is 1.3.

Therefore, the mole fraction of Ar (xAr) is calculated below.

  Molefraction=Masspercentage100=1.3100=0.013

Therefore, the entropy of mixing 1mol of gases is calculated below.

    ΔSmix=1mol×8.314JK1mol1×((0.755×ln0.755)+(0.232×ln0.232)+(0.013×ln0.013))=1mol×8.314JK1mol1×((0.2122)+(0.3389)+(0.05645))=1mol×8.314JK1mol1×(0.6075)=5.052JK1_

Therefore, the entropy of mixing of a gaseous mixture with mass percentage composition 75.5(N2), 23.2(O2), and 1.3(Ar), is 5.052JK1_.

(b)

Interpretation Introduction

Interpretation:

The change in entropy from part (a) when air is taken as a mixture with mass percentage composition 75.52(N2), 23.15(O2), 1.28(Ar), and 0.046(CO2), has to be calculated.

Concept Introduction:

Refer to (a)

(b)

Expert Solution
Check Mark

Answer to Problem 5A.3P

The change in entropy from part (a) when air is taken as a mixture with mass percentage composition 75.52(N2), 23.15(O2), 1.28(Ar), and 0.046(CO2) is 0.02J K-1_.

Explanation of Solution

The entropy of mixing of perfect gases (ΔSmix) is calculated by the formula given below.

    ΔSmix=nR(xN2lnxN2+xO2lnxO2+xArlnxAr+xCO2lnxCO2)(2)

Where,

n is the number of moles.

R is the gas constant. (8.314JK1mol1).

xN2, xO2, xAr, xCO2 are the mole fractions of N2, O2, Ar, and CO2 .

The mass percentage of N2 is 75.52.

Therefore, the mole fraction of N2 (xN2) is calculated below.

  Molefraction=Masspercentage100=75.52100=0.7552

The mass percentage of O2 is 23.15.

Therefore, the mole fraction of O2 (xO2) is calculated below.

  Molefraction=Masspercentage100=23.15100=0.2315

The mass percentage of Ar is 1.28.

Therefore, the mole fraction of Ar (xAr) is calculated below.

  Molefraction=Masspercentage100=1.28100=0.0128

The mass percentage of CO2 is 0.046.

Therefore, the mole fraction of CO2 (xCO2) is calculated below.

  Molefraction=Masspercentage100=0.046100=0.00046

Therefore, the entropy of mixing 1mol of gases is calculated below.

    ΔSmix=1mol×8.314JK1mol1×((0.7552×ln0.7552)+(0.2315×ln0.2315)+(0.0128×ln0.0128)+(0.00046×ln0.00046))=1mol×8.314JK1mol1×((0.2120)+(0.3387)+(0.05579)+(0.000353))=1mol×8.314JK1mol1×(0.61)=5.072JK1

Therefore, the entropy of mixing of a gaseous mixture with mass percentage composition 75.5(N2), 23.2(O2), 1.3(Ar), and 0.046(CO2) is 5.072JK1.

Therefore, the change in entropy from part (a) is calculated below.

    Changeinentropy=5.072JK15.052JK1=0.02JK1_

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Chapter 5 Solutions

ATKINS' PHYSICAL CHEMISTRY

Ch. 5 - Prob. 5A.4DQCh. 5 - Prob. 5A.5DQCh. 5 - Prob. 5A.1AECh. 5 - Prob. 5A.1BECh. 5 - Prob. 5A.2AECh. 5 - Prob. 5A.2BECh. 5 - Prob. 5A.3AECh. 5 - Prob. 5A.3BECh. 5 - Prob. 5A.4AECh. 5 - Prob. 5A.4BECh. 5 - Prob. 5A.5AECh. 5 - Prob. 5A.5BECh. 5 - Prob. 5A.6AECh. 5 - Prob. 5A.6BECh. 5 - Prob. 5A.7AECh. 5 - Prob. 5A.7BECh. 5 - Prob. 5A.8AECh. 5 - Prob. 5A.8BECh. 5 - Prob. 5A.9AECh. 5 - Prob. 5A.9BECh. 5 - Prob. 5A.10AECh. 5 - Prob. 5A.10BECh. 5 - Prob. 5A.11AECh. 5 - Prob. 5A.11BECh. 5 - Prob. 5A.1PCh. 5 - Prob. 5A.3PCh. 5 - Prob. 5A.4PCh. 5 - Prob. 5A.5PCh. 5 - Prob. 5A.6PCh. 5 - Prob. 5A.7PCh. 5 - Prob. 5B.1DQCh. 5 - Prob. 5B.2DQCh. 5 - Prob. 5B.3DQCh. 5 - Prob. 5B.4DQCh. 5 - Prob. 5B.5DQCh. 5 - Prob. 5B.6DQCh. 5 - Prob. 5B.7DQCh. 5 - Prob. 5B.1AECh. 5 - Prob. 5B.1BECh. 5 - Prob. 5B.2AECh. 5 - Prob. 5B.2BECh. 5 - Prob. 5B.3AECh. 5 - Prob. 5B.3BECh. 5 - Prob. 5B.4AECh. 5 - Prob. 5B.4BECh. 5 - Prob. 5B.5AECh. 5 - Prob. 5B.5BECh. 5 - Prob. 5B.6AECh. 5 - Prob. 5B.6BECh. 5 - Prob. 5B.7AECh. 5 - Prob. 5B.7BECh. 5 - Prob. 5B.8AECh. 5 - Prob. 5B.8BECh. 5 - Prob. 5B.9AECh. 5 - Prob. 5B.9BECh. 5 - Prob. 5B.10AECh. 5 - Prob. 5B.10BECh. 5 - Prob. 5B.11AECh. 5 - Prob. 5B.11BECh. 5 - Prob. 5B.12AECh. 5 - Prob. 5B.12BECh. 5 - Prob. 5B.1PCh. 5 - Prob. 5B.2PCh. 5 - Prob. 5B.3PCh. 5 - Prob. 5B.4PCh. 5 - Prob. 5B.5PCh. 5 - Prob. 5B.6PCh. 5 - Prob. 5B.9PCh. 5 - Prob. 5B.11PCh. 5 - Prob. 5B.13PCh. 5 - Prob. 5C.1DQCh. 5 - Prob. 5C.2DQCh. 5 - Prob. 5C.3DQCh. 5 - Prob. 5C.1AECh. 5 - Prob. 5C.1BECh. 5 - Prob. 5C.2AECh. 5 - Prob. 5C.2BECh. 5 - Prob. 5C.3AECh. 5 - Prob. 5C.3BECh. 5 - Prob. 5C.4AECh. 5 - Prob. 5C.4BECh. 5 - Prob. 5C.1PCh. 5 - Prob. 5C.2PCh. 5 - Prob. 5C.3PCh. 5 - Prob. 5C.4PCh. 5 - Prob. 5C.5PCh. 5 - Prob. 5C.6PCh. 5 - Prob. 5C.7PCh. 5 - Prob. 5C.8PCh. 5 - Prob. 5C.9PCh. 5 - Prob. 5C.10PCh. 5 - Prob. 5D.1DQCh. 5 - Prob. 5D.2DQCh. 5 - Prob. 5D.1AECh. 5 - Prob. 5D.1BECh. 5 - Prob. 5D.2AECh. 5 - Prob. 5D.2BECh. 5 - Prob. 5D.3AECh. 5 - Prob. 5D.3BECh. 5 - Prob. 5D.4AECh. 5 - Prob. 5D.4BECh. 5 - Prob. 5D.5AECh. 5 - Prob. 5D.5BECh. 5 - Prob. 5D.6AECh. 5 - Prob. 5D.1PCh. 5 - Prob. 5D.2PCh. 5 - Prob. 5D.3PCh. 5 - Prob. 5D.4PCh. 5 - Prob. 5D.5PCh. 5 - Prob. 5D.6PCh. 5 - Prob. 5D.7PCh. 5 - Prob. 5E.1DQCh. 5 - Prob. 5E.2DQCh. 5 - Prob. 5E.3DQCh. 5 - Prob. 5E.4DQCh. 5 - Prob. 5E.1AECh. 5 - Prob. 5E.1BECh. 5 - Prob. 5E.2AECh. 5 - Prob. 5E.2BECh. 5 - Prob. 5E.3AECh. 5 - Prob. 5E.3BECh. 5 - Prob. 5E.4AECh. 5 - Prob. 5E.4BECh. 5 - Prob. 5E.5AECh. 5 - Prob. 5E.5BECh. 5 - Prob. 5E.1PCh. 5 - Prob. 5E.2PCh. 5 - Prob. 5E.3PCh. 5 - Prob. 5F.1DQCh. 5 - Prob. 5F.2DQCh. 5 - Prob. 5F.3DQCh. 5 - Prob. 5F.4DQCh. 5 - Prob. 5F.5DQCh. 5 - Prob. 5F.1AECh. 5 - Prob. 5F.1BECh. 5 - Prob. 5F.2AECh. 5 - Prob. 5F.2BECh. 5 - Prob. 5F.3AECh. 5 - Prob. 5F.3BECh. 5 - Prob. 5F.4AECh. 5 - Prob. 5F.4BECh. 5 - Prob. 5F.5AECh. 5 - Prob. 5F.5BECh. 5 - Prob. 5F.6AECh. 5 - Prob. 5F.6BECh. 5 - Prob. 5F.7AECh. 5 - Prob. 5F.7BECh. 5 - Prob. 5F.8AECh. 5 - Prob. 5F.8BECh. 5 - Prob. 5F.1PCh. 5 - Prob. 5F.2PCh. 5 - Prob. 5F.3PCh. 5 - Prob. 5F.4PCh. 5 - Prob. 5.1IACh. 5 - Prob. 5.2IACh. 5 - Prob. 5.3IACh. 5 - Prob. 5.4IACh. 5 - Prob. 5.5IACh. 5 - Prob. 5.6IACh. 5 - Prob. 5.8IACh. 5 - Prob. 5.9IACh. 5 - Prob. 5.10IA
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Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY